Go Math Grade 1 Answer Key Chapter 3 Addition Strategies

Go Math Grade 1 Chapter 3 Answer Key Addition Strategies

Curious George

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 1
There are 4 fish in the tank. If you doubled the number of fish, how many would there be?

Addition Strategies Show What You Know

Model Addition

Use Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 to show each number. Draw the cubes. Write how many in all.

Question 1.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 3

Answer:

2         +           3         =                 5

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 +Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 =Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 2

Therefore, there are 5 in total

Use Symbols to Add
Use the picture. Write the addition sentence.

Question 2.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 4

Answer: 2 + 4 = 6

Explanation:

The number of red bugs =  2

The number of green bugs = 4

Total : 2 + 4 = 6

Therefore, the total number of bug pictures are 6.

Question 3.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 5

Answer: 4

Explanation:

The number of purple butterflies = 3

The number of orange butterflies = 1

Total butterflies : 3 + 1 = 4

Therefore, there are 4 butterflies

Add in Any Order

Use Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6 and Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 to add. Color to match. Write each sum.

Question 4.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 8
1 + 4 = __

Answer:

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6 + Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 = Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7

Total : 1 + 4 = 5

Question 5.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 9
4 + 1 = ___

Answer:

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6 + Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7 = Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 6Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 7

Total : 4 + 1 = 5

Addition Strategies Vocabulary Builder

Visualize It
Write the addends and the sum for the addition sentence

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 10

Answer:

Understand Vocabulary
Use a review word to complete the sentence.

Question 1.
4 and 3 in 4 + 3 = 7 are ___

Answer:

In 4 + 3 = 7 , 4 and 3 are addends.

Question 2.
4 + 3 = 7 is an ___

Answer:

4 + 3 = 7 is an Addition sentence.

Question 3.
4 cubes and 3 cubes are put together to __ the groups.

Answer:

4 cubes and 3 cubes are put together to 7 the groups.

Addition Strategies Game: Ducky Sums

Ducky Sums

Materials
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 11

Play with a partner.
1. Put your Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 11.1 on START.
2. Toss the Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 12. Move your Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 12.1 that number of spaces.
3. Use a Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 13 a Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 14, and a Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 15 to make the spinner. Spin.
4. Add the number you spin and the number your Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16 is on. Your partner checks the sum.
5. Take turns. The winner is the first person to get to END.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.1

Addition Strategies Vocabulary Game

Going Places with GOMATH! words

Concentration

Materials
2 sets of word cards
How to Play
Play with a partner.

  1. Mix the cards. Put the cards in rows blank side up.
  2. Turn over two cards.
    • If the words match, keep the pair of cards.
    • If the words do not match, turn the cards blank side up again.
  3. The other player takes a turn.
  4. Find all the pairs. The player with more pairs wins.

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.2

The Write Way

Reflect
Choose one idea. Draw and write about it.
• Mary writes the addition sentence 4 + 5.
Rose writes the addition sentence 5 + 4.
Will they get the same answer? Tell how you know.
• Think about the ways you learned to add. Tell about your favorite way to add. Explain why.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.3

Lesson 3.1 Algebra • Add in Any Order

Essential Question
What happens if you change the order of the addends when you add?

Listen and Draw

Use Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4 and Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5. Color to model the problem. Write the addition sentence.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 17

Answer:

Addition sentence :

7 + 8 = 15

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4 + Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5

= Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5

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Use Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4 and Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5. Color to change the order. Write the addition sentence.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 18

Answer:

Addition sentence ( changing the order )  :

8 + 7 = 15

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5 + Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4 =

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.5Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 16.4

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Math Talk
MATHEMATICAL PRACTICES

Describe how knowing the fact 7 + 8 helps you find 8 + 7.

Model and Draw

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 20

Answer:

5 + 6 = 11

Fact : Even after changing the order of addends , the sum is same .

Share and Show MATH BOARD

Add. Change the order of the addends. Add again.

Question 1.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 22

Answer:

8 + 9 = 17

Addition sentence after changing the order of addends =

9 + 8 = 17

Question 2.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 23

Answer:

6 + 7 = 1 3

Addition sentence after changing the order of addends =

7 + 6 = 13

Question 3.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 24

Answer:

7 + 5 = 12

Addition sentence after changing the order of addends =

5 + 7 = 12

Question 4.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 25

Answer:

2 + 8 = 10

Addition sentence after changing the order of addends =

8 + 2 = 10

Question 5.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 26

Answer:

9 + 2 = 11

Addition sentence after changing the order of addends =

2 + 9 = 11

Question 6.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 27

Answer:

8 + 4 = 12

Addition sentence after changing the order of addends =

4 + 8 = 12

On Your Own

MATHEMATICAL PRACTICE
Attend to Precision Add. Change the order of the addends. Add again.

Question 7.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 28

Answer:

9 + 6 = 15

Addition sentence after changing the order of addends =

6 + 9 = 15

Question 8.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 29

Answer: 0 + 6 = 6

Addition sentence after changing the order of addends =

6 + 0 = 6

Question 9.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 30

Answer:

8 + 3 = 11

Addition sentence after changing the order of addends =

3 + 8 = 11

Question 10.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 31

Answer:

5 + 9 = 14

Addition sentence after changing the order of addends =

9 + 5 = 14

Question 11.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 32

Answer:

4 +5 = 9

Addition sentence after changing the order of addends =

5 + 4 = 9

Question 12.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 33

Answer:

8 + 5 = 13

Addition sentence after changing the order of addends =

5 + 8 = 13

Question 13.
THINK SMARTER
Nina uses the number sentence 3 + 7 = 10 to tell about her toy trucks. What other number sentence could Nina write to tell about her trucks using the same addends?
__ = __ + __

Answer:

Given,

Addition sentence : 3 + 7 = 10

Other Addition sentence : 10 = 7 + 3

Question 14.
GO DEEPER
Explain If Adam knows 4 + 7 = 11, what other addition fact does he know? Write the new fact in the box. Tell how Adam knows the new fact.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 34
________________________
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
________________________

Answer:

Given ,

Addition sentence = 4 + 7 = 11

Other addition fact : 7 + 5 = 11

Problem Solving • Applications WRITE Math

Write two addition sentences you can use to solve the problem. Write the answer.

Question 15.
Roy sees 4 big fish and 9 small fish. How many fish does Roy see?
__ fish
__ + __ = __
__ + __ = __

Answer: 13

Explanation:

Given,

Number of big fishes = 4

Number of small fishes = 9

Total number of fishes = 4 + 9 = 13

Therefore, Roy sees 13 fishes

Addition sentences:

4 + 9 = 1 3

9 + 3 = 13

Question 16.
THINK SMARTER
Justin has 6 toys. He gets 8 more toys. How many toys does he have now?
__ toys
__ + __ = __
__ + __ = __

Answer: 14

Explanation:

Given,

Justin has 6 toys.

Number of toys he gets more = 8

Total : 6 + 8 = 14

Therefore, total number of toys = 14

Addition sentences:

6 + 8 = 14

8 + 6 = 14

Question 17.
THINK SMARTER
Anna has two groups of pennies. She has 10 pennies in all. When she changes the order of the addends, the addition sentence is the same. What sentence can Anna write?
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 35
__ = __ + __

Answer: 5 + 5 = 10

Explanation :

Given,

Anna has 2 groups of pennies and also total number of pennies = 10

The addition sentence can be written with same addends as

5 + 5 = 10

Question 18.
THINK SMARTER
Write the addends in a different order.
3 + 4 = 7
__ + __ = 7

Answer:

4 + 3 = 7

TAKE HOME ACTIVITY
• Ask your child to explain what happens to the sum when you change the order of the addends.

Algebra • Add in Any Order Homework & practice 3.1

Add. Change the order of the addends. Add again.

Question 1.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 36

Answer:

7 + 3 =10

3 + 7 = 10

Question 2.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 37

Answer:

4 + 7 = 11

7 + 4 = 11

Question 3.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 38

Answer:

9 + 8 = 17

8 + 9 = 17

Problem Solving

Write two addition sentences you can use to solve the problem.

Question 4.
Camila has 5 shells. Then she finds 4 more shells. How many shells does she have now?
__ + __ = __
__ + __ = __

Answer: 9

Explanation :

Given that,

The number of shells Camila have = 5

The number of shells she finds more = 4

Total : 5+4= 9

Therefore, the number of shells she have now = 9

Addition sentences :

5 + 4 = 9

4 + 5 = 9

Question 5.
WRITE Math
Use pictures or words to explain how you would use the sum of 13 to show how to add in any order.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 39

13 can be shown as the sum of 6 and 7

6 + 7 = 13

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42 + Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

=Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Lesson Check

Question 1.
What is another way to write 7 + 6 = 13?
6 + 7 = __

Answer:

6 + 7 = 13

Question 2.
What is another way to write 6 + 8 = 14?
8 + 6 = __

Answer:

8 + 6 = 14

Spiral Review

Question 3.
What is the sum? Write the number.

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 40

Answer:

4 + 3 = 7

Question 4.
How many nests are there? Write the number.
2 nests and 1 more nest __ nests

Answer: 3

Explanation :

2 + 1 = 3

Therefore, total number of nests = 3

Lesson 3.2 Count On

Essential Question
How do you count on 1, 2, or 3?

Listen

Start at 9. How can you count on to add? Add 1.

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 41

Add 1.  9 + 1 =10

Add 2.  9 + 2 = 11

Add 3.  9 + 3 = 12

Math Talk
MATHEMATICAL PRACTICES
Analyze How is counting on 2 like adding 2?

Share and Show MATH BOARD

Circle the greater addend. Draw Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42 to count on 1, 2, or 3. Write the sum.

Question 1.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 43

Answer:

Greater addend : 6

6 + 2 = 8

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 44 Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

 

Question 2.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 44

6 + 3 = __

Answer:

Greater addend : 6

6 + 3 = 9

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 44Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Question 3.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 45
__ = 1 + 6

Answer:

Greater addend = 6

6 + 1 = 7

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 45Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Question 4.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 46
__ = 7 + 1

Answer:

Greater addend  : 7

7 + 1 = 8

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 46Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Question 5.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 47
2 + 7 = __

Answer:

Greater addend = 7

7 + 2 = 9

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 47Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Question 6.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 48
__ = 7 + 3

Answer:

Greater addend : 7

7 + 3 = 10

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 48Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

On Your Own

MATHEMATICAL PRACTICE
Circle the greater addend. Count on to find the sum.

Question 7.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 49

Answer:

Greater addend : 9

1 + 9 = 10

Question 8.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 50

Answer:

Greater addend  : 8

3 + 8 = 11

Question 9.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 51

Answer:

Greater addend : 8

1 + 8 = 9

Question 10.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 52

Answer:

Greater addend : 6

1 + 6 = 7

Question 11.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 53

Answer :

Greater addend : 9

9 + 3 =12

Question 12.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 54

Answer:

Greater addend : 7

7 + 2 = 9

Question 13.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 55

Answer:

Greater addend : 6

2 + 6 = 8

Question 14.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 56

Answer:

Greater addend : 5 + 3 = 8

Question 15.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 57

Answer:

Greater addend : 7

7 + 1 = 8

Question 16.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 58

Answer:

Greater addend = 7

3 + 7 =10

Question 17.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 59

Answer:

Greater addend : 9

9 + 2 = 11

Question 18.
Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 60

Answer:

Greater addend : 4

3 + 4 = 7

Question 19.
GO DEEPER
Adam has 6 hats. Molly has 3 hats. They stack all their hats. Then Blake puts 2 more hats on the stack. How many hats are on the stack?
__ + __ = ___ hats
__ + __ = ___ hats

Answer:

6 + 3 = 9

9 + 2 = 11

Question 20.
MATHEMATICAL PRACTICE
Explain Terry added 3 and 7. He got a sum of 9. His answer is not correct. Describe how Terry can find the correct sum.
_______________
_ _ _ _ _ _ _ _ _ _ _
_______________
_______________
_ _ _ _ _ _ _ _ _ _ _
_______________

Answer:

No, he is wrong .

The sum of 3 and 7 is 10

Addition sentence :  3 + 7 = 10

Starts at 7 and count by 3 , means 8 , 9 , 10

So, the answer is 0.

Problem Solving • Applications WRITE Math

Draw to solve. Write the addition sentence.

Question 21.
THINK SMARTER
Cindy and Joe pick 8 oranges. Then they pick 3 more oranges. How many oranges do they pick?
__ + __ = ___ oranges
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61

Answer: 11

Explanation :

Given that,

The number of oranges Cindy and Joe pick = 8

The number of  more oranges they pick = 3

Total : 8 + 3 = 11 oranges

Therefore, the number of total oranges = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61+      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61

________________________________________________________________________________________________________

= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 61

1                  2                3                 4                5              6                  7                  8               9               10              11

Which three numbers can you use to complete the problem?

Question 22.
THINK SMARTER
Jennifer has __ stamps. She gets __ more stamps. How many stamps does she have now?
__ + __ = __ stamps
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 62

Answer: 10

Explanation :

Jennifer has  7 stamps

Number of more stamps = 3

Addition sentence :

7 + 3 = 10

Question 23.
THINK SMARTER
Count on from 3. Write the number that shows 2 more in the box below
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 63

Answer: 5

TAKE HOME ACTIVITY
• Have your child tell you how to count on to find the sum for 6 + 3.

Count On Homework & Practice 3.2

Circle the greater addend. Count on to find the sum.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 64

Answer:

Greater addend : 8

8 + 2 = 10

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 65

Answer:

Greater addend : 7

1 + 7 = 8

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 66

Answer:

Greater addend : 9

3 + 9 = 12

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 67

Answer:

Greater addend : 5

5 + 3 = 8

Problem Solving

Draw to solve. Write the addition sentence.

Question 5.
Jon eats 6 crackers. Then he eats 3 more crackers. How many crackers does he eat?
__ + __ = __ crackers

Answer:  9

Explanation :

Number of crackers Jon eat = 6

Number of crackers he ate more = 3

Total : 6 + 3 = 9 Crackers

Therefore, Jon ate 9 crackers .

Question 6.
WRITE Math
Use pictures or words to explain how you can find 9 + 3 by counting on.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 68

Answer:

9 + 3 = 12

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42  +  Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42 = Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

1      2      3        4      5       6      7       8     9              10     11    12        Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Lesson Check

Question 1.
Count on to solve 5 + 2. Write the sum
__

Answer:

5 + 2 = 7

Question 2.
Count on to solve 1 + 9. Write the sum.
__

Answer:

1 + 9 = 10

Spiral Review

Question 3.
What does the model show?
__ + __ = ___

Answer:

3 + 3 = 6

Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42 +Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42 = Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42Go Math Grade 1 Answer Key Chapter 3 Addition Strategies 42

Question 4.
4 ducks swim in the pond. 2 more ducks join them. How many ducks are in the pond now? Complete the model and the number sentence.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 69
4 + 2 = __

Answer:

4 + 2 = 6

Lesson 2.3 Add Doubles

Essential Question
What are doubles facts?

Listen and Draw

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70 to solve. Write the addition sentence.

__ + __ = __

Answer:

2 + 2 = 4

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70

Math Talk
MATHEMATICAL PRACTICES
Use Tools Describe how your model shows a doubles fact.

Model and Draw

Why are these doubles facts?
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 71

Answer:                                                             Answer:

1 + 1 = 2                                                           2 + 2 = 4

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70                                      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70  + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70   =Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70

Share and Show MATH BOARD

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72 to show your work. Write the sum.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 74

Answer:

3 + 3 = 6

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

_____________________________

=Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 75

Answer:

4 + 4 = 8

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

___________________________

= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 76

Answer:

5 + 5 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

_______________________________________________

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 77

Answer: 6 + 6 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

____________________________________________________________________________

=Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

On Your Own

GO DEEPER
Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 78. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 78 to show your work. Write the sum.

Question 5.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 80

Answer:

7 + 7 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

______________________________________________________________________________

= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Question 6.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 81

Answer:

8 + 8 = 16

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

________________________________________________________________________

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Question 7.
MATHEMATICAL PRACTICE
Look for a Pattern Look back at Exercises 1 – 6. Write the fact that would be next in the pattern. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 82 to show your work.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 82.1

Add.

Question 8.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 83

Answer:

5 + 5 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70

________________________________________________________________

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70

= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 70

 

Question 9.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 84

Answer:

7 + 7 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

______________________________________________________________________________

= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

 

Question 10.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 85

Answer: 6 + 6 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

____________________________________________________________________________

=Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

 

Question 11.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 86

Answer:

10 + 10 = 20

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72  + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72  = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Question 12.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 87

Answer:

4 + 4 = 8

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

___________________________

= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

 

Question 13.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 88

Answer:

8 + 8 = 16

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

+    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

________________________________________________________________________

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 72

 

Problem Solving • Applications WRITE Math

Write a doubles fact to solve.

Question 14.
THINK SMARTER
Meg and Paul each put 8 apples into a basket. How many apples are in the basket?
__ + ___ = __ apples
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 89

Answer : 16

Explanation :

The number of apples meg kept in a basket = 8

The number of apples Paul kept in a basket = 8

Total : 8 + 8 = 16 apples

Therefore, total number of apples in the basket = 16

Question 15.
THINK SMARTER
There are 18 people at the party. Some are boys and some are girls. The number of boys is the same as the number of girls.
__ = __ + __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 90

Answer : 9 + 9 = 18

Explanation :

Given,

Total number of people in the party = 18

Also given , The number of boys is the same as the number of girls.

So, 9 + 9 = 18.

Question 16.
THINK SMARTER
The cubes show a doubles fact. Choose the doubles fact and the sum.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 91

Answer:

The number sentence with doubles fact = 4 + 4 = 8

 

TAKE HOME ACTIVITY
• Have your child choose a number from 1 to 10 and use that number in a doubles fact. Repeat with other numbers.

Add Doubles Homework & Practice 3.3

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 to show your work. Write the sum.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 94

Answer:

4 + 4 = 8

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92= Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 95

Answer:

6 + 6 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 96

Answer:

3 + 3 = 6

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 =Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Question 4.

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 97

Answer:

8 + 8 = 16

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92  +  Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92                                                                                                                                                                                       Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Problem Solving

Write a doubles fact to solve.

Question 5.
There are 16 crayons in the box. Some are green and some are red. The number of green crayons is the same as the number of red crayons.
__ = __ + __

Answer: 16 = 8 + 8

Explanation:

Given ,

The total number of crayons in the box = 16

Also give, the number of green crayons is the same as the number of red crayons.

Now, By using doubles fact we can write number sentence as:

8 + 8 = 16

Question 6.
WRITE Math
Use pictures or words to explain how you could find the sum of 7 + 7.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 98

Answer:

7 + 7 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92 =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 92

Lesson Check

Question 1.
Write a doubles fact with the sum of 18.
__ + __ = 18

Answer:

A doubles fact with the sum of 18 =

9 + 9 = 18

Question 2.
Write a doubles fact with the sum of 12.
__ + __ = 12

Answer:

A doubles fact with the sum of 12 =

6 + 6 = 12

Spiral Review

Question 3.
What is the sum of 3 and 2? Draw the Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99. Write the sum.
__+ __ = __

Answer: 5

Explanation:

The sum of 3 and 2 is 5

Number sentence :

3 + 2 = 5

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99 =Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 99

Question 4.
Draw circles to show the numbers. Write the sum.
__ + __ = __

Lesson 3.4 Use Doubles to Add

Essential Question
How can you use doubles to help you add?

Listen and Draw

Draw to show the problem. Write the number of fish.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 100

There are __ fish.

Math Talk
MATHEMATICAL PRACTICES

Look for Structure How does knowing 3 + 3 help you solve the problem?

Model and Draw

How can a doubles fact help you solve 6 + 7?
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 101
So, 6 + 7 = __

Answer:

6 + 7 =  13

By using doubles fact ,

13 can be shown as 6 + 6 + 1 = 13

Share and Show MATH BOARD

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 102 to model. Make doubles. Add.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 103

Answer :

9 + 8 = 17

8 + 8 + 1 = 17

So, 9 + 8 = 17

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 104

Answer:

5 + 5 + 1 = 11

So, 5 + 6 = 11

On Your Own

MATHEMATICAL PRACTICE
Use a Concrete Model
Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 105. Make doubles. Add.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 106

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 107

Answer:

7 + 8 = 15

By using doubles fact ,

7 + 7 + 1 = 15

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 108

Answer :

5 + 4 = 9

By using doubles fact ,

4 + 4 + 1 = 9

Question 5.
THINK SMARTER
Mandy has the same number of red and yellow leaves. Then she finds one more yellow leaf. There are 17 leaves in all. How many leaves are red? How many leaves are yellow?
__ red leaves __ yellow leaves
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 109

Answer :

Explanation :

Given , the total number of leaves = 17

Also given , Mandy has the same number of red and yellow leaves

And also find another yellow flower

By using doubles plus one effect , we can show addition sentence as :

8 + 8 + 1 = 17

Therefore,

Number of red flowers = 8

Number of yellow flowers = 9.

GO DEEPER
Explain Would you use count on or doubles to solve? Why?

Question 6.
3 + 4
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 110

Answer: 3 + 3 + 1 = 7

Doubles plus one fact.

Question 7.
3 + 9
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 111

Answer :

9 and count on 3

Problem Solving • Applications

Question 8.
THINK SMARTER
Use what you know about doubles to complete the Key. Write the missing sums.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 112

Answer :

Question 9.
THINK SMARTER
There are 7 red cubes. There are 8 yellow cubes. How many cubes are there in all? Use a double to add. Write the missing numbers.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 113

Answer: 15

Explanation:

The number of red cubes = 7

The number of yellow cubes = 8

Total : 7 + 8 = 15

Number sentence with double fact = 7 + 7+ 1 = 15

So, 7 + 8 = 15

TAKE HOME ACTIVITY
• Ask your child to show you how to use what he or she knows about doubles to help solve 6 + 5.

Use Doubles to Add Homework & Practice 3.4

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 114. Make doubles. Add.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 115

Answer:

5 + 6 = 11

The number sentence with doubles fact =

5 + 5 + 1 = 11

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 116
So, 9 + 8 = __

Answer:

9 + 8 = 17

The number sentence with doubles fact =

8 + 8 + 1 = 17

Use doubles to help you add.

Question 3.
8 + 7 = __

Answer:

Doubles plus one fact .

Explanation :

8 + 7 = 15

The number sentence with doubles fact :

7 + 7 + 1 = 15

Question 4.
6 + 5 = __

Answer:

Doubles plus one fact .

Explanation :

6 + 5 = 11

The number sentence with doubles plus one fact :

5 + 5 + 1 = 11

Question 5.
7 + 6 = __

Answer:

Doubles plus one fact .

Explanation :

7 + 6 = 13

The number sentence with doubles plus one fact :

6 + 6 + 1 = 13

Problem Solving

Solve. Draw or write to explain.

Question 6.
Bo has 6 toys. Mia has 7 toys. How many toys do they have?
__ toys

Answer: 13

Explanation:

The number of toys Bo have = 6

The number of toys Mia have = 7

Total : 6 +7 = 13

Therefore, the total number of toys they both have = 13

 

Question 7.
WRITE Math
Draw and label a picture to show how knowing 7 + 7 helps you find 7 + 8.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 117

Lesson Check

Question 1.
Use doubles to find the sum of 7 + 8. Write the number sentence.
__ + __ + __ = __

Answer:

The sum of 7 + 8 = 15

The number sentence with doubles plus one fact :

7 + 7 + 1 = 15

Spiral Review

Question 2.
Circle the part you take from the group. Cross it out. Then write the difference.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 118
8 – 6 = __

Answer:

8 – 6 = 2

Question 3.
There are 7 gray kittens. 2 kittens are black. How many fewer black kittens are there than gray kittens?
Use the bar model to solve. Then write the number sentence.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 119
__ – __ = __

Answer:

7 – 2 = 5

Therefore, there are 5 fewer black kittens are there than gray kittens.

Lesson 3.5 Doubles Plus 1 and Doubles Minus 1

Essential Question
How can you use what you know about doubles to find other sums?

Listen and Draw

How can you use the doubles fact, 4 + 4, to solve each problem? Draw to show how. Complete the addition sentence.

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 120

Answer :

  1. 4 + 4 + 1 = 9 – Doubles plus one fact
  2.  4 + 4 – 1 = 7 – Doubles minus one fact.

 

Math Talk
MATHEMATICAL PRACTICES

Explain what happens to the doubles fact when you increase one addend by one or decrease one addend by one.

Share and Show MATH BOARD

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 121 to add. Solve the doubles fact. Then use doubles plus one or doubles minus one. Circle + or − to show how you solved each one.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 122

Answer:

2 + 2 = 4

2+ 2+ 1 = 5  doubles plus one

2 + 2 – 1 = 3 – doubles minus one.

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 123

Answer :

3 + 3 = 6

3 + 3 + 1 = 7 – Doubles plus one

3 + 3 – 1 = 5 – Doubles minus one

 

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 124

Answer :

4 + 4 = 8

4 + 4 + 1 = 9 – Doubles plus one

4 + 4 – 1 = 7 – Doubles minus one

On Your Own

MATHEMATICAL PRACTICE
Make Connections Add. Write the doubles fact you used to solve the problem.

Question 4.
8 + 9 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 125

Answer:

Doubles plus one fact .

Explanation :

8 + 9 = 17

The number sentence with doubles plus one fact :

8 + 8 + 1 = 17

 

Question 5.
2 + 3 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 126

Answer:

Doubles plus one fact .

Explanation :

2 + 3 = 5

The number sentence with doubles plus one fact :

2 + 2 + 1 = 5

Question 6.
7 + 6 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 127

Answer:

Doubles plus one fact .

Explanation :

7 + 6 = 13

The number sentence with doubles plus one fact :

6 + 6 + 1 = 15

Question 7.
6 + 5 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 127.1

Answer:

Doubles plus one fact .

Explanation :

6 + 5 = 11

The number sentence with doubles plus one fact :

5 + 5 + 1 = 15

Question 8.
3  + 4 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 127

Answer:

Doubles plus one fact .

Explanation :

3 + 4 = 7

The number sentence with doubles plus one fact :

3 + 3 + 1 = 7

Question 9.
4 + 5 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 127

Answer:

Doubles plus one fact .

Explanation :

4 + 5 = 9

The number sentence with doubles plus one fact :

4 + 4 + 1 = 15

Question 10.
THINK SMARTER
Brianna has 6 toy ducks. Ian has the same number of toy ducks and a toy fish. How many toys do Brianna and Ian have?
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 128

Answer : 13

Explanation :

The number of toy ducks Brianna

THINK SMARTER
Add. Write the doubles plus one fact. Write the doubles minus one fact.

Question 11.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 129

Answer :

6 + 6 = 12

6 + 6 + 1 = 1 3

6 + 6 – 1 = 11

Problem Solving • Applications

Question 12.
GO DEEPER
Grace wants to write the sums for the doubles plus one and doubles minus one facts. She has started writing sums. Help her find the rest of the sums.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 130

Question 13.
THINK SMARTER
Choose all the doubles facts that can help you solve 4 + 5.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 131

Answer :

9 + 9 = 18 – No

5 + 5 = 10 -Yes

4 + 4 = 8 – Yes

TAKE HOME ACTIVITY
• Have your child explain how to use a doubles fact to solve the doubles plus one fact 4 + 5 and the doubles minus one fact 4 + 3.

Doubles Plus 1 and Doubles Minus 1 Homework & Practice 3.5

Add. Write the doubles fact you used to solve the problem.

Question 1.
8 + 7 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 132

Answer:

Doubles plus one fact .

Explanation :

8 + 7 = 15

The number sentence with doubles plus one fact :

7 + 7 + 1 = 15

Question 2.
6 + 7 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 132

Answer:

Doubles plus one fact .

Explanation :

6 + 7 = 13

The number sentence with doubles plus one fact :

6 + 6 + 1 = 13

Question 3.
4 + 3 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 132

Answer:

Doubles plus one fact .

Explanation :

4 + 3 = 7

The number sentence with doubles plus one fact :

3 + 3 + 1 = 7

Question 4.
2 + 1 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 132

Answer:

Doubles plus one fact .

Explanation :

2 + 1 = 3

The number sentence with doubles plus one fact :

1 + 1 + 1 = 3

Question 5.
8 + 9 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 132

Answer:

Doubles plus one fact .

Explanation :

8 + 9 = 17

The number sentence with doubles plus one fact :

8 + 8 + 1 = 17

Question 6.
3 + 2 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 132

Answer:

Doubles plus one fact .

Explanation :

3 + 2 = 5

The number sentence with doubles plus one fact :

2 + 2 + 1 = 5

Problem Solving

Question 7.
Andy writes an addition fact. One addend is 9. The sum is 17. What is the other addend? Write the addition fact.
__ + ___ = 17

Answer :

9 + 8 = 17

Therefore, other addend = 8

Question 8.
WRITE Math
Use pictures or words to explain how you would use doubles plus one to solve 4 + 5.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 133

Answer :

4 + 5 = 9

By using doubles plus one fact ,

4 + 4 + 1 = 9

Lesson Check

Question 1.
Use the picture. Write a doubles plus one number sentence.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 134
__ + __ + __

Answer:

The number sentence with double plus one :

2 + 2 + 1 = 5

Question 2.
Which doubles fact helps you solve 8 + 7 = 15? Write the number sentence.
__ + __ + __

Answer:

8 + 7 = 15

The number sentence with doubles plus one fact  :

7 + 7 + 1 = 15

Spiral Review

Question 3.
There are 7 large dogs and 2 small dogs. How many dogs are there?
Use ○ to solve. Draw to show your work. Write the number sentence and how many.
__ dogs
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 135

Answer :

7 + 2 = 9

Therefore, total number of dogs = 9

Question 4.
What is the sum of 2 and 1 more? Draw the Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136. Write the sum.
__ 2 + 1 = __

Answer : 2 + 1 = 3

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136

Lesson 3.6 Practice the Strategies

Essential Question
What strategies can you use to solve addition fact problems?

Listen and Draw

Think of different addition strategies. Write or draw two ways you can solve 4 + 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 136.1

Math Talk
MATHEMATICAL PRACTICES
Look for Structure Why is the sum the same when you use different strategies?

Model and Draw

These are the ways you have learned to find sums.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 137

Share and Show MATH BOARD

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 138

Answer :

4 + 1 =5

5+ 1 = 6

6+ 1= 7

7 + 1 = 8

 

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 139

5+ 2= 7

6 + 2 = 8

7 + 2 = 9

8 + 2 = 10

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 140

Answer :

6 + 3 = 9

7 + 3 = 10

8 + 3 = 11

9 + 3 = 1 2

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 141

Answer :

7 + 7 = 14

8 + 8 = 16

9 + 9 =18

10 + 10 = 20

 

Question 5.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 142

Answer :

Doubles plus one :

5 + 5 + 1 = 11

6 + 6 + 1 = 13

Doubles minus one :

7 + 7 – 1 = 15

8 + 8 -1 = 17

On Your Own

MATHEMATICAL PRACTICE
Apply Add. Color doubles facts Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143. Color count on facts Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144. Color doubles plus one or doubles minus one facts Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.

Question 6.
9 + 9 = __

Answer :

Doubles fact Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143

9 + 9 = 18

Question 7.
7 + 1 = __

Answer :

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

7 + 1 = 8

Question 8.
5 + 3 = __

Answer :

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

5 + 3 = 8

Question 9.
2 + 9 = __

Answer :

2 + 9 = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 10.
7 + 3 = __

Answer :

7 + 3 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 11.
7 + 7 = __

Answer :

7 + 7 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143

Question 12.
6 + 5 = __

Answer :

6 + 5 = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145

Question 13.
2 + 8 = __

Answer :

8 + 2 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 14.
8 + 8 = __

Answer :

8 + 8 = 16

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143

Question 15.
8 +9 = __

Answer :

8 + 9 = 17

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145

Question 16.
9 + 3 = __

Answer :

9 + 3 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 17.
7 + 8 = __

Answer :
7 + 8 = 15

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145

THINK SMARTER
Make a counting on problem. Write the missing numbers.

Question 18.
__ birds were in a tree. __ more birds flew there. How many birds are in the tree now?
__ birds
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.1

Answer :

The birds on a tree = 7

The number of birds flew there = 3

Total : 7 + 3 = 10

TAKE HOME ACTIVITY
• Have your child point out a doubles fact, a doubles plus one fact, a doubles minus one fact, and a fact he or she solved by counting on. Have him or her describe how each strategy works.

Practice the Strategies Homework & Practice 3.6

Add. Color doubles facts Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.2. Color count on facts Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.2. Color doubles plus one or doubles minus one facts Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.2.

Question 1.
8 + 8 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer:

8 + 8 = 16

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143

Question 2.
8 + 1 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

8 + 1 = 9

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 3.
1 + 7 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

1 + 7 = 8

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 4.
8 + 3 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

8  + 3 = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 5.
5 +5 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

5 + 5 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143

Question 6.
8 + 7 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

8 + 7 = 15

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145

Question 7.
8 + 9 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

8 + 9 = 17

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145

Question 8.
6 + 3 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

6 + 3 = 9

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 144

Question 9.
6 + 6 = __
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 145.3

Answer :

6 + 6 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 143

Problem Solving

Make a counting on problem. Write the missing numbers.

Question 10.
__ apples are in a bag. __ more apples are put in the bag. How many apples are in the bag now?
__ apples

Answer :

5 apples are in a bag.

2 more apples are put in the bag.
Total number of apples = 5 + 2 = 7

Question 11.
WRITE Math
Use pictures or words to explain a strategy you would use to find 8 + 9.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 146

Answer :

8 + 9 = 17

Lesson Check

Question 1.
Which strategy would you use to find 2 + 8? Explain how you decided.
_________________________
_________________________

Answer :

Count on.

8 + 2 = 10

Question 2.
What is the sum of 9 + 9? Write the number.
__

Answer: 18

The sum of 9 + 9= 18

Spiral Review

Question 3.
What is the sum of 5 + 2 or 2 + 5? Why is the sum the same?
_________________________
_________________________

Answer: The sum of 5 + 2 or 2 + 5 = 7

The sum is same because

Question 4.
How many flowers are there? Write the number.
3 flowers and 3 more flowers __ flowers

Answer : 6

3 + 3 = 6

Addition Strategies Mid-Chapter Checkpoint

Concepts and Skills

Add. Change the order of the addends. Add again.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 147

Answer :

4 + 8 = 12

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 148

Answer :

9 + 7 = 16

Circle the greater addend. Count on to find the sum.

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 149

Answer:

Greater addend : 8

1 + 8 = 9

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 150

Answer:

Greater addend : 7

3 + 7 = 10

Question 5.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 151

Answer:

Greater addend : 9

9 + 2 = 11

Question 6.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 152

Answer :

Greater addend  : 6

6 + 3 = 9

Question 7.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 153

Answer :

Greater addend  : 7

7 + 1 = 8

Question 8.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 154

Answer :

Greater addend  : 8

2 + 8 = 10

Use doubles to help you add.

Question 9.
7 + 8 = __

Answer :

7 + 8 = 15

Doubles fact :

7 + 7 + 1 = 15

Question 10.
6 + 7 = __

Answer :

6 +7 = 13

Doubles fact  :

6 + 6 + 1 = 13

Question 11.
9 + 8 = __

Answer :

9 + 8 = 17

Doubles fact :

8 + 8 +1 = 17

Question 12.
THINK SMARTER
Write a count on 1 fact to show a sum of 8. Then write a doubles fact to show a sum of 8.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 155

Answer :

Count on fact to show sum of 8 =

7 +1 = 8

Doubles fact plus one = 4+ 4+ 1 = 8

Lesson 3.7 Add 10 and More

Essential Question
How can you use a ten frame to add 10 and some more?

Listen and Draw

What is 10 + 5? Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 156 and the ten frame. Model and draw to solve.

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 156.1

Answer:

10+ 5 = 15

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

 

 

Math Talk
MATHEMATICAL PRACTICES
Reasoning Explain how your model shows 10 + 5.

Model and Draw
You can use a ten frame to add 10 + 6.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 156.2

Share and Show MATH BOARD

Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157 to show 10. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158 to show the other addend. Write the sum.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 159

Answer:

10 + 3 = 13

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

+ Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 160

Answer :

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

+ Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

 

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 161

Answer :

10+ 1 = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

+ Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

 

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 162

Answer :

10 + 2 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

+ Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

 

Question 5.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 163

Answer :

10+ 4 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

+ Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

 

Question 6.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 164

Answer :

10+ 7 = 17

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

+ Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 157

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 158

On Your Own

MATHEMATICAL PRACTICE
Represent a Problem
Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 to show 10. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 to show the other addend. Write the sum.

Question 7.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165.2

Answer:

10 + 8 = 18

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 8.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166.2

Answer:

10 + 2 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 9.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 167

Answer:

10 + 6 = 16

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166   =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 10.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 168

Answer:

10+9 = 19

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Add.

Question 11.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 169

Answer:

10+ 1 = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165     Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Question 12.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 170

Answer :

4 + 10 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165             Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

+    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Question 13.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 171

Answer :

5 + 10 = 15

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165             Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

+           Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 =

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Question 14.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 172

Answer :

10 + 3 = 13

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165     Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +

=

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Question 15.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173

0 + 10 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Question 16.
THINK SMARTER
Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1 to show 10. Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2 to show the missing addend. Write the missing addend.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 174

Answer:

10+ 4 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2=Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.2

Problem Solving • Applications

GO DEEPER
Draw Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 175. Write the addition sentence. Explain your model.

Question 17.
Marina has 10 crayons. She gets 7 more crayons. How many crayons does she have now?
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 175.1

Answer:

1 7

Explanation :

The number of crayons Marina have = 10

The number of crayons she gets more = 7

Addition sentence : 1 0 + 7 = 17

Therefore, total crayons Marina have = 17

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

 

Question 18.
THINK SMARTER
Match the models to the number sentences.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 176

Answer :

TAKE HOME ACTIVITY
• Have your child choose a number between 1 and 10 and then find the sum of 10 and that number. Repeat using other numbers.

Add 10 and More Homework & Practice 3.7

Draw red ○ to show 10. Draw ○ yellow to show the other addend. Write the sum.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 177

Answer :

10 + 7 = 17

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 178

Answer :

10+ 5 = 15

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Problem Solving

Draw red and yellow ○ to solve. Write the addition sentence.

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 179

Question 3.
Linda has 10 toy cars. She gets 6 more cars. How many toy cars does she have now?
__ + __ = __ toy cars

Answer: 16

Explanation:

The number of toys Linda have = 10

The number of toys she gets more = 6

Total : 10 + 6 = 16

Therefore, the total number of toys she have now = 16

Question 4.
WRITE Math
Use pictures or words to explain how you can solve 10 + 6.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 180

Lesson Check

Question 1.
Draw more Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 181 to show the addition fact. Then solve.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 181

Question 2.
What number sentence does this model show? Write the number sentence.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 183

__ + __ = __

Answer:

The addition sentence :

10 + 8 = 18

Spiral Review

Question 3.
Show three different ways to make 10. Write the number sentences.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 184

Answer:

  1. 2 + 8 = 10

2. 5 + 5 = 10

3. 8 + 2 = 10

Question 4.
There are 3 large turtles and 1 small turtle. How many turtles are there? Write the number sentence and how many.
__ turtles
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 185

Answer : 4

Explanation :

The number of large turtles : 3

The number of small turtles = 1

Number sentence : 3 + 1 = 4

Therefore, total number of turtles = 4

Lesson 3.8 Make a 10 to Add

Essential Question
How do you use the make a ten strategy to add?

Listen and Draw
What is 9 + 6? Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 185.1 and the ten frame. Model and draw to solve.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 186

Answer:

9 + 6 = 15

9 + 1 = 10

10 + 5 = 15

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Math Talk
MATHEMATICAL PRACTICES
Use Tools Why do you start by putting 9 counters in the ten frame?

Model and Draw

Why do you show 8 in the ten frame to find 4 + 8?
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 187

Share and Show MATH BOARD
Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 188 and a ten frame. Show both addends. Draw to make a ten. Then write the new fact. Add.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 189

Answer :

9 + 5 = 14

9 + 1 = 10

So, 10 + 4 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165  + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165                                            Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 190

Answer :

4 + 7 = 11

7 + 3 = 10

10 + 1 = 11

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165                                                          Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 3.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 191

Answer :

9 + 8 = 17

9 + 1 = 10

10+ 7 = 17

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165  + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165            Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166                    Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

On Your Own

MATHEMATICAL PRACTICE
Use a Concrete Model

Use Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 192 and a ten frame. Show both addends. Draw to make a ten. Then write the new fact. Add.

Question 4.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 192.1

Answer :

8 + 2 = 10

10 + 3 = 13

So, 5 + 8 = 13

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165                                                          Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Draw to make a ten. Write the missing number.

Question 5.
THINK SMARTER
Andrew has visited Cedar Hill State Park 7 times. Cathy has visited the same park 9 times. How many times did they both visit the park.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 193
__ visits

Answer : 16 times

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165                          Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166          Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 6.
THINK SMARTER
What strategy would you choose to solve 7 + 8? Why?
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 194

Problem Solving • Applications WRITE Math

Solve.

Question 7.
10 + 8 has the same sum as 9 + __.

Answer:

10+ 8 = 18

9 + 9 = 18

Question 8.
10 + 7 has the same sum as 8 + __.

Answer:

10 + 7 = 17

8 +  9 = 17

Question 9.
10 + 5 has the same sum as 6 + __.

Answer :

10 +5 = 15

6 + 9 = 15

Question 10.
GO DEEPER
Write the numbers 6, 8, or 10 to complete the sentence __ + __ has the same sum as __ + 8

Answer:

10+ 6 = 18 has the same sum of 8 + 8.

Question 11.
THINK SMARTER
The model shows 7 + 4 = 11. Write the 10 fact that has the same sum.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 195

Answer:

7 + 3 = 10

10 + 1 = 11

So, 7 + 4 = 11

TAKE HOME ACTIVITY
• Cut off 2 cups from an egg carton or draw a 5-by-2 grid on a sheet of paper to create a ten frame. Have your child use small objects to show how to make a ten to solve 8 + 3, 7 + 6, and 9 + 9.

Make a 10 to Add Homework & Practice 3.8

Use red and yellow ○ and a ten frame. Show both addends. Draw to make a ten. Then write the new fact. Add.

Question 1.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 196

Answer:

5 + 5 = 10

10 + 2 = 12

So, 5 + 7 = 12

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Question 2.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 197

Answer:

9 + 1 = 10

10 + 4 = 14

So, 9 + 5 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Problem Solving

Solve.
Question 3.
10 + 6 has the same sum as 7 + __.

Answer: 9

10 + 6 = 16

7 + 9 = 16

Question 4.
WRITE Math Use pictures or words to explain how you would use the make a ten strategy to solve 5 + 7.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 198

Answer:

5 + 7 = 12

5 + 5 = 10

10 + 2 = 12

Lesson Check

Question 1.
What sum does this model show? Write the number.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 199

Answer:

Number sentence :

10 + 5 = 15

Question 2.
What addition sentence does this model show? Write the number sentence.
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 200

Answer:

Number sentence :

10 + 4 = 14

Spiral Review

Question 3.
What is the sum of 4 + 6? Write the sum.
__

Answer:

4 + 6 = 8

The sum of 4 + 6 = 8

Question 4.
There are 2 big flowers and 4 small flowers. How many flowers are there? Write the number sentence and how many.
__ flowers
Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 201

Answer: 6

Explanation :

The number of big flowers = 2

The number of small flowers = 4

Number sentence : 2 + 4 = 6

Therefore, Total number of flowers = 6

 

Lesson 3.9 Use Make a 10 to Add

Essential Question
How can you make a ten to help you add?

Listen and Draw

Draw to show the addends. Then draw to show how to make a ten. Write the sum.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 202

Answer:

6 + 7 = 13

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165        Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165                               =  Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166                      Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 166

 

Math Talk
MATHEMATICAL PRACTICES
Represent Describe how the drawings show how to make a ten to solve 6 + 7.

Model and Draw

What is 9 + 6?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 203

Share and Show MATH BOARD

Write to show how you make a ten. Then add.

Question 1.
What is 8 + 4?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 204
So, 8 + 4 = __

Answer:

8 + 2 + 2 = 12

10 + 2 = 12

So, 8 + 4 = 12

Question 2.
What is 5 + 7?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 205
So, 5 + 7 = __

Answer:

7 + 3 = 10

10 + 2 = 12

So, 5 + 7 = 12

On Your Own

THINK SMARTER
Write to show how you make a ten. Then add.

Question 3.
What is 7 + 8?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 206
So, 7 + 8 = __

Answer:

8 + 2 = 10

10 + 5 = 15

So, 7 + 8 = 15

Question 4.
What is 9 + 8?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 207
So, 9 + 8 = __

Answer:

9 + 1 = 10

1 0 + 7 = 17

So, 9 + 8 = 17 .

MATHEMATICAL PRACTICE
Use Models

THINK SMARTER
Use the model. Write to show how you make a ten. Then add.

Question 5.
Joe has 8 green balls of clay. Leah has 6 blue balls of clay. How many balls of clay do they have?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 208

Answer:

8 + 2 + 4

10 + 4 = 14

So, 8 + 6 = 14

Therefore, There are  14 balls of clay.

Problem Solving • Applications WRITE Math

Use the clues to solve. Draw lines to match.

Question 6.
Han, Luis, and Mike buy apples. Mike buys 10 red apples and 4 green apples. Luis and Mike buy the same number of apples. Match each person to his apples.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 209

Answer:

Question 7.
GO DEEPER
Look at Exercise 6. Han eats one apple. Now he has the same number of apples as Luis and Mike. How many red and green apples could he have?
__ red apples and __ green apples

Answer :

7 red apples and 7 green apples

Question 8.
THINK SMARTER
Does the addition show how to make a ten to add? Choose Yes or No.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 210

Answer:

  1. 8 + 2 + 2 = Yes
  2. 5 + 4 + 3 = No
  3. 6 + 7 + 3 = Yes

Use Make a 10 to Add Homework & Practice 3.9

Write to show how you make a ten. Then add.

Question 1.
What is 9 + 7?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 211
__ + __ = __
So, 9 + 7 = __

Answer :

9 + 1 + 6

9 + 1 = 10

So, 9 + 6 = 15

Question 2.
What is 5 + 8?
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 212
__ + __ = __
So, 5 + 8 = __

Answer :

8 + 2 + 3

8 + 2= 10

So, 5 + 8 = 13

Problem Solving

Use the clues to solve. Draw lines to match.

Question 3.
Ann eats 10 green grapes and 6 red grapes. Gia eats the same number of grapes as Ann. Match each person to her grapes.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 213

Answer:;

Ann – 10 green grapes and 6 red grapes .

Gia – 7 green grapes and 9 red grapes .

Question 4.
WRITE Math
Draw to explain how you would make a ten to find 5 + 8.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 214

Lesson Check

Question 1.
Write the number sentence. Make a ten to find 8 + 4. Write the number sentence.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 215
__ + __ + __ = __

Answer :

8 + 2 = 10

10 + 2 = 14

Spiral Review

Question 2.
What is the difference? Complete the subtraction sentence.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 216

Answer:

5 – 5 = 0

Question 3.
What is the difference? Write the difference.
Go Math Answer Key Grade 1 Chapter 3 Addition Strategies 217

Answer :

8 – 2 = 6

Lesson 3.10 Algebra • Add 3 Numbers

Essential Question
How can you add three addends?

Listen and Draw

Use Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 218 to model the problem. Draw to show your work.
__ birds

Math Talk
MATHEMATICAL PRACTICES
Apply Which two addends did you add first? Explain.

Share and Show MATH BOARD

Use Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 219 to change which two addends you add first. Complete the addition sentences.

Question 1.
5 + 2 + 3 = __
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 220

Answer:

5 + 2 + 3 = 10

Addition sentences :

  1.  7 + 3 = 10
  2.  5 + 5 = 10

Question 2.
3 + 4 + 6 = __
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 221

Answer:

3 + 4 + 6  = 13

Addition sentences :

  1.  3 + 10 = 13
  2.  7 + 6 =   13

On Your Own

MATHEMATICAL PRACTICE

Compare Models
Look at the Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 222. Complete the addition sentences showing two ways to find the sum.

Question 3.
7 + 3 + 1 = __
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 222.1

Answer:

7 + 3 + 1 = 11

Addition sentences :

  1. 7   + 4 = 11
  2. 10 + 1 = 11

Question 4.
3 + 6 + 3 = __
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 223

Answer:

3 + 6 + 3 = 12

  1.  6 + 6 = 12
  2.  3 + 9 = 12

GO DEEPER
Solve both ways.

Question 5.
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 224

Answer:

2 + 3 + 7 = 12

  1. 2 + 10 = 12
  2. 5 +  7  = 1 2

Question 6.
THINK SMARTER
I used Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 226.1 to model 3 addends. Use my model. Write the 3 addends.
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 226

Answer:

Addition sentence :

2 + 2 + 3 = 7

Problem Solving • Applications

Question 7.
THINK SMARTER
Choose three numbers from 1 to 6. Write the numbers in an addition sentence. Show two ways to find the sum.
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 227

Answer:

6 = 4 + 2

6 = 5 + 1

Question 8.
THINK SMARTER
Write each addition sentence in the box that shows the sum.
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 228

Answer:

The addition sentences shows the sum 12 are :

1. 2 + 2 + 8 = 12

2.  6 + 0 + 6 = 12

The addition sentences shows the sum 13 are :

  1.  3 + 5 + 5 = 13
  2.  4 + 4 +5 = 13

TAKE HOME ACTIVITY
• Have your child draw to show two ways to add the numbers 2, 4, and 6.

Algebra • Add 3 Numbers Homework & Practice 3.10

Look at the Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 229. Complete the addition sentences showing two ways to find the sum.

Question 1.
5 + 4 + 2 = __
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 230

Answer: 11

Explanation:

The two addition sentences  to find the sum are

  1.  5 + 6 = 11
  2.  9 + 2 = 11

Problem Solving

Question 2.
Choose three numbers from 1 to 6. Write the numbers in an addition sentence. Show two ways to find the sum.

Question 3.
WRITE Math
Use pictures or words to explain how you can find the sum for 3 + 5 + 2.
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 231

Answer:

3 + 5 + 2 = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165  + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Lesson Check

Question 1.
What is the sum of 3 + 4 + 2? Write the sum.
__

Answer: 9

Explanation:

The sum of 3 + 4 + 2 = 9

Therefore, sum = 9

Question 2.
What is the sum of 5 + 1 + 4? Write the sum.
__

Answer: 10

Explanation:

The sum of 5 + 1 + 4 = 10

Therefore, sum = 10

Spiral Review

Question 3.
What is the sum of 3 and 7?
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 232

Answer: 10

Explanation:

The sum of 3 and 7:

3 + 7 =10

Question 4.
4 cows are in the barn. 2 more cows join them. How many cows are in the barn now?
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 233
Complete the model and the number sentence.
__ cows 4 + 2 = __

Answer:

Number sentence :

4 + 2 = 6

Therefore, the number of cows in the barn = 6

Lesson 3.11 Algebra • Add 3 Numbers

Essential Question
How can you group numbers to add three addends?

Listen and Draw

Listen to the problem. Show two ways to group and add the numbers.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 234

Math Talk
MATHEMATICAL PRACTICES

Apply Describe the two ways you grouped the numbers to add.

Share and Show MATH BOARD

Choose a strategy. Circle two addends to add first. Write the sum. Then find the total sum. Then use a different strategy and add again.

Question 1.
Go Math 1st Grade Answer Key Chapter 3 Addition Strategies 235

Answer:

Two addends : 6 , 4

Sum : 6 + 4 = 10

Total sum : 6 + 4 + 2 = 12

Question 2.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 236

Answer:

Two addends : 3 , 4

Sum : 3 + 4 = 7

Total sum : 3 + 4 + 4 = 11

Question 3.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 237

Answer:

Two addends : 2 , 5

Sum : 2 + 5 = 7

Total sum : 2 + 5 + 0 = 7

Question 4.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 238

Answer:

Two addends : 5 , 5

Sum : 5 + 5 = 10

Total sum : 5 + 4 + 5 = 14

On Your Own

MATHEMATICAL PRACTICE
Use Repeated Reasoning Choose a strategy. Circle two addends to add first. Write the sum.

Question 5.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 239

Answer:

Two addends : 8 , 2

Sum : 8 + 2 + 2 =12

Question 6.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 240

Answer :

Two addends : 6, 0

Sum : 6 + 0 + 8 = 14

Question 7.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 241

Answer :

Two addends : 3 , 4

Sum : 3 + 4 + 6 = 13

Question 8.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 242

Answer:

Two addends : 2 , 3

Sum : 2 + 3 + 7 = 12

Question 9.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 243

Answer:

Two addends : 7 , 7

Sum : 7 + 7 + 2 = 16

Question 10.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 244

Answer:

Two addends : 1 , 1

Sum : 1 + 9 + 1 = 11

Question 11.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 245

Answer:

Two addends : 4 , 4

Sum : 5 + 4 + 4 = 13

Question 12.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 246

Answer :

Two addends : 5 , 5

Sum: 5 + 5+ 5 = 15

Question 13.
THINK SMARTER
Susan has 7 shells. Kai has 3 shells. Zach has 5 shells. How many shells do they have?
__ + __ + __ + __ = __ shells

Answer: 15

Explanation :

The number of shells Susan has = 7

The number of Shells Kai have = 3

The number of shells Zach have = 5

Total : 7 + 3 + 5 = 15

Therefore, the number of  shells they have = 7+3+5 = 15

THINK SMARTER
Write the missing addends. Add.

Question 14.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 247

Answer:

The possible addend be 0,8 ; 1,7; 2,6; 3,5; 4,4 ; 5,3 ; 6,2 ; 7,1 ; 8,0

The sum of the number sentence is 11

Question 15.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 248

Answer:

Possible addends are : 0,7; 1,6; 2,5; 3,4; 4,3; 5,2; 2,6; 1,7; 7,0.

The sum of number sentence = 10

Problem Solving • Applications

Draw a picture. Write the number sentence.

Question 16.
Maria has 3 cats. Jim has 2 cats. Cheryl has 5 cats. How many cats do they have?
__ + __ + __ = __ cats
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 249

Answer: 10

Explanation:

The number of cats Maria have = 3

The number of cats Jim have = 2

The number of cats Cheryl have = 5

Total : 3 + 2 + 5= 10

Therefore, the total number of cats Maria have = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Question 17.
Tony sees 5 small turtles. He sees 0 medium turtles. He sees 4 big turtles. How many turtles does he see?
__ + __ + __ = __ turtles
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 250

Answer:  9

Explanation:

The number of small turtles = 5

The number of medium turtles = 0

The number of big turtles = 4

Total : 5 + 0 + 4 = 9

Therefore, the number of turtles tony have = 9

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + 0 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165.

Question 18.
GO DEEPER
Kathy sees 13 fish in the tank. 6 fish are gold. The rest are blue or red. How many of each could she see?
__ + __ + 6 = 13 fish
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 251

Answer:

The sum of  blue and red fish is 7

Total number of fishes in the tank = 13

Number of gold fishes = 6

Let, the number of blue fishes = 3

the number of red fishes = 4

Total red and blue fishes = 3 + 4 = 7

Number sentence : 3 + 4 + 6 = 14

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165                               Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Question 19.
THINK SMARTER
Write two ways to group and add 2 + 3 + 4.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 252

Answer:

2 + 3 + 4 = 9

  1. 6 + 3 = 9
  2. 2 + 7 = 9

TAKE HOME ACTIVITY
• Have your child look at Exercise 18. Have your child tell you how he or she decided which numbers to use. Have him or her tell you two new numbers that would work.

Algebra • Add 3 Numbers Homework & practice 3.11

Choose a strategy. Circle two addends to add first. Write the sum.

Question 1.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 253

Answer:

Two addends : 7, 3

Sum : 7 + 3 + 3 = 13

Question 2.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 254

Two addends : 2 , 2

Sum : 2 + 2+ 6 = 10

Question 3.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 255

Answer:

Two addends : 6 , 6

Sum : 6 + 6 +3 = 15

Question 4.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 256

Answer:

Two addends : 0 , 8

Total : 2 + 0 + 8 = 10

Problem Solving

Draw a picture. Write the number sentence.

Question 5.
Don has 4 black dogs. Tim has 3 small dogs. Sue has 3 big dogs. How many dogs do they have?
__ + __ + __ = __ dogs

Answer: 10

Explanation :

The number of dogs Don have = 4

The number of dogs Tim have = 3

The number of dogs Sue have = 3

Total : 4 + 3 +3 = 10

Therefore, the number of dogs they have = 10

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 +Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 165

Question 6.
WRITE Math
Use pictures or words to explain how you would find 6 + 4 + 4.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 257

Answer :

6 + 4 + 4 = 14

Lesson Check

Question 1.
What is the sum of 4 + 4 + 2?
__

Answer:

4 + 4 + 2 = 10

Question 2.
Circle two addends to add first. Find the sum. Explain your strategy.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 258
_______________
_______________

Answer:

7 + 3 = 10

Total sum : 7 + 3 + 2 = 12

Spiral Review

Question 3.
Write a doubles plus one fact for the sum of 7.
__ + __ = __

Answer:

A doubles plus one fact for the sum of 7 :

3 + 4 = 7

3 + 3 + 1 = 7

Question 4.
What addition sentence does this model show? Write the number sentence.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 259

__ + __ = __

Answer:

Number sentence  :

8 + 3 = 11

Lesson 3.12 Problem Solving • Use Addition Strategies

Essential Question
How do you solve addition word problems by drawing a picture?

Megan put 8 fish in the tank. Tess put in 2 more fish. Then Bob put in 3 more fish. How many fish are in the tank now?

Unlock the Problem

What do I need to find?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 260
What information do I need to use?
Megan put in 8 fish.
Tess put in 2 fish.
Bob put in 3 fish.

Show how to solve the problem.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 261
__ fish

Answer :

Explanation :

Given,

Megan put in 8 fish.
Tess put in 2 fish.
Bob put in 3 fish.

Total : 8 + 2 + 3 = 13

Therefore, total number of fishes = 13

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1   + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1 + Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1 = Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1                                                              Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1

Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1Go Math Grade 1 Chapter 3 Answer Key Pdf Addition Strategies 173.1

 

HOME CONNECTION
• Your child will continue to use this chart throughout the year to help him or her unlock the problem. In this lesson, your child used the strategy draw a picture to solve problems.

Try Another Problem

Draw a picture to solve.

Question 1.
Mark has 9 green toy cars. He has 1 yellow toy car. He also has 5 blue toy cars. How many toy cars does he have?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 262

Answer : 15

Explanation :

The number of green toy cars = 9

The number of yellow toy cars = 1

The number of blue toy cars = 5

Total :

9 + 1 + 5 = 15

Therefore, the total number of toy cars he have now = 15

Math Talk
MATHEMATICAL PRACTICES
Reasoning Explain how using make a ten helps you solve the problem.

Share and Show MATH BOARD

MATHEMATICAL PRACTICE

Write an Equation Draw a picture to solve.

Question 2.
Ken put 5 marbles in a jar. Lou put in 0 marbles. Mae put in 5 marbles. How many marbles did they put in the jar?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 263

Answer: 10

Explanation :

The number of marbles Ken kept in a jar = 5

The number of Lou kept in a jar = 0

The number of marbles Mae kept in jar = 5

Total : 5 + 0 + 5 = 10 marbles

Therefore, there are 10 marbles in the jar.

Question 3.
Ava has 3 kites. Lexi has 3 kites. Fred has 5 kites. How many kites do they have?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 264

Answer : 11

Explanation :

The number of kites Ava have = 3

The number of kites Lexi have = 3

The number of kites Fred have = 5

Total : 3 + 3 + 5 = 11 Kites

Therefore, the number of kites they have = 11

 

Question 4.
Al got 8 books at the library. Ryan got 7 books. Dee got 1 book. How many books do they have?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 265

Answer : 16

Explanation :

The number of books Al got from library = 8

The number of books Ryan got from library = 7

The number of books Dee got from library = 1

Total : 8 + 7 + 1 = 16 books

Therefore, the number of books they have = 16

Question 5.
Pete sends 4 letters. Then he sends 3 more letters. Then he sends 2 more letters. How many letters did Pete send?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 266

Answer: 9

Explanation :

The number of letters Pete sends = 4

The number of letters he sends more = 3

Then also the number of letters he sends more = 2

Total : 4 + 3 + 2 = 9 letters

Therefore, the number of letters Pete sends = 9

On Your Own WRITE Math

Solve. Draw or write to show your work.

Question 6.
Kevin has 15 baseball cards. He gives away 8 baseball cards. How many baseball cards does he have?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 267
__ baseball cards

Answer : 7

Explanation :

Total number of basketball cards = 15

The number of cards he gives away = 8

Now, the number basketball he have now =

15 – 8 = 7

Therefore, the number of cars he have now = 7

 

Question 7.
There are 14 pencils. Haley has 6 pencils. Mac has 4 pencils. Sid has some pencils. How many pencils does Sid have?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 268
___ pencils

Answer: 4

Explanation :

Total number of pencils = 14

The number of pencils Haley have = 6

The number of pencils Mac have = 4

Total : 6 + 4 = 10

The number of pencils Sid have = 14 – 10 = 4

Therefore, Sid have 4 pencils.

 

Question 8.
GO DEEPER
12 marbles are in a bag. Shelly takes 3 marbles. Dan puts in 4. How many marbles are in the bag now?
__ marbles

Answer:

Explanation :

The total number of marbles in a bag = 12

The number of marbles Shelly takes out =

1 2 – 3 = 9

The number of marbles Dan puts in = 4

9 + 4 = 13

Therefore, total number of marbles in the bag = 13

Question 9.
THINK SMARTER
Eric has 4 pencils. Sandy gives Eric 3 pencils. Tracy gives Eric 5 more pencils. How many pencils does Eric have in all?
Eric has ☐ pencils in all.

Answer: 12

Explanation :

The number of pencils Eric have = 4

The umber of pencils Sandy gives Eric  = 3

The number of pencils Tracy gives Eric  = 5

Total : 4 + 3 + 5 = 12

Therefore, the total number pencils Eric have = 12.

 

TAKE HOME ACTIVITY
• Ask your child to look at Exercise 8 and tell how he or she found the answer.

Problem Solving • Use Addition Strategies Homework & Practice 3.12

Draw a picture to solve.

Question 1.
Franco has 5 crayons. He gets 8 more crayons. Then he gets 2 more crayons. How many crayons does he have now?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 269

Answer : 15

Explanation :

The number of crayons Franco have = 5

The number of crayons he gets more = 8

The number of crayons he gets more = 2

Total number of crayons = 5 + 8 + 2 = 15

Therefore, the number of crayons he have now = 15

Question 2.
Jackson has 6 blocks. He gets 5 more blocks. Then he gets 3 more blocks. How many blocks does he have now?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 270

Answer : 14

The number of blocks Jackson have = 6

The number of blocks Jackson gets more = 5 + 3 = 8

Total : 6 + 8 = 14

Therefore, Jackson have 14 blocks.

Question 3.
Ava has 7 gifts. Then he gets 2 more gifts. Then he gets 3 more gifts. How many gifts does Ava have now?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 271

Answer : 12

Explanation :

The number of gifts Ava have = 7

The number of gifts Ava gets more = 2 + 3 = 5

Total : 7 + 5 = 12

Therefore, Ava have 12 gifts.

Question 4.
WRITE Math
Draw a picture to show how you would solve this problem.
Jeb has 4 large rocks. He has 4 medium rocks. He has 7 small rocks. How many rocks does Jeb have?
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 272

Answer: 15

Explanation :

The number of large rocks Jeb have = 4

The number of medium rocks Jeb have = 4

The number of small rocks Jeb have = 7

Total : 4 + 4 + 7 = 15

Therefore, Jeb have 15 rocks.

Lesson Check

Question 1.
Lila has 3 gray stones. She has 4 black stones. She also has 7 white stones. How many stones does she have? Write the number sentence.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 273

Answer:

The total number of stones = 14

Number sentence: 3 + 4 + 7 = 14

Explanation :

The number of gray stones = 3

The number of black stones = 4

The number of white stones = 7

Total :

3 + 4 + 7 = 14

 

Question 2.
Patrick has 3 red stickers, 6 pink stickers, and 8 green stickers. How many stickers does Patrick have? Write the number sentence.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 274

Answer:

The Total number of stickers = 17

Number sentence: 3 + 6 + 8 = 17

Explanation:

The number of red stickers = 3

The number of pink stickers = 6

The number of green stickers = 8

Total: 3 + 6 + 8 = 17

Spiral Review

Question 3.
What is the sum of 2 + 4 or 4 + 2? Write the number.
____

Answer: 6

Explanation:

The sum of of 2+4 or 4+2 = 6

Question 4.
There are 6 black pens. There are 3 blue pens. How many pens are there? Write the number.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 275

Answer:

The total number of pens = 9

Number sentence : 6 + 3 =9

Explanation:

The number of black pens = 6

The number of blue pens = 3

Total : 6 + 3 = 9

 

Addition Strategies Chapter 3 Review/Test

Question 1.
Write the addends in a different order.
5 + 4 = 9
__ + __ = __

Answer:

The addition sentence with the addends in a different order =

4 + 5 = 9

Question 2.
Count on from 4. Write the number that shows 1 more.
__

Answer: 5

Question 3.
The cubes show a doubles fact. Choose the doubles fact and the sum.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 276

Answer:

The addition sentence with doubles fact :

5 + 5 = 10

Therefore, The sum of cubes = 10

Question 4.
THINK SMARTER
There are 3 red leaves. There are 4 yellow leaves. How many leaves in all? Use a double to add. Write the missing numbers.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 277

Answer:

3 + 4 = 3 + 3 +1

So, 3 + 4 = 7

Question 5.
Choose all the doubles facts that can help you solve 8 + 7.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 278

Answer:

  1.   7 + 7 = 14
  2.   8 + 8 = 14

Question 6.
GO DEEPER
Write a count on 2 fact to show a sum of 10. Then write a doubles fact to show a sum of 10.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 279

Answer:

Addition sentence on 2 fact to show a sum of 10

8 + 2 = 10

Addition sentence with doubles fact to show a sum of 10.

5 + 5 = 10

Question 7.
Match the models to the number sentences.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 280

Answer :

Question 8.
The model shows 8 + 5 = 13.
Write the 10 fact that has the same sum.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 281

Answer: 10 + 3 = 13

Explanation:

Given,

The model shows 8 + 5 = 13.

Addition sentence with 10 fact that has same sum =

10 + 3 = 13

Question 9.
Does the addition show how to make a ten to add? Choose Yes or No.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 282

Answer:

1.  7 + 3 + 2 –  Yes

2.   7 + 5 +5  –  Yes

3.  5 + 4 + 7 –   No

Question 10.
Look at the 1st Grade Go Math Answer Key Chapter 3 Addition Strategies 283. Complete the addition sentence to show the sum. Choose the missing number and the sum.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 283.1

Answer:

2 + 3 + 4 = 9

Explanation :

The number of red cubes  = 2

The number of blue cubes = 3

The number of yellow cubes = 3

Now, The addition sentence :

2 + 3 + 4 = 9

Total sum of cubes = 9

 

Question 11.
Write two ways to group and add 4 + 2 + 5.
__ + __ = __
__ + __ = __

Answer:

4 + 2 + 5 = 11

The two ways are:

6 + 5 = 11

8 + 3 = 11

Question 12.
Beth sees 4 red birds. She sees 2 yellow birds. She sees 4 blue birds. Draw a picture of the birds.
1st Grade Go Math Answer Key Chapter 3 Addition Strategies 284

Answer:

4 + 2 + 4 = 10

Beth sees = 10 birds

 

 

 

 

 

Go Math Grade 2 Answer Key Chapter 7 Money and Time

Go Math Grade 2 Chapter 7 Answer Key Money and Time Concepts

Money and Time Show What You Know

Order Numbers to 100 on a Number Line
Write the number that is just before, between, or just after.
Question 1.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.1
Answer:

57

Question 2.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.2
Answer:

25

Skip Count by Fives and Tens
Question 3.
Count by fives. Write how many in all.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.3
_________ _____ paints in all
Answer:

5,10,15,20,25

5+5+5+5+5= 25

So, there are 25 paints in all

Question 4.
Count by tens. Write how many in all.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.4
________ ____ paints in all
Answer:

10, 20, 30, 40, 50

10+10+10+10+10=50

So, there are 50 paints in all

Time to the Hour
Write the time shown on the clock.
Question 5.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.5
Answer: 11’o clock

Explanation:

The hours hand is on 11 and the minutes hand is on  12.

So, the time is 11:00

Question 6.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.6
Answer: 5’o clock

Explanation :

The hours hand is on the 5 and the minutes hand is on 12.

So, the time is 5:00

 

Money and Time Vocabulary Builder

2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.7
Visualize It
Fill in the graphic organizer.
Show ways to count on.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 1.8

Answer:

Understand Vocabulary
Write the missing numbers in each counting pattern.
Question 1.
Count by ones. 40, ____, ____, ____, 44, ___, 46, ____
Answer:

40, 41, 42, 43, 44, 45, 46, 47

Question 2.
Count by fives. 10, 15, ___, ___, ___, 35, ___, ____
Answer:

10, 15, 20, 25, 30, 35, 40, 45.

Question 3.
Count by tens. 20, ___, ___, 50, ___, ____, 80, ___
Answer:

20, 30, 40, 50, 60, 70, 80, 90.

Money and Time Game 5 and 10 Count

Materials
2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.1
Play with a partner.

  1. Spin the pointer on 2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.2 for your starting number. Put your cube on that number.
  2. Spin the pointer. Count on by that number two times.
  3. Take turns. The first player to get to 100 wins. Play again.
    2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.3

Money and Time Vocabulary Cards

2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.4
2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.5

Money and Time Vocabulary Game

2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.6

DIRECTIONS 2 to 4 players. Take turns to play.

  • To take a turn, toss the numbered cube. Move that many spaces
  • Follow the directions for the space you land.
  • First player to reach FINISH wins.

MATERIALS 1 connecting cube per player

  • 1 number cube
  • 1 set of clue cards
    2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.7
    2nd Grade Go Math Answer Key Chapter 7 Money and Time 2.8

The Write Way
Reflect
Choose one idea. Write about it in the space below.

  • Write and draw to explain the following amount as if you were talking to a young child. Use another sheet of paper for your drawing.
    $1.36
  • What time is it now? Use at least three of these terms in your answer.
    A.M. midnight minute noon P.M. quarter past
    Write at least three things you know about money.
    _________________________
    _________________________

Lesson 7.1 Dimes, Nickels, and Pennies

Essential Question How do you find the total value of a group of dimes, nickels, and pennies?

Listen and Draw

Sort the coins. Then draw the coins.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 1

Answer:

– Dime = 10 cents

– Nickle = 5 cents

– penny = 1 cent

 

MATHEMATICAL PRACTICES
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 2
A nickel has the same value as how many pennies? Explain.
Answer:5

1 nickel = 5 cents

1 penny = 1 cents

Which means

1+1+1+1+1 = 5

5 cents = 5 pennies

So, A nickel has the same value a 5 pennies.

Share and Show

Count on to find the total value.
Question 1.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 3
Answer: 26¢

Explanation:

1 nickel = 5 cents

Which means, 4 nickels = 5+5+5+5= 20 cents

5 cents and 1 cent

Total value of coins = 20+5+1=26 ¢

Question 2.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 4
Answer:35¢

Explanation:

1 dime = 10 cents

Which means , 2 dimes = 20 cents

1 nickel = 5 cents

Which means, 2 nickels = 10cents

and 5 cents

Therefore, Total value of coins = 20+10+5=35¢

On Your Own

2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 5
Count on to find the total value.
Question 3.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 6
Answer:60¢

Explanation:

1 dime = 10 cents

Which means , 6 dimes = 10+10+10+10+10+10=60¢

Therefore, the total value is 60¢.

Question 4.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 7
Answer: 30¢

Explanation:

1 nickel = 5 cents

Which means 5 nickels = 5+5+5+5+5= 25¢

and 5 cents

Total = 25¢+5¢

Therefore, the total value is 30¢

 

Question 5.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 8
Answer:45¢

Explanation:

1dime = 10 cents

Which means , 3 dimes = 30 cents

1 nickels = 5 cents

Which means , 2 nickels = 10 cents

and 5 cents

Total : 30+10+5 = 45

Therefore, the total value = 45¢

Question 6.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 9
Answer: 18¢

Explanation:

1 nickel = 5 cents

Which means , 2 nickels = 10cents

3 pennies = 3 cents and 5 cents

Total : 10+3+5 = 18

Therefore, the total value is 18¢

Question 7.
THINK SMARTER
Maggie had 5 nickels. She gave 2 nickels to her sister. What is the total value of the nickels that Maggie has now?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 10
Answer: 3 nickels

Explanation:

Given , Maggie had 5 nickels

The number of nickels she gave to her sister =2

Leftover nickel coins = 5-2 = 3

1 nickel = 5 cents

Which means , 3 nickels = 5+5+5= 15

The total value of the nickels that Maggie has now =15¢

Problem Solving • Applications

Solve. Write or draw to explain.
Question 8.
MATHEMATICAL PRACTICE Analyze
Jackson has 4 pennies and 3 dimes. How much money does Jackson have?
Answer: 34¢

Explanation :

4 pennies = 4 cents

1 dime = 10 cents

Which means , 3 dimes = 30 cents

Total : 4 +30 = 34

The total money Jackson have = 34¢

Question 9.
MATHEMATICAL PRACTICE
Use Models Draw two ways to show 25¢. You can use dimes, nickels, and pennies.
Answer:

25cents can be shown as

2 dimes , 1 nickel

1 dime = 10 cents

Which means , 2 dimes = 10+10= 20

and 1 nickel = 5 cents

Total : 20+5 = 25 ¢

Another way :

25 cents can be shown as 1 dime 2 nickel and 5 pennies

1 dime = 10 cents

1 nickel = 5 cents

Which means , 2 nickels= 5+5= 10

5 pennies = 5 cents

Total value : 10+10+5 = 25

Question 10.
THINK SMARTER
Sue has 40¢. Circle coins to show this amount.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 11
Answer:

 

TAKE HOME ACTIVITY • Draw pictures of five coins, using dimes, nickels, and pennies. Ask your child to find the total value.

Dimes, Nickels, and Pennies Homework & Practice 7.1

Count on to find the total value.
Question 1.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 12
Answer: 24 cents

Explanation :

1 dime = 10 cents

Which means , 2 dimes = 20 cents

and 4 cents

Total : 20+4 = 24

Therefore, the total value = 24cents

Question 2.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 13
Answer: 13 cents

Explanation :

2 five cents and 3 cents

5+5+3 = 13 cents

Therefore, the total value of coins = 13 cents

Question 3.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 14
Answer:28 cents

Explanation:

1 dime = 10 cents

Which means, 2 dimes = 20 cents

1 nickel = 5 cents

and 3 cents

Total : 20+5+3 = 28

Therefore, the total value of coins is 28 cents

Problem Solving
Solve. Write or draw to explain.
Question 4.
Aaron has 5 dimes and 2 nickels. How much money does Aaron have?
________
Answer:60 cents

Explanation :

1 dime = 10 cents

Which means, 5 dimes = 50 cents

1 nickel = 5 cents

Which means , 2 nickels = 10 cents

Total : 50+10 = 60 cents

Therefore, the total value of coins = 60 cents

 

Question 5.
WRITE
Draw three dimes, 1 nickel, and 2 pennies. Describe how to count on to find the total value of this group of coins.
_____________
Answer:37 cents

Explanation:

1 dime = 10 cents

Which means , 3 dimes = 10+10+10= 30 cents

1 nickel = 5 cents

and 2 pennies = 2 cents

Total value of coins = 30+5+2 = 37 cents

 

Lesson Check
Question 1.
What is the total value of this group of coins?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.1 15 ______
Answer:21 cents

Explanation :

1 dime = 10 cents

2 five cents = 5+5= 10

and 1 cent

Total value of coins = 10+10+1 = 21 ¢

Spiral Review
Question 2.
Hayden is building toy cars. Each car needs 4 wheels. How many wheels will Hayden use to build 3 toy cars?
______ wheels
Answer:

Total number of toy cars = 3

Given that, each car needs 4 wheels

Now, The number of wheels Hayden use to 3 toy cars =

4+4+4 = 12

Therefore , Hayden will use 12 wheels to build 3 toy cars.

Question 3.
What is the value of the underlined digit?
429 _____
Answer:

Four hundred twenty nine

The underlined digit is 4

The value of 4 is hundred

Question 4.
Lillian is counting by fives. What numbers did she say next?
40, ___, ____, ___, ____
Answer:

40, 45, 50, 55, 60.

Question 5.
Sophie has 12 grapes in her lunch bag. She shared 7 grapes with her sister. How many grapes does she have?
12 – 7 = ______
Answer:

Given , the number of grapes in her lunch bag = 12

The number grapes she shared =7

12 – 7 = 5

Therefore, the number of games she have = 5

Lesson 7.2 Quarters

Essential Question How do you find the total value of a group of coins?

Listen and Draw

Sort the coins. Then draw the coins.

MATHEMATICAL PRACTICES
Describe how the value of a quarter is greater than the value of a dime.
Answer:

1 quarter = 25 cents

1 dime = 10 cents

The value of a quarter is 15 cents greater than the value of a dime

Share and Show

Count on to find the total value.
Question 1.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 1
Answer: 50 cents

Explanation:

1 quarter = 25 cents

Which means , 2 quarters = 25+25= 50

Therefore , the total value of coins = 50 ¢

Question 2.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 2
Answer:86 cents

Explanation:

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75

1 dime = 10 cents and

1 penny = 1 cent

Total : 75+10+1 = 86

Therefore ,the total value of the coins = 86 ¢

 

Question 3.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 3
Answer: 57¢

Explanation:

1 quarter =  25 cents

Which means , 2 quarters = 25 +25 = 50

1 nickel = 5 cents and

2 pennies = 2 cents

Total : 50+5+2 = 57

Therefore, the total value of the coins = 57¢

On Your Own

Count on to find the total value.
Question 4.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 4

Answer: 100 ¢

Explanation:

1 quarter = 25 cents

4quarters = 25+25+25+25= 100

Therefore, the total value of coins = 100¢

Question 5.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 5
Answer: 65¢

Explanation :

1 quarter = 25 cents

1 dime = 10 cents

Which mean , 4 dimes = 10+10+10+10= 40

Total : 25+40= 65

Therefore, the total value of the coins = 65¢

Question 6.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 6
Answer: 70¢

Explanation :

1 quarter = 25 cents

Which means , 2 quarters = 25+25 = 50

1 nickel = 5 cents

Which means , 4 nickels = 5+5+5+5+ = 20

Total : 50+20= 70

Therefore, the total value of the coins = 70¢

Question 7.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 7
Answer: 38¢

Explanation :

1 quarter = 25 cents

1 nickel = 5 cents

Which means , 2 nickels= 5+5+ = 10

3 pennies = 3 cents

Total : 25+10+3 = 38

Therefore, the value of coins = 38¢

Draw and label a coin to solve.
Question 8.
THINK SMARTER
Ed’s coin has the same value as a group of 5 pennies and 4 nickels. What is his coin?
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 8
Answer: Quarter coin

Explanation :

1 penny = 1 cent

Which means , 5 pennies = 5 cents

1 nickel = 5 cents

Which means, 4 nickels = 5+5+5+5= 20

Total : 5 +20

We know that , 1 quarter = 25 cents

Therefore, His coin is a quarter coin .

Problem Solving • Applications

MATHEMATICAL PRACTICE Make Connections
Read the clue. Choose the name of a coin from the box to answer the question.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 9
Question 9.
I have the same value as 5 pennies.
What coin am I?
________
Answer:

5 pennies =5 cents

We know that , 1 nickel = 5 cents

So, I am a nickel coin.

Question 10.
I have the same value as 25 pennies.
What coin am I?
_________
Answer:

1 penny = 1 cent

25 pennies = 2 cents

We know that , 1 quarter = 25 cents

Therefore , I am a quarter coin .

Question 11.
I have the same value as 2 nickels.
What coin am I?
_________
Answer:

1 nickel = 5 cents

Which means 2 nickels = 5+5 = 10

We know that 1 dime = 10 cents

Therefore, I am a dime coin.

Question 12.
I have the same value as a group of 5 nickels.
What coin am I?
_________
Answer: Quarter

1 nickel = 5 cents

Which means, 5 nickels = 5+5+5+5+5 =25 cents

1 quarter = 25 cents

So , I am a quarter coin .

Question 13.
THINK SMARTER
Tom gives these coins to his brother.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 10
Answer: 65 cents

Explanation:

1 quarter = 25 cents

Which means , 2 quarters = 25+25 = 50

3 dimes = 5+5+5= 15

Total: 50+15 = 65 cents

Therefore, tom gives his brother 65 cents

TAKE HOME ACTIVITY • Have your child draw two quarters, two dimes, and two nickels, and then find the total value.

Quarters Homework & Practice 7.2

Count on to find the total value.
Question 1.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 11
Answer:55¢

Explanation:

1 quarter = 25 cents

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

1 nickel = 5 cents

Which means, 2 nickels = 5+5 =10

Total :

25+20+10 = 55

Therefore, the value of coins = 55¢

Question 2.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 12
Answer: 66¢

Explanation :

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 nickel = 5 cents

Which means, 3 nickels = 5+5+5=15

1 penny = 1 cent

Total: 50+15+1=66

Therefore, the total value of coins = 66¢

Problem Solving
Read the clue. Choose the name of a coin from the box to answer the question.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 13
Question 3.
I have the same value as a group of 2 dimes and 1 nickel. What coin am I?
_________
Answer: Quarter

Explanation:

1 dime = 10 cents

Which means , 2 dimes = 20 cents

1 nickel = 5 cents

Total value of coins = 20 +5 = 25 cents

We know that , 1 quarter = 25 cents

So , I am a quarter coin.

Question 4.
WRITE
Draw 1 quarter, 1 dime, and 4 pennies. Describe how to count to find the total value of this group of coins.
Answer: 39 cents

Explanation :

1 quarter = 25 cents

1 dime = 10 cents

4 pennies = 4 cents

Total value of coins = 25+10+4 = 39

Lesson Check
Question 1.
What is the total value of this group of coins?
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.2 14
_________
Answer: 61¢

Explanation:

1 quarter = 25 cents

Which means, 2 quarters = 25+25 =50

1 nickel = 5 cents

Which means, 2 nickels = 5+5= 10

1 penny = 1 cent

Total : 50+10+1

Therefore, the total value of coins = 61¢

Spiral Review
Question 2.
Circle the odd number.
8 14 17 22
Answer:

8 , 14 are the odd numbers

Question 3.
Kai scored 4 points and Gail scored 7 points. How many points did they score altogether?
4 + 7 = ______ points
Answer:

Given,

The points scored by kai = 4

The points scored by Gail = 7

The points that they score altogether = 4+7 = 11 points.

Question 4.
There were 382 chairs in the music hall. Write a number greater than 382.
_______
Answer: 383

383 is greater than 382.

Question 5.
Write the number 61 using words.
________
Answer: sixty one

Lesson 7.3 Count Collections

Essential Question How do you order coins to help find the total value of a group of coins?

Listen and Draw

Line up the coins from greatest value to least value. Then draw the coins in that order.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 1

>>>

Quarter> dime> nickel > penny

MATHEMATICAL PRACTICES
Describe how the values of the different kinds of coins compare.
Answer:

The values of the coins are compared with cents

1 quarter = 25 cents

1 dime = 10 cents

1 nickel= 5 cents and

1 penny = 1 cent

 

Model and Draw

Order the coins from greatest value to least value.
Then find the total value.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 2
Answer:

1 quarter = 25 cents

Which means, 2 quarters = 25 +25 = 50

1 dime = 10 cents

and 2 pennies = 2 cents

Total: 50 + 10 +2 = 32

Therefore, the total value of coins 62¢

Share and Show

Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 3
Draw and label the coins from greatest to least value. Find the total value.
Question 1.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 4
_______
Answer:

1 quarter = 25 cents

1 nickel = 5 cents

Which means , 2 nickels = 5+5 = 10

2 pennies = 2 cents

Total : 25 +10+2 = 37

Therefore, the total value of the coins = 37 ¢

Question 2.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 5
_______
Answer:

1 quarter = 25 cents

1 dime = 10 cents

Which means , 2 dimes = 20 cents

1 nickel = 5 cents

and 1 penny = 1 cent

Total : 25+20+5+1 = 51

Therefore ,  the total value of coins = 51¢

Question 3.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 6
_______
Answer:

1quarter = 25 cents

Which means ,2 quarters = 25+25 = 50

1 dime = 10 cents and

1 nickel = 5 cents

Total: 50+10+5 =65

Therefore the total value of coins = 65¢

On Your Own

Draw and label the coins from greatest to least value. Find the total value.
Question 4.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 7
______
Answer:

1 quarter = 25 cents

1 dime = 10 cents

Which means , 3 dimes = 10+10+10= 30

Total :25+30 = 55

Therefore, the total value of coins  55¢

Question 5.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 8
_______
Answer:

1 quarter = 25 cents

Which means ,2 quarters = 25+25 = 50 cents

1 dime = 10 cents

and 2 pennies

Total : 50+10+2 = 62

Therefore, the total value of coins = 62¢

Question 6.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 9
_______
Answer:

1 quarter = 25 cents

1 dime = 10 cents

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

Total : 25+10+10 =45 cents

Therefore, the total value of coins =45¢

Question 7.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 10
_______
Answer:

1 quarter = 25 cents

Which means, 2 quarters = 25+25= 50

1 dime = 10 cents

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

Total : 50+10+10 = 70

Therefore, the total value of the coins = 70¢

Question 8.
GO DEEPER
Andy has only quarters and nickels. The total value of his coins is 75¢. What coins could Andy have?
______ quarters _____ nickels
Answer:

75¢ can be shown by using only quarters and nickels as 2 quarters 5 nickels

Explanation :

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50 cents

1 nickel = 5 cents

Which means , 5 nickels = 5+5+5+5+5= 25

Total : 50+25 = 75 ¢

Therefore, Andy has 2 quarters and 5 nickels .

Problem Solving • Applications

Solve. Write or draw to explain.
Question 9.
THINK SMARTER
Paulo had these coins.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 11
He spent 1 quarter. How much money does he have now?
______
Answer: 21¢

Explanation :

Total value of coins = 46¢

After spending a quarter

1 quarter = 25 cents

1 nickel = 5 cents

Which means, 4 nickels = 5+5+5+5= 20 cents

and 1 penny = 1 cent

46 – 25 = 21¢

Therefore, the leftover money = 21¢

Question 10.
Rachel has 2 quarters, 3 dimes, and 1 nickel in her bank. How much money is in Rachel’s bank?
_______
Answer: 85¢

Explanation:;

1 quarter = 25 cents

Which means 2 quarters =25+25 =50

1 dime = 10 cents

Which means, 3 dimes = 10+10+10 = 30

1 nickel = 5 cents

Total : 50+30+5 = 85¢

Therefore, the amount Rachels have in his bank = 85¢

Question 11.
GO DEEPER
Blake has only nickels and dimes. He has double the number of nickels as dimes. The total value of his coins is 60¢. What coins does Blake have?
______ nickels _____ dimes
Answer: 6 nickels and 3 dimes

Explanation :

Given , Rachel has double the number of nickels and dimes

1 nickel = 5 cents

Which means, 6 nickels = 5+5+5+5+5+5= 30 cents

1 dime = 10 cents

Which means , 3 dimes = 10+10+10 = 30 cents

Total : 30+30= 60¢

Therefore , Rachel has 3 dimes and 6 nickels .

Question 12.
THINK SMARTER
Malik has these coins in his pocket. What is the total value of the coins?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 12
Answer: 50¢

Explanation :

1 dime = 10 cents

Which means, 4 dimes = 10+10+10+10 = 40 cents

1 nickel = 5 cents

Which means , 2 nickels = 5+5 = 10 cents

Total : 40+10 = 50

Therefore , the total value of coins Malik had = 50¢

TAKE HOME ACTIVITY • Have your child draw and label coins with a total value of 32¢.

Count Collections Homework & Practice 7.3

Draw and label the coins from greatest to least value. Find the total value.
Question 1.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 13
________
Answer: 22¢

1 dime = 10 cents

1 nickel = 5 cents Which means, 2 nickels = 5+5 = 10 cents and

2 pennies = 2 cents

Total : 10+10+2 = 22

Therefore, the total value f the coins = 22¢

 

Question 2.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 14
_________
Answer: 75¢

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

and 1 nickel = 5 cents

Total : 50+20+5 = 75

Therefore, total value of the coins = 75¢

Problem Solving
Solve. Write or draw to explain.
Question 3.
Rebecca has these coins. She spends 1 quarter. How much money does she have left?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 15
Answer:

Firstly, Rebecca has 2 quarters , 1 dime, 1 nickel and 1 penny

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

1 nickel = 5 cents and

1 penny = 1 cent

Total : 50+10+5+1 = 66¢

Given that,

she spends 1 quarter which means, 25 cents

Now, 66¢ – 25¢ = 41¢

Therefore, the leftover money is 41¢

Question 4.
WRITE
Draw 2 dimes, 1 nickel, and 2 quarters. Describe how to order and then count to find the total value of the coins.
_______
Answer:

Explanation:

We have to order coins on the basis of cent values

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50 cents

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20 cents

and 1 nickel = 5 cents

Total : 50+20+5 = 75¢

Therefore, the total value of the coins = 75¢

Lesson Check
Question 1.
What is the total value of this group of coins?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 16
_______
Answer:

Explanation:

1 quarter = 25 cents

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20 cents

and 2 pennies = 2 cents

Total : 25 + 20 +2 = 47

Therefore, the total value of the coins is 47¢

 

Spiral Review
Question 2.
What number is 100 more than 562?
_____
Answer:

The number 100 more than 562 is 662

Question 3.
Describe 58 as a sum of tens and ones.
______
Answer: 58 can be shown in sum of tens and once as

50 + 8  = 58

Question 4.
Pete helps his grandmother gather pecans. He finds 6 pecans on his left and 3 on his right. How many pecans did Pete find altogether?
6 + 3 = _____ pecans
Answer:

The number of pecans Pete finds on his left side = 6

The number of Pecans Pete finds on his right side = 3

Total number of Pecans Pete find altogether =

6 + 3 = 9

Therefore, total number of pecans = 9

Question 5.
What number do the blocks show?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.3 17
______
Answer:

The first and second figures have 100 blocks

The 3 single line contains 10+10+10 = 30 blocks

and 4 blocks

Total number of blocks = 100+100+10+10+10+1+1+1+1 = 234

Therefore, Total number of blocks = 234

Lesson 7.4 Show Amounts in Two Ways

Essential Question How do you choose coins to show a money amount in different ways?

Listen and Draw

Show the amount with coins. Draw the coins.
Write the amount.
Answer: 26¢

1 quarter = 25 cents and 1 penny = 1 cent

Total value = 26¢

Another way:

26¢ can also be shown as 2 dimes 1 nickel and 1 penny

1 dime = 10 cents

Which means , 2 dimes = 10+10 = 20

1 nickel = 5 cents

and 1 penny = 1 cent

Total : 20+1 = 26 ¢

 

MATHEMATICAL PRACTICES
Can you show 10¢ with 3 coins? Explain how you know.
Answer:

No, it is not possible to show 10¢ with 3 coins because

Share and Show

Use coins. Show the amount in two ways.
Draw and label the coins.
Question 1.
61¢
Answer:

61 ¢ can be shown as 2 quarters 1 dime and 1 penny

1 quarter = 25 cents

Which means, 2 quarters = 25+25= 50

1  dime = 10 cents

and 1 penny = 1 cent

Total : 50+10+1= 61¢

Another way:

61¢ can also be shown as 6 dimes and 1 penny

1 dime = 10 cents

and 6 dimes = 10+10+10+10+10+10 = 60

and 1 penny = 1 cent

Total : 60+1 = 61¢

Question 2.
36¢
Answer:

36 cents be shown as 3 dimes 1 nickel and 1 penny

1 dime = 10 cents

Which means, 3 dimes = 10+10+10

1 nickel = 5 cents

and 1 penny = 1 cent

Total : 30+5+1 = 36¢

Another way :

36¢ can also be shown as 1 quarter 1 dime and 1 penny

1 quarter = 25 cents

1 dime = 10 cents and 1 penny = 1 cent

Total : 25+10+1 = 36¢

 

On Your Own

Use coins. Show the amount in two ways.
Draw and label the coins.
Question 3.
55¢
Answer:

55 cents can be shown as 2 quarters and 1 nickel

1 quarter = 25 cents

Which means, 25+25 = 50 cents

and 1 nickel = 5 cents

Total : 50+5 = 55 cents

Another way:

55 cents can also be shown as 5 dimes and 1 nickel

1 dime = 10 cents

Which means, 5 dimes = 10+10+10+10+10 = 50

and 1 nickel = 5 cents

Total : 50+5 = 55

Question 4.
90¢
Answer:

90 cents can be shown as 3 quarters 1 dime and 1 nickel

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75

1 dime = 10 cents

and 1 nickel = 5 cents

Total : 75+10+5 = 90

Another way:

90 can also be shown as 2 quarters and 4 dimes

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

and 4 dimes = 10+10+10+10 = 40

Total: 50+40 = 90 cents

Question 5.
75¢
Answer:

75¢ can be shown as 3 quarters

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75 cents

Another way :

75 cents can also be shown as 2 quarters 2 dimes and 1 nickel

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50 cents

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

and 1 nickel = 5 cents

Total: 50+20+5 = 75 cents

 

Question 6.
THINK SMARTER
Teresa has 42¢. She has no dimes. Draw to show what coins she might have.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.4 1
Answer:

42¢ can be shown as 1 quarter 3 nickels and 2 pennies

1 quarter = 25 cents

1 nickel = 5 cents

Which means , 3 nickels = 15 cents

2 pennies = 2 cents

Total value of coins = 25 +15+ 2 = 42

Problem Solving • Applications

MATHEMATICAL PRACTICE Model with Mathematics
Use coins to solve.
Question 7.
Lee buys a pen for 50¢. Draw coins to show two different ways to pay 50¢.
Answer:

50 cents can be shown as:

2 quarters

1 quarter = 25cents

Which means , 2 quarters = 25 +25 = 50 ¢

Another way :

50 cents can be shown as 5 dimes

1 dime = 10 cents

Which means ,

5 dimes = 10+10+10+10+10 = 50

Question 8.
MATHEMATICAL PRACTICE Make Sense of Problems
Delia used 4 coins to buy a book for 40¢. Draw coins to show two ways to pay 40¢ with 4 coins.
Answer:

40¢ can be shown as 4 dimes

1 dime = 10 cents

Which means ,

4 dimes = 10+10+10+10 = 40

Another way :

40 cents can also be shown in 4 coins as 1 quarter 3 nickel

1 quarter = 25 cents

1 nickel = 5 cents

Which means ,

3 nickels = 15 cents

Total value = 25 +15 = 40 cents

Question 9.
THINK SMARTER
Fill in the bubble next to all the groups of coins with a total value of 30¢.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.4 2
Answer:

2 nickels and 2 dimes

1 nickel = 5 cents

Which means , 2 nickels = 5+5 = 10cents

1 dime =10 cents

Which means , 2 dimes = 10+10 = 20 cents

Total value = 10 +20 = 30 ¢.

 

TAKE HOME ACTIVITY • With your child, take turns drawing different collections of coins to show 57¢.

Show Amounts in Two Ways Homework & Practice 7.4

Use coins. Show the amounts in two ways.
Draw and label the coins.
Question 1.
39¢
Answer:

39 cents be shown as 1 quarter 1 dime 4 pennies .

1 quarter = 25 cents

1 dime = 10 cents

4 pennies = 4 cents

Total value = 25+10+4 = 39 cents

 

Another way :

39 cents can be shown as 3 dimes , 1nickel and 4 pennies

1 dime = 10 cents

Which means ,

3 dimes = 10 +10+10= 30

1 nickel = 5 cents

4 pennies = 4 cents

Total value = 30+5+4 = 39 cents

Question 2.
70¢
Answer:

70 cents can be shown as 2 quarters and 2 dimes

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

Total : 50+20= 70 cents

Another way:

70 cents can also be shown as 1 quarter 4 dimes and 1 nickel

1 quarter = 25 cents

1 dime = 10 cents

Which means, 4 dimes = 10+10+10+10 = 40 cents

and 1 nickel = 5 cents

Total : 25 + 40 + 5 = 70 cents

Problem Solving
Question 3.
Madeline uses fewer than 5 coins to pay 60¢. Draw coins to show one way she could pay 60¢.
Answer:

60¢ can be shown in fewer than 5 coins 2 quarters and 1 dime

1 quarter = 25 cents

Which means, 2 quarters = 25+25  50

and 1 dime = 10 cents

Total : 50 + 10 = 60¢

Question 4.
WRITE
Draw coins in two ways to show 57¢. Describe how to choose the coins for each way.
Answer:

57 ¢ can be shown as 2 quarters 1 nickel and  2 pennies

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 nickel = 5 cents

and 2 pennies = 2 cents

Total : 50+5+2 = 57 ¢

Another way :

57 cents can also be shown as 5 dimes, 1 nickel and 2 pennies

1 dime = 10 cents

Which means, 5 dimes = 10+10+10+10+10 = 50

1 nickel = 5 cents and

2 pennies = 2 cents

Lesson Check
Question 1.
Circle the group of coins that has the same total value.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.4 3

 

Answer:

 

 

Spiral Review
Question 2.
Write the number 31 as a sum of tens and ones.
2 tens ____ ones
Answer:

The number 31 can be shown as a sum of tens and once as

3 tens and 1 once or 30+1 = 31

Question 3.
Write 13 tens as a sum of hundreds and tens.
_____ hundreds ______ tens

Answer: 13 x 10 = 130

can be written as 100+30

1 hundreds and 3 tens

Question 4.
What is the value of the underlined digit?
28 _____
Answer: The value of underlined digit is once

The underlined digit in once place is 8

Question 5.
Baylie’s softball team scored 5 runs in the first inning and 6 runs in the second inning. How many runs did her team score?
5 + 6 = ______ runs
Answer:

5+6 = 11

Therefore, the total number of runs her team score = 11

Lesson 7.5 One Dollar

Essential Question How can you show the value of one dollar with coins?

Listen and Draw

Draw the coins. Write the total value.
________
Answer:

$1 is equal to 100 cents

100¢ can be shown as 3 quarters 2 dimes and 1 nickel

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75 ¢

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20¢

and 1 nickel = 5 cents

Total : 75+20+5 = 100¢

Therefore, 3 quarters , 2 dimes and 1 nickel have the value of $1.00

MATHEMATICAL PRACTICES
How many pennies have the same value as 80¢? Explain.
Answer: 80

Explanation:

1 penny = 1 cent

So, 80 pennies = 80¢

Therefore, 80 pennies have the same value as 80¢

Share and Show

Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 1
Draw the coins to show $1.00. Write the total value.
Question 1.
nickels
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 2
Answer: 20

Explanation :

$1 = 100 cents

1 nickel = 5 cents

20 nickels = 100 cents

Question 2.
quarters
________
Answer: 4

Explanation :

$1 = 100 cents

1 quarter = 25 cents

Which means ,

4 quarters = 25+25+25+25 = 100

Question 3.
dimes
________
Answer: 10

Explanation:

$1 = 100 cents

1 dime = 10 cents

Which means , 10 dimes = 100 cents

On Your Own

Circle coins to make $1.00.
Cross out the coins you do not use.
Question 4.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 3
Answer:

Question 5.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 4
Answer:

Question 6.
GO DEEPER
Warren shows $1.00 using only two kinds of coins. Draw and label coins he could use.
Answer:

$1 = 100 cents

100 cents can be shown in two kinds of coins as 3 quarters and 5 nickels

1 quarter =25 cents

Which means , 3 quarters = 25+25+25 = 75

1 nickel = 5 cents

Which means ,5 nickels = 5+5+5+5+5= 25

Total value = 75+25 = 100

 

Question 7.
THINK SMARTER
Sara has these coins.
Draw more coins to show $1.00.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 5
Answer:

The given coins are 1 quarter , 1 dime and 1 nickel

1 quarter = 25 cents

1 dime = 10 cents

and 1 nickel = 5 cents

Total : 25+10+5 = 35

$1 is equals to 100 cents

100 cents – 35 cents = 65 cents

We need 65 cents more to make $1

65 cents can be shown as 2 quarters  1 dime and 1 nickel

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

and 1 nickel = 5 cents

Total : 50+10+5 = 65

Therefore, we need 2 quarters  1 dime and 1 nickel more to make $1

TAKE HOME ACTIVITY • Have your child draw a group of coins to show $1.00.

One Dollar Homework & Practice 7.5

Circle coins to make $1.00.
Cross out the coins you do not use.
Question 1.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 6
Answer:

 

Question 2.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 7
Answer:

Problem Solving
Question 3.
Draw more coins to show $1.00 in all.

Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 8
Answer:

$ 1 = 100 cents

2 quarters = 25+25= 50 cents

We need 50¢ more to make $1.00

50¢ can be shown as  5 dimes

Question 4.
WRITE
Draw coins to show one way to make $1.00 using only nickels and quarters.
Answer:

$1 = 100 cents

$1 can be shown by using only nickels and quarters =3 quarters and 5 nickels

1 quarter is equal to 25 cents

Which means , 3 quarters = 25+25+25+= 75

1 nickel= 5 cents

Which means ,  5 nickels = 5+5+5+5+5= 25

Total value = 75+25= 100

 

Lesson Check
Question 1.
Which group of coins has a value of $1.00?
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 9
Answer:

$1 = 100 cents

1 quarter = 25 cents

Which means , 4 quarters = 25+25+25+25= 100 cents

Spiral Review
Question 2.
Write 692 using words.
________
Answer:

692= Six hundred ninety two

Question 3.
Keith ate 7 almonds, then ate 7 more. Is the total number of almonds even or odd?
7 + 7 = ______ almonds
_______
Answer:

7+7 = 14

Therefore , the total number of almonds is even

Question 4.
What is the total value of 1 quarter and 3 nickels?
_________
Answer: 40 cents

Explanation:

1 quarter = 25 cents

1 nickel = 5 cents

Which means , 3 nickels = 5+5+5= 15

Total value = 25+ 15= 40 cents

Therefore, the value of 1 quarter and 3 nickels = 40 cents.

Question 5.
Kristin is counting by tens. What numbers does she say next?
230, _____, _____, ______
Answer:

230, 240, 250, 260.

Money and Time Mid-Chapter Checkpoint

Concepts and Skills

Count on to find the total value.
Question 1.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 10
Answer: 28¢

Explanation:

1 dime= 10 cents

Which means, 2 dimes = 10+10= 20

1 nickel = 5 cents and

3 pennies = 3 cents

Total :20+5+3

Therefore, the total value of the coins = 28¢

 

Question 2.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 11
Answer: 76¢

Explanation:

1 quarter = 25 cents

Which means , 2 quarters = 25+25= 50

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

1 nickel = 5 cents and

1 penny = 1 cent

Total : 50+20+5+1 = 76

Therefore, the total value of the coins = 76¢

Use coins. Show the amount in two ways.
Draw and label the coins.
Question 3.
31¢
Answer:

31cents can be shown as 1 quarter, 1 nickel and 1 penny

1 quarter = 25 cents

1 nickel = 5 cents and

1 penny =1 cent

Total value = 25+5+1=31

Another way :

31 cents can be shown as 3 dimes , 1 penny

1 dime = 10 cents

Which means, 3 dimes = 10+10+10=30

and 1 penny = 1 cent

Total : 30 =1 = 31 cents

Question 4.
THINK SMARTER
Mary used these coins to buy a folder. What is the total value of these coins?
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.5 12
Answer: 60¢

Explanation:

1 quarter = 25 cents

Which means , 2 quarters = 25+25=50

1 nickel= 5 cents

Which means, 2 nickels= 5+5 = 10

Total : 50+10 =60

Therefore, the total value of the money = 60¢

 

Lesson 7.6 Amounts Greater Than $1

Essential Question How do you show money amounts greater than one dollar?

Listen and Draw

Draw and label the coins.
Write the total value.
_______
total value
Answer:

MATHEMATICAL PRACTICES
Use Repeated Reasoning
Explain how you found the total value of the coins in the coin bank.
Answer:

Share and Show

Circle the money that makes $1.00. Then
write the total value of the money shown.
Question 1.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 1
__________
Answer:

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 nickel = 5 cents ,

1 penny = 1 cent

and one $1 bill

Total value : $1+50+5+1 = $1.56

Question 2.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 2
________
Answer:

1 quarter = 25 cents

Total value : 4 quarters = 25+25+25+25 = 100

1 dime = 10 cents

Which means, 3 dimes = 10+10+10 = 30

Total : 100+30 = 130

Therefore, total value of money = 130

On Your Own

Circle the money that makes $1.00. Then write the total value of the money shown.
Question 3.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 3
_______
Answer:

The given coins are 4 quarters 1 dime and 1 nickel

1 quarter = 25 cents

Which means, 4 quarters = 25+25+25+25 = 100

1 dime = 10 dime

and 1 nickel = 5 cents

Total : 100+10+5 = $1.15

Therefore, the total value of coins = $1.15

Question 4.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 4
Answer:

$1 bill= 100 cents

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50 cents

1 dime = 10 cents

Which means, 3 dimes = 10+10+10 = 30

1 nickel = 5 cents and

1 penny = 1 cents

Total value : 100+ 50+30+5+1 = $1.86

Question 5.

2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 5
_______
Answer:

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75

1 dime = 10 cents

Which means, 3 dimes = 10+10+10 = 30

and 1 nickel = 5 cents

Total : 75 + 30 + 5 $1.10

Therefore, the total amount of money = $1.10

Question 6.
THINK SMARTER
Martin used 3 quarters and 7 dimes to pay for a kite. How much money did he use?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 6
_______
Answer:

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75

1 dime = 10 cents

Which means, 7 dimes= 10+10+10+10+10+10+10 = 70

Total value = 75+70 = 145 cent

Therefore, the total value of money = $1.45

Problem Solving • Applications

Question 7.
GO DEEPER
Pam has fewer than 9 coins. The coins have a total value of $1.15. What coins could she have? Draw the coins. Then write a list of her coins.
Answer:

2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 3

$1.15 can be shown in fewer than 9 coins as

4 quarters , 1 dime and 1 nickel

1 quarter = 25 cents

Which means, 4 quarters = 25+25+25+25 = 100

1 dime = 10 dime

and 1 nickel = 5 cents

Total : 100+10+5 = $1.15

Therefore, the total value of coins = $1.15

 

Question 8.
THINK SMARTER+
Jason put this money in his bank.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 7
Answer:

Jason put a total of $1.35 in his bank

TAKE HOME ACTIVITY • With your child, take turns drawing coins or a $1 bill and coins with a total value of $1.23.

Amounts Greater Than $1 Homework & Practice 7.6

Circle the money that makes $1.00. Then write the total value of the money shown.
Question 1.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 8
_________
Answer:

The total value of the money shown is

1 quarter = 25 cents

1 $1 bill , 1 dime = 10 cents

and 1 nickel = 5 cents

Total value : $1 +25+10+5 = $1.40

Question 2.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 9
__________
Answer:

The total value of money shown :

1 quarter = 25 cents

Which means, 25+25+25 = 75

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

and 1 cent

Total : 75+20+10+1 = 166  cents or $1.16

Problem Solving
Solve. Write or draw to explain.
Question 3.
Grace found 3 quarters, 3 dimes, and 1 nickel in her pocket. How much money did she find?
________
Answer:

$1.10

Explanation:

1 quarter= 25 cents

Which means, 3 quarters = 75 cents

1 dime = 10 cents

Which means, 3 dimes = 30 cents

and 1 nickel = 5 cents

Total :75+30+5=110

So the total value of the money = $1.10

Question 4.
WRITE
Write about how to use the dollar sign and decimal point to show the total value of 5 quarters.
Answer:

1 quarter = 25 cents

Which means, 5 quarters = 25+25+25+25+25 = 125 cents

125 cents can be written as $1.25

Lesson Check
Question 1.
Julie has this money in her bank. What is the total value of this money?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 10
________
Answer:

1 quarter = 25 cents

1 dime = 10 cents and

one $1 bill

Total value : $1 +25+10 = $1.35

Therefore, total value in Julie bank = $1.35

Spiral Review
Question 2.
There are 79 squash plants and 42 pepper plants in Julia’s garden. How many vegetable plants are in Julia’s garden altogether?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 11
Answer:

7 9

+  4 2

__________

= 1 2 1

Therefore , the total number of plants in Julia’s garden are 121

Question 3.
What is the difference?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.6 12
Answer:

6 1

–  2 7

_________________

3 4

 

Question 4.
What number is 100 less than 694?
_________
Answer:

594

Question 5.
Write an addition fact that has the same sum as 6 + 5.
10 + ______
Answer:

10+1 =11

Lesson 7.7 Problem Solving • Money

Essential Question How does acting it out help when solving problems about money?

Kendra gave 2 dimes, 2 nickels, 1 quarter, and two $1 bills to her sister. How much money did Kendra give her sister?

Unlock the Problem
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.7 1

Show how to solve the problem.
Draw to show the money that Kendra used.
Kendra gave her sister _________
Answer:

1 dime = 10 cents

Which means , 2 dimes = 10+10 = 20

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

1 quarter = 25 cents

and two $1 bills

Total value of money = $1+$1 +20+10+25 = $2.55

Therefore, Kendra gave her sister = $2.25

HOME CONNECTION • Your child used play money to act out the problem. Representing problems with materials can be a useful strategy for children to use to solve problems.

Try Another Problem

Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.7 2
Use play coins and bills to solve.
Draw to show what you did.
Question 1.
Jacob has two $1 bills, 2 dimes, and 3 pennies in his pocket. How much money does Jacob have in his pocket?
_________
Answer:

two $1 bills = $1+$1 = $2

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

and 3 pennies = 3 cents

Total value : $2+20+3 = $2.23

Question 2.
Amber used 2 quarters, 1 nickel, 1 dime, and three $1 bills to buy a toy. How much money did Amber use to buy the toy?
__________
Answer:

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 nickel = 5 cents

1 dime = 10 cents and

three $1 bills = $1+$1+$1 = $3

Total value : 50+5+10+$3 = $3.65

MATHEMATICAL PRACTICES
Explain how you found the amount of money in Jacob’s pocket.
Answer: The total value of the money is find in his pockets

Share and Show

Use play coins and bills to solve.
Draw to show what you did.
Question 3.
Val used 3 quarters, 2 nickels, 2 pennies, and one $1 bill to buy a book. How much money did Val use to buy the book?
__________
Answer:

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75 cents

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

2 pennies = 2 cents

and $1 bill

Total value of money = 75+10+2 +$1= $1.87

Therefore, the amount Val use to buy the book = $1.87

Question 4.
Derek has two $1 bills, 2 quarters, and 6 dimes. How much money does he have?
___________
Answer:

Two $1 bills = $1 +$1 = $2

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

Which means, 6 dimes = 10+10+10+10+10+10 = 60 cents

Total value = $2 +50+60 = $3.10

Question 5.
THINK SMARTER
Katy has 3 quarters, 2 nickels, 2 dimes, and 3 pennies. How many more pennies does she need to have $1.10?
Go Math 2nd Grade Answer Key Chapter 7 Money and Time 7.7 3
________ more pennies
Answer:

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

1 dime = 10 cents

Which means, 2 dimes = 10+10 = 20

and 3 pennies = 3 cents

Total value = 75+10+20+3 = $1.08

To make $1.10 we need two more pennies

Problem Solving • Applications

Question 6.
MATHEMATICAL PRACTICE Make Sense of Problems
Victor has some dollar bills, some quarters, and some nickels. Draw and label dollar bills, quarters, and nickels to show $2.25.
Answer:

one $1 bill ,

1 quarter = 25 cents

Which means , 4 quarters = 25+25+25+25 = 100

1 nickel = 5 cents

Which means, 5 nickels = 5+5+5+5+5= 25

Total value : $1+100+20 = $2.25

Question 7.
THINK SMARTER
Ross used 3 quarters, 4 dimes, 3 nickels, and 5 pennies to buy a card. How much money did Ross use to buy the card? Draw to show how you solve the problem.
Answer:

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 =75 cents

1 dime = 10 cents

Which means, 4 dimes = 10+10+10+10 = 40

1 nickel = 5 cents

Which means, 3 nickels = 5+5+5 =15

and 5 pennies = 5 cents

Total : 75+40+15+5 = 135 cents

Therefore, the total value of money Ross used to buy the card = $1.35

TAKE HOME ACTIVITY • Ask your child to explain how he or she solved one problem in this lesson.

Problem Solving • Money Homework & Practice 7.7

Use play coins and bills to solve.
Draw to show what you did.
Question 1.
Sara has 2 quarters, 1 nickel, and two $1 bills.
How much money does Sara have?
_________
Answer:

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 nickel = 5 cents

and two $1 bills = $1+$1 = $2

Total value : $1+$1 +50+5 = $2.55

Question 2.
Brad has one $1 bill, 4 dimes, and 2 nickels in his bank. How much money does Brad have in his bank?
__________
Answer:

$1 bill,

1 dime = 10 cents

Which means, 4 dimes = 10+10+10+10 = 40

and 1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

Total : $1+40+10 = $1.50

Question 3.
Mr. Morgan gives 1 quarter, 3 nickels, 4 pennies, and one $1 bill to the clerk. How much money does Mr. Morgan give the clerk?
__________
Answer:

1 quarter = 25 cents

1 nickel = 5 cents

Which means, 3 nickels = 5+5+5 15

4 pennies = 4 cents

and $1 bill

Total : $1 + 25+15+4 = $1.44

Question 4.
WRITE
Write or draw to explain how you would find the total value of two $1 bills and 3 quarters.
Answer:

Two $1 bills = $1+$1 = $2

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25 = 75

Total = $2 +75 = $2.75

Therefore, the total value of money =$2.75

Lesson Check
Question 1.
Lee has two $1 bills and 4 dimes. How much money does Lee have?
________
Answer:

4 dimes = 40 cents and $1 bill

The total value of money = $1.40

Question 2.
Dawn has 2 quarters, 1 nickel, and one $1 bill. How much money does Dawn have?
__________
Answer:

1 quarter = 25 cents

Which means, 2 quarters = 50 cents

1 nickel = 5 cents

and $1 bill

Total value of money is $1.55

Spiral Review
Question 3.
What is the value of the underlined digit?
56 ______
Answer:

The value of underlined digit 6 is once

Question 4.
Cecilia collected 342 pennies for her class’s penny drive. Marked collected 243 pennies. Use <, >, or = to compare. Who collected more?
342 ____ 243
________ collected more.
Answer:

342 > 243

Cecilia have collected more .

Question 5.
Brooke’s dog has 15 treats. Then he ate 8 of them. How many treats does he have left?
15 – 8 = _______
Answer:

15 – 8 = 7

Therefore, the number of treats he left = 7

Question 6.
What is the next number in this pattern?
225, 325, 425, 525, ______
Answer: 625

Lesson 7.8 Time to the Hour and Half Hour

Essential Question How do you tell time to the hour and half hour on a clock?

Listen and Draw

Draw the hour hand to show each time.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 1

             

2 hours 30 minutes                        2 hours 50 minutes

       

5 hours 5 minutes                                    7 hours 25 minutes

MATHEMATICAL PRACTICES
Communicate Describe where the hour hand points to show half past 4:00.
Answer:

The time half past 4:00 is 4: 30

The hours hand is between 4 and 5 nd the minutes hand is on 6 , Which means 30 minutes.

The time is 4:30

Share and Show

Look at the clock hands. Write the time.
Question 1.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 2
Answer:

The time is 1:00

As the hours hand on 1 and the minutes hand is on 12 .

So, the time is 1 :00

 

Question 2.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 3
Answer:

8 hours 30 minutes

As the hours hand is between 8 and 9 and the minutes hand is on 6 .Which means, 30 minutes

So, the time is 8 hours 30 minutes

 

Question 3.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 4
Answer:

3 hours 30 minutes

As the hours hand is between 3 and 4 and the minutes hand is on 6 .Which means, 30 minutes

So, the time is 3 hours 30 minutes

 

On Your Own

Look at the clock hands. Write the time.
Question 4.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 5
Answer: 9 hours 30 minutes

As the hours hand is between 9 and 10 and the minutes hand is on 6 .Which means, 30 minutes

So, the time is 9 hours 30 minutes

 

Question 5.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 6
Answer: 12 hours 30 minutes

Explanation:

The  hours hand is between 12 and 1 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 12 hours 30 minutes

 

Question 6.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 7
Answer: 6’o clock

Explanation:

The  hours hand is on 6 and the minutes hand is on 12

So, the time is 6 hours or 6:00

 

Question 7.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 8
Answer: 4’o clock

Explanation:

The  hours hand is on 4 and the minutes hand is on 12

So, the time is 4 hours or 4:00

Question 8.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 9

Answer: 11 hours 30 minutes

Explanation:

The  hours hand is between 11 and 12 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 11 hours 30 minutes

 

Question 9.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 10

Answer: 5 hours 30 minutes

Explanation:

The  hours hand is between 5 and 6 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 5 hours 30 minutes

 

Question 10.
THINK SMARTER
Look at the time. Draw the hour hand and the minute hand to show the same time.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 11
Answer:

7:30                         2:00                           11:00

Problem Solving • Applications

Question 11.
MATHEMATICAL PRACTICE Make Connections
Allie eats lunch when the hour hand points halfway between the 11 and the 12, and the minute hand points to the 6. When does Allie eat lunch? Show the time on both clocks.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 12
How do you know what time to write in the digital clock? Explain.
_______________________
_________________________
Answer:

The  hour hand points halfway between the 11 and the 12, and the minute hand points to the 6.

The time Allie eat lunch is 11:30

Question 12.
THINK SMARTER
Match the clocks that show the same time.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 13
Answer:

TAKE HOME ACTIVITY • Have your child describe what he or she knows about a clockface.

Time to the Hour and Half Hour Homework & Practice 7.8

Look at the clock hands. Write the time.
Question 1.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 14
Answer:  3’o clock

Explanation:

The  hours hand is on 3 and the minutes hand is on 12

So, the time is 3 hours or 3:00

Question 2.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 15
Answer: 10 hours 30 minutes

Explanation:

The  hours hand is between 10 and 11 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 10 hours 30 minutes

 

Question 3.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 16
Answer: 4’o clock

Explanation:

The  hours hand is on 4 and the minutes hand is on 12

So, the time is 4 hours or 4:00

Problem Solving
Question 4.
Amy’s music lesson begins at 4:00. Draw hands on the clock to show this time.
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 17
Answer:

The hours hand is on 4 and the minutes hand is on 12

So, the time is 4:00

 

Question 5.
WRITE
Draw a clock to show the time as 2:30. Describe how you decided where the clock hands should point.
______________________
______________________
Answer:

The hours hand is between 2 and 3 and the minutes hand is on 6 .Which means, 30 minutes

So, the time is 2 hours 30 minutes .

Lesson Check
Question 1.
What is the time on this clock?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 18
_____
Answer: 3 hours 30 minutes

Explanation:

The  hours hand is between 3 and 4 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 3 hours 30 minutes

 

Question 2.
What is the time on this clock?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 19
______

Answer: 6’o clock

Explanation:

The  hours hand is on 6 and the minutes hand is on 12

So, the time is 6 hours or 6:00

 

Spiral Review
Question 3.
Rachel has one $1 bill, 3 quarters, and 2 pennies. How much money does Rachel have?
________
Answer: $1.77

Explanation:

1 $1 bill ,

1 quarter = 25 cents

Which means, 3 quarters = 25+25+25=75

and 2 pennies = 2 cents

Total $1 +75+2 = $1.77

Therefore, the total value of the money = $1.77

 

Question 4.
Write <, >, or = to compare 260 and 362.
260 _____ 362
Answer:

260 < 362

Question 5.
What number is shown with these blocks?
Go Math Answer Key Grade 2 Chapter 7 Money and Time 7.8 20
______
Answer:

Question 6.
Circle any even numbers.
1 3 4 5
Answer:

4 – Even number

Lesson 7.9 Time to 5 Minutes

Essential Question How do you tell and show time to five minutes?

Listen and Draw

Draw the hour hand and the minute hand to show the time.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 1
Answer:

MATHEMATICAL PRACTICES
Describe where the minute hand points to show half past the hour.
Answer: The minutes hand points to 6 Which means , 30 minutes or half past

Share and Show

Look at the clock hands. Write the time.
Question 1.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 2

Answer: 3 hours 25 minutes

Explanation:

The  hours hand is between 3 and 4 and the minutes hand is on 5 , which means, 25 minutes

So, the time is 3 hours 25 minutes

Question 2.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 3

Answer: 9 hours 5 minutes

Explanation:

The  hours hand is between 9 and 10 and the minutes hand is on 1 , which means, 5 minutes

So, the time is 9 hours 5 minutes

 

Question 3.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 4
Answer:

Answer: 11 hours 40 minutes

Explanation:

The  hours hand is between 11 and 40 and the minutes hand is on 8 , which means, 40 minutes

So, the time is 11 hours 40 minutes

 

Question 4.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 5
Answer:

Answer: 12 hours 50 minutes

Explanation:

The  hours hand is between 12 and 1 and the minutes hand is on 10 , which means, 50 minutes

So, the time is 12 hours 50 minutes

 

Question 5.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 6

Answer: 7 hours 15 minutes

Explanation:

The  hours hand is between 7 and 8 and the minutes hand is on 3 , which means, 15 minutes

So, the time is 7 hours 15 minutes

 

Question 6.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 7

Answer: 5 hours 55 minutes

Explanation:

The  hours hand is between 5 and 6 and the minutes hand is on 11 , which means, 55 minutes

So, the time is 5 hours 55 minutes

 

On Your Own

Look at the clock hands. Write the time.
Question 7.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 8

Answer: 6 hours 40 minutes or 6:40

Explanation:

The  hours hand is between 6 and 7 and the minutes hand is on 8 , which means, 40 minutes

So, the time is 6 hours 40 minutes

 

Question 8.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 9
Answer: 9 hours 20 minutes or 9: 20

Explanation:

The  hours hand is between 9 and 10 and the minutes hand is on 4 , which means, 20 minutes

So, the time is 9 hours 20 minutes

Question 9.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 10
Answer: 4 hours 30 minutes or 4:30

Explanation:

The  hours hand is between 4 and 5 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 4 hours 30 minutes

Question 10.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 11
Answer:

2 hours 45 minutes or 2: 45

Explanation:

The  hours hand is between 2 and 3 and the minutes hand is on 9 , which means, 45 minutes

So, the time is 2 hours 45 minutes

Question 11.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 12
Answer:

10 hours 10 minutes or 10 :10

Explanation:

The  hours hand is between 10 and 11 and the minutes hand is on 2 , which means, 10 minutes

So, the time is 10 hours 10 minutes

Question 12.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 13
Answer:

8 hours 35 minutes or 8:35

Explanation:

The  hours hand is between 8 and 9 and the minutes hand is on 7 , which means, 35 minutes

So, the time is 8 hours 35 minutes

MATHEMATICAL PRACTICE Use Models
Look at the time. Draw the minute hand to show the same time.
Question 13.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 14
Answer:

Question 14.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 15
Answer:

Question 15.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 16
Answer:

Problem Solving • Applications

Draw the clock hands to show the time.
Then write the time.
Question 16.
THINK SMARTER
My hour hand points between the 8 and the 9. In 35 minutes it will be the next hour. What time is it?
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 17
Answer:

The hours hand is between 8 and 9 and the minutes hand is on 5. Which means, 25 minutes

Also in 35 minutes , it will be 1 hour

So, the time is 8 hours 25 minutes

Question 17.
GO DEEPER
Mr. Brady fixes broken computers. Look at the start and finish times for his work on one computer. How many minutes did he work on the computer?
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 18
_______ minutes
Answer:

60 minutes

Explanation:

Given that , Mr. Brady fixes broken computers.

The time he starts  the work = 4:00

The time he finishes the work = 5: 00

The time difference between 1 hour is 60 minutes

Therefore, the time he took to fix the computer = 60 minutes

Question 18.

THINK SMARTER
Angel eats lunch at 12:45. Angel spent 10 minutes eating lunch. Draw the minute hand on the clock to show when Angel finished eating. Write the time.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 19
___ : ____
Answer:

Given that, Angel eats lunch at 12 :45

She spend 10 minutes to eat

10 minutes after 12 :45 means 12:55

Therefore, Angel completes her lunch at 12:55

TAKE HOME ACTIVITY • Have your child draw a large blank clock face and use two pencils as clock hands to show some different times.

Time to 5 Minutes Homework & Practice 7.9

Look at the clock hands. Write the time.
Question 1.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 20
Answer:

8 hours 15 minutes or 8: 15

Explanation:

The  hours hand is between 8 and 9 and the minutes hand is on 3 , which means, 15 minutes

So, the time is 8 hours 15 minutes

Question 2.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 21
Answer:

2 hours 40 minutes or 2:40

Explanation:

The  hours hand is between 2 and 3 and the minutes hand is on 8 , which means, 40 minutes

So, the time is 2 hours 40 minutes

Question 3.
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 22
Answer:

5 hours 55 minutes or 5: 55

Explanation:

The  hours hand is between 5 and 6 and the minutes hand is on 11 , which means, 55 minutes

So, the time is 5 hours 55 minutes

Problem Solving
Draw the minute hand to show the time.
Then write the time.
Question 4.
My hour hand points between the 4 and the 5. My minute hand points to the 9. What time do I show?
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 23
Answer:

The time is 4: 45

The hours hand is between 4 and 5 and the minutes hand is on 9 . Which means, 45 minutes

So, the time is 4 hours 45 minutes.

Question 5.
WRITE
Draw a clock showing 2:50. Explain how you know where the clock hands point.
Answer:

The clock shows the time 2 : 50

The hours hand is between 2 and 3 and the minutes hand is on 10 .Which means, 50 minutes

The time is 2 hours 50 minutes .

Lesson Check
Question 1.
What is the time on this clock?
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 24
Answer:

8 hours 5 minutes or 8: 05

Explanation:

The  hours hand is between 8 and 9 and the minutes hand is on 1 , which means, 5 minutes

So, the time is 8 hours 5 minutes

Question 2.
What is the time on this clock?
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 25
Answer:

7 hours 20 minutes or 7:20

Explanation:

The  hours hand is between 7 and 8 and the minutes hand is on 4 , which means, 20 minutes

So, the time is 7 hours 20 minutes

Spiral Review
Question 3.
What is the sum?
1 + 6 + 8 = _____
Answer: 15

Question 4.
Which number has the same value as 30 tens?
______
Answer: 300

Question 5.
Steven has 3 rows of toys. There are 4 toys in each row. How many toys are there?
_____ toys
Answer:

Given that , Steven has 3 rows of toys

There are 4 toys in each row

Total number of toys 4+4+4 = 12

Therefore, the total number of toys = 12

Question 6.
Jill has 14 buttons. She buys 8 more buttons. How many buttons does Jill have?
Go Math Grade 2 Answer Key Chapter 7 Money and Time 7.9 26
Answer:

Given,

The total number of buttons Jill have =  14

Also, she has 8 more buttons

Now, total buttons =

14 + 8 = 22

Therefore, the total number of buttons = 120

Lesson 7.10 Practice Telling Time

Essential Question What are the different ways you can read the time on a clock?

Listen and Draw

Write the times on the digital clocks.
Then label the clocks with the children’s names.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 1

  1. 11 hours 45 minutes – 11: 45
  2. 3 hours 25 minutes – 3 :25
  3. 7 hours 15 minutes – 7:15
  4. 6 hours 10 minutes – 6:10

MATHEMATICAL PRACTICES
Where would the minute hand point to show 15 minutes after the hour? Explain.
Answer:

The minutes hand should point on 3 to show 15 minutes after the hour.

Share and Show

Draw the minute hand to show the time. Write the time.
Question 1.
15 minutes after 1
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 2

Answer:

15 minutes after 1

The time is 1 hours 15 minutes

The hours hand is between 1 and 2 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 1 hours 15 minutes

 

Question 2.
half past 9
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 3
Answer:

Half past 9 means 30 minutes after 9

The time is 9 hours 30 minutes

The hours hand is between 9 and 10 and the minutes hand is on 6. Which means, 30 minutes

So, the time is 9 hours 30 minutes

Question 3.
quarter past 5
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 4
Answer:

Quarter past 5 means 15 minutes after 5

The time is 5 hours 15 minutes

The hours hand is between 5 and 6 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 5 hours 15 minutes

 

Question 4.
quarter past 10
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 5
Answer:

Quarter past 10 means 15 minutes after 10

The time is 10 hours 15 minutes

The hours hand is between 10 and 11 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 10 hours 15 minutes

 

Question 5.
40 minutes after 3
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 6
Answer:

40 minutes after 3 means, 3:40

The time is 3 hours 40 minutes

The hours hand is between 3 and 4 and the minutes hand is on 8. Which means, 40 minutes

So, the time is 3 hours 40 minutes

 

Question 6.
half past 7
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 7
Answer:

Half past 7 means 30 minutes after 7

The time is 7 hours 30 minutes

The hours hand is between 7 and 8 and the minutes hand is on 6. Which means, 30 minutes

So, the time is 7 hours 30 minutes

 

On Your Own

Draw the minute hand to show the time.
Write the time.
Question 7.
15 minutes after 11
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 8
Answer:

15 minutes after 11 means, 11:15

The time is 11 hours 15 minutes

The hours hand is between 11 and 12 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 11 hours 15 minutes

 

Question 8.
quarter past 4
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 9
Answer:

Quarter past 4 means 15 minutes after 4

The time is 4 hours 15 minutes

The hours hand is between 4 and 5 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 4 hours 15 minutes

 

Question 9.
25 minutes after 8
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 10
Answer:

25 minutes after 8

The time is 8 hours 25 minutes

The hours hand is between 8 and 9 and the minutes hand is on 5. Which means, 25 minutes

So, the time is 8 hours 25 minutes

 

Question 10.
10 minutes after 6
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 11
Answer:

10 minutes after 6 means 6: 10

The time is 6 hours 10 minutes

The hours hand is between 6 and 7 and the minutes hand is on 2. Which means, 10 minutes

So, the time is 6 hours 10 minutes

 

Question 11.
half past 2
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 12
Answer:

Half past 2 means 30 minutes after 2

The time is 2 hours 30 minutes

The hours hand is between 2 and 3 and the minutes hand is on 6. Which means, 30 minutes

So, the time is 2 hours 30 minutes

 

Question 12.
45 minutes after 3
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 13
Answer:

45 minutes after 4 means,  4: 45

The time is 4 hours 45 minutes

The hours hand is between 4 and 5 and the minutes hand is on 9. Which means, 45 minutes

So, the time is 4 hours 45 minutes

 

Question 13.
5 minutes after 7
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 14
Answer:

5 minutes after 7 means 7:05

The time is 7 hours 5 minutes

The hours hand is between7 and 8 and the minutes hand is on 1. Which means, 5 minutes

So, the time is 7 hours 5 minutes

 

Question 14.
30 minutes after 12
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 15
Answer:

30 minutes after 12

The time is 12 hours 30 minutes

The hours hand is between 12 and 1 and the minutes hand is on 6. Which means, 30 minutes

So, the time is 12 hours 30 minutes

Question 15.
quarter past 10
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 16
Answer:

Quarter past 10 means 15 minutes after 10

The time is 10 hours 15 minutes

The hours hand is between 10 and 11 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 10 hours 15 minutes

Problem Solving • Applications

Question 16.
THINK SMARTER
Lily eats lunch at quarter past 12. Meg eats lunch at 12:30. Katie eats lunch at 12:15. Which girls eat lunch at the same time?
_______ and _______
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 17
Answer:

Given that, Lily eats lunch at quarter past 12 which also means 30 minutes after 12, 12: 30

Meg eats lunch at 12: 30

and Katie eats lunch at 12: 15

Therefore, Lily and Meg eat lunch at the same time .

Question 17.
MATHEMATICAL PRACTICE
Explain Soccer practice starts at 4:30. Gabe arrives at soccer practice at 4:15. Does he arrive before or after practice starts? Explain.
____________________
____________________
Answer:

Given that, Soccer practice starts at 4: 30

and Gabi arrives practice at 4: 15

So, Gabi arrives 15 minutes before the soccer practice starts.

Question 18.
THINK SMARTER
What time is shown on the clock? Fill in the bubble next to all the ways to write or say the time.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 18
Answer:

The hours hand is between 3 and 4 and the minutes hand is on 5 . Which means, 25 minutes

So the time is 3 : 25 or 25 minutes after 3

  1. 3: 25
  2. 25 minutes after 3

 

TAKE HOME ACTIVITY • Name a time to 5 minutes. Ask your child to describe where the clock hands point at this time.

Practice Telling Time Homework & Practice 7.10

Draw the minute hand to show the time. Write the time.
Question 1.
quarter past 7
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 19
Answer:

 

Quarter past 7 means 15 minutes after 7

The time is 7 hours 15 minutes

The hours hand is between 7 and 8 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 7 hours 15 minutes

Question 2.
half past 3
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 20
Answer:

Half past 3 means 30 minutes after 3

The time is 3 hours 30minutes

The hours hand is between 3 and 4 and the minutes hand is on 6. Which means, 30 minutes

So, the time is 3 hours 30 minutes

Question 3.
50 minutes after 1
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 21
Answer:

50 minutes after 1 means  1 hour 50 minutes

The time is 1 hours 50minutes

The hours hand is between 1 and 2 and the minutes hand is on 10. Which means, 50 minutes

So, the time is 1 hours 50 minutes

Question 4.
quarter past 11
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 22

Answer:

 

Quarter past 11 means 15 minutes after 11

The time is 11 hours 15 minutes

The hours hand is between 11 and 12 and the minutes hand is on 3. Which means, 15 minutes

So, the time is 11 hours 15 minutes

 

Problem Solving
Draw hands on the clock to solve.
Question 5.
Josh got to school at half past 8. Show this time on the clock.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 23
Answer:

Half past 8 means, 30 minutes after 8

The time Josh got to school is 8 hours 30 minutes

As the hours hand is between 8 and 9 and the minutes hand is on 6. Which means, 30 minutes

So, the time is 8 hours 30 minutes

Question 6.
WRITE
Write the time 8:30. Then write this time in two other ways, using words.
______________
________________
Answer:

The time is

8:30 = Half past 8

8 :30 – 8 hours 30 minutes

Lesson Check
Question 1.
Write the time on this clock using words.
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 24
Answer: The time is six hours fifteen minutes or quarter past 6

Spiral Review
Question 2.
What is the value of this group of coins?
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 25
Answer: 46 cents

Explanation:

1 quarter = 25 cents

1 dime = 10 cents

Which means , 2 dimes = 10+10=20

1 penny = 1 cents

Total : 25+20+1 = 46 cents

Therefore, the total value of the coins = 46 cents

Question 3.
What time is shown on this clock?
Go Math Grade 2 Chapter 7 Answer Key Pdf Money and Time Concepts 7.10 26
________
Answer: 7:15

Explanation :

The hours hand is between 7 and the minutes hand is on 3 ,Which means 15 minutes

So ,the time is 7 hours 15 hours.

 

Question 4.
Write six hundred forty-seven using numbers.
______________
Answer:

647 – Six hundred forty-seven

Lesson 7.11 A.M. and P.M.

Essential Question How do you use A.M. and P.M. to describe times?

Listen and Draw

Draw the clock hands to show each time.
Then write each time.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 1
Answer:

The time is 7 hours 15 minutes a.m.

As the hours hand is between 7 and 8 and the minutes hand is on 3 Which means, 15 minutes

So, the time is 7 hours 15 minutes a.m.

The time is 3 hours 45 minutes p.m.

As the hours hand is between 3 and 4 and the minutes hand is on 9 .Which means, 45 minutes

So, the time is 3 hours 45 minutes .

MATHEMATICAL PRACTICES
Describe some activities that you do in both the morning and in the evening.
Answer:

The activities that we do in both the morning and in the evening are

  1. Reading
  2. Exercise/meditation
  3. Brushing teeth
  4. Taking bath

Share and Show

Write the time. Then circle a.m. or p.m.
Question 1.
eat breakfast
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 2
Answer:

7:15 a.m.

7 hours 15 minutes

Explanation:

The  hours hand is between 7 and 8 and the minutes hand is on 3 , which means, 15 minutes

So, the time is 7 hours 15 minutes

 

Question 2.
go to art class
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 3
Answer:

1:40 p.m.

1 hours 40 minutes

Explanation:

The  hours hand is between 1 and 2 and the minutes hand is on 8 , which means, 40 minutes

So, the time is 1 hours 40 minutes

 

Question 3.
do homework
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 4
Answer:

4:30 p.m.

4 hours 30 minutes

Explanation:

The  hours hand is between 4 and 5 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 4 hours 30 minutes

 

Question 4.
arrive at school
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 5
Answer:

8: 25 a.m.

8 hours 25 minutes

Explanation:

The  hours hand is between 8 and 9 and the minutes hand is on 5 , which means, 25 minutes

So, the time is 8 hours 25 minutes

 

On Your Own

Write the time. Then circle a.m. or p.m.
Question 5.
go to the library
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 6
Answer:

2:50 p.m.

2 hours 50 minutes

Explanation:

The  hours hand is between 2 and 3 and the minutes hand is on 10 , which means, 50 minutes

So, the time is 2 hours 50 minutes

 

Question 6.
go to science class
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 7
Answer:

10:30 a.m.

10 hours 30 minutes

Explanation:

The  hours hand is between 10 and 11 and the minutes hand is on 6 , which means, 30 minutes

So, the time is 10 hours 30 minutes

 

Question 7.
eat lunch
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 8
Answer:

11:45 a.m.

11 hours 45 minutes

Explanation:

The  hours hand is between 11 and 12 and the minutes hand is on 9 , which means, 45 minutes

So, the time is 11 hours 45 minutes

 

Question 8.
look at the moon
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 9
Answer:

8: 15 p.m.

8 hours 15 minutes

Explanation:

The  hours hand is between 8 and 9 and the minutes hand is on 3 , which means, 15 minutes

So, the time is 8 hours 15 minutes

 

Question 9.
THINK SMARTER
Use the times in the list to complete the story.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 10
Don got to school at _______. His class went to the library at _____. After school, Don read a book at _____.
Answer:

Don got to school at 8 : 30 a.m. His class went to the library at 10:15a.m.  After school, Don read a book at 3:20p.m.

Problem Solving • Applications

Question 10.
GO DEEPER
Some times are shown on this time line. Write a label for each dot that names something you do at school during that part of the day.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 11
At what times would you say the dots are placed on the time line?
_____ and _____
Answer:

8:00 a.m. – Yoga class

10:00 a.m. – Project work

Noon – Eat lunch

2:00 p.m. – Music class

4: 00 – Games

  1. 9: 00 a.m.
  2. 1:00 p.m.

Question 11.
THINK SMARTER+
The clock shows the time Jane goes to recess. Write the time. Then circle a.m. or p.m.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 12
Recess lasted one hour. Write the time recess was over. Write a.m. or p.m.
_______________
Answer:

The clock sows the time is 11: 30 a.m.

As the hours hand is between 11 and 12 and the minutes hand is on 6 Which means, 30 minutes

So , the time is 11 hours 30 minutes a.m.

Given that, Recess lasted one hour .

So, the time recess was over is 12 : 30 p.m.

TAKE HOME ACTIVITY • Name some activities and times. Have your child say a.m. or p.m. for the times.

A.M. and P.M. Homework & Practice 7.11

Write the time. Then circle A.M. or P.M.
Question 1.
walk the dog
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 13
Answer: 4 hours 40 minutes p.m.

Explanation:

The hours hand is between 4 and 5 and the minutes hand is on 8 .Which means 40 minutes

So the time is 4:40 p.m.

Question 2.
finish breakfast
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 14
Answer:

7 hours 30 minutes A.m.

Explanation :

The hours hand is between 7 and 8 and the minutes hand is on 6 . Which means , 30 minutes

So, the time is 7 : 30 a.m.

Problem Solving
Use the list of times. Complete the story.
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 15
Question 3.
Jess woke up at _____. She got on the bus at _____ and went to school. She left school at _______.
Answer:

Jess woke up at 7:00 A.m. She got on the bus at 8:30 A.m. and went to school . She left the school at 3:15 p.m.

Question 4.
WRITE
List two school activities that you do in the morning and two school activities that you do in the afternoon. Write times for these activities using A.M. and P.M.
___________________
__________________
Answer:

Morning activities:

a. Yoga class – 8:30 a.m.

b. Art class – 11:45 a.m.

Evening activities

a. Drawing time : 3:30 p.m.

b. Music class : 5:00 p.m.

Lesson Check
Question 1.
The clock shows when the soccer game ended. What time was it?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 16
_______
Answer:

4 hours 50 minutes

Explanation:

The hours hand is between 4 and 5 and the minutes hand is on 10 ,Which means 50 minutes

So, the time is 4 hours 50 minutes

Question 2.
The clock shows when Dad gets up for work. What time is it?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 17
_______
Answer:

6 hours 10 minutes

Explanation :

The hours hand is on 6 and the minutes hand is on 2 .Which means, 10 minutes

So, the time is 6 hours 10 minutes .

Spiral Review
Question 3.
What coin has the same value as 25 pennies? Draw your answer.
Answer:

Quarter

Explanation:

1 penny = 1 cent

Which means , 25 pennies = 25 cents

We know that 1 quarter = 25 cents

So, the coin is a quarter coin.

 

Question 4.
Describe 72 as a sum of tens and ones.
____ + ____
Answer:

72 can be shown in tens as 70 and once as 2

Total : 70 + 2 = 72

 

Question 5.
At the beginning of the school year there were 437 2nd graders at Woods Elementary. Over the course of the year,24 students joined. How many 2nd graders are there at the end of the year?
2nd Grade Go Math Answer Key Chapter 7 Money and Time 7.11 18
Answer:

4 3 7

+    2 4

_______________

4 6 1

Therefore, the number of  2nd graders are there at the end of the year = 461

Question 6.
What time is quarter past 3?
_______
Answer:

3 : 15

quarter past means 15 minutes after 3

So the time is 3 hours 15 minutes.

Money and Time Review/Test

Question 1.
Andrea pays $2.15 for a jump rope. Fill in the bubble next to all the ways that show $2.15.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 1
Answer:

  1. two $1 bills , 1 dime and 1 nickel = $2.15
  2. one $1 bill, 3 quarters and 4 dimes = $2.15

Question 2.
The clock shows the time Michael eats breakfast.
Write the time. Circle a.m. or p.m.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 2
Tell how you knew whether to select a.m. or p.m.
Answer: 7:10 a.m.

The time is 7 hours 10 minutes as the hours hand is on 7 and the minutes hand is on 2, Which means 10 minutes

And also breakfast is consumed in morning time

So, the time is 7: 10 a.m.

Question 3.
Fill in the bubble next to all the groups of coins with a total of 60¢.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 3
Answer:

  1. Yes, because 2 quarters = 50¢ , 1 dime= 10¢ ; Total = 60¢
  2. Yes, because 1 quarter = 25 ¢,2 dimes = 20¢ and 3 nickels = 15¢ ;Total = 60¢
  3. No, because 5 dimes = 50¢, 1 nickel = 5¢ 6 pennies = 6 cents ;Total = 61¢
  4. No, because 4 nickels = 20¢ , 20 pennies = 20 cents ;Total = 40¢

Question 4.
GO DEEPER
Tess gave Raul these coins. Tess says she gave Raul $1.00. Is Tess correct? Explain.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 4
_______________________
______________________
Answer:  No, she is wrong

Explanation:

$1 = 100 cents

The value of the coins she gave =

1 quarter = 25 cents

Which means, 2 quarters= 25+25 = 50

and 1 dime = 10 cents

Total :

50 + 10 = 60

So, the value of coins she gave to Raul = 60¢

Therefore, Tess is wrong .

Question 5.
Write the time that is shown on this clock.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 5
Answer: 8 hours 5 minutes

Explanation :

The hours hand is on 8 and the minutes hand is on 1 which means, 5 minutes

So, the time is 68 hours 5 minutes

Question 6.
What time is shown on the clock? Fill in the bubble next to all the ways to write or say the time.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 6
Answer: The time in the given clock is 4 hours 35 minutes

It can be shown as 4 : 35

It can also be written as 35 minutes past 4

Question 7.
THINK SMARTER+
Alicia has this money in her pocket.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 7
Answer: $1.70

Explanation:

1 dollar = 100 cents

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 dime = 10 cents

Which means , 2 dimes = 10+10 = 20

Total value of money = 100+50+20 = 170 cents

Is can be shown as $1 .70

Therefore, Alicia has a total of $1.70 in her pocket.

Question 8.
Kate’s father gave her these coins. Write the value of the coins. Explain how you found the the total value.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 8
_________________________
_________________________
Answer:31 ¢

Explanation:

1 quarter = 25 cents

Which means, 2 quarters = 25+25 = 50

1 nickel = 5 cents

Which means, 2 nickels = 5+5 = 10

and 1 penny = 1 cent

Total value of coins = 50+10+1 = 61¢

 

Question 9.
Write the times the clocks show.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 9
Answer:

  1. 4 hours 30 minutes
  2. 7 hours 30 minutes
  3. 5 ‘o clock

Question 10.
Ben has 30¢. Circle coins to show this amount.
Go Math 2nd Grade Answer Key Chapter 7 Money and Time rt 10
Answer:

Question 11.
Mia buys apples that costs 76¢.
Draw and label coins to show a total value of 76¢
Answer:

76¢ can be shown as 3 quarters and 1 penny

1 quarter = 25 cents

Which means , 3 quarters = 25+25+25 = 75

And 1 penny = 1 cents

Total : 75+1 = 76¢

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles

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Big Ideas Math Book Geometry Answer Key Chapter 5 Congruent Triangles

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Congruent Triangles Maintaining Mathematical Proficiency

Find the coordinates of the midpoint M of the segment with the given endpoints. Then find the distance between the two points.

Question 1.
P(- 4, 1) and Q(0, 7)
Answer:
The given points are:
P (-4, 1), Q (0, 7)
We know that,
The midpoint M of the segment with the 2 endpoints is:
( \(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\) )
Let the give points are:
(x1, y1) and (x2, y2)
So,
By comparing the given poits,
We will get
x1 = -4, x2 = 0, y1 = 1, y2 = 7
Hence,
The midpoint M = ( \(\frac{-4 + 0}{2}\), \(\frac{1 + 7}{2}\) )
= ( \(\frac{-4}{2}\), \(\frac{8}{2}\) )
= (-2, 4)
Hence, from the above,
We can conclude that the midpoint M of the segment with the given endpoints is: (-2, 4)

Question 2.
G(3, 6) and H(9, – 2)
Answer:
The given points are:
G (3, 6), H (9, -2)
We know that,
The midpoint M of the segment with the 2 endpoints is:
( \(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\) )
Let the give points are:
(x1, y1) and (x2, y2)
So,
By comparing the given poits,
We will get
x1 = 3, x2 = 9, y1 = 6, y2 = -2
Hence,
The midpoint M = ( \(\frac{3 + 9}{2}\), \(\frac{6 – 2}{2}\) )
= ( \(\frac{12}{2}\), \(\frac{4}{2}\) )
= (6, 2)
Hence, from the above,
We can conclude that the midpoint M of the segment with the given endpoints is: (6, 2)

Question 3.
U(- 1, – 2) and V(8, 0)
Answer:
The given points are:
U (-1, -2), V (8, 0)
We know that,
The midpoint M of the segment with the 2 endpoints is:
( \(\frac{x1 + x2}{2}\), \(\frac{y1 + y2}{2}\) )
Let the give points are:
(x1, y1) and (x2, y2)
So,
By comparing the given poits,
We will get
x1 = -1, x2 = 8, y1 = -2, y2 = 0
Hence,
The midpoint M = ( \(\frac{-1 + 8}{2}\), \(\frac{-2 + 0}{2}\) )
= ( \(\frac{7}{2}\), \(\frac{-2}{2}\) )
= ( \(\frac{7}{2}\), -1 )
Hence, from the above,
We can conclude that the midpoint M of the segment with the given endpoints is: ( \(\frac{7}{2}\), -1 )

Solve the equation.

Question 4.
7x + 12 = 3x
Answer:
The given equation is:
7x + 12 = 3x
So,
7x – 3x = 12
4x = 12
x = \(\frac{12}{4}\)
x = 3
Hence, from the above,
We can conclude that the value of x is: 3

Question 5.
14 – 6t = t
Answer:
The given equation is:
14 – 6t = t
So,
14 = 6t + t
7t = 14
t = \(\frac{14}{7}\)
t = 2
Hence, from the above,
We can conclude that the value of t is: 2

Question 6.
5p + 10 = 8p + 1
Answer:
The given equation is:
5p + 10 = 8p + 1
So,
5p – 8p = 1 – 10
-3p = -9
3p = 9
p = \(\frac{9}{3}\)
p = 3
Hence, from the above,
We can conclude that the value of p is: 3

Question 7.
w + 13 = 11w – 7
Answer:
The given equation is:
w + 13 = 11w – 7
So,
w – 11w = -7 – 13
-10w = -20
10w = 20
w = \(\frac{20}{10}\)
w = 2
Hence, from the above,
We can conclude that the value of w is: 2

Question 8.
4x + 1 = 3 – 2x
Answer:
The given equation is:
4x + 1 = 3 – 2x
So,
4x + 2x = 3 – 1
6x = 2
x = \(\frac{2}{6}\)
x = \(\frac{1}{3}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{1}{3}\)

Question 9.
z – 2 = 4 + 9z
Answer:
The given equation is:
z – 2 = 4 + 9z
So,
z – 9z = 4 + 2
-8z = 6
z = –\(\frac{6}{8}\)
z = –\(\frac{3}{4}\)
Hence, from the above,
We can conclude that the value of z is: –\(\frac{3}{4}\)

Question 10.
ABSTRACT REASONING
Is it possible to find the length of a segment in a coordinate plane without using the Distance Formula? Explain your reasoning.
Answer:
Yes, it is possible to find the length of a segment in a coordinate plane without using the distance formula
Since the segment is a portion of a line, we can use the graph to calculate the distance of a segment even though it would not provide accurate results.
Hence,
We use the distance formula to find the length of a segment in a coordinate plane

Congruent Triangles Mathematical Practices

Monitoring Progress

Classify each statement as a definition, a postulate, or a theorem. Explain your reasoning.

Question 1.
In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is – 1.
Answer:
The given statement is:
In a coordinate plane, two non-vertical lines are perpendicular if and only if the product of their slopes is – 1.
We know that,
According to the “parallel and perpendicular lines theorem”, two non-vertical lines are perpendicular if and only if the product of their slopes is -1
Hence, from the above,
We can conclude that the given statement is a Theorem

Question 2.
If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.
Answer:
The given statement is:
If two lines intersect to form a linear pair of congruent angles, then the lines are perpendicular.
We know that,
According to the “Linear pair perpendicular theorem”,
When two straight lines intersect at a point and form a linear pair of congruent angles, then the lines are perpendicular
Hence, from the above,
We can conclude that the given statement is a Theorem

Question 3.
If two lines intersect to form a right angle. then the lines are perpendicular.
Answer:
The given statement is:
If two lines intersect to form a right angle. then the lines are perpendicular.
We know that,
According to the “Perpendicular lines theorem”,
When two lines intersect to form a right angle, the lines are perpendicular
Hence, from the above,
We can conclude that the given statement is a Theorem

Question 4.
Through any two points, there exists exactly one line.
Answer:
The given statement is:
Through any two points, there exists exactly one line
We know that,
Between two points, only one line can be drawn and we don’t need any proof to prove the above statement
We know that,
The statement that is true without proof to prove is called “Postulate”
Hence, from the above,
We can conclude that the given statement is a Postulate

5.1 Angles of Triangles

Exploration 1

Writing a Conjecture

Work with a partner.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 1

a. Use dynamic geometry software to draw any triangle and label it ∆ABC.
Answer:
By using the dynamic geometry software, the triangle drawn is:

b. Find the measures of the interior angles of the triangle.
Answer:
From part (a),
We can observe that the vertices of the triangle are: A, B, and C
Let the interior angles of the vertices A, B, and C be α, β, and γ respectively
Hence,
The measures of the given triangle are:

Hence, from the above,
The measures of the interior angles are:
α = 62.1°, β = 64.1°, and γ = 53.8°

c. Find the sum of the interior angle measures.
Answer:
From part (b),
The measures of the interior angles are:
α = 62.1°, β = 64.1°, and γ = 53.8°
Hence,
The sum of the interior angles = 62.1° + 64.1° + 53.1° = 180°
Hence, from the above,
We can conclude that the sum of the interior angle measures is: 180°

d. Repeat parts (a)-(c) with several other triangles. Then write a conjecture about the sum of the measures of the interior angles of a triangle.
Answer:
The representation of the 3 different triangles and their internal angle measures is:

Hence, from the above,
We can conclude that the conjecture about the sum of the measures of the interior angles of a triangle is:
The sum of the internal angle measures of a triangle is always: 180°

CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data and write conjectures.
Answer:
Inductive reasoning:
Inductive reasoning is the process of arriving at a conclusion based on a set of observations.
Inductive reasoning is used in geometry in a similar way.
Conjecture:
A statement you believe to be true based on inductive reasoning.

Exploration 2

Writing a Conjecture

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 2

Work With a partner.

a. Use dynamic geometry software to draw any triangle and label it ∆ABC.
Answer:
The triangle drawn by using the dynamic geometry software is:

Hence, from the above,
We can conclude that the vertices of the triangle are: A, B, and C

b. Draw an exterior angle at any vertex and find its measure.
Answer:
From part (a),
The vertices of the triangle are: A, B, and C
Let the external angle measures of the triangle are: α, β, and γ
Hence,
The representation of the external angle measures of the triangle are:

Hence,
From the above,
We can conclude that
The external angle measures of the triangle are:
α = 310.7°, β = 299.3°, and γ = 290°

c. Find the measures of the two nonadjacent interior angles of the triangle.
Answer:
From part (b),
The external angle measures of the triangle are:
α = 310.7°, β = 299.3°, and γ = 290°
Hence,
The representation of the non-adjacent interior angles and the external angle measures of the triangle are:

Hence, from the above,
The angle measures of two non-adjacent sides are:
α = 70°, β = 60.7°, and γ = 49.3°

d. Find the sum of the measures of the two nonadjacent interior angles. Compare this sum to the measure of the exterior angle.
Answer:
From part (b),
The external angle measures of the triangle are:
α = 310.7°, β = 299.3°, and γ = 290°
From part (c),
The measures of the two non-adjacent interior angles are:
α = 70°, β = 60.7°, and γ = 49.3°
Now,
The sum of the measures of the external angles of the triangle are:
α + β + γ = 310.7° + 299.3°+ 290°
= 900.0°
The sum of the measures of the two non-adjacent interior angles is:
α + β + γ = 70° + 60.7° + 49.3°
= 180.0
Hence, from the above,
We can conclude that the sum of the measures of the external angles is 5 times the sum of the measures of the two non-adjacent interior angles

e. Repeat parts (a)-(d) with several other triangles. Then write a conjecture that compares the measure of an exterior angle with the sum of the measures of the two nonadjacent interior angles.
Answer:

Hence, from the above,
We can conclude that
The external angle measure of a vertex for a given triangle = 360° – (Internal angle measure of a vertex that we are finding the external angle measure)
The sum of the internal angle measures of the triangle is: 180°

Communicate Your Answer

Question 3.
How are the angle measures of a triangle related?
Answer:
The angle measures of a triangle are related as shown below:
The external angle measure of a vertex for a given triangle = 360° – (Internal angle measure of a vertex that we are finding the external angle measure)
The sum of the internal angle measures of the triangle is: 180°

Question 4.
An exterior angle of a triangle measures 32° What do you know about the measures of the interior angles? Explain your reasoning.
Answer:
It is given that an exterior angle of a triangle measures 32°
We know that,
The external angle measure of a vertex for a given triangle = 360° – (Internal angle measure of a vertex that we are finding the external angle measure)
So,
32° = 360° – (The internal angle measure of 32°)
The internal angle measure of 32° = 360° – 32°
The interior angle measure of 32° = 328°
Hence, from the above,
We can conclude that the interior angle measure of a triangle for an external angle measure of 32° is: 328°

Lesson 5.1 Angles of Triangles

Monitoring Progress

Question 1.
Draw an obtuse isosceles triangle and an acute scalene triangle.
Answer:
The figures of an obtuse isosceles triangle and an acute triangle are as follows:

Question 2.
∆ABC has vertices A(0, 0), B(3, 3), and C(- 3, 3), Classify the triangle by its sides. Then determine whether it is a right triangle.
Answer:
The given points are:
A (0, 0), B (3, 3), and C (-3, 3)
and the triangle is ΔABC
We know that,
To find whether the given triangle is a right-angled triangle or not,
We have to prove,
AC² = AB² + BC²
Where,
AC is the distance between A and C points
AB is the distance between A and B points
BC is the distance between B and C points
We know that,
The distance between 2 points = √(x2 – x1)² + (y2 – y1)²
Now,
Let the given points be considered as A(x1, y1), B(x2, y2), and C( x3, y3)
So,
AB = √(3 – 0)² + (3 – 0)² = √3² + 3²
= √9 + 9 = √18
BC = √(-3 – 3)² + (3 – 3)²
= √(-6)² + 0²
= √6² = 6
AC = √(-3 – 0)² + (3 – 0)²
= √(-3)² + 3²
= √9 + 9 = √18
Now,
AC² = AB² + BC²
(√18)² = (√18)² + 6²
18 = 18 + 36
18 ≠54
Hence, from the above,
We can conclude that the given triangle is not a right-angled triangle

Question 3.
Find the measure of ∠1
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 3
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 3
We know that,
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles
From the given triangle,
The exterior angle is: (5x – 10)°
The interior angles are: 40°, 3x°, ∠1
So,
(5x – 10)° = 40° + 3x°
5x° – 3x° = 40° + 10°
2x° = 50°
x = 50° ÷ 2
x = 25°
So,
The interior angles are 40°, 3 (25)°, ∠1
= 40°, 75°, ∠1
We know that,
The sum of the interior angles of a triangle is: 180°
So,
40° + 75° + ∠1 = 180°
115° + ∠1 = 180°
∠1 = 180° – 115°
∠1 = 65°
Hence, from the above,
We can conclude that the value of ∠1 is: 65°

Question 4.
Find the measure of each acute angle.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 36
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 36
We know that,
The sum of the interior angles in a triangle is: 180°
From the given figure,
The interior angles of the right-angled triangle are: 90°, 2x°, and (x – 6)°
So,
90° + 2x° + (x – 6)° = 180°
84°+ 3x° = 180°
3x° = 180° – 84°
3x° = 96°
x = 96° ÷ 3°
x = 32°
So,
The measure of each acute angle is 90°, 2x°, (x – 6)°
= 90°, 2(32)°, (32 – 6)°
= 90°, 64°, 26°
Hence, from the above,
We can conclude that,
The measure of each acute angle is 90°, 64°, and 26°

Exercise 5.1 Angles of Triangles

Vocabulary and Core Concept Check

Question 1.
WRITING
Can a right triangle also be obtuse? Explain our reasoning.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 1

Question 2.
COMPLETE THE SENTENCE
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two ____________ interior angles.
Answer:
The given statement is:
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two ____________ interior angles.
Hence,
The completed form of the given statement is:
The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, classify the triangle by its sides and by measuring its angles.

Question 3.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 4
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 3

Question 4.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5
We know that,
“|” represents the “Congruent” or “Equal” in geometry
So,
From the given figure,
We can observe that all three sides of the given triangle are equal
We know that,
If a triangle has all the sides equal, then the triangle is called an “Equilateral triangle”
Hence, from the above,
We can conclude that the ΔLMN is an “Equilateral triangle”

Question 5.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 6
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 5

Question 6.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 7
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 7
We know that,
If any side is not equal to each other in the triangle, then the triangle is called a “Scalene triangle”
The angle greater than 90° is called as “Obtuse angle”
An angle less than 90° is called an “Acute angle”
Hence, from the above,
We can conclude that ΔABC is an “Acute scalene triangle”

In Exercises 7-10, classify ∆ABC by its sides. Then determine whether it is a right triangle.

Question 7.
A(2, 3), B(6, 3), (2, 7)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 7

Question 8.
A(3, 3), B(6, 9), (6, – 3)
Answer:
The given points are:
A (3, 3), B(6, 9), and C (6, -3)
We know that,
To find whether the given triangle is a right angle or not,
We have to prove,
AC² = AB² + BC²
Where,
AC is the distance between points A and C
AB is the distance between points A and B
BC is the distance between points B and C
The slope of any one side must be equal to -1
Now,
Let the given points be
A (x1, y1), B(x2, y2), and C (x3, y3)
So,
A (x1, y1)= (3, 3), B (x2, y2) = (6, 9), and C (x3, y3) = (6, -3)
We know that,
The distance between 2 points = √(x2 – x1)² + (y2 – y1)²
So,
AB = √(6 – 3)² + (9 – 3)²
= √3² + 6²
= √9 + 36 = √45
BC = √6 – 6)² + (-3 – 9)²
= √0 + 12²
= √12² = 12
AC = √(6 – 3)² + (-3 – 3)²
= √(3)² + (-6)²
= √9 + 36 = √45
So,
From the length of the sides,
We can say that the given triangle is an Isosceles triangle,
We know that,
Slope (m) = \(\frac{y2 – y1} {x2 – x1}\)
So,
Slope of AB = \(\frac{9 – 3} {6 – 3}\)
= \(\frac{6} {3}\)
= 2
Slope of BC = \(\frac{-9 – 3} {6 – 6}\)
= \(\frac{-12} {0}\)
= Undefined
Slope of AC = \(\frac{-3 – 3} {6 – 3}\)
= \(\frac{-6} {3}\)
= -2
Hence, from the above,
We can conclude that the given triangle is not a right triangle

Question 9.
A(1, 9), B(4, 8), C(2, 5)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 9

Question 10.
A(- 2, 3), B(0, – 3), C(3, – 2)
Answer:
The given points are:
A (-2, 3), B(0, -3), and C (3, -2)
We know that,
To find whether the given triangle is a right angle or not,
We have to prove,
AC² = AB² + BC²
Where,
AC is the distance between points A and C
AB is the distance between points A and B
BC is the distance between points B and C
The slope of any one side must be equal to -1
Now,
Let the given points be
A (x1, y1), B(x2, y2), and C (x3, y3)
So,
A (x1, y1)= (-2, 3), B (x2, y2) = (0, -3), and C (x3, y3) = (3, -2)
We know that,
The distance between 2 points = √(x2 – x1)² + (y2 – y1)²
So,
AB = √(0 – [-2])² + (3 – 3)²
= √2² + 0²
= √4 + 0 = 2
BC = √3 – 0)² + (-2 -[-3] )²
= √9 + 1²
= √10
AC = √(3 – [-2])² + (-2 – 3)²
= √(5)² + (-5)²
= √25 + 25 = √50
Now,
AC² = AB² + BC²
50 = 10 + 4
50 ≠ 14
So,
From the length of the sides,
We can say that the given triangle is a scalene triangle since all the lengths of the sides are different
We know that,
Slope (m) = \(\frac{y2 – y1} {x2 – x1}\)
So,
Slope of AB = \(\frac{9 – 3} {6 – 3}\)
= \(\frac{6} {3}\)
= 2
Slope of BC = \(\frac{-9 – 3} {6 – 6}\)
= \(\frac{-12} {0}\)
= Undefined
Slope of AC = \(\frac{-3 – 3} {6 – 3}\)
= \(\frac{-6} {3}\)
= -2
Hence, from the above,
We can conclude that the given triangle is not a right triangle

In Exercises 11 – 14. find m∠1. Then classify the triangle by its angles

Question 11.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 8
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 11

Question 12.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 9
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 9
We know that,
The sum of interior angles in a triangle is: 180°
So,
From the above,
The interior angles of the given triangle are: 40°,  30°,  ∠1
Now,
40° + 30° + ∠1 = 180°
70 + ∠1 = 180°
∠1 = 180° – 70°
∠1 = 110°
We know that,
The angle greater than 90° is called an “Obtuse angle”
Hence, from the above,
We can conclude that the given triangle is an “Obtuse angled triangle”

Question 13.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 10
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 13

Question 14.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 11
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 11
We know that,
The sum of interior angles in a triangle is: 180°
So,
From the above,
The interior angles of the given triangle are: 60°,  60°,  ∠1
Now,
60° + 60° + ∠1 = 180°
120 + ∠1 = 180°
∠1 = 180° – 120°
∠1 = 60°
We know that,
An angle less than 90° is called an “Acute angle”
The triangle that all the angles 60° is called an “Equilateral triangle”
Hence, from the above,
We can conclude that the given triangle is an “Equilateral triangle”

In Exercises 15-18, find the measure of the exterior angle.

Question 15.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 12
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 15

Question 16.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 13
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 13
We know that,
An exterior angle is equal to the sum of the two non-adjacent interior angles in a triangle
So,
(2x – 2)° = x° + 45°
2x° – x° = 45° + 2°
x = 47°
Hence,
The measure of the exterior angle is: (2x – 2)°
= (2 (47) – 2)°
= (94 – 2)°
= 92°
Hence, from the above,
We can conclude that the measure of the exterior angle is: 92°

Question 17.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 14
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 17

Question 18.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 15
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 15
We know that,
An exterior angle is equal to the sum of the two non-adjacent interior angles in a triangle
So,
(7x – 16)° = (x + 8)° + 4x°
7x° – 5x° = 16° + 8°
2x = 24°
x = 24° ÷ 2
x = 12°
Hence,
The measure of the exterior angle is: (7x – 16)°
= (7 (12) – 16)°
= (84 – 16)°
= 68°
Hence, from the above,
We can conclude that the measure of the exterior angle is: 68°

In Exercises 19-22, find the measure of each acute angle.

Question 19.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 16
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 19

Question 20.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 17
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 17
From the given figure,
We can observe that one angle is 90° and the 2 sides are perpendicular
So,
We can say that the given triangle is a right-angled triangle
We know that,
The sum of interior angles of a triangle is: 180°
So,
x° + (3x + 2)° + 90° = 180°
4x° + 2° + 90° = 180°
4x° = 180° – 90° – 2°
4x° = 88°
x = 88° ÷ 4°
x = 22°
So,
The 2 acute angle measures are: x° and (3x + 2)°
= 22° and (3(22) + 2)°
= 22° and (66 + 2)°
= 22° and 68°
Hence, from the above,
We can conclude that the 2 acute angle measures are: 22° and 68°

Question 21.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 18
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 21

Question 22.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 19
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 19
From the given figure,
We can observe that one angle is 90° and the 2 sides are perpendicular
So,
We can say that the given triangle is a right-angled triangle
We know that,
The sum of interior angles of a triangle is: 180°
So,
(19x – 1)° + (13x – 5)° + 90° = 180°
32x° – 6° + 90° = 180°
32x° = 180° – 90° – 6°
4x° = 84°
x = 84° ÷ 4°
x = 21°
So,
The 2 acute angle measures are: (19x – 1)° and (13x – 5)°
= (19 (21) – 1)° and (13(21) – 5)°
= 398° and (273 – 5)°
= 398° and 268°
Hence, from the above,
We can conclude that the 2 acute angle measures are: 398° and 268°

In Exercises 23-26. find the measure of each acute angle in the right triangle.

Question 23.
The measure of one acute angle is 5 times the measure of the other acute angle.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 23

Question 24.
The measure of one acute angle is times the measure of the other acute angle.
Answer:

Question 25.
The measure of one acute angle is 3 times the sum of the measure of the other acute angle and 8.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 25

Question 26.
The measure of one acute angle is twice the difference of the measure of the other acute angle and 12.
Answer:
The given statement is:
The measure of one acute angle is twice the difference of the measure of the other acute angle and 12.
So,
x° + [2 (x – 12)]° = 90°
x° + 2x° – 2(12)° = 90°
3x° – 24° = 90°
3x° = 90° + 24°
3x° = 114°
x = 114° ÷ 3
x = 38°
So,
The 2 acute angle measures are: x°, 2 (x – 12)°
= 38°, 2 (38 – 12)°
= 38°, 2(26)°
= 38° , 52°
Hence, from the above,
We can conclude that the acute angle measures are: 38°, 52°

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in finding m∠1.

Question 27.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 20
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 27

Question 28.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 21
Answer:
We know that,
The exterior angle of a triangle is equal to the sum of the non-adjacent interior angles of a triangle
So,
From the figure,
The external angle is: ∠1
The interior angles are 80°, 50°
So,
∠1 = 80° + 50°
∠1 = 130°
Now,
The interior angle measure of ∠1= 180° – (External angle measure of 130°)
= 180° – 130°
= 50°
Hence, from the above,
The internal angle measure of ∠1 is: 50°

In Exercises 29-36, find the measure of the numbered angle.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 22

Question 29.
∠1
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 29

Question 30.
∠2
Answer:
We know that,
The external angle measure is equal to the sum of the non-adjacent interior angles
So,
∠2 = 90° + 40°
∠2 = 130°

Question 31.
∠3
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 31

Question 32.
∠3
Answer:
From the above figure,
∠2 = ∠4
Hence, from the above,
We can conclude that
∠2 = ∠4 = 130°

Question 33.
∠5
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 33

Question 34.
∠6
Answer:
The external angle measure is equal to the sum of the non-adjacent interior angles
So,
∠6 = 90° + ∠3
∠6 = 90° + 50°
∠6 = 140°

Question 35.
∠7
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 35

Question 36.
∠8
Answer:
The external angle measure is equal to the sum of the non-adjacent interior angles
So,
∠8 = 90° + ∠1
∠6 = 90° + 50°
∠6 = 140°

Question 37.
USING TOOLS
Three people are standing on a stage. The distances between the three people are shown in the diagram. Classify the triangle by its sides and by measuring its angles.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 23
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 37

Question 38.
USING STRUCTURE
Which of the following sets of angle measures could form a triangle? Select all that apply.
(A) 100°, 50°, 40°
Answer:
The given angles are: 100°, 50°, 40°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 100° + 50° + 40°
= 100° + 90°
= 190°
Hence, from the above,
We can conclude that the given angles do not form a triangle

(B) 96°, 74°, 10°
Answer:
The given angles are: 96°, 74°, 10°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 96° + 74° + 10°
= 96° + 84°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

(C) 165°, 113°, 82°
Answer:
The given angles are: 165°, 113°, 82°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 165° + 113° + 82°
= 165° + 195°
= 360°
But,
We know that,
The sum of exterior angles of a triangle is: 360°
Hence, from the above,
We can conclude that the given angles forms a triangle

(D) 101°, 41°, 38°
Answer:
The given angles are: 101°, 41°, 38°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 101° + 38° + 41°
= 101° + 79°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

(E) 90°, 45°, 45°
Answer:
The given angles are: 90°, 45°, 45°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 90° + 45° + 45°
= 90° + 90°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

(F) 84°, 62°, 34°
Answer:
The given angles are: 84°, 62°, 34°
We know that,
The sum of the angles of a triangle should be equal to 180°
So,
The sum of the given angles = 84° + 62° + 34°
= 84° + 96°
= 180°
Hence, from the above,
We can conclude that the given angles forms a triangle

Question 39.
MODELING WITH MATHEMATICS
You are bending a strip of metal into an isosceles triangle for a sculpture. The strip of metal is 20 inches long. The first bend is made 6 inches from one end. Describe two ways you could complete the triangle.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 39

Question 40.
THOUGHT-PROVOKING
Find and draw an object (or part of an object) that can be modeled by a triangle and an exterior angle. Describe the relationship between the interior angles of the triangle and the exterior angle in terms of the object.
Answer:

From the above figure,
We can say that
The sum of the interior angles of a given triangle is: 180°
The sum of the exterior angles of a given triangle is: 360°
The relation between the interior angles and the exterior angles is:
The exterior angle measure = Sum of the two non-adjacent interior angles

Question 41.
PROVING A COROLLARY
Prove the Corollary to the Triangle Sum Theorem (Corollary 5. 1).
Given ∆ABC is a right triangle
Prove ∠A and ∠B are complementary
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 24
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 41

Question 42.
PROVING A THEOREM
Prove the Exterior Angle Theorem (Theorem 5.2).
Given ∆ABC, exterior ∠ACD
Prove m∠A + m∠B = m∠ACD

Answer:
It is given that
In ΔABC, the exterior angle is ∠ACD
We have to prove that
m∠A + m∠B = m∠ACD
Proof:

Hence, from the above,
We can conclude that
m∠A + m∠B = m∠ACD is proven

Question 43.
CRITICAL THINKING
Is it possible to draw an obtuse isosceles triangle? obtuse equilateral triangle? If so, provide examples. If not, explain why it is not possible.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 43

Question 44.
CRITICAL THINKING
Is it possible to draw a right isosceles triangle? right equilateral triangle? If so, provide an example. If not, explain why it is not possible.
Answer:
It is possible to draw a right isosceles triangle but it is not possible to draw a right equilateral triangle
We know that,
In a triangle, if the length of the 2 sides are equal and one angle is a right-angle, then, it is called an “Right Isosceles triangle”
In a triangle, if the length of all the sides are equal and each angle is 60°, then it is an “Equilateral triangle”
Hence,
From the above definitions,
We can observe that it is possible to draw right isosceles triangle but it is not possible to dran a right equilateral triangle

Question 45.
MATHEMATICAL CONNECTIONS
∆ABC is isosceles.
AB = x, and BC = 2x – 4.
a. Find two possible values for x when the perimeter of ∆ABC is 32.
b. How many possible values are there for x when the perimeter of ∆ABC is 12?
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 45

Question 46.
HOW DO YOU SEE IT?
Classify the triangles, in as many ways as possible. without finding any measurements.
a. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 26
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 26
From the figure,
We can observe that all the length of the sides of the triangle are equal
We know that,
The triangle that has the length of all the sides equal is called an “Equilateral triangle”
Hence, from the above,
We can conclude that the given triangle is an “Equilateral triangle”

b. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 27
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 27
From the figure,
We can observe that the lengths of all the 3 sides are different
We know that,
The triangle that has all the different side lengths is called a “Scalene triangle”
Hence, from the above,
We can conclude that the given triangle is called a “Scalene triangle”

c. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 28
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 28
From the figure,
We can observe that the length of all the 3 sides are different and 1 angle is obtuse i.e., greater than 90°
We know that,
The triangle that has any angle obtuse is called an “Obtuse angled triangle”
Hence, from the above,
We can conclude that the given triangle is an “Obtuse angled scalene triangle”

d. Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 29
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 29
From the figure,
We can observe that 1 angle is 90° and the 2 sides are perpendicular to each other
We know that,
The triangle that has an angle of 90° and the slope -1 is called a “Right-angled triangle”
Hence, from the above,
We can conclude that the given triangle is called a “Right-angled triangle”

Question 47.
ANALYZING RELATIONSHIPS
Which of the following could represent the measures of an exterior angle and two interior angles of a triangle? Select all that apply.
A) 100°, 62°, 38°
(B) 81°, 57°, 24°
(C) 119°, 68°, 49°
(D) 95°, 85°, 28°
(E) 92°, 78°, 68°
(F) 149°, 101°, 48°
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 47

Question 48.
MAKING AN ARGUMENT
Your friend claims the measure of an exterior angle will always be greater than the sum of the nonadjacent interior angle measures. Is your friend correct? Explain your reasoning.
Answer:
No, your friend is not correct

Explanation:
We know that,
According to the exterior angle theorem,
The external angle measure is always equal to the sum of the non-adjacent internal angle measures
But,
According to your friend,
The external angle measure will always be greater than the sum of the non-adjacent interior angle measures
Hence, from the above,
We can conclude that your friend is not correct

MATHEMATICAL CONNECTIONS
In Exercises 49-52, find the values of x and y.

Question 49.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 30
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 49

Question 50.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 31
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 31
From the figure,
We have to obtain the values of x and y
Now,
By using the alternate angles theorem,
x = 118°
Now,
By using the exterior angle theorem,
x = y + 22°
y = x – 22°
y = 118° – 22°
y = 96°
Hence, from the above,
We can conclude that the values of x and y are: 118° and 96° respectively

Question 51.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 32
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 51

Question 52.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 33
Answer:
The given figure is:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 33
From the above figure,
We have to find the values of x and y
Now,
By using the sum of interior angle measures,
x° + 64° + 90° = 180°
x° + 154° = 180°
x° = 180° – 154°
x° = 26°
Now,
By using the exterior angle theorem,
y° = x° + 64°
y° = 26° + 64°
y° = 90°
Hence, from the above,
We can conclude that the values of x and y are: 26° and 90° respectively

Question 53.
PROVING A THEOREM
Use the diagram to write a proof of the Triangle Sum Theorem (Theorem 5. 1). Your proof should be different from the proof of the Triangle Sum Theorem shown in this lesson.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 34
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 53.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 53.2

Maintaining Mathematical Proficiency

Use the diagram to find the measure of the segment or angle.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 35

Question 54.
m∠KHL
Answer:
From the given figures,
We can observe that
∠ABC = ∠GHK
∠KHL = ∠GHK / 2
So,
(6x + 2)° = (3x + 1)° + (5x – 27)°
6x – 3x – 5x = 1 – 27 – 2
6x – 8x = -27 – 1
-2x = -28
2x = 28
x = 28 ÷ 2
x = 14
So,
∠KHL = ∠GHK / 2
= [(3 (14) + 1)° + (5 (14) – 27)°] / 2
= [43° + 43°] / 2
= 86° / 2
= 43°
Hence, from the above,
We can conclude that
∠KHL = 43°

Question 55.
m∠ABC
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 55

Question 56.
GH
Answer:
From the given figures,
We can observe that
AB = GH
So,
3y = 5y – 8
3y – 5y = -8
-2y = -8
2y = 8
y = 8 ÷ 2
y = 4
So,
The value of GH = 3y = 3 (4) = 12
Hence, from the above,
We can conclude that the value of GH is: 12

Question 57.
BC
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.1 a 57

5.2 Congruent Polygons

Exploration 1

Describing Rigid Motions

Work with a partner: of the four transformations you studied in Chapter 4, which are rigid motions? Under a rigid motion. why is the image of a triangle always congruent to the original triangle? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 37
Answer:
Rigid motion occurs in geometry when an object moves but maintains its shape and size, which is unlike non-rigid motions, such as dilation, in which the object’s size changes. All rigid motion starts with the original object, called the pre-image, and results in the transformed object, called the image.
There are 4 types of rigid motion. They are:
a. Translation
b. Rotation
c. Reflection
d. Glide reflection
We know that,
Rotation only occurs in terms of 90° or 180°
Now,
The given transformations are:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 37
So,
From the above figure,
The first figure and the second figure are different
The second figure and the third figure are the same in shape
The first figure and the fourth figure are the same in shape
So,
We can say that the first and the fourth figures are rigid motions
W can say that the second and the third figures are rigid motions
In the second and the third figures,
The “Rotation” takes place i.e., the second figure is rotated 180° keeping the original shape
In the first and the fourth figures,
The “Reflection” takes place i.e., the first figure is reflected keeping the original shape
Now,
The image of the triangle is always congruent to the original triangle because of the “Translation” i.e., the original triangle and the image of the triangle have the same sides and the same angles but not in the same position.

Exploration 2

Finding a Composition of Rigid Motions

Work with a partner. Describe a composition of rigid motions that maps ∆ABC to ∆DEF. Use dynamic geometry software to verify your answer.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.

a. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 38
Answer:

b. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 39
Answer:

c. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 40
Answer:

d. ∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 41
Answer:

Communicate Your Answer

Question 3.
Given two congruent triangles. how can you use rigid motions to map one triangle to the other triangle?
Answer:
It is given that there are two congruent triangles
So,
Let the one triangle be the original or pre-image triangle and another triangle be the image triangle
Now,
We can use rigid motions on the original triangle to create the image triangle as the rigid motions won’t alter the size and shape of the figure and it maps each part of the triangle with the corresponding part of its image

Question 4.
The vertices of ∆ABC are A(1, 1), B(3, 2), and C(4, 4). The vertices of ∆DEF are D(2, – 1), E(0, 0), and F(- 1, 2). Describe a composition of rigid motions that maps ∆ABC to ∆DEF.
Answer:

Lesson 5.2 Congruent Polygons

Monitoring Progress

In the diagram, ABGH ≅ CDEF.

Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 42

Question 1.
Identify all pairs of congruent corresponding parts.

Answer:
∠A = ∠C, ∠B = ∠D, ∠H = ∠F, ∠G = ∠E
AB = CD, BG= DE, HG = ED, AH = CF

Question 2.
Find the value of x.

Answer:
x = 25

Explanation:
∠H = ∠F
(4x + 5) = 105
4x = 100
x = 25

Question 3.
In the diagram at the left. show that ∆PTS ≅ ∆RTQ.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 43

Answer:
PS = QR, ST = TQ, PT = TR
So, ∆PTS ≅ ∆RTQ

Use the diagram.

Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 44

Question 4.
Find m∠DCN.

Answer:
m∠DCN = 75°

Explanation:
∆CDN ≅ SRN
So, m∠DCN = m∠SRN
m∠DCN = 75°

Question 5.
What additional information is needed to conclude that ∆NDC ≅ ∆NSR?

Answer:
The sides RC = RS, DN = NS and m∠DNC = m∠RNS

Exercise 5.2 Congruent Polygons

Question 1.
WRITING
Based on this lesson. what information do you need to prove that two triangles are congruent? Explain your reasoning.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 45

Is ∆JKL ≅ ∆RST?
Is ∆KJL ≅ ∆SRT?
Is ∆JLK ≅ ∆STR?
Is ∆LKJ ≅ ∆TSR?

Answer:
∆JLK ≅ ∆STR is different

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4. identify all pairs of congruent corresponding parts. Then write another congruence statement for the polygons.

Question 3.
∆ABC ≅ ∆DEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 46
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 3

Question 4.
GHJK ≅ QRST
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 47
Answer:
Corresponding angles: ∠G = ∠Q, ∠H = ∠R, ∠K = ∠T, ∠J = ∠T
Correspoding sides: ST = JK, TQ = KG, QR = GH, RS = HJ
So, GHJK ≅ QRST

In Exercises 5-8, ∆XYZ ≅ ∆MNL. Copy and complete the statement.

Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 48

Question 5.
m∠Y = ______
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 5

Question 6.
m∠M = ______

Answer:
m∠M = m∠X, so m∠M = 33°

Question 7.
m∠Z = _______
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 7

Question 8.
XY= _______

Answer:
XY = MN, so, XY = 8

In Exercises 9 and 10. find the values of x and y.

Question 9.
ABCD ≅ EFGH
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 49
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 9

Question 10.
∆MNP ≅ ∆TUS
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 50

Answer:
x = 32°, y = 51

Explanation:
Sum of angles = 180°
142 + 24 + 2x – 50 = 180
116 + 2x = 180
2x = 64
x = 32°
NP = TU
2x – y = 13
2(32) – y = 13
64 – 13 = y
y = 51

In Exercises 11 and 12. show that the polygons are congruent. Explain your reasoning.

Question 11.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 51
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 11

Question 12.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 52

Answer:
∠X = ∠Z, ∠XWY = ∠ZWY, ∠XYW = ∠ZWY
WX = YZ, XY = WZ
All corresponding parts of polygons are congruent
So, △ XWZ ≅ △ ZWY

In Exercises 13 and 14, find m∠1.

Question 13.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 53
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 13

Question 14.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 54

Answer:
∠1 = 180 – (80 + 45) = 180 – 125
∠1 = 55°

Question 15.
PROOF
Triangular postage stamps, like the ones shown, are highly valued by stamp collectors. Prove that ∆AEB ≅ ∆CED.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 55
Given \(\overline{A B}\) || \(\overline{D C}\), \(\overline{A B}\) ≅ \(\overline{D C}\) is the midpoint of \(\overline{A C}\) and \(\overline{B D}\)
Prove ∆AEB ≅ ∆CED
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 15

Question 16.
PROOF
Use the information in the figure to prove that ∆ABG ≅ ∆DCF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 56

Answer:
Proved

Explanation:
FE = EG, BE = EC, AF = GD, AB = CD
∠B = ∠C, ∠A = ∠D
All the corresponding pairs are similar So, ∆ABG ≅ ∆DCF.

ERROR ANALYSIS
In Exercises 17 and 18, describe and correct the error.

Question 17.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 57
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 17

Question 18.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 58

Answer:
∠N = ∠S, ∠M = ∠R, MN ≠ RS
The corresponding sides are not congruent. So, ∆MNP is not similar to ∆RSP.

Question 19.
PROVING A THEOREM
Prove the Third Angles Theorem (Theorem 5.4) by using the Triangle Sum Theorem (Theorem 5. 1).
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 19.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 19.2

Question 20.
THOUGHT PROVOKING
Draw a triangle. Copy the triangle multiple times to create a rug design made of congruent triangles. Which property guarantees that all the triangles are congruent?

Answer:
Suppose the carpet is in rectangle shape. We will draw one triangle and copy it several times
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.2 1
The interior angles of these triangles are equal because one pair of angles is equal because they are cross angles and the other pairs are equal because the angles are on the transversal.
When the angles are equal, then the sides of these triangles are also equal that means these triangles are congruent.

Question 21.
REASONING
∆JKL is congruent to ∆XYZ Identify all pairs of congruent corresponding parts.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 21

Question 22.
HOW DO YOU SEE IT?
In the diagram, ABEF ≅ CDEF
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 59
a. Explain how you know that \(\overline{B E}\) ≅ \(\overline{D E}\) and ∠ABE ≅∠CDE.
Answer:

b. Explain how you know that ∠GBE ≅ ∠GDE.
Answer:
GF is the altitude of triangle AGC.
∠GED = 90 so, ∠GEB = 90
That means ∠GBE ≅ ∠GDE

c. Explain how you know that ∠GEB ≅ ∠GED.
Answer:
GF is the altitude of triangle AGC. So, ∠GEB ≅ ∠GED

d. Do you have enough information to prove that ∠BEG ≅ ∠DEG? Explain.
Answer:

MATHEMATICAL CONNECTIONS
In Exercises 23 and 24, use the given information to write and solve a system of linear equations to find the values of x and y.

Question 23.
∆LMN ≅ ∆PQR. m∠L = 40°, m∠M = 90° m∠P = (17x – y)°. m∠R (2x + 4y)°
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 23

Question 24.
∆STU ≅ ∆XYZ, m∠T = 28°, m∠U = (4x + y)°, m∠X = 130°, m∠Y = (8x – 6y)°

Answer:
x = 5, y = 2

Explanation:
m∠Y = m∠T
28 = (8x – 6y)°
4x – 3y = 14 —- (I)
m∠T + m∠S + m∠U = 180°
28°+ 130° + (4x + y)° = 180°
158 + 4x + y = 180
4x + y = 22 —- ( II)
Subtract (I) from (II)
4x – 3y – 4x – y = 14 – 22
-4y = -8
y = 2
Put y = 2 in (II)
4x + 2 = 22
4x = 20
x = 5

Question 25.
PROOF
Prove that the criteria for congruent triangles in this lesson is equivalent to the definition of congruence in terms of rigid motions.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 25

Maintaining Mathematical Proficiency

What can you conclude from the diagram?

Question 26.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 60
Answer:
∠Z = ∠W

Question 27.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 61
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 27

Question 28.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 62

Answer:
JK = KM
∠J = ∠M

Question 29.
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 63
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.2 a 29

5.3 Proving Triangle Congruence by SAS

Exploration 1

Drawing Triangles

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 64

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 65

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 66

Work with a partner.

Use dynamic geometry software.
a. Construct circles with radii of 2 units and 3 units centered at the origin. Construct a 40° angle with its vertex at the origin. Label the vertex A.
Answer:

b. Locate the point where one ray of the angle intersects the smaller circle and label this point B. Locate the point where the other ray of the angle intersects the larger circle and label this point C. Then draw ∆ABC.
Answer:

c. Find BC, m∠B, and m∠C.
Answer:

d. Repeat parts (a)-(c) several times. redrawing the angle indifferent positions. Keep track of your results by copying and completing the table below. What can you conclude?
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to help visualize the results of varying assumptions, explore consequences, and compare predictions with data.
Answer:

Communicate Your Answer

Question 2.
What can you conclude about two triangles when you know that two pairs of corresponding sides and the corresponding included angles are congruent?
Answer:

Question 3.
How would you prove your conclusion in Exploration 1(d)?
Answer:

Lesson 5.3 Proving Triangle Congruence by SAS

Monitoring Progress

In the diagram, ABCD is a square with four congruent sides and four right
angles. R, S, T, and U are the midpoints of the sides of ABCD. Also, \(\overline{R T}\) ⊥ \(\overline{S U}\) and \(\overline{S V}\) ≅ \(\overline{V U}\).
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 67
Question 1.
Prove that ∆SVR ≅ ∆UVR.

Answer:
Given that,
ABCD is a square with four congruent sides and four right angles. R, S, T, and U are the midpoints of the sides of ABCD.
So, SV = VU, RS = RU, ∠S = ∠U
All the corresponding sides and angles are congruent. So, ∆SVR ≅ ∆UVR.

Question 2.
Prove that ∆BSR ≅ ∆DUT.

Answer:
∠B = ∠D, ∠S = ∠U, ∠R = ∠T and BS = UD, BR = TD, SR = TU
So, ∆BSR ≅ ∆DUT

Question 3.
You are designing the window shown in the photo. You want to make ∆DRA congruent to ∆DRG. You design the window so that \(\overline{D A}\) ≅ \(\overline{D G}\) and ∠ADR ≅ ∠GDR. Use the SAS Congruence Theorem to prove ∆DRA ≅ ∆DRG.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 68

Answer:
When you rotate ∆DRG 90° towards left, then ∆DRG coincides ∆DRA.
so, according to the Side Angle Side Congruence theorem, ∆DRA ≅ ∆DRG

Exercise 5.3 Proving Triangle Congruence by SAS

vocabulary and core concept check

Question 1.
WRITING
What is an included angle?
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 1

Question 2.
COMPLETE THE SENTENCE
If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then __________ .

Answer:
If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then two triangles are congruent according to the SAS Congruence Theorem.

Monitoring progress and Modeling with Mathematics

In Exercises 3-8, name the included an1e between the pair of sides given.

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 69

Question 3.
\(\overline{J K}\) and \(\overline{K L}\)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 3

Question 4.
\(\overline{P K}\) and \(\overline{L K}\)

Answer:
∠KLP is the included angle between \(\overline{P K}\) and \(\overline{L K}\)

Question 5.
\(\overline{L P}\) and \(\overline{L K}\)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 5

Question 6.
\(\overline{J L}\) and \(\overline{J K}\)

Answer:
∠KJL is the included angle between \(\overline{J L}\) and \(\overline{J K}\)

Question 7.
\(\overline{K L}\) and \(\overline{J L}\)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 7

Question 8.
\(\overline{K P}\) and \(\overline{P L}\)

Answer:
∠KPL is the included angle between \(\overline{K P}\) and \(\overline{P L}\)

In Exercises 9-14, decide whether enough information is given to prove that the triangles are congruent using the SAS Congruence Theorem (Theorem 5.5). Explain.

Question 9.
∆ABD, ∆CDB
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 70
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 9

Question 10.
∆LMN, ∆NQP
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 71

Answer:
Yes, two pairs of sides and the included angles are congruent.

Question 11.
∆YXZ, ∆WXZ
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 72
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 11

Question 12.
∆QRV, ∆TSU
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 73

Answer:
Yes, two pairs of sides and the included angles are congruent.

Question 13.
∆EFH, ∆GHF
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 74
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 13

Question 14.
∆KLM, ∆MNK
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 75

Answer:
No, one of the congruent angles is not the included angle.

In Exercises 15 – 18, write a proof.

Question 15.
Given \(\overline{P Q}\) bisects ∠SPT, \(\overline{S P}\) ≅ \(\overline{T P}\)
Prove ∆SPQ ≅ ∆TPQ
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 76
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 15

Question 16.
Given \(\overline{A B}\) ≅ \(\overline{C D}\), \(\overline{A B}\) || \(\overline{C D}\)
Prove ∆ABC ≅ ∆CDA
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 77

Answer:
Since AB || CD
∠BAC = ∠ACD (Alternative Interior Angle)
∠DAC = ∠BCA (Alternative Interior Angle)
AC = AC (Reflexive property of congruence)
∆ABC ≅ ∆CDA by Angle Side Angle Congruence Theorem.

Question 17.
Given C is the midpoint of \(\overline{A E}\) and \(\overline{B D}\)
Prove ∆ABC ≅ ∆EDC
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 78
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 17

Question 18.
Given \(\overline{P T}\) ≅ \(\overline{R T}\), \(\overline{Q T}\) ≅ \(\overline{S T}\)
Prove ∆PQT ≅ ∆RST
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 79

Answer:
\(\overline{P T}\) ≅ \(\overline{R T}\)
\(\overline{Q T}\) ≅ \(\overline{S T}\)
In a parallelogram, diagonals bisect at 90 degrees.
∠PTQ = ∠STR (Vertical Angles Congruence Theorem)
∆PQT ≅ ∆RST (SAS Congruence Theorem)

In Exercises 19-22, use the given information to name two triangles that are congruent. Explain your reasoning.

Question 19.
∠SRT ≅ ∠URT, and R is the center of the circle.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 80
Answer:

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 19

Question 20.
ABCD is a square with four congruent sides and four congruent angles.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 81

Answer:
BC ≅ AD, AB ≅ CD as four sides are congruent
BD ≅ BD by the reflexive property of congruence.
∠A = ∠C, ∠DBA = ∠CBD
All the corresponding sides and angles are congruent
So, △BAD ≅ △BCD

Question 21.
RSTUV is a regular pentagon.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 82
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 21

Question 22.
\(\overline{M K}\) ⊥ \(\overline{M N}\), \(\overline{K L}\) ⊥ \(\overline{N L}\), and M and L are centers of circles.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 83
Answer:
∠M = ∠L = 90°
MK = LN
NK = NK by the reflexive property of congruence
So, △ MKN ≅ △LKN

CONSTRUCTION
In Exercises 23 and 24, construct a triangle that is congruent to ∆ABC using the SAS Congruence Theorem (Theorem 5.5).

Question 23.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 84
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 85
Answer:
Construct side DE which is congruent to AC. Construct ∠D with vertex D and side DE so that it is congruent to ∠A.
Construct DF that is congruent to AB. Draw △DFE. By SAS congruence theorem △ABC ≅ △DFE

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding the value of x.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 86
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 25

Question 26.
HOW DO YOU SEE IT?
What additional information do you need to prove that ∆ABC ≅ ∆DBC?
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 87

Answer:
One included congruent angle is needed to prove that ∆ABC ≅ ∆DBC.

Question 27.
PROOF
The Navajo rug is made of isosceles triangles. You know ∠B ≅∠D. Use the SAS Congruence Theorem (Theorem 5.5 to show that ∆ABC ≅ ∆CDE. (See Example 3.)
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 88
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 27

Question 28.
THOUGHT PROVOKING
There are six possible subsets of three sides or angles of a triangle: SSS, SAS, SSA, AAA, ASA, and AAS. Which of these correspond to congruence theorems? For those that do not, give a counterexample.

Answer:
SSS, SAS, ASA and AAS are congruence theorems.
SSA does not prove two traingles congruent. For example, two triangles have the same angle and two common sides, but they are not congruent.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.3 1
AAA does not prove two triangles congruent. For example, these two triangles have the same angles but are not similar.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.3 2

Question 29.
MATHEMATICAL CONNECTIONS
Prove that
∆ABC ≅ ∆DEC
Then find the values of x and y.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 89
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 29

Question 30.
MAKING AN ARGUMENT
Your friend claims it is possible to Construct a triangle congruent to ∆ABC by first constructing \(\overline{A B}\) and \(\overline{A C}\), and then copying ∠C. Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 90

Answer:
My friend is wrong. Because to copy the ∠C, you need to construct the vertex C first. But by constructing \(\overline{A B}\) and \(\overline{A C}\), you will get vertex A.

Question 31.
PROVING A THEOREM
Prove the Reflections in Intersecting Lines Theorem (Theorem 4.3).
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.2
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.3
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 31.4

Maintaining Mathematical Proficiency

Classify the triangle by its sides and by measuring its angles.

Question 32.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 91

Answer:
Two sides are equal and one angle is the right angle. So, the triangle is right angles isosceles triangle

Question 33.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 92
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 33

Question 34.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 93

Answer:
Three sides of the triangle are equal. So, it is an equilateral triangle.

Question 35.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 94
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.3 a 35

5.4 Equilateral and Isosceles Triangles

Exploration 1

Writing a Conjecture about Isosceles Triangles

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 95

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 96

Work with a partner: Use dynamic geometry software.

a. Construct a circle with a radius of 3 units centered at the origin.
Answer:

b. Construct ∆ABC so that B and C are on the circle and A is at the origin.
Answer:

c. Recall that a triangle is isosceles if it has at least two congruent sides. Explain why ∆ABC is an isosceles triangle.
Answer:

d. What do you observe about the angles of ∆ABC?
Answer:

e. Repeat parts (a)-(d) with several other isosceles triangles using circles of different radii. Keep track of your observations by copying and completing the table below. Then write a conjecture about the angle measures of an isosceles triangle.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to make conjectures and build a logical progression of statements to explore the truth of your conjectures.
Answer:

f. Write the converse of the conjecture you wrote in part (e). Is the converse true?
Answer:

Communicate Your Answer

Question 2.
What conjectures can you make about the side lengths and angle measures of an
isosceles triangle?
Answer:

Question 3.
How would you prove your conclusion in Exploration 1 (e)? in Exploration 1(f)?
Answer:

Lesson 5.4 Equilateral and Isosceles Triangles

Monitoring Progress

Copy and complete the statement.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 97

Question 1.
If \(\overline{H G}\) ≅ \(\overline{H K}\), then ∠ _______ ≅ ∠ _______ .

Answer:
∠G = ∠K by using the base angles theorem.

Question 2.
If ∠KHJ ≅∠KJH, then ______ ≅ ______ .

Answer:
\(\overline{K H}\) ≅ \(\overline{K J}\)

Question 3.
Find the length of \(\overline{S T}\) of the triangle at the left.

Answer:
As per Collary to the Converse of the Base Angles Theorem, ST = SU = TU
So, \(\overline{S T}\) = 5

Question 4.
Find the value of x and y in the diagram.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 98

Answer:
As all sides are equal, angles are also equal.
So, x° = 60°
y° = 90° – 60° = 30°

Question 5.
In Example 4, show that ∆PTS ≅ ∆QTR

Answer:
TS ≅ TR,
PS ≅ QR
and ∠TPS = ∠TQR
So, ∆PTS ≅ ∆QTR by SAS theorem.

Exercise 5.4 Equilateral and Isosceles Triangles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Describe how to identify the vertex angle of an isosceles triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 1

Question 2.
WRITING
What is the relationship between the base angles of an isosceles triangle? Explain.

Answer:
The base angle of an isosceles triangle is congruent.

Explanation:
An isosceles triangle consists of two equal legs and one base. There are two equal angles on the base. These are called base angles. So, the base angles are opposite to the congruent sides. Using the base angle theorem, we can conclude that the base angles of an isosceles triangle are congruent.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6. copy and complete the statement. State which theorem you used.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 99

Question 3.
If \(\overline{A E}\) ≅ \(\overline{D E}\) then ∠_____ ≅ ∠_____ .
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 3

Question 4.
If \(\overline{A B}\) ≅ \(\overline{E B}\) then ∠_____ ≅ ∠_____ .

Answer:
If \(\overline{A B}\) ≅ \(\overline{E B}\) then ∠A ≅ ∠E by the Base Angles Theorem.

Question 5.
If ∠D ≅ ∠CED, then _______ ≅ _______ .
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 5

Question 6.
If ∠EBC ≅ ∠ECB, then _______ ≅ _______ .

Answer:
If ∠EBC ≅ ∠ECB, then EC ≅ EB by the converse of the base angles theorem.

In Exercises 7-10. find the value of x.

Question 7.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 100
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 101

Answer:
x = 16

Explanation:
As ∠L, ∠N are 60 degrees so ∠M is also 60 degrees
So, LM = LN = MN
x = 16

Question 9.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 102
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 103

Answer:
x = 20°

Explanation:
All the sides are equal so angles are also equal
3x + 3x + 3x = 180
9x = 180
x = 20°

Question 11.
MODELING WITH MATHEMATICS
The dimensions of a sports pennant are given in the diagram. Find the values of x and y.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 104
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 11

Question 12.
MODELING WITH MATHEMATICS
A logo in an advertisement is an equilateral triangle with a side length of 7 centimeters. Sketch the logo and give the measure of each side.

Answer:
The remaining sides of the logo are 7 cm

Explanation:
An equilateral triangle has three sides of equal length. So, the remaining two sides are also 7 cm.
The logo is Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 1

In Exercises 13-16, find the values of x and y.

Question 13.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 105
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 106

Answer:

Question 15.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 107
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 108

Answer:
x = 7.416, y = 5.25

Explanation:
3x – 5 = 5y – 4
3x – 5y + 4 = 0 —- (i)
y + 12 = 3x – 5
3x – y – 17 = 0 —- (ii)
Subtract (i) from (ii)
3x – y – 17 – 3x + 5y – 4 = 0
4y – 21 = 0
4y = 21
y = 5.25
Put y = 5.25 in (ii)
3x – 5.25 – 17 = 0
3x – 22.25 = 0
3x = 22.25
x = 7.416

CONSTRUCTION
In Exercises 17 and 18, construct an equilateral triangle whose sides are the given length.

Question 17.
3 inches
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 17

Question 18.
1.25 inches

Answer:
Construct the base leg with length 1.25 inches and draw an arc with center at one of the endpoints and a radius 1.25 in. Draw another arc that intersects the first arc with the same radius and from another endpoint of the base leg. Connect intersection of arcs to two endpoints of the base to form an equilateral triangle.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 2

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding the length of \(\overline{B C}\).
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 109
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 19

Question 20.
PROBLEM SOLVING
The diagram represents part of the exterior of the Bow Tower in Calgary. Alberta, Canada, In the diagram. ∆ABD and ∆CBD arc congruent equilateral triangles.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 110

a. Explain why ∆ABC is isosceles.

Answer:
AB and BC have the same length.
So, ∆ABC is isosceles.

b. Explain ∠BAE ≅ ∠BCE.

Answer:
AC is a diagonal that cuts the angles of the opposite vertex into two half. So, ∠BAE ≅ ∠BCE.

c. Show that ∆ABE and ∆CBE are congruent.

Answer:
AB ≅ BC, BE ≅ BE and ∠BAE ≅ ∠BCE
By SAS theorem, ∆ABE and ∆CBE are congruent.

d. Find the measure of ∠BAE.

Answer:
∠BAE ≅ ∠BCE.

Question 21.
FINDING A PATTERN
In the pattern shown. each small triangle is an equilateral triangle with an area of 1 square unit.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 111
a. Explain how you know that an triangle made out of equilateral triangles is equilateral.
b. Find the areas of the first four triangles in the pattern.
c. Describe any patterns in the areas. Predict the area of the seventh triangle in the pattern. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 21

Question 22.
REASONING
The base of isosceles ∆XYZ is \(\overline{Y Z}\). What
can you prove? Select all that apply.
(A) \(\overline{X Y}\) ≅ \(\overline{X Z}\)
(B) ∠X ≅ ∠Y
(C) ∠Y ≅ ∠Z
(D) \(\overline{Y Z}\) ≅ \(\overline{Z X}\)

Answer:
(A) \(\overline{X Y}\) ≅ \(\overline{X Z}\)
(C) ∠Y ≅ ∠Z

Explanation:
If, \(\overline{Y Z}\) is base of isosceles ∆XYZ.
Then, let us take \(\overline{X Y}\) ≅ \(\overline{X Z}\)
If \(\overline{X Y}\) ≅ \(\overline{X Z}\), then ∠Y ≅ ∠Z.

In Exercises 23 and 24, find the perimeter of the triangle.

Question 23.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 112
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 23

Question 24.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 113

Answer:
Perimeter = 39

Explanation:
21 – x = 2x – 3
21 + 3 = 2x + x
24 = 3x
x = 8
Perimeter = 13 + 13 + 13 = 39

MODELING WITH MATHEMATICS
In Exercises 25 – 28. use the diagram based on the color wheel. The 12 triangles in the diagram are isosceles triangles with congruent vertex angles.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 114

Question 25.
Complementary colors lie directly opposite each other on the color wheel. Explain how you know that the yellow triangle is congruent to the purple triangle.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 25

Question 26.
The measure of the vertex angle of the yellow triangle is 30°. Find the measures of the base angles.

Answer:
The measures of base angles are 75°, 75°.

Explanation:
2x° + 30° = 180°
2x = 180 – 30 = 150°
x° = 75°

Question 27.
Trace the color wheel. Then form a triangle whose vertices are the midpoints of the bases of the red. yellow. and blue triangles. (These colors are the primary colors.) What type of triangle is this?
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 27

Question 28.
Other triangles can be brined on the color wheel that are congruent to the triangle in Exercise 27. The colors on the vertices of these triangles are called triads. What are the possible triads?
Answer:

Question 29.
CRITICAL THINKING
Are isosceles triangles always acute triangles? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 29

Question 30.
CRITICAL THINKING
Is it possible for an equilateral triangle to have an angle measure other than 60°? Explain your reasoning.

Answer:
No, because an equilateral triangle has all the angles of equal measurement and the sum of angles in a triangle must be equal to 180 degrees. So, it is not possible for an angle measure other than 60° in an equilateral triangle.

Question 31.
MATHEMATICAL CONNECTIONS
The lengths of the sides of a triangle are 3t, 5t – 12, and t + 20. Find the values of t that make the triangle isosceles. Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 31

Question 32.
MATHEMATICAL CONNECTIONS
The measure of an exterior angle of an isosceles triangle is x°. Write expressions representing the possible angle measures of the triangle in terms of x.

Answer:
If x is the exterior angle of a traingle then the interior angle of the triangle that is supplementary with this angle is equal to 180° – x.
We know that the exterior angle of a triangle is equal to the sum of 2 opposite interior angles on the triangle. Based on this, it can be said that the 2 other interior angles can be 0.25x and 0.75x

Question 33.
WRITING
Explain why the measure of the vertex angle of an isosceles triangle must be an even number of degrees when the measures of all the angles of the triangle are whole numbers.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 33

Question 34.
PROBLEM SOLVING
The triangular faces of the peaks on a roof arc congruent isosceles triangles with vertex angles U and V.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 115
a. Name two angles congruent to ∠WUX. Explain your reasoning.
b. Find the distance between points U and V.

Answer:
a. ∠WUX ≅ ∠XVY
∠WUX ≅ ∠UXV
b. The distance between thee points U and V is 8 m.

Question 35.
PROBLEM SOLVING
A boat is traveling parallel to the shore along \(\vec{R}\)T. When the boat is at point R, the captain measures the angle to the lighthouse as 35°. After the boat has traveled 2.1 miles, the captain measures the angle to the lighthouse to be 70°.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 116
a. Find SL. Explain your reasoning.
b. Explain how to find the distance between the boat and the shoreline.
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 35

Question 36.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry, all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry, do all equiangular triangles have the same angle measures? Justify your answer.
Answer:

Question 37.
PROVING A COROLLARY
Prove that the Corollary to the Base Angles Theorem (Corollary 5.2) follows from the Base Angles Theorem (Theorem 5.6).
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 37

Question 38.
HOW DO YOU SEE IT?
You are designing fabric purses to sell at the school fair.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 117
a. Explain why ∆ABE ≅ ∆DCE.
b. Name the isosceles triangles in the purse.
c. Name three angles that are congruent to ∠EAD.

Answer:
a. AB ≅ CD, AE ≅ ED and ∠A ≅ ∠D
So, ∆ABE ≅ ∆DCE by SAS theorem.
b. ∠EAD ≅ ∠AED ≅ ∠EDA

Question 39.
PROVING A COROLLARY
Prove that the Corollary to the Converse of the Base Angles Theorem (Corollary 5.3) follows from the Converse of the Base Angles Theorem (Theorem 5.7)
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 39

Question 40.
MAKING AN ARGUMENT
The coordinates of two points are T(0, 6) and U(6, 0) Your friend claims that points T, U, and V will always be the vertices of an isosceles triangle when V is any point on the line y = x. Is your friend correct? Explain your reasoning.

Answer:
My friend is wrong.

Explanation:
We know that the coordinates of two points are T(0, 6) and U (6, 0)
We also consider that V is any point on the line y = x
The coordinates of V be x = 3, y = 3. So V(3, 3)
draw a graph.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 3
When we connect the points T, V, U we can see that we have not obtained an isosceles triangle.

Question 41.
PROOF
Use the diagram to prove that ∆DEF is equilateral.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 118
Given ∆ABC is equilateral
∠CAD ≅ ∠ABE ≅ ∠BCF
Prove ∆DEF is equilateral
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 41.1
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 41.2

Maintaining Mathematical Proficiency

Use the given property to complete the statement.

Question 42.
Reflexive Property of Congruence (Theorem 2. 1): ________ ≅ \(\overline{S E}\)

Answer:
\(\overline{S E}\) ≅ \(\overline{S E}\)

Question 43.
Symmetric Property of Congruence (Theorem 2.1): If ________ ≅ ________, then \(\overline{R S}\) ≅ \(\overline{J K}\)
Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.4 a 43

Question 44.
Transitive Property of Congruence (Theorem 2.1): If \(\overline{E F}\) ≅ \(\overline{P Q}\), and \(\overline{P Q}\) ≅ \(\overline{U V}\) ________ ≅ ________.

Answer:
Transitive Property of Congruence (Theorem 2.1): If \(\overline{E F}\) ≅ \(\overline{P Q}\), and \(\overline{P Q}\) ≅ \(\overline{U V}\), then \(\overline{U V}\) ≅ \(\overline{E F}\).

5.1 to 5.4 Quiz

Find the measure of the exterior angle.

Question 1.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 119

Answer:
x° = 110°

Explanation:
x° = 180° – (180° – (30° + 80°))
= 180° – (180° – 110°)
= 180° – 70°
= 110°

Question 2.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 120

Answer:
y° = 138°

Explanation:
90° + 5x + 2 + 6x = 180°
11x + 92 = 180°
11x = 180 – 92
11x = 88
x = 8°
y° = 180° – (5(8) + 2)
= 180° – 42
y° = 138°

Question 3.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 121

Answer:
(15x + 34)° = 109°

Explanation:
29 + 12x + 26 + third_angle = 180
third_angle = 180 – 55 – 12x = 125 – 12x
15x + 34 = 180 – (125 – 12x)
15x + 34 = 180 – 125 + 12x
15x – 12x = 55 – 34
3x = 21
x = 7
15x + 34 = 15(7) + 34
= 105 + 34 = 139

Identify all pairs of congruent corresponding parts. Then write another congruence statement for the polygons.

Question 4.
∆ABC ≅ ∆DEF
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 122

Answer:
∠C = ∠F, ∠B = ∠E, ∠A = ∠D

Question 5.
QRST ≅ WXYZ
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 123

Answer:
∠Q = ∠W, ∠R = ∠X, ∠S = ∠Y, ∠T = ∠Z

Decide whether enough information is given to prove that the triangles are congruent using the SAS Congruence Theorem (Thm 5.5). If so, write a proof. If not, explain why.

Question 6.
∆CAD, ∆CBD
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 124

Answer:
∠A ≅ ∠B, AD ≅ BD, CD ≅ CD (Reflexive Property)
∆CAD ≅ ∆CBD by the SAS Congruence Theorem.

Question 7.
∆GHF, ∆KHJ
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 125

Answer:
GH ≅ HK, FH ≅ HJ and ∠GHF ≅ ∠JHK
∆GHF ≅ ∆KHJ by the SAS Congruence Theorem.

Question 8.
∆LWP, ∆NMP
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 126

Answer:
LM ≅ MN, MP ≅ MP by reflexive property
∠LMP ≅ ∠PMN
So, ∆LWP ≅ ∆NMP by the SAS Congruence Theorem.

Copy and complete the statement. State which theorem you used.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 127

Question 9.
If VW ≅ WX, then ∠______ ≅ ∠ ________.

Answer:
If VW ≅ WX, then ∠V ≅ ∠X.

Question 10.
If XZ ≅ XY. then∠______ ≅ ∠ ________.

Answer:
If XZ ≅ XY. then∠Y ≅ ∠Z.

Question 11.
If ∠ZVX ≅∠ZXV, then ∠_______ ≅ ∠_______.

Answer:
If ∠ZVX ≅∠ZXV, then ∠ZV ≅ ∠ ZX.

Question 12.
If ∠XYZ ≅∠ZXY, then ∠______ ≅ ∠ ________.

Answer:
If ∠XYZ ≅∠ZXY, then ∠XY ≅ ∠ ZX.

Find the values of x and y.

Question 13.
∆DEF ≅ ∆QRS
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 128

Answer:
x° = 14°, y = 9

Explanation:
EF ≅ SR, ED ≅ RQ, DF ≅ SQ
5y – 7 = 38
5y = 45
y = 9
123 + 29 + 2x + 2 = 180
154 + 2x = 180
2x = 28
x° = 14°

Question 14.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 129

Answer:
x = 5, y° = 20°

Explanation:
5x – 1 = 24
5x = 25
x = 5
2 • 6y° = 360° – (60° + 60°) = 240
y° = 20°

Question 15.
In a right triangle, the measure of one acute angle is 4 times the difference of the measure of the other acute angle and 5. Find the measure ol each acute angle in the triangle. (Section 5.1)

Answer:
The first angle = 22°
The second angle = 68°

Explanation:
Assuming the measure of the first acute angle equals x
The measure of the second acute angle = 4(x – 5)
x + 4(x – 5) = 90
5x – 20 = 90
5x = 110
x = 22
The first angle = 22°
The second angle = 90 – 22 = 68°

Question 16.
The figure shows a stained glass window. (Section 5.1 and Section 5.3)

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 130

a. Classify triangles 1 – 4 by their angles.

Answer:
Triangle 1 is a right-angled trinagle, 2 is an acute angle triangle, 3 is an obtuse angle triangle and 4 is an equilateral triangle.

b. Classify triangles 4 – 6 by their sides.

Answer:
Triangle 4 is an equilateral triangle, 5 is a scalene triangle, 6 is an isosceles triangle.

c. Is there enough information given to prove that ∆7 ≅ ∆8? If so, label the vertices
and write a proof. If not, determine what additional information is needed.

Answer:
Yes, two congruent sides and one congruent angle.
By Side Angle Side Congruent Theorem, ∆7 ≅ ∆8.

5.5 Proving Triangle Congruence by SSS

Exploration 1

Drawing Triangles

Work with a partner.
Use dynamic geometry software.

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 131

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 132

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 133

a. Construct circles with radii of 2 units and 3 units centered at the origin. Label the origin A. Then draw \(\overline{B C}\) of length 4 units.
Answer:

b. Move \(\overline{B C}\) so that B is on the smaller circle and C is on the larger circle. Then draw ∆ABC.
Answer:

c. Explain why the side lengths of ∆ABC are 2, 3, and 4 units.
Answer:

d. Find m∠A, m∠B, and m∠C.
Answer:

e. Repeat parts (b)and (d) several times, moving \(\overline{B C}\) to different locations. Keep track of ‘our results by copying and completing the table below. What can you conclude?
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technology to help visualize the results of varying assumptions, explore consequences, and compare predictions with data.
Answer:

Communicate Your Answer

Question 2.
What can you conclude about two triangles when you know the corresponding sides are congruent?
Answer:

Question 3.
How would you prove your conclusion in Exploration 1(e)?
Answer:

Lesson 5.5 Proving Triangle Congruence by SSS

Monitoring Progress

Decide whether the congruence statement is true. Explain your reasoning.

Question 1.
∆DFG ≅ ∆HJK
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 134

Answer:
True

Explanation:
DG ≅ HK, DF ≅ HJ, FG ≅ JK
According to SSS Congruence Theorem, ∆DFG ≅ ∆HJK.

Question 2.
∆ACB ≅ ∆CAD
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 135

Answer:
True

Explanation:
AB ≅ CD, AD ≅ BC, AC ≅ AC by reflexive property
So, ∆ACB ≅ ∆CAD by SSS Congruence Theorem

Question 3.
∆QPT ≅ ∆RST
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 136

Answer:
True

Explanation:
QT ≅ TR, PQ ≅ SR, PT ≅ ST
So, ∆QPT ≅ ∆RST by SSS Congruence Theorem

Determine whether the figure is stable. Explain your reasoning.

Question 4.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 137
Answer:
The figure is not stable. Because it doesn’t have a triangle. By the SSS Congruence Theorem, those triangles cannot change shape.

Question 5.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 138

Answer:
The figure is stable as the diagonal forms the triangle. By the SSS Congruence Theorem, these triangles cannot change the shape, so the figure is stable.

Question 6.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 139

Answer:
The figure is stable as the diagonal forms the triangle. By the SSS Congruence Theorem, these triangles cannot change the shape, so the figure is stable.

Use the diagram.

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 140

Question 7.
Redraw ∆ABC and ∆DCB side by side with corresponding parts in the same position.

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.5 1

Question 8.
Use the information in the diagram to prove that ∆ABC ≅ ∆DCB.

Answer:
AC ≅ BD, BC ≅ BC by reflexive property of congruence, and AB ≅ CD
So, ∆ABC ≅ ∆DCB by SSS Congruence Theorem.

Exercise 5.5 Proving Triangle Congruence by SSS

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The side opposite the right angle is called the __________of the right triangle.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 1

Question 2.
WHICH ONE DOESNT BELONG?
Which triangles legs do not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 142

Answer:
The second triangle legs do not belong with the other three.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, decide whether enough information is given to prove that the triangles are congruent using the SSS Congruence Theorem (Theorem 5.8). Explain.

Question 3.
∆ABC, ∆DBE
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 141
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 3

Question 4.
∆PQS, ∆RQS
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 143

Answer:
No, PQ ≅ QR, QS ≅ QS by reflexive property.
We need one more congruence property to prove those triangles are congruent.

In Exercises 5 and 6, decide whether enough information is given to prove that the triangles are congruent using the HL Congruence Theorem (Theorem 5.9). Explain.

Question 5.
∆ABC, ∆FED
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 144
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 5

Question 6.
∆PQT, ∆SRT
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 145

Answer:
Yes

Explanation:
PT ≅ TS, RT ≅ TQ, ∠PTQ ≅ ∠RTS
∆PQT ≅ ∆SRT by SAS congruence theorem.

In Exercises 7-10. decide whether the congruence statement is true. Explain your reasoning.

Question 7.
∆RST ≅ ∆TQP
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 146
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 7

Question 8.
∆ABD ≅ ∆CDB
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 147

Answer:
Yes

Explanation:
AB ≅ CD, AD ≅ BC
BD ≅ BD by reflexive property of congruence
So, ∆ABD ≅ ∆CDB by SSS Congruence theorem.

Question 9.
∆DEF ≅ ∆DGF
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 148
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 9

Question 10.
∆JKL ≅ ∆LJM
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 149

Answer:
Yes

Explanation:
KJ ≅ LM, KL ≅ JM
JL ≅ JL by reflexive property of congruence
So, ∆JKL ≅ ∆LJM by SSS Congruence Theorem.

In Exercises 11 and 12, determine whether the figure is stable. Explain your reasoning.

Question 11.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 150
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 11

Question 12.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 151

Answer:
No, the figure has no diagonals. So the figure is not stable.

In Exercises 13 and 14, redraw the triangles so they are side by side with corresponding parts in the same position. Then write a proof.

Question 13.
Given \(\overline{A C}\) ≅ \(\overline{B D}\)
\(\overline{A B}\) ⊥ \(\overline{A D}\)
\(\overline{C D}\) ⊥ \(\overline{A D}\)
Prove ∆BAD ≅ ∆CDA
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 152
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 13

Question 14.
Given G is the midpoint of \(\overline{E H}\), \(\overline{F G}\) ≅ \(\overline{G I}\), ∠E and ∠H are right angles.
Prove ∆EFG ≅ ∆HIG
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 153

Answer:
EG ≅ GH as G is the midpoint of \(\overline{E H}\)
\(\overline{F G}\) ≅ \(\overline{G I}\), ∠E ≅ ∠H
∆EFG ≅ ∆HIG by SAS congruence theorem.

In Exercises 15 and 16. write a proof.

Question 15.
Given \(\overline{L M}\) ≅ \(\overline{J K}\), \(\overline{M J}\) ≅ \(\overline{K L}\)
Prove ∆LMJ ≅ ∆JKL
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 154
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 15

Question 16.
Given \(\overline{W X}\) ≅ \(\overline{V Z}\), \(\overline{W Y}\) ≅ \(\overline{V Y}\), \(\overline{Y Z}\) ≅ \(\overline{Y X}\)
Prove ∆VWX ≅ ∆WVZ
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 155

Answer:
XY ≅ YZ, WY ≅ VY, WX ≅ VZ
∆VWX ≅ ∆WVZ by SSS Congruence theorem

CONSTRUCTION
In Exercises 17 and 18, construct a triangle that is congruent to ∆QRS using the SSS Congruence Theorem Theorem 5.8).

Question 17.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 156
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 17

Question 18.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 157

Answer:
At first, construct a side that is congruent to QS. Draw an arc with the compass with Q as center and radius as QR. Draw another arc that intersects the first arc with S as center and radius as SR. Join the point to Q and S to form a circle that is congruent to ∆QRS.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.5 2

Question 19.
ERROR ANALYSIS
Describe and correct the error in identifying congruent triangles.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 158
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in determining the value of x that makes the triangles congruent.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 159

Answer:
4x + 4 = 6x
2x = 4
x = 2

Question 21.
MAKING AN ARGUMENT
Your friend claims that in order to use the SSS Congruence Theorem (Theorem 5.8) Lo prove that two triangles are congruent, both triangles must be equilateral triangles. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 21

Question 22.
MODELING WITH MATHEMATICS
The distances between consecutive bases on a softball field are the same. The distance from home plate to second base is the same as the distance from first base to third base. The angles created at each base are 90°. Prove
∆HFS ≅ ∆FST ≅ ∆STH
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 160

Answer:
Given that,
HS ≅ FT
FS ≅ FS by reflexive property of congruence
∠HFS ≅ ∠TFS
So, ∆HFS ≅ ∆FST by SAS congruence theorem
ST ≅ ST by reflexive property of congruence
HS ≅ FT
∠FST ≅ ∠TSH
So, ∆FST ≅ ∆STH by SAS congruence theorem
Therefore, ∆HFS ≅ ∆FST ≅ ∆STH

Question 23.
REASONING
To support a tree you attach wires from the trunk of the tree to stakes in the ground, as shown in the diagram.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 161
a. What additional information do you need to use the HL Congruence Theorem (Theorem 5.9) to prove that ∆JKL ≅ ∆MKL?
b. Suppose K is the midpoint of JM. Name a theorem you could use to prove that ∆JKL ≅ ∆MKL. Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 23

Question 24.
REASONING
Use the photo of the Navajo rug, where \(\overline{B C}\) ≅ \(\overline{D E}\) and \(\overline{A C}\) ≅ \(\overline{C E}\)
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 162
a. What additional intormation do you need to use the SSS Congruence Theorem (Theorem 5.8) to prove that ∆ABC ≅ ∆CDE?
b. What additional information do you need to use the HL Congruence Theorem (Theorem 5.9) to prove that ∆ABC ≅ ∆CDE?

Answer:
a. AB ≅ CD is required o use the SSS Congruence Theorem (Theorem 5.8) to prove that ∆ABC ≅ ∆CDE
b. One angle should be the right angle in those triangles to use the HL Congruence Theorem (Theorem 5.9) to prove that ∆ABC ≅ ∆CDE

In Exercises 25-28. use the given coordinates to determine whether ∆ABC ≅ ∆DEF.

Question 25.
A(- 2, – 2), B(4, – 2), C(4, 6), D(5, 7), E(5, 1), F(13, 1)
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 25

Question 26.
A(- 2, 1), B(3, – 3), C(7, 5), D(3, 6), E(8, 2), F( 10, 11)

Answer:
∆ABC ≇ ∆DEF

Explanation:
AB = √(3 + 2)² + (-3 – 1)²
= √25 + 16 = √41
BC = √(7 – 3)² + (5 + 3)²
= √16 + 64 = √80 = 8.9
AC = √(7 + 2)² + (5 – 1)²
= √81 + 16 = √97 = 9.8
DE = √(8 – 3)² + (2 – 6)²
= √25 + 16 = √41
EF = √(10 – 8)² + (11 – 2)²
= √4 + 81 = √85 = 9.2
DF = √(10 – 3)² + (11 – 6)²
= √49 + 25 = √74 = 8.6
AB ≅ DE, BC ≅ EF, AC ≇ DF. So, ∆ABC ≇ ∆DEF

Question 27.
A(0, 0), B(6, 5), C(9, 0), D(0, – 1), E(6, – 6), F(9, – 1)
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 27

Question 28.
A(- 5, 7), B(- 5, 2), C(0, 2), D(0, 6), E(0, 1), F(4, 1)

Answer:
∆ABC ≅ ∆DEF

Explanation:
AB = √(-5 + 5)² + (2 – 7)² = √0 + 25 = 5
BC = √(0 + 5)² + (2 – 2)² = √0 + 25 = 5
AC = √(0 + 5)² + (2 – 7)² = √25 + 25 = √50
DE = √(0 – 0)² + (1 – 6)² = √0 + 25 = 5
EF = √(0 – 4)² + (1 – 1)² = √0 + 16 = 4
DF = √(4 – 0)² + (1 – 6)² = √16 + 25 = √41
AB = DE, BC ≅ EF, AC ≅ DF. So, ∆ABC ≅ ∆DEF by SSS Congruence theorem.

Question 29.
CRITICAL THINKING
You notice two triangles in the tile floor of a hotel lobby. You want to determine whether the triangles are congruent. but you only have a piece of string. Can you determine whether the triangles are congruent? Explain.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 29

Question 30.
HOW DO YOU SEE IT?
There are several theorems you can use to show that the triangles in the “square” pattern are congruent. Name two of them.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 163
Answer:

Question 31.
MAKING AN ARGUMENT
Your cousin says that ∆JKL is congruent to ∆LMJ by the SSS Congruence Theorem (Thm. 5.8). Your friend says that ∆JKL is congruent to ∆LMJ by the HL Congruence Theorem (Thm. 5.9). Who is correct? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 164
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 31

Question 32.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry. all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. In spherical geometry. do you think that two triangles are congruent if their corresponding sides are congruent? Justify your answer.
Answer:

USING TOOLS
In Exercises 33 and 34, use the given information to sketch ∆LMN and ∆STU. Mark the triangles with the given information.

Question 33.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 165
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 33

Question 34.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 166
Answer:

Question 35.
CRITICAL THINKING
The diagram shows the light created by two spotlights, Both spotlights are the same distance from the stage.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 167
Answer:
a. Show that ∆ABD ≅ ∆CBD. State which theorem or postulate you used and explain your reasoning.
b. Are all four right triangles shown in the diagram Congruent? Explain your reasoning.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 35

Question 36.
MATHEMATICAL CONNECTIONS
Find all values of x that make the triangles congruent. Explain.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 168

Answer:
The possible values of x are 3, 6, 5.

Explanation:
5x = 4x + 3
5x – 4x = 3
x = 3
5x – 2 = 3x + 10
2x = 12
x = 6
5x – 2 = 4x + 3
x = 5
5x = 3x + 10
2x = 10
x = 5
The possible values of x are 3, 6, 5.

Maintaining Mathematical proficiency

Use the congruent triangles.

Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 169

Question 37.
Name the Segment in ∆DEF that is congruent to \(\overline{A C}\).
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 37

Question 38.
Name the segment in ∆ABC that is congruent to \(\overline{E F}\).

Answer:
BC is congruent to EF.

Question 39.
Name the angle in ∆DEF that is congruent to ∠B.
Answer:
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 5.5 a 39

Question 40.
Name the angle in ∆ABC that is congruent to ∠F.

Answer:
∠C is congruent to ∠F.

5.6 Proving Triangle Congruence by ASA and AAS

Exploration 1

Determining Whether SSA Is Sufficient

Work with a partner.
a. Use dynamic geometry software to construct ∆ABC. Construct the triangle so that vertex B is at the origin. \(\overline{A B}\) has a length of 3 units. and \(\overline{B C}\) has a length of 2 units.
Answer:

b. Construct a circle with a radius of 2 units centered at the origin. Locate point D where the circle intersects \(\overline{A C}\). Draw \(\overline{B D}\).
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 170
Answer:

c. ∆ABC and ∆ABD have two congruent sides and a non included congruent angle.
Name them.
Answer:

d. Is ∆ABC ≅ ∆ABD? Explain your reasoning.
Answer:

e. Is SSA sufficient to determine whether two triangles are congruent? Explain your reasoning.
Answer:

Exploration 2

Determining Valid Congruence Theorems

Work with a partner. Use dynamic geometry software to determine which of the following are valid triangle congruence theorems. For those that are not valid. write a counter example. Explain your reasoning.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to recognize and use counterexamples.

Possible Congruence Theorem Valid or not valid?
SSS
SSA
SAS
AAS
ASA
AAA

Answer:

Communicate Your Answer

Question 3.
What information is sufficient to determine whether two triangles are congruent?
Answer:

Question 4.
Is it possible to show that two triangles are congruent using more than one congruence theorem? If so, give an example.
Answer:

Lesson 5.6 Proving Triangle Congruence by ASA and AAS

Monitoring Progress

Question 1.
Can the triangles be proven congruent with the information given in the diagram? If so, state the theorem you would use.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 171

Answer:
WX ≅ YZ, XY ≅ WZ, and ∠1 ≅ ∠3
So, WXY ≅ WYZ by the AAS congruence theorem.

Question 2.
In the diagram, \(\overline{A B}\) ⊥ \(\overline{A D}\), \(\overline{D E}\) ⊥ \(\overline{A D}\), and \(\overline{A C}\) ≅ \(\overline{D C}\) . Prove ∆ABC ≅ ∆DEF.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 172

Answer:
\(\overline{A C}\) ≅ \(\overline{D C}\), ∠A ≅ ∠D
So, the given information is not enough to prove that ∆ABC ≅ ∆DEF.

Question 3.
In the diagram, ∠S ≅ ∠U and \(\overline{B D}\)\(\overline{B D}\) . Prove that ∆RST ≅ ∆VYT
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 173

Answer:
∠S ≅ ∠U, RS ≅ UV
So, the given information is not enough to prove that ∆RST ≅ ∆VYT

Lesson 5.6 Proving Triangle Congruence by ASA and AAS

Vocabulary and Core Concept Check

Question 1.
WRITING
How arc the AAS Congruence Theorem (Theorem 5. 11) and the ASA Congruence
Theorem (Theorem 5.10) similar? How are they different?
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 1

Question 2.
WRITING
You know that a pair of triangles has two pairs of congruent corresponding angles. What other information do you need to show that the triangles are congruent?

Answer:
In order to use the AAS or ASA Congruence theorem, you need to know that two pairs of corresponding angles and one more pair of corresponding sides are congruent.

Monitoring Progress and Modeling with Mathematics

In Exercises 3-6, decide whether enough information is given to prove that the triangles are congruent. If so, state the theorem you would use.

Question 3.
∆ABC, ∆QRS
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 174
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 3

Question 4.
∆ABC, ∆DBC
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 175

Answer:
No, two angles are not sufficient to determine congruence.

Question 5.
∆XYZ, ∆JKL
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 176
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 5

Question 6.
∆RSV, ∆UTV
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 177

Answer:
No, one side and one included angle is not sufficient to determine congruence.

In Exercises 7 and 8, state the third congruence statement that is needed to prove that ∆FGH ≅ ∆LMN the given theorem.

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 178
Question 7.
Given \(\overline{G H}\) ≅ \(\overline{M N}\), ∠G ≅ ∠M, _______ = ________
Use the AAS Congruence Theorem (Thm. 5.11).
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 7

Question 8.
Given \(\overline{F G}\) ≅ \(\overline{L M}\), ∠G ≅ ∠M, _______ = ________
Use the ASA Congruence Theorem (Thm. 5.10).

Answer:
\(\overline{F G}\) ≅ \(\overline{L M}\), ∠G ≅ ∠M, ∠F ≅ ∠L

In Exercises 9 – 12. decide whether you can use the given information to prove that ∆ABC ≅ ∆DEF Explain your reasoning.

Question 9.
∠A ≅ ∠G, ∠C ≅∠F, \(\overline{A C}\) ≅ \(\overline{D F}\)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 9

Question 10.
∠C ≅ ∠F, \(\overline{A B}\) ≅ \(\overline{D E}\), \(\overline{B C}\) ≅ \(\overline{E F}\)

Answer:
Yes, ∆ABC ≅ ∆DEF by SAS congruence theorem

Question 11.
∠B ≅ ∠E, ∠C ≅∠F, \(\overline{A C}\) ≅ \(\overline{D E}\)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 11

Question 12.
∠A ≅ ∠D, ∠B ≅∠E, \(\overline{B C}\) ≅ \(\overline{E F}\)

Answer:
Yes, ∆ABC ≅ ∆DEF by AAS theorem.

CONSTRUCTION
In Exercises 13 and 14, construct a triangle that is congruent to the given triangle using the ASA Congruence Theorem (Theorem 5.10). Use a compass and straightedge.

Question 13.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 179
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 13

Question 14.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 180

Answer:
Construct a side that is similar to JK. With J as center and JL as radius, draw an arc. With K as center and KL as radius draw another arc that intersects the first arc. Label the intersection of arcs as L. Connect LK and LJ.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.6 1

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error.

Question 15.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 181
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 15

Question 16.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 182

Answer:
△QRS ≅ △VWX by ASA congruence theorem.

PROOF
In Exercises 17 and 18, prove that the triangles are congruent using the ASA Congruence Theorem (Theorem 5.10).

Question 17.
Given M is the midpoint of \(\overline{N L}\).
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 183
Prove ∆NQM ≅ ∆MPL
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 184
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 17.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 17.2

Question 18.
Given \(\overline{A J}\) ≅ \(\overline{K C}\) ∠BJK ≅ ∠BKJ, ∠A ≅ ∠C
Prove ∆ABK ≅ ∆CBJ
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 185

Answer:
AJ ≅ KC, ∠K ≅ ∠J, ∠A ≅ ∠C
So, ∆ABK ≅ ∆CBJ by ASA congruence theorem.

PROOF
In Exercises 19 and 20, prove that the triangles are congruent using the AAS Congruence Theorem (Theorem 5.11).

Question 19.
Given \(\overline{V W}\) ≅ \(\overline{U W}\), ∠X ≅ ∠Z
Prove ∆XWV ≅ ∆ZWU
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 186
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 19

Question 20.
Given ∠NKM ≅∠LMK, ∠L ≅∠N
Prove ∆NMK ≅ ∆LKM
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 187

Answer:
Given ∠NKM ≅∠LMK, ∠L ≅∠N
KM ≅ KM ≅ by reflexive property of congruence
∆NMK ≅ ∆LKM by AAS congruence theorem.

PROOF
In Exercises 21-23, write a paragraph proof for the theorem about right triangles.

Question 21.
Hypotenuse-Angle (HA) Congruence Theorem
If an angle and the hypotenuse of a right triangle are congruent to an angle and the hypotenuse of a second right triangle, then the triangles are congruent.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 21

Question 22.
Leg-Leg (LL) Congruence Theorem
If the legs of a right triangle are congruent to the legs of a second right triangle, then the triangles are congruent.

Answer:
You are given that egs of a right triangle are congruent to the legs of a second right triangle so triangles are congruent by LL congruence theorem.

Question 23.
Angle-Leg (AL) Congruence Theorem
If an angle and a leg of a right triangle are congruent to an angle and a leg of a second right triangle, then the triangles are Congruent.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 23

Question 24.
REASONING
What additional in information do you need to prove ∆JKL ≅ ∆MNL by the ASA Congruence Theorem (Theorem 5. 10)?
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 188
(A) \(\overline{K M}\) ≅ \(\overline{K J}\)
(B) \(\overline{K H}\) ≅ \(\overline{N H}\)
(C) ∠M ≅ ∠J
(D) ∠LKJ ≅ ∠LNM

Answer:

Question 25.
MATHEMATICAL CONNECTIONS
This toy contains △ABC and △DBC. Can you conclude that △ABC ≅ △DBC from the given angle measures? Explain
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 189
m∠ABC = (8x – 32)°
m∠DBC = (4y – 24)°
m∠BCA = (5x + 10)°
m∠BCD = (3y + 2)°
m∠CAB = (2x – 8)°
m∠CDB = (y – 6)°
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 25.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 25.2

Question 26.
REASONING
Which of the following congruence statements are true? Select all that apply.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 190
(A) \(\overline{B D}\) ≅ \(\overline{B D}\)
(B) ∆STV ≅ ∆XVW
(C) ∆TVS ≅ ∆VWU
(D) ∆VST ≅ ∆VUW
Answer:
(C) ∆TVS ≅ ∆VWU by ASA congruence theorem.
∠T ≅ ∠V, TS ≅ VU, ∠S ≅ ∠U
(D) ∆VST ≅ ∆VUW by ASA congruence theorem.
∠T ≅ ∠V, TS ≅ VU, ∠S ≅ ∠U

Question 27.
PROVING A THEOREM
Prove the Converse of the Base Angles Theorem (Theorem 5.7). (Hint: Draw an auxiliary line inside the triangle.)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 27.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 27.2

Question 28.
MAKING AN ARGUMENT
Your friend claims to be able Lo rewrite any proof that uses the AAS Congruence Theorem (Thin. 5. 11) as a proof that uses the ASA Congruence Theorem (Thin. 5.10). Is this possible? Explain our reasoning.

Answer:
No your friend is not correct. This is because the congruent side in AAS is not between the congruent angles and ASA has the congruent side between the two angles. So they are not interchangeable.

Question 29.
MODELING WITH MATHEMATICS
When a light ray from an object meets a mirror, it is reflected back to your eye. For example, in the diagram, a light ray from point C is reflected at point D and travels back to point A. The law of reflection states that the angle of incidence, ∠CDB. is congruent to the angle of reflection. ∠ADB.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 191
a. Prove that ∆ABD is Congruent to ∆CBD.
Given ∠CBD ≅∠ABD
DB ⊥ AC
Prove ∆ABD ≅ ∆CBD
b. Verify that ∆ACD is isosceles.
c. Does moving away from the mirror have an effect on the amount of his or her reflection a person sees? Explain.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 29.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 29.2

Question 30.
HOW DO YOU SEE IT?
Name as man pairs of congruent triangles as you can from the diagram. Explain how you know that each pair of triangles is congruent.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 192

Answer:

Question 31.
CONSTRUCTION
Construct a triangle. Show that there is no AAA congruence rule by constructing a second triangle that has the same angle measures but is not congruent.
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 31

Question 32.
THOUGHT PROVOKING
Graph theory is a branch of mathematics that studies vertices and the way they are connected. In graph theory. two polygons are isomorphic if there is a one-to-one mapping from one polygon’s vertices to the other polygon’s vertices that preserves adjacent vertices. In graph theory, are any two triangles isomorphic? Explain your reasoning. second triangle that has the same angle measures but is not congruent.

Answer:
Yes, any two congruent triangles can show isomorphism because the corresponding vertices can be mapped on to each other to completely overlap each other as required for isomorphism.

Question 33.
Mathematical Connections
Six statements are given about ∆TUV and ∆XYZ
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 193
a. List all combinations of three given statements that could provide enough information to prove that ∆TUV is congruent to ∆XYZ.
b. You choose three statements at random. What is the probability that the statements you choose provide enough information to prove that the triangles are congruent?
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 33.1
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 33.2

Maintaining Mathematical proficiency

Find the coordinates of the midpoint of the line segment with the given endpoints.

Question 34.
C(1, 0) and D(5, 4)

Answer:
Midpoint = (\(\frac { 1 + 5 }{ 2 } \), \(\frac { 0 + 4 }{ 2 } \)) = (\(\frac { 6 }{ 2 } \), \(\frac { 4 }{ 2 } \)) = (3, 3)

Question 35.
J(- 2, 3) and K(4, – 1)
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 35

Question 36.
R(- 5, – 7) and S(2, – 4)

Answer:
Midpoint = (\(\frac { -5 + 2 }{ 2 } \), \(\frac { -7 – 4 }{ 2 } \)) = (\(\frac { -3 }{ 2 } \), \(\frac { -11 }{ 2 } \))

Copy and angle using a compass and straightedge.

Question 37.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 194
Answer:
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 5.6 a 37

Question 38.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 195

Answer:
Draw a segment. Label a point D on the segment. Draw an arc with center B, and label the intersection points A and C. Using the same radius, draw an arc with center D. Label the point of intersection of the arc with radius BC with center D. Label the intersection F. Draw DF. So, ∠B ≅ ∠FDE.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.6 2

5.7 Using Congruent Triangles

Exploration 1

Measuring the Width of a River

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 219

Work with a partner:
The figure shows how a surveyor can measure the width of a river by making measurements on only one side of the river.

a. Study the figure. Then explain how the surveyor can find the width of the river.
Answer:

b. Write a proof to verify that the method you described in part (a) is valid.
Given ∠A is a right angle, ∠D is a right angle, \(\overline{A C}\) ≅ \(\overline{C D}\)
Answer:

c. Exchange Proofs with your partner and discuss the reasoning used.
CRITIQUING THE REASONING OF OTHERS
To be proficient in math, you need to listen to or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments.
Answer:

Exploration 2

Measuring the Width of a River

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 196

Work with a partner. It was reported that one of Napoleon’s offers estimated the width of a river as follows. The officer stood on the hank of the river and lowered the visor on his cap until the farthest thin visible was the edge of the bank on the other side. He then turned and rioted the point on his side that was in line with the tip of his visor and his eye. The officer then paced the distance to this point and concluded that distance was the width of the river.

a. Study the figure. Then explain how the officer concluded that the width of the river is EG.
Answer:

b. Write a proof to verify that the conclusion the officer made is correct.
Given ∠DEG is a right angle, ∠DEF is a right angle, ∠EDG ≅ ∠EDF
Answer:

c. Exchange proofs with your partner and discuss the reasoning used.
Answer:

Communicate Your Answer

Question 3.
How can you use congruent triangles to make an indirect measurement?
Answer:

Question 4.
Why do you think the types of measurements described in Explorations 1 and 2 are called indirect measurements?
Answer:

Lesson 5.7 Using Congruent Triangles

Monitoring Progress

Question 1.
Explain how you can prove that ∠A ≅ ∠C.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 197

Answer:
If you can prove that △ABD ≅ △CBD, then ∠A ≅ ∠C.
AB ≅ BC, AD ≅ CD
BD ≅ BD by reflexive property of congruence
So, △ABD ≅ △CBD by SSS congruence theorem.

Question 2.
In Example 2, does it mailer how far from point N you place a stake at point K? Explain.

Answer:

Question 3.
Write a plan to prove that ∆PTU ≅ ∆UQP.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 198

Answer:
TU ≅ PQ
∠PTU ≅ ∠UQP
PU ≅ PU by reflexive property of congruence
∆PTU ≅ ∆UQP by SAS congruence theorem.

Question 4.
Use the construction of an angle bisector on page 42. What segments can you assume are congruent?

Answer:

Exercise 5.7 Using Congruent Triangles

Vocabulary and core concept check

Question 1.
COMPLETE THE SENTENCE
_____________ parts of congruent triangle are congruent.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 1

Question 2.
WRITING
Describe a situation in which you might choose to use indirect measurement with
congruent triangles to find a measure rather than measuring directly.

Answer:
Indirect measurements are calculations based on known lengths
When two triangles are similar using AAS, SSS, ASA or SAS indirect measurement can be used to find missing measurements and angles.
If you were given a triangle with two angles marked and a similar triangle with two sides marked you would be able to find measurements of the angles indirectly.

Monitoring Progress and Modeling With Mathematics

In Exercise 3-8, explain how to prove that the statement is true.

Question 3.
∠A ≅ ∠D
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 199
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 3

Question 4.
∠Q ≅∠T
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 200

Answer:
QP ≅ PT, RP ≅ SP, QR ≅ ST
All pairs of sides are congruent by SSS congruence theorem. △QPR ≅ △STP. Because corresponding parts of congruent triangles are congruent, ∠Q ≅∠T.

Question 5.
\(\overline{J M}\) ≅ \(\overline{L M}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 201
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 5

Question 6.
\(\overline{A C}\) ≅ \(\overline{D B}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 202

Answer:
\(\overline{A C}\) ≅ \(\overline{D B}\),
\(\overline{A D}\) ≅ \(\overline{A D}\) by reflexive property of congruence
∠C ≅∠B
△ACD ≅ △BDC by SAS congruence theorem.

Question 7.
\(\overline{G K}\) ≅ \(\overline{H J}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 203
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 7

Question 8.
\(\overline{Q W}\) ≅ \(\overline{V T}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 204

Answer:
VW ≅ RT, ∠Q ≅ ∠S, ∠W ≅ ∠T
△QVW ≅ △VRT by the AAS congruence theorem
So, \(\overline{Q W}\) ≅ \(\overline{V T}\)

In Exercises 9-12, write a plan to prove that ∠1 ≅∠2.

Question 9.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 205
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 9

Question 10.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 206
Answer:
∠A ≅ ∠D, AB ≅ CD and ∠BEA ≅ ∠CED
So, △ABE ≅ △EDC by AAS congruence theorem
∠ABE ≅ ∠DCE
Use the congruent complements theorem to prove that ∠1 ≅ ∠2

Question 11.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 207
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 11

Question 12.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 208

Answer:
AF ≅ CD, ∠AEF ≅ ∠CED, ∠FAE ≅ ∠ECD
So, △AFE ≅ △CDE by ASA congruence thorem
Then all parts of the triangles are congruent.
So, ∠1 ≅ ∠2

In Exercises 13 and 14. write a proof to verify that the construction is valid.

Question 13.
Line perpendicular to a line through a point not on the line
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 209
Plan for proof ∆APQ ≅ ∆BPQ by the congruence Theorem (Theorem 5.8). Then show the ∆APM ≅ ∆BPM using the SAS Congruence Theorem (Theorem 5.5). Use corresponding parts of congruent triangles to show that ∠AMP and ∠BMP are right angles.
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 13.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 13.2

Question 14.
Line perpendicular to a line through a p0int on the line
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 210
Plan for Proof Show that ∆APQ ≅ ∆BPQ by the SSS Congruence Theorem (Theorem 5.8) Use corresponding parts of congruent triangles to show that ∠QPA and ∠QPB are right angles.
Answer:

In Exercises 15 and 16, use the information given in the diagram to write a proof.

Question 15.
Prove \(\overline{F L}\) ≅ \(\overline{H N}\)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 211
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 15.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 15.2

Question 16.
Prove ∆PUX ≅ ∆QSY
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 212
Answer:

Question 17.
MODELING WITH MATHEMATICS
Explain how to find the distance across the canyon.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 213
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 17

Question 18.
HOW DO YOU SEE IT?
Use the tangram puzzle.
Answer:

Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 214

a. Which triangle(s) have an area that is twice the area of the purple triangle?
b. How man times greater is the area of the orange triangle than the area of the purple triangle?
Answer:

Question 19.
PROOF
Prove that the green triangles in the Jamaican flag congruent if \(\overline{A D}\) || \(\overline{B C}\) and E is the midpoint of \(\overline{A C}\).
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 215
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 19.1
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 19.2

Question 20.
THOUGHT PROVOKING
The Bermuda Triangle is a region in the Atlantic Ocean in which many ships and planes have mysteriously disappeared. The vertices are Miami. San Juan. and Bermuda. Use the Internet or some other resource to find the side lengths. the perimeter, and the area of this triangle (in miles). Then create a congruent triangle on land using cities as vertices.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 216
Answer:

Question 21.
MAKING AN ARGUMENT
Your friend claims that ∆WZY can be proven congruent to ∆YXW using the HL Congruence Theorem (Thm. 5.9). Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 217
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 21

Question 22.
CRITICAL THINKING
Determine whether each conditional statement is true or false. If the statement is false, rewrite it as a true statement using the converse, inverse, or contrapositive.
a. If two triangles have the same perimeter, then they are congruent.
b. If two triangles are congruent. then they have the same area.

Answer:
a. The statement is false
we will rewrite this statement suing the converse.
If two triangles are congruent, then they have the same perimeter.
b. The statement is true.

Question 23.
ATTENDING TO PRECISION
Which triangles are congruent to ∆ABC? Select all that apply.
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 218
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 23

Maintaining Mathematical Proficiency

Find the perimeter of the polygon with the given vertices.

Question 24.
A(- 1, 1), B(4, 1), C(4, – 2), D(- 1, – 2)

Answer:
Perimeter of the polygon = 16

Explanation:
AB = √(4 + 1)² + (1 – 1)² = √5² + 0² = 5
BC = √(4 – 4)² + (-2 – 1)² = √0² + (-3)² = 3
CD = √(-1 – 4)² + (-2 + 2)² = √(-5)² + 0² = 5
AD = √(-1 + 1)² + (-2 – 1)² = √0² + (-3)² = 3
Perimeter of the polygon = AB + BC + CD + AD
= 5 + 3 + 5 + 3 = 16

Question 25.
J(- 5, 3), K(- 2, 1), L(3, 4)
Answer:
Big Ideas Math Answers Geometry Chapter 5 Congruent Triangles 5.7 a 25

5.8 Coordinate Proofs

Exploration 1

Writing a coordinate Proof

Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 218

Work with a partner.

a. Use dynamic geometry software to draw \(\overline{A B}\) with endpoints A(0, 0) and B(6, 0).
Answer:

b. Draw the vertical line x = 3.
Answer:

c. Draw ∆ABC so that C lies on the line x = 3.
Answer:

d. Use your drawing to prove that ∆ABC is an isosceles triangle.
Answer:

Exploration 2

Writing a Coordinate proof

Work with a partner.

a. Use dynamic geometry software to draw \(\overline{A B}\) with endpoints A(0, 0) and B(6, 0).

b. Draw the vertical line x = 3.

c. Plot the point C(3, 3) and draw ∆ABC. Then use your drawing to prove that ∆ABC is an isosceles right triangle.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 219

d. Change the coordinates of C so that C lies below the x-axis and ∆ABC is an isosceles right triangle.
Answer:

e. Write a coordinate proof to show that if C lies on the line x = 3 and ∆ABC is an isosceles right triangle. then C must be the point (3, 3) or the point found in part (d).
CRITIQUING THE REASONING OF OTHERS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
Answer:

Communicate Your Answer

Question 3.
How can you use a coordinate plane to write a proof?
Answer:

Question 4.
Write a coordinate proof to prove that ∆ABC with vertices A(0, 0), 8(6, 0), and C(3, 3√3) is an equilateral triangle.
Answer:

Lesson 5.8 Coordinate Proofs

Monitoring Progress

Question 1.
Show another way to place the rectangle in Example 1 part (a) that is convenient
for finding side lengths. Assign new coordinates.

Answer:
Place the base of the rectangle anywhere you want on the coordinate plane.
From there take height and draw the rectangle.
Mark the points so that it is easy to find the distance between vertices.
The new coordinates are A(1, 1), B (5, 1), C (5, 4), D (1, 4)
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 1

Question 2.
A square has vertices (0, 0), (m, 0), and (0, m), Find the fourth vertex.

Answer:
The fourth vertex is (m, m)

Explanation:
By drawing the square on a graph. We can find the fourth vertex.
The fourth vertex is (m, m)

Question 3.
Write a plan for the proof.
Given \(\vec{G}\)J bisects ∠OGH.
Proof ∆GJO ≅ ∆GJH
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 220

Answer:
GJ ≅ GJ by the reflexive property of congruence
OG = √(3 – 0)² + (7 – 0)² = √9 + 49 = √58
GH = √(6 – 3)² + (0 – 7)² = √9 + 49 = √58
OG ≅ GH
OJ = √(3 – 0)² + (2 – 0)² = √9 + 4 = √13
JH = √(3 – 6)² + (2 – 0)² = √9 + 4 = √13
OJ ≅ JH
So, ∆GJO ≅ ∆GJH by SAS congruence theorem.

Question 4.
Graph the points 0(0, 0), H(m, n), and J(m, 0). Is ∆OHJ a right triangle? Find the side lengths and the coordinates of the midpoint of each side.

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 2
Side lengths are OA = √(m – 0)² + (n – 0)² = √m² + n²
OB = √(m – 0)² + (0 – 0)² = √m² = m
AB = √(m – m)² + (n – 0)² = √n² = n
By using the Pythagorean theorem,
OA² = OB² + AB²
m² + n² = m² + n²
So, the trinagle is a right-angled triangle.

Question 5.
Write a coordinate proof.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 221
Given Coordinates of vertices of ∆NPO and ∆NMO
Prove ∆NPO ≅ ∆NMO

Answer:
NP = √(h – 0)² + (h – 2h)² = √h² + h² = h√2
OP = √(0 – 0)² + (0 – 2h)² = √4h² = 2h
ON = √(0 – h)² + (0 – h)² = √h² + h² = h√2
NM = √(h – 2h)² + (h – 0)² = √h² + h² = h√2
ON = √(0 – h)² + (0 – h)²  = √h² + h² = h√2
OM = √(2h – 0)² + (0 – 0)² = √4h² = 2h
NP ≅ NM, OP ≅ OM, ON ≅ ON
So, ∆NPO ≅ ∆NMO by SSS congruence theorem

Exercise 5.8 Coordinate Proofs

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
How is a coordinate proof different from other types of proofs you have studied?
How is it the same?
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 1

Question 2.
WRITING
Explain why it is convenient to place a right triangle on the grid as shown when writing a coordinate proof.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 222

Answer:
Because the right triangle has the base and another leg on the same line in the coordinate plane.

Maintaining Progress and Modeling with Mathematics

In Exercises 3-6, place (he figure in a coordinate plane in a convenient way. Assign coordinates to each vertex. Explain the advantages of your placement.

Question 3.
a right triangle with leg lengths of 3 units and 2 units
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 3

Question 4.
a square with a side length of 3 units

Answer:
Place the sides on the x-axis, y-axis
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 3
It is easy to find the lengths of horizontal and vertical segments and distances from the origin.

Question 5.
an isosceles right triangle with leg length p
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 5

Question 6.
a scalene triangle with one side length of 2m

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 4

In Exercises 7 and 8, write a plan for the proof.

Question 7.
Given Coordinates of vertices of ∆OPM and ∆ONM Prove ∆OPM and ∆ONM are isosceles triangles.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 223
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 7

Question 8.
Given G is the midpoint of \(\overline{H F}\).
Prove ∆GHJ ≅ ∆GFO
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 224

Answer:
The coordinates of G are (3, 2)
OG = √(3 – 0)² + (2 – 0)² = √9 + 4 = √13
OF = √(5 – 0)² + (0 – 0)² = 5
GF = √(5 – 3)² + (0 – 2)² = √2² + 2² = √8
GH = √(3 – 1)² + (2 – 4)² = √2² + 2² = √8
HJ = √(6 – 1)² + (4 – 4)² = √5² = 5
GJ = √(6 – 3)² + (4 – 2)² = √9 + 4 = √13
OG ≅ GJ, OF ≅ HJ, GF ≅ GH
All the sides are congruent. So, ∆GHJ ≅ ∆GFO by SSS congruence theorem.

In Exercises 9-12, place the figure in a coordinate plane and find the indicated length.

Question 9.
a right triangle with leg lengths of 7 and 9 units; Find the length of the hypotenuse.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 9

Question 10.
an isosceles triangle with a base length of 60 units and a height of 50 units: Find the length of one of the legs.

Answer:
The length of one of the legs = 58.31

Explanation:
The coordinates of the vertices A(0, 0), B(60, 0), C(30, 50)
The length of one leg AC = √(50 – 0)² + (30 – 0)² = √2500 + 900 = √3400 = 58.31
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 5

Question 11.
a rectangle with a length o! 5 units and a width of 4 units: Find the length of the diagonal.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 11

Question 12.
a square with side length n: Find the length of the diagonal.

Answer:
The length of the diagonal is n√2

Explanation:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 6
The length of diagonal = √(n – 0)² + (n – 0)² = √n² + n² = n√2

In Exercises 13 and 14, graph the triangle with the given vertices. Find the length and the slope of each side of the triangle. Then find the coordinates of the midpoint of each side. Is the triangle a right triangle? isosceles? Explain. Assume all variables are positive and in m ≠ n.)

Question 13.
A(0, 0), B(h, h), C(2h, 0)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 13.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 13.2

Question 14.
D(0, n), E(m, n), F(m, 0)

Answer:
Side lengths of △DEF
DE = √(m – 0)² + (n – n)² = m
EF = √(m – m)² + (0 – n)² = n
DF = √(m – 0)² + (0 – n)² = √m² + n²
Slopes of △DEF
Slope of DE = \(\frac { n – n }{ n – 0 } \) = 0
Slope of EF = \(\frac { 0 – n }{ m – m } \) = undefined
Slope of DF = \(\frac { m – 0 }{ 0 – n } \) = \(\frac { -m }{ n } \)
Midpoint of DE = (\(\frac { 0 + m }{ 2 } \), \(\frac { n + n }{ 2 } \)) = (\(\frac { m }{ 2 } \), n)
Midpoint of EF = (\(\frac { m + m }{ 2 } \), \(\frac { n + 0 }{ 2 } \)) = (m, \(\frac { n }{ 2 } \))
Midpoint of DF = (\(\frac { 0 + m }{ 2 } \), \(\frac { n + 0 }{ 2 } \)) = (\(\frac { m }{ 2 } \), \(\frac { n }{ 2 } \))

In Exercises 15 and 16, find the coordinates of any unlabeled vertices. Then find the indicated length(s).

Question 15.
Find ON and MN.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 225
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 15

Question 16.
Find OT.
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 226
Answer:

PROOF
In Exercises 17 and 18, rite a coordinate proof.

Question 17.
Given Coordinates of vertices of ∆DEC and ∆BOC
Prove ∆DEC ≅ ∆BOC
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 227
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 17

Question 18.
Given Coordinates of ∆DEA, H is the midpoint of \(\overline{D A}\), G is the mid point of \(\overline{E A}\)
Prove \(\overline{D G}\) ≅ \(\overline{E H}\)
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 228

Answer:
G is the mid point of \(\overline{E A}\)
So, G = (\(\frac { 2h + 0 }{ 2 } \), \(\frac { 0 + 2k }{ 2 } \)) = (h, k)
H is the midpoint of \(\overline{D A}\)
So, H = (\(\frac { -2h + 0 }{ 2 } \), \(\frac { 2k + 0 }{ 2 } \)) = (-h, k)
\(\overline{D G}\) = √(h + 2h)² + (k – 0)² = √9h² + k²
\(\overline{E H}\) = √(-h – 2h)² + (0 – k)² = √9h² + k²

Question 19.
MODELING WITH MATHEMATICS
You and your cousin are camping in the woods. You hike to a point that is 500 meters cast and 1200 meters north of the Campsite. Your cousin hikes to a point that is 1000 meters cast of the campsite. Use a coordinate proof to prove that the triangle formed by your Position, your Cousin’s position. and the campsite is isosceles. (See Example 5.)
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 229
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 19.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 19.2

Question 20.
MAKING AN ARGUMENT
Two friends see a drawing of quadrilateral PQRS with vertices P(0, 2), Q(3, – 4), R(1, – 5), and S(- 2, 1). One friend says the quadrilateral is a parallelogram but not a rectangle. The other friend says the quadrilateral is a rectangle. Which friend is correct? Use a coordinate proof to support your answer.

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 5.8 7
It can be seen that PS || QR and PQ || SR. If PQRS is a rectangle then PS ⊥ PQ, therefore find slopes of PS and PQ.
Slope of PS = \(\frac { 1 – 2 }{ -2 – 0 } \) = \(\frac { 1 }{ 2 } \)
Slope of PQ = \(\frac { -4 – 2 }{ 3 – 0 } \) = -2
It can be seen that slope of PS is negative reciprocal of slope of PQ.
So, PS is perpendicular to PQ and PQRS is a rectangle.

Question 21.
MATHEMATICAL CONNECTIONS
Write an algebraic expression for the coordinates of each endpoint of a line segment whose midpoint is the origin.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 21

Question 22.
REASONING
The vertices of a parallelogram are (w, 0), (0, v), (- w, 0), and (0, – v). What is the midpoint of the side in Quadrant III?
(a) \(\left(\frac{w}{2}, \frac{v}{2}\right)\)
(b) \(\left(-\frac{w}{2},-\frac{v}{2}\right)\)
(c) \(\left(-\frac{w}{2}, \frac{v}{2}\right)\)
(d) \(\left(\frac{w}{2},-\frac{v}{2}\right)\)

Answer:
(b) \(\left(-\frac{w}{2},-\frac{v}{2}\right)\)

Explanation:
The midpoint of the side in Quadrant III = (\(\frac { -w + 0 }{ 2 } \), \(\frac { 0 – v }{ 2 } \))
= (\(\frac { -w }{ 2 } \), \(\frac { -v }{ 2 } \))

Question 23.
REASONING
A rectangle with a length of 3h and a width of k has a vertex at (- h, k), Which point cannot be a vertex of the rectangle?
(A) (h, k)
(B) (- h, 0)
(c) (2h, 0)
(D) (2h, k)
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 23

Question 24.
THOUGHT PROVOKING
Choose one of the theorems you have encountered up to this point that you think would be easier to prove with a coordinate proof than with another type of proof. Explain your reasoning. Then write a coordinate proof.

Answer:

Question 25.
CRITICAL THINKING
The coordinates of a triangle are (5d – 5d), (0, – 5d), and (5d, 0). How sh
would the coordinates be changed to make a coordinate proof easier to complete?
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 25

Question 26.
HOW DO YOU SEE IT?
without performing any calculations, how do you know that the diagonals of square TUVW are perpendicular to each oilier? How can you use a similar diagram to show that the diagonals of any square are perpendicular to each other?
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 230

Answer:
As the vertices are in the opposite sign, the diagonals of square TUVW are perpendicular to each other.

Question 27.
PROOF
Write a coordinate proof for each statement.
a. The midpoint o! the hypotenuse of a right triangle is the same distance from each vertex of the triangle.
b. Any two congruent right isosceles triangles can be combined to form a single isosceles triangle.
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 27.1
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 27.2

Maintaining Mathematical proficiency

\(\vec{Y}\)W bisects ∠XYZ such that m∠XYW = (3x – 7)° and m∠WYZ = (2x + 1)°.

Question 28.
Find the value of x.

Answer:
x = 8

Explanation:
(3x – 7)° = (2x + 1)
3x – 2x = 1 + 7
x = 8

Question 29.
Find m∠XYZ
Answer:
Big Ideas Math Geometry Answer Key Chapter 5 Congruent Triangles 5.8 a 29

Congruent Triangles Chapter Review

5.1 Angles of Triangles

Question 1.
Classify the triangle at the right by its sides and by measuring its angles.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 231

Answer:
The trinagle have two congruent sides. So, the triangle is an isosceles triangle. Therefore, the measure of angles is less than 90 degrees. So these are obtuse angles.

Find the measure of the exterior angle.

Question 2.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 232

Answer:
x° = 132°

Explanation:
Sum of angles = 180
46 + 86 + y = 180
y = 48
x° = 180° – 48° = 132°

Question 3.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 234

Answer:
x = 9

Explanation:
y = 180 – (9x + 9)
Sum of interior angles = 180
45 + 5x + y = 180
45 + 5x + 180 – (9x + 9) = 180
45 – 4x – 9 = 0
4x = 36
x = 9

Find the measure of each acute angle.

Question 4.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 235

Answer:
x = 6

Explanation:
Sum of interior angles = 180
90 + 8x + 7x = 180
15x = 90
x = 6

Question 5.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 236

Answer:
x = 7

Explanation:
Sum of interior angles = 180
7x + 6 + 90 + 6x – 7 = 180
13x – 1 = 90
13x = 91
x = 7

5.2 Congruent Polygons

Question 6.
In the diagram. GHJK ≅ LMNP. Identify all pairs of congruent corresponding parts. Then write another congruence statement for the quadrilaterals.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 237

Answer:
Corresponding sides: GH ≅ LM, KJ ≅ PN
Corresponding angles: ∠G ≅ ∠H ≅ ∠L ≅ ∠M, ∠K ≅ ∠J ≅ ∠P ≅ ∠N

Question 7.
Find m ∠ V.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 238

Answer:
m ∠ V = 16

Explanation:
∠T ≅ ∠S
Sum of interior angles = 180
90 + 74 + m ∠ V =180
164 + m ∠ V = 180
m ∠ V = 16

5.3 Proving Triangle Congruence by SAS

Decide whether enough information is given to prove that ∆WXZ ≅ ∆YZX using the SAS Congruence Theorem (Theorem 5.5). If so, write a proof. If not, explain why.

Question 8.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 239

Answer:
Yes

Explanation:
∠W ≅ ∠Y
WZ ≅ XY
XZ ≅ XZ by the reflexive property of congruence
△ XZW ≅ △XYZ by SAS Congruence Theorem

Question 9.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 240

Answer:
Yes

Explanation:
WX ≅ ZY
WZ ≅ XY
XZ ≅ XZ by the reflexive property of congruence
△ XZW ≅ △XYZ by SAS Congruence Theorem

5.4 Equilateral and Isosceles Triangles

Copy and Complete the statement.

Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 241

Question 10.
If \(\overline{Q P}\) ≅ \(\overline{Q R}\), then ∠ ______ ≅ ∠ ______ .

Answer:
If \(\overline{Q P}\) ≅ \(\overline{Q R}\), then ∠ P ≅ ∠ R

Question 11.
If ∠TRV ≅ ∠TVR, then ______ ≅ ______ .

Answer:
If ∠TRV ≅ ∠TVR, then TR ≅ TV

Question 12.
If \(\overline{R Q} \cong \overline{R S}\), then ∠ ______ ≅ ∠ ______ .

Answer:
If \(\overline{R Q} \cong \overline{R S}\), then ∠ Q ≅ ∠ S .

Question 13.
If ∠SRV ≅ ∠SVR, then ______ ≅ ______ .

Answer:
If ∠SRV ≅ ∠SVR, then SR ≅ VR .

Question 14.
Find the values of x and y in the diagram.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 242

Answer:

5.5 Proving Triangle Congruence by SSS

Question 15.
Decide whether enough information is given to prove that ∆LMP ≅ ∆NPM using the SSS Congruence Theorem (Thin. 5.8). If so, write a proof. If not, explain why.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 243

Answer:
LM ≅ NP, MP ≅ MP by reflexive property of congruence.
We need one more side to be congruent to prove that ∆LMP ≅ ∆NPM using the SSS Congruence Theorem

Question 16.
Decide whether enough information is given to prove that ∆WXZ ≅ ∆YZX using the HL Congruence Theorem (Thm. 5.9). If so, write a proof. If not, explain why.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 244

Answer:
XZ ≅ XZ by reflexive property of congruence.
The hypotenuse leg of ∆WXZ and ∆YZX are congruent. So, ∆WXZ ≅ ∆YZX

5.6 Proving Triangle Congruence by ASA and AAS

Question 17.
∆EFG, ∆HJK
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 245

Answer:
∠E ≅ ∠H, ∠F ≅ ∠J, FG ≅ KJ
So, ∆EFG ≅ ∆HJK by AAS Congruence Theorem.

Question 18.
∆TUS, ∆QRS
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 246

Answer:
∠T ≅ ∠Q, TU ≅ QR
The given information is not enough to prove that ∆TUS ≅ ∆QRS using AASCongruence Theorem.

Decide whether enough information is given to prove that the triangles are congruent using the ASA Congruence Theorem (Thm. 5.10). If so, write a proof, If not, explain why.

Question 19.
∆LPN, ∆LMN
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 247

Answer:
∠PNL ≅ ∠MNL, ∠PLN ≅ ∠MLN
LN ≅ LN by reflexive property of congruence
So, ∆LPN ≅ ∆LMN using ASA Congruence Theorem.

Question 20.
∆WXZ, ∆YZX
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 248

Answer:
WZ ≅ XY
XZ ≅ XZ by reflexive property of congruence theorem
So, the given information is not enough to prove that ∆WXZ ≅ ∆YZX using ASA Congruence Theorem.

5.7 Using Congruent Triangles

Question 21.
Explain how to prove that ∠K ≅∠N.

Answer:
HJ ≅ ML
KJ ≅ NM
∠J ≅ ∠M
So, ∆HKJ ≅ ∆MNL using SAS congruence theorem.
Therefore, ∠K ≅∠N

Question 22.
Write a plan to prove that ∠1 ≅ ∠2

Answer:

5.8 Coordinate Proofs

Question 23.
Write a coordinate proof.
Given Coordinates of vertices of quadrilateral OPQR
Prove ∆OPQ ≅ ∆QRO
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 251

Answer:
Find the distance of all sides
OP = √(h – 0)² + (k – 0)² = √h² + k²
OQ = √(h -0)² + (k + j – 0 )² = √h² + (k + j)²
PQ = √(h – h)² + (k – k – j)² = √0² + j² = j
OR = √(0 – 0)² + (0 – j)² = √j² = j
OQ = √(h – 0)² + (k + j – 0)² = √h² + (k + j)²
QR = √(h – 0)² + (k + j – j)² = √h² + k²
OP ≅ QR, OQ ≅ OQ, PQ ≅ OR
So, ∆OPQ ≅ ∆QRO using SSS Congruence Theorem.

Question 24.
Place an isosceles triangle in a coordinate plane in a way that is convenient for finding side lengths. Assign coordinates to each vertex.

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 1

Question 25.
A rectangle has vertices (0, 0), (2k, 0), and (0, k), Find the fourth vertex.

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 2
From the image, the fourth vertex is (2k, k).

Congruent Triangles Test

Write a Proof.

Question 1.
Given \(\overline{C A} \cong \overline{C B} \cong \overline{C D} \cong \overline{C E}\)
Prove ∆ABC ≅ ∆EDC
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 252

Answer:
Given \(\overline{C A} \cong \overline{C B} \cong \overline{C D} \cong \overline{C E}\)
As ∆ABC and ∆EDC are isosceles triangles.
So, ∆ABC ≅ ∆EDC

Question 2.
Given \(\overline{J K}\|\overline{M L}, \overline{M J}\| \overline{K L}\)
Prove ∆MJK ≅ ∆KLM
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 253

Answer:
\(\overline{J K}\|\overline{M L}, \overline{M J}\| \overline{K L}\)
MK ≅ MK by reflexive property of congruence
So, ∆MJK ≅ ∆KLM using SSS congruence theorem.

Question 3.
Given \overline{Q R} \cong \overrightarrow{R S}\(\), ∠P ≅ ∠T
Prove ∆SRP ≅ ∆QRT
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 254

Answer:
QR ≅ RS, ∠P ≅ ∠T
∠R ≅ ∠R by reflexive property of congruence
So, ∆SRP ≅ ∆QRT using AAS Congruence Theorem

Question 4.
Find the measure of each acute angle in the figure at the right.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 255

Answer:
x = 12

Explanation:
90 + 4x – 2 + 3x + 8 = 180
96 + 7x = 180
7x = 180 – 96
7x = 84
x = 12

Question 5.
Is it possible to draw an equilateral triangle that is not equiangular? If so, provide an example. If not, explain why.

Answer:
No

Explanation:
All equilateral triangles will be equiangular

Question 6.
Can you use the Third Angles Theorem (Theorem 5.4) to prove that two triangles are congruent? Explain your reasoning.

Answer:
Big Ideas Math Geometry Answers Chapter 5 Congruent Triangles 3
∠B ≅ ∠E, ∠C ≅ ∠F
Then ∠A ≅ ∠D using third angles theorem.

Write a plan through that ∠1 ≅∠2

Question 7.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 256

Answer:

Question 8.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 257

Answer:

Question 9.
Is there more than one theorem that could be used to prove that ∆ABD ≅ ∆CDB? If so, list all possible theorems.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 258
Answer:
Yes

Explanation:
The first theorem is HL Congruence Theorem
AB ≅ CD
The second theorem is SAS Congruence Theorem
AB ≅ CD
DB ≅ DB by reflexive property of congruence
∠D ≅ ∠B

Question 10.
Write a coordinate proof to show that the triangles created b the keyboard stand are congruent.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 259

Answer:
The coordinates of T(3, 0), S(21, 0), R (12, 15), P(3, 30), Q(21, 30)
PQ = √(21 – 3)² + (30 – 30)² = 18
PR = √(3 – 12)² + (30 – 15)² = √9² + 15² = √306
RQ = √(21 – 12)² + (30 – 15)² = √9² + 15² = √306
TR = √(12 – 3)² + (15 – 0)² = √9² + 15² = √306
TS = √(21 – 3)² + (0 – 0)² = √18² = 18
RS = √(21 – 12)² + (0 – 15)² = √9² + 15² = √306
PQ ≅ TS, PR ≅ TR, RQ ≅ RS
So, △PQR ≅ △TRS using SSS congruence theorem

Question 11.
The picture shows the Pyramid of Cestius. which is located in Rome, Italy. The measure of the base for the triangle shown is 100 Roman feet. The measures of the other two sides of the triangle are both 144 Roman feet.
Big Ideas Math Geometry Solutions Chapter 5 Congruent Triangles 260
a. Classify the triangle shown by its sides.
Answer:
The triangle is an isosceles triangle.

b. The measure of ∠3 is 40° What are the measures of ∠1 and ∠2? Explain your reasoning.
Answer:
The measures of ∠1 and ∠2 are 70°

Explanation:
∠3 + 2x = 180
40 + 2x = 180
2x = 140
x = 70

Congruent Triangles Cumulative Assessment

Question 1.
Your friend claims that the Exterior Angle Theorem (Theorem 5.2) can be used to prove the Triangle Sum Theorem (Theorem 5, 1). Is your friend correct? Explain your reasoning.

Answer:
Yes

Question 2.
Use the steps in the construction to explain how you know that the line through point P is parallel to line m.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 261

Answer:
From step 4 the red line is parallel to m and passes through the pint P. So, point P is parallel to line m.

Question 3.
The coordinate plane shows ∆JKL and ∆XYZ
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 262
a. Write a composition of transformations that maps ∆JKL to ∆XYZ
Answer:
The coordinates of J(-3, 2), L (0, 2) K(-2, 4), X(1, -2), Y(2, -4), Z(4, -2)
JL = √(0 + 3)² + (2 – 2)² = 3
XZ = √(4 – 1)² + (-2 + 2)² = 3
JK = √(-2 + 3)² + (4 – 2)² = √5
XY = √(2 – 1)² + (-4 + 2)² = √5
KL = √(-2 – 0)² + (4 – 2)² = √8
YZ = √(4 – 2)² + (-2 + 4)²= √8
JK ≅ XY, JL ≅ XZ, KL ≅ YZ

b. Is the composition a congruence transformation? If so, identify all congruent corresponding parts.
Answer:
△JKL ≅ △XYZ using the SSS Congruence theorem.

Question 4.
The directed line segment RS is shown. Point Q is located along \(\overline{R S}\) so that the ratio of RQ to QS is 2 to 3. What are the coordinates of point Q?
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 263
(A) Q(1.2, 3)
(B) Q(4, 2)
(C) Q(2, 3)
(D) Q(-6, 7)

Answer:
(B) Q(4, 2)

Explanation:
RQ : RS = 2 : 3
The coordinates of R(-2, 5), S(8, 0)
Let us validate answer with Q(2, 3)
RQ = √(2 + 2)² + (3 – 5)² = √16 + 4 = √20 = 4.47
RS = √(8 + 2)² + (0 – 5)² = √125 = 11.18
If Q(4, 2)
RQ = √(4 + 2)² + (2 – 5)² = √45
Now, RQ : RS = 2 : 3
So, Q(4, 2)

Question 5.
The coordinate plane shows that ∆ABC and ∆DEF
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 264
a. Prove ∆ABC ≅ ∆DEF using the given information.
Answer:
∠B ≅ ∠E, BC ≅ EF
So, ∆ABC ≅ ∆DEF

b. Describe the composition of rigid motions that maps ∆ABC to ∆DEF
Answer:

Question 6.
The vertices of a quadrilateral are W(0, 0), X(- 1, 3), )(2, 7), and Z(4, 2). Your friend claims that point W will not change after dilatinig quadrilateral WXYZ by a scale factor of 2. Is your friend correct? Explain your reasoning.
Answer:

Question 7.
Which figure(s) have rotational symmetry? Select all that apply.
(A) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 265
(B) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 266
(C) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 267
(D) Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 268
Answer:
(D)
Because by rotating the symbol, we get the same image.

Question 8.
Write a coordinate proof.
Given Coordinates of vertices of quadrilateral ABCD
Prove Quadrilateral ABCD is a rectangle.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 269

Answer:
The vertices of A(2, 5), B(4, 7), C(7, 4), D(5, 2)
The slope of AB = \(\frac { 4 – 2 }{ 7 – 5 } \) = \(\frac { 2 }{ 2 } \) = 1
The slope of AD = \(\frac { 5 – 2 }{ 2 – 5 } \) = \(\frac { 3 }{ -3 } \) = -1
The slope of AB and AD are negative reciprocals
So, AB is perpendicular to AD
So, ABCD is a rectangle.

Question 9.
Write a proof to verify that the construction of the equilateral triangle shown below is valid.
Big Ideas Math Answer Key Geometry Chapter 5 Congruent Triangles 270

Answer:
At step 1, we need to draw the base AB
With A as centre and AB as radius draw an arc and with B as centre and AB as radius draw another arc intersecting the first arc at C.
Name that point as C. Now formed triangle ABC is an equilateral triangle.

Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions

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Big Ideas Math Book Algebra 1 Answer Key Chapter 3 Graphing Linear Functions

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Graphing Linear Functions Maintaining Mathematical Proficiency

Plot the point in a coordinate plane. Describe the location of the point.

Question 1.
A(3, 2)
Answer:
The given point is: A (3,2)
Compare the given point with (x, y)
So,
x = 3 and y = 2
Hence,
The representation of the point in the coordinate plane is:

Question 2.
B(-5, 1)
Answer:
The given point is: B (-5,1)
Compare the given point with (x, y)
So,
x = -5 and y = 1
Hence,
The representation of the given point in the coordinate plane is:

Question 3.
C(0, 3)
Answer:
The given point is: c (0,3)
Compare the given point with (x,y)
So,
x = 0 and y = 3
Hence,
The representation of the given point in the coordinate plane is:

Question 4.
D(-1, -4)
Answer:
The given point is: D (-1,-4)
Compare the given point with (x,y)
So,
x = -1 and y = -4
Hence,
The representation of the given point in the coordinate plane is:

Question 5.
E(-3, 0)
Answer:
The given point is: E (-3,0)
Compare the given point with (x,y)
So,
x = -3 and y = 0
Hence,
The representation of the given point in the coordinate plane is:

Question 6.
F(2, -1)
Answer:
The given point is: F (2,-1)
Compare the given point with (x,y)
So,
x = 2 and y = -1
Hence,
The representation of the given point in the coordinate plane is:

Evaluate the expression for the given value of x.

Question 7.
3x – 4; x = 7
Answer:
The value of the expression for the given value of x is: 17

Explanation:
The given expression is:
3x – 4 with x = 7
Hence,
The value of the expression is:
3 (7) – 4 = 21 – 4 = 17
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 17

Question 8.
-5x + 8; x = 3
Answer:
The value of the expression for the given value of x is: -7

Explanation:
The given expression is:
-5x + 8 with x = 3
Hence,
The value of the expression is:
-5 (3) + 8 = -15 + 8 = -7
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: -7

Question 9.
10x + 18; x = 5
Answer:
The value of the expression for the given value of x is: 68

Explanation:
The given expression is:
10x + 18 with x = 5
Hence,
The value of the expression is:
10 (5) + 18 = 50 + 18 = 68
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 68

Question 10.
-9x – 2; x = -4
Answer:
The value of the expression for the given value of x is: 34

Explanation:
The given expression is:
-9x – 2 with x = -4
Hence,
The value of the expression is:
-9 (-4) – 2 = 36 – 2 = 34
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 34

Question 11.
24 – 8x; x = -2
Answer:
The value of the expression for the given value of x is: 40

Explanation:
The given expression is:
24 – 8x with x = -2
Hence,
The value of the expression is:
24 – 8 (-2) = 24 + 16 = 40
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: 40

Question 12.
15x + 9; x = -1
Answer:
The value of the expression for the given value of x is: -6

Explanation:
The given expression is:
15x + 9 with x = -1
Hence,
The value of the expression is:
15 (-1) + 9 = -15 + 9 = -6
Hence, from the above,
We can conclude that the value of the expression for the given value of x is: -6

Question 13.
ABSTRACT REASONING
Let a and b be positive real numbers. Describe how to plot (a, b), (-a, b), (a, -b), and (-a, -b).
Answer:
It is given that a and b are positive real numbers
The given points are (a, b), (-a, b), (a, -b) and (-a, -b)
Let the names of the points be:
A (a, b), B (-a, b), C (a, -b), and D (-a, -b)
We know that,
The coordinate plane is divided into 4 parts. These parts are called  “Quadrants”
So,
The representation of a and b in the 4 quadrants are:
1st Quadrant: (a, b)
2nd Quadrant: (-a, b)
3rd Quadrant: (-a, -b)
4th Quadrant: (a, -b)
Hence,
The representation of the given points in the coordinate plane is:

Graphing Linear Functions Mathematical Practices

Monitoring Progress

Determine whether the viewing window is square. Explain.

Question 1.
-8 ≤ x ≤ 7, -3 ≤ y ≤ 7

Question 2.
-6 ≤ x ≤ 6, -9 ≤ y ≤ 9

Question 3.
-18 ≤ x ≤ 18, -12 ≤ y ≤ 12

Use a graphing calculator to graph the equation. Use a square viewing window.

Question 4.
y = x + 3
Answer:
The given equation is:
y = x + 3
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 5.
y = -x – 2
Answer:
The given equation is:
y = -x – 2
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 6.
y = 2x – 1
Answer:
The given equation is:
y = 2x – 1
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 7.
y = -2x + 1
Answer:
The given equation is:
y = -2x + 1
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 8.
y = –\(\frac{1}{3}\)x – 4
Answer:
The given equation is:
y = –\(\frac{1}{3}\)x – 4
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 9.
y = \(\frac{1}{3}\)x + 2
Answer:
The given equation is:
y = \(\frac{1}{3}\)x + 2
Now,
We can find the values of x and y by putting the values 0, 1, 2…..
Hence,
The representation of the given equation in the coordinate plane is:

Question 10.
How does the appearance of the slope of a line change between a standard viewing window and a square viewing window?
Answer:
A typical graphing calculator screen has a height to width ratio of 2 to 3. This means that when you use the standard viewing window of -10 to 10 ( on each axis ), the graph will not be in its true perspective.
To see a graph in its true perspective, you need to use a square viewing window, in which the tick marks on the x-axis are spaced the same as the tick marks on the y-axis.

Lesson 3.1 Functions

Essential Question

A relation pairs inputs with outputs. When a relation is given as ordered pairs, the x-coordinates are inputs and the y-coordinates are outputs. A relation that pairs each input with exactly one output is a function.

EXPLORATION 1
Describing a Function
Work with a partner.
Functions can be described in many ways.

  • by an equation
  • by an input-output table
  • using words
  • by a graph
  • as a set of ordered pairs

a. Explain why the graph is shown represents a function.
Answer:
The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because that x value has more than one output. A function has only one output value for each input value.

b. Describe the function in two other ways.
Answer:
The function can be described in 4 ways. They are:
a. A function can be represented verbally.
Example:
The circumference of a square is four times one of its sides.
b. A function can be represented algebraically.
Example:
3 x + 6 .
c. A function can be represented numerically.
d. A function can be represented graphically.

EXPLORATION 2

Identifying Functions
Work with a partner. Determine whether each relation represents a function. Explain your reasoning.

Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 1.1
e. (-2, 5), (-1, 8), (0, 6), (1, 6), (2, 7)
f. (-2, 0), (-1, 0), (-1, 1), (0, 1), (1, 2), (2, 2)
g. Each radio frequency x in a listening area has exactly one radio station y.
h. The same television station x can be found on more than one channel y.
i. x = 2
j. y = 2x + 3
Answer:
We know that,
Functions can be described in different ways. They are:
a. By an equation
b. By an input-output table
c. Using words
d. By a graph
e. As a set of ordered pairs
We know that,
A function has only a single input for each or multiple outputs
Hence,
The given Exercises above i.e., a, e, g,  j are the relations that represent a function and the remaining relations are not functions

Communicate Your Answer

Question 3.
What is a function? Give examples of relations, other than those in Explorations 1 and 2, that (a) are functions and (b) are not functions.
Answer:
Definition of Function:
A relation from a set of inputs to a set of possible outputs where each input is related to exactly one output is known as ” Function”
Examples of relation that is a function:
a. y = x + 3
b. y = -x² + 1
Example of relation that is not a function:

We know that,
A relation is a set of inputs for specific outputs.
But from the above,
A single input has multiple outputs

3.1 Lesson

Monitoring Progress

Determine whether the relation is a function. Explain.

Question 1.
(-5, 0), (0, 0), (5, 0), (5, 10)
Answer:
The given relation is:
(-5, 0), (0, 0), (5, 0), (5, 10)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
But,
When we observe the given relation, there are 2 outputs for a single input
Hence, from the above,
We can conclude that the given relation is not a function

Question 2.
(-4, 8), (-1, 2), (2, -4), (5, -10)
Answer:
The given relation is:
(-4, 8), (-1, 2), (2, -4), (5, -10)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
Hence, from the above,
We can conclude that the given relation is a function

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 2
Answer:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 2
The representation of the table in the form of relation is:
(2, 2.6), (4, 5.2), (6, 7.8)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
Hence, from the above,
We can conclude that the given relation is a function

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 3
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 3
The representation of the given figure in the form of relation is:
(1, -2), (1, 0), (2, 4)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
But,
When we observe the given relation, there are 2 outputs for a single input
Hence, from the above,
We can conclude that the given relation is not a function

Monitoring Progress

Determine whether the graph represents a function. Explain.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 4
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 4
The representation of the points from the graph is:
(0, 3), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3)
From the given relation,
We know that,
x represents the input
y represents the output
We know that,
For a function, each input should correspond with only one output
We have to remember that the inputs must be different but outputs may be the same or different
Hence, from the above,
We can conclude that the given graph is a function

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 5
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 5
From the graph,
The representation of the points are:
(2, 1), (1, 2), (1, 3), (1, 4), (0.5, 5), (0.5, 6), (3, 0), (4, 1), (4.2, 2), (4.8, 3), (5, 4), (5.1, 5), (5.1, 6)
From the points,
We can observe that the output is different for the same input
Hence, from the above,
We can conclude that the given graph is not a function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 6
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 6
From the given graph,
We can observe that for the same value of x, there are different values of y
Where,
x represents the input
y represents the output
We know that,
For a function, each input must be matched with a single output
Hence, from the above,
We can conclude that the given graph is not a function

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 7
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 7
From the graph,
We can observe that there are multiple outputs for a single input
Hence, from the above,
We can conclude that the given graph is not a function

Monitoring Progress

Find the domain and range of the function represented by the graph.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 8

Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 8
We know that,
“Domain” is defined as the set of all values present in the x-axis
“Range” is defined as the set of all values present in the y-axis
Hence,
The domain of the given graph is: -2, -1, 0, 1, 2
The range of the given graph is: 1, 2, 3,  4

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 9
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 9
We know that,
“Domain” is defined as the set of all values present in the x-axis
“Range” is defined as the set of all values present in the y-axis
Hence,
The domain of the given graph is: 1, 2, 3, 4, 5
The range of the given graph is: 0, 1, 2, 3,  4

Monitoring Progress

Question 11.
The function a = -4b + 14 represents the number ‘a’ of avocados you have left after making b batches of guacamole.
a. Identify the independent and dependent variables.
Answer:
The “Independent variable” represents the input values of a function and can be any value in the domain.
The dependent variable represents the output values of the function and depends on the value of the independent variable
Hence, from the above,
The independent variable is: b (batches of guacamole)
The dependent variable is: a ( The number of avocados )

b. The domain is 0, 1, 2, and 3. What is the range?
Answer:
The given function is:
a = -4b + 14
In the given function,
Input: b
Output: a
We know that,
The “Domain” is defined as the set of all values present in the input or the x-axis
The given Domain is: 0, 1, 2, and 3
So,
To find the range, we have to find the values of b for each value present in the domain
So,
a = -4 (0) + 14 = 14
a = -4 (1) + 14 = 10
a = -4 (2) + 14 = 6
a = -4 (3) + 14 = 2
Hence, from the above,
We can conclude that the range for the given equation is: 2,  6, 10, and 14

Question 12.
The function t = 19m + 65 represents the temperature t (in degrees Fahrenheit) of an oven after preheating for m minutes.
a. Identify the independent and dependent variables.
Answer:
The “Independent variable” represents the input values of a function and can be any value in the domain.
The dependent variable represents the output values of the function and depends on the value of the independent variable
Hence, from the above,
The independent variable is: Minutes
The dependent variable is: Temperature

b. A recipe calls for an oven temperature of 350°F. Describe the domain and range of the function.
Answer:
The given function is:
t = 19m + 65
Compare the given function with
y = mx + c
It is given that a recipe calls for an oven temperature of 350°F
So,
350 = 19m + 65
19m = 350 – 65
19m = 285
m = 285 / 19
m = 15 minutes
So,
t = 19 (15) + 350
t = 285 + 65
t = 350°F
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ m ≤ 15 [ Since the minutes will not be -ve ]
The range of the given function is: 65 ≤ t ≤ 350 [ Since the  minimum temperature is the value of c ]

Functions 3.1 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
How are independent variables and dependent variables different?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 9.1
Answer:
The given statements are:
a. Find the range of the function represented by the table?
b. Find the inputs of the function represented by the table?
c. Find the x values of the function represented by (-1, 7), (0, 5), and (1, -1)?
d. Find the domain of the function represented by (-1, 7), (0, 5), and (1, -1)?
Now,
From the given table,
The values of x are: -1, 0, 1
The values of y are: 7, 5, -1
We know that,
For a function,
The x values represent the input and the domain
The y values represent the output and the range
So,
a.
The range of the function represented by the table is: 7, 5, -1
b.
The inputs of the function represented by the table are: -1, 0, 1
c.
The x values of the function represented by the given points are: -1, 0, 1
d.
The domain of the function represented by the given points are: -1, 0, 1
Hence, from the above,
We can conclude that all the given four are the same

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, determine whether the relation is a function. Explain.

Question 3.
(1, -2), (2, 1), (3, 6), (4, 13), (5, 22)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 3

Question 4.
(7, 4), (5, -1), (3, -8), (1, -5), (3, 6)
Answer:
The given points are:
(7, 4), (5, -1), (3, -8), (1, -5), (3, 6)
We know that,
For a function, each input has only a single output
But from the above,
We can observe that 3 has multiple outputs
Hence, from the above,
We can conclude that the given relation is not a function

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 11
Answer:
The given relation is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 11
We know that,
For a function, a single input has a single output
Hence, from the above,
We can say that each input has only 1 output
Hence, from the above,
We can conclude that the given relation is a function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 13
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 13
From the given table,
We can observe that each input has a single output
Hence, from the above,
We can conclude that the given relation is a function

In Exercises 9–12, determine whether the graph represents a function. Explain.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 14
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 15
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 15
From the given graph,
We can observe that the vertical line can be drawn through more than one point on the graph i.e., input 2 has repeated 2 times i.e., (2, 1) and (2, 5)
Hence, from the above,
We can conclude that the given graph is not a function

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 17
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 17
From the given graph,
We can observe that each input corresponds to a single output
Hence, from the above,
We can conclude that the given graph is a function

In Exercises 13–16, find the domain and range of the function represented by the graph.

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 18
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 19
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 19
From the given graph,
The ordered pairs are: (0, 4), (-2, 4), (2, 4), (4, 4)
Hence,
The domain of the given graph is: 0, -2, 2, 4
The range of the given graph is: 4

Question 15.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 20
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 15

Question 16.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 21
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 21
From the given graph,
Identify the x and y values
So,
The values of x range from 2 to 7 excluding 2 and 7
The values of y range from 1 to 6 excluding 1 and 6
Hence, from the above,
We can conclude that
The domain of the given graph is: 2 < x < 7
The range of the given graph is: 1 < y < 6

Question 17.
MODELING WITH MATHEMATICS
The function y = 25x + 500 represents your monthly rent y (in dollars) when you pay x days late.
a. Identify the independent and dependent variables.
b. The domain is 0, 1, 2, 3, 4, and 5. What is the range?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 17

Question 18.
MODELING WITH MATHEMATICS
The function y = 3.5x + 2.8 represents the cost y (in dollars) of a taxi ride of x miles.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 22
a. Identify the independent and dependent variables.
Answer:
The given function is:
y = 3.5x + 2.8
From the above function,
The independent variable is: x which represents the number of miles
The dependent variable is: y which represents the cost in dollars

b. You have enough money to travel at most 20 miles in a taxi. Find the domain and range of the function.
Answer:
The given function is:
y = 3.5x + 2.8
It is given that you have enough money to travel at most 20 miles i.e., the value of x in a taxi i.e.,
x ≥ 0 and x ≤ 20
So,
The value of x ranges from 0 ≤ x ≤ 20
Now,
y = 3.5 (20) + 2.8
y = 7 + 2.8
y = $9.8
y = 3.5 (0) + 2.8
y = 0 + 2.8
y = $2.8
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ x ≤ 20 miles
The range of the given function is: $2.8 ≤ y ≤ $9.8

ERROR ANALYSIS
In Exercises 19 and 20, describe and correct the error in the statement about the relation shown in the table.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 23

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 24
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 19

Question 20.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 25
Answer:
It is given that the relation is a function and the range is 1, 2, 3, 4, and 5
We know that,
The relation is a function only when a single input pairs with an output
For the function, the domain, and the range exist
From the given table,
The domain is: 1, 2, 3, 4, 5
The range is: 6, 7, 8, 6, 9
Hence, from the above,
We can conclude that the given statement is not correct

ANALYZING RELATIONSHIPS
In Exercises 21 and 22, identify the independent and dependent variables.

Question 21.
The number of quarters you put into a parking meter affects the amount of time you have on the meter.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 21

Question 22.
The battery power remaining on your MP3 player is based on the amount of time you listen to it.
Answer:
The given statement is:
The battery power remaining on your MP3 player is based on the amount of time you listen to it.
Hence, from the above,
We can conclude that
The Independent variable: Amount of time
The dependent variable: Battery power

Question 23.
MULTIPLE REPRESENTATIONS
The balance y (in dollars) of your savings account is a function of the month x.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 26
a. Describe this situation in words.
b. Write the function as a set of ordered pairs.
c. Plot the ordered pairs in a coordinate plane.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 23

Question 24.
MULTIPLE REPRESENTATIONS
The function 1.5x + 0.5y = 12 represents the number of hardcover books x and softcover books y you can buy at a used book sale.
a. Solve the equation for y.
Answer:
The given function is:
1.5x + 0.5y = 12
So,
0.5y = 12 – 1.5x
y = \(\frac{12 – 1.5x}{0.5}\)
y = \(\frac{12}{0.5}\) – \(\frac{1.5x}{0.5}\)
y = 24 – 3x
Hence, from the above,
We can conclude that the equation for y is:
y = 24 – 3x

b. Make an input-output table to find ordered pairs for the function.
Answer:
From part (a),
The equation for y is:
y = 24 – 3x
Now,
Put the values of 0, 1, 2, 3 in x
So,
y = 24 – 3(0) = 24
y = 24 – 3(1) = 21
y = 24 – 3(2) = 18
y = 24 – 3 (3) = 15
Hence,
The input-output table for the given equation of y is:

c. Plot the ordered pairs in a coordinate plane.
Answer:
From part (b),
The table is:

We know that,
The ordered pair is in the form of (x, y)
From the table,
The ordered pairs are:
(0, 24), (1, 21), (2, 18), (3, 15)
Hence,
The representation of the ordered pairs in the coordinate plane is:

Question 25.
ATTENDING TO PRECISION
The graph represents a function. Find the input value corresponding to an output of 2.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 27
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 25

Question 26.
OPEN-ENDED
Fill in the table so that when t is the independent variable, the relation is a function, and when t is the dependent variable, the relation is not a function.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 28
Answer:
Let the function in terms of t and v  such as t is the independent variable and v is the dependent variable is:
v = t + 4 ———–(1)
Let the function in terms of t and v  such as v is the independent variable and t is the dependent variable is:
t = 4v ———–(2)
Now,
place the values 0, 1, 2, 3 …….. in the independent variables of both functions
Now,
In equation (1),
v = 0 + 4 = 4
v = 1 + 4 = 5
v = 2 + 4 = 6
v = 3 + 4 = 7
In equation (2),
t = 4(0) = 0
t = 4(1) = 4
t = 4(2) = 8
t = 4(3) = 12
Hence,
     
Hence, from the above,
We can conclude that equation (1) is a function and equation (2) is not a function

Question 27.
ANALYZING RELATIONSHIPS
You select items in a vending machine by pressing one letter and then one number.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 29
a. Explain why the relation that pairs letter-number combinations with food or drink items is a function.
b. Identify the independent and dependent variables.
c. Find the domain and range of the function
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 27

Question 28.
HOW DO YOU SEE IT?
The graph represents the height h of a projectile after t seconds.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 30
a. Explain why h is a function of t.
b. Approximate the height of the projectile after 0.5 seconds and after 1.25 seconds.
c. Approximate the domain of the function.
d. Is t a function of h? Explain.
Answer:
a.
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 30
From the graph,
We can observe that when we draw the vertical lines, each vertical line corresponds to only 1 value.
Hence, from the above,
We can conclude that h is a function of t by using the vertical line method

b.
From the given graph,
We can observe that the height of the projectile after 0.5 seconds increases and after some time, the height of the projectile decreases
We can observe that the height of the projectile after 1.25 seconds decreases steadily
Hence, from the above,
We can conclude that
The approximate maximum height after 0.5 seconds is: 30 feet
The approximate maximum height after 1.25 seconds is: 25 feet

c.
From the given graph,
The values of t in the x-axis vary from 0 to 2.5
Hence,
The domain of the given graph is: 0 ≤ t ≤ 2.5 seconds

d.
t is not a function of h
Reason:
When we observe the graph,
We can see that for a single value of t, there are multiple values of h.
We know that a relation can be considered as a function only when a single input pairs with a single output
hence, from the above,
We can conclude that that t is not a function of h

Question 29.
MAKING AN ARGUMENT
Your friend says that a line always represents a function. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 29

Question 30.
THOUGHT-PROVOKING
Write a function in which the inputs and/or the outputs are not numbers. Identify the independent and dependent variables. Then find the domain and range of the function.
Answer:
It is given that the inputs and/or the outputs will not be numbers.
So,
The given function in which the inputs and/or the outputs are not numbers is:
°C = 32 + °F
From the given function,
The independent variable is: °F
The dependent variable is: °C
Now,
From the given function,
We can observe that the values of °F vary from -∞ to ∞
Now,
Place the values of -∞ to ∞ in the place of °F
So,
°C = 32 + 0 = 32°F
°C = 32 + 1 = 33°F
°C = 32 – 1 = 31°F
°C = 32 – 40 = -8°F
Hence, from the above values,
We can observe that the values of °C vary from -∞ to ∞
Hence, from the above,
We can conclude that
The domain of the function is: -∞ to ∞
The range of the function is: -∞ to ∞

ATTENDING TO PRECISION In Exercises 31–34, determine whether the statement uses the word function in a way that is mathematically correct. Explain your reasoning.

Question 31.
The selling price of an item is a function of the cost of making the item.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 31

Question 32.
The sales tax on a purchased item in a given state is a function of the selling price.
Answer:
The given statement is:
The sales tax on a purchased item in a given state is a function of the selling price
We know that,
The sales tax is a percentage applied to the selling price
Hence, from the above,
We can conclude that the given statement uses the word function in a way that is mathematically correct.

Question 33.
A function pairs each student in your school with a homeroom teacher.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 33

Question 34.
A function pairs each chaperone on a school trip with 10 students.
Answer:
The given statement is:
A function pairs each chaperone on a school trip with 10 students.
We know that,
Each chaperone on a school trip pairs with more than 1 student i.e., the number may be 2 or ∞ but not exactly 10
Hence, from the above,
We can conclude that the given statement does not use the word function in a way that is mathematically correct.

REASONING
In Exercises 35–38, tell whether the statement is true or false. If it is false, explain why.

Question 35.
Every function is a relation.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 35

Question 36.
Every relation is a function.
Answer:
The given statement is false

Explanation:
We know that,
Every function is a relation that has only 1 output for a single input
But we can not say that every relation has a single output for a single input
Hence, from the above,
We can conclude that the given statement is false

Question 37.
When you switch the inputs and outputs of any function, the resulting relation is a function.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 37

Question 38.
When the domain of a function has an infinite number of values, the range always has an infinite number of values.
Answer:
The given statement is false

Explanation:
We know that,
The domain is defined as the set of all the values of x
The range is defined as the set of all the values of y
Now, consider an example
Let the input be x
Let the output be a constant
Now,
The domain of the input can vary from -∞ to ∞
But the range of the output is only a constant even though we put any value of x
Hence, from the above,
We can conclude that the given statement is false

Question 39.
MATHEMATICAL CONNECTIONS
Consider the triangle shown.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 31
a. Write a function that represents the perimeter of the triangle.
b. Identify the independent and dependent variables.
c. Describe the domain and range of the function. (Hint: The sum of the lengths of any two sides of a triangle is greater than the length of the remaining side.)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 39

REASONING
In Exercises 40–43, find the domain and range of the function.

Question 40.
y = | x |
Answer:
The given function is:
y = | x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
We can put the values of x from -∞ to ∞
So,
The values of x vary from 0 to ∞ since x can’t be negative
Hence, from the above,
We can conclude that
The domain of the given function is: -∞ to ∞
The range of the given function is: 0 to ∞

Question 41.
y = – | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 41

Question 42.
y = | x | – 6
Answer:
The given function is:
y = | x | – 6
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
We can vary the values of x from -∞ to ∞
The values of y vary from -6 to ∞
Hence, from the above,
We can conclude that
The domain of the given function is: -∞ to ∞
The range of the given function is: y ≥ -6

Question 43.
y = 4 – | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 43

Maintaining Mathematical Proficiency

Write the sentence as an inequality. (Section 2.1)

Question 44.
A number y is less than 16.
Answer:
The given worded form is:
A number y is less than 16
Hence,
The representation of the given worded form in the form of inequality is:
y < 16

Question 45.
Three is no less than a number x.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 45

Question 46.
Seven is at most the quotient of a number d and -5.
Answer:
The given worded form is:
Seven is at most the quotient of a number d and -5
Hence,
The representation of the given worded form in the form of inequality is:
7 ≤ d ÷ (-5)

Question 47.
The sum of a number w and 4 is more than -12.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 47

Evaluate the expression.

Question 48.
112
Answer:
The product of 11² is:
11² = 11 × 11 = 121

Question 49.
(-3)4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 49

Question 50.
-52
Answer:
The product of -5² is:
-5² = -5 × -5 = 25 [ We know that – × – = + ]

Question 51.
25
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.1 Question 51

Lesson 3.2 Linear Functions

Essential Question

How can you determine whether a function is linear or non-linear?
Answer:
Simplify the equation as closely as possible to the form of y = mx + c.
Check to see if your equation has exponents.
If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.

EXPLORATION 1
Finding Patterns for Similar Figures

Work with a partner. Copy and complete each table for the sequence of similar figures. (In parts (a) and (b), use the rectangle shown.) Graph the data in each table. Decide whether each pattern is linear or nonlinear. Justify your conclusion.
a. perimeters of similar rectangles
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 32
Answer:
We know that,
The perimeter of the rectangle (P) = 2 (Length + Width)
P = 2 (x + 2x)
P = 2 (3x)
P = 6x
Hence,
The completed table for the perimeters of similar rectangles is:

The representation of the perimeters of the similar rectangles in the coordinate plane is:

b. areas of similar rectangles
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 32
Answer:
We know that,
Area of the rectangle (A) = Length × Width
A = x × (2x)
A = 2x²
Hence,
The complete table for the area of the similar rectangles is:

The representation of the areas of the similar rectangles in the coordinate plane is:

c. circumferences of circles of radius r
Answer:
We know that,
The circumference of circle = 2πr
Take the value of π as 3
Hence,
The complete table for the circumferences of circles of radius r is:

The representation of the circumferences of the circles in the coordinate plane is:

d. areas of circles of radius r
Answer:
We know that,
The area of the circle = πr²
Take the value of π as 3
Hence,
The complete table for the areas of the similar circles is:


The representation of the areas of the similar circles is:

Communicate Your Answer

Question 2.
How do you know that the patterns you found in Exploration 1 represent functions?
Answer:
In Exploration 1,
From the graphs,
We can observe that from the vertical test, only one point passes through each vertical line i.e., each input hs only 1 output
Our observation coincides with the definition of function
Hence, from the above,
We can conclude that the given patterns in Exploration 1 represent functions

Question 3.
How can you determine whether a function is linear or nonlinear?
Answer:
Simplify the equation as closely as possible to the form of y = mx + c.
Check to see if your equation has exponents.
If it has exponents, it is nonlinear. If your equation has no exponents, it is linear.

Question 4.
Describe two real-life patterns: one that is linear and one that is nonlinear. Use patterns that are different from those described in Exploration 1.
Answer:
The real-life pattern that is linear is:
The distance you travel when you go for a jog, you can graph the function and make some assumptions with only two points. The slope of a function is the same as the rate of change for the dependent variable (y), For instance, if you’re graphing distance Vs
The real-life pattern that is non-linear is:
Triangulation of GPS signals
Example:
A device like your cellphone receives signals from GPS satellites, which have known orbital positions around the Earth.

3.2 Lesson

Monitoring Progress

Does the graph or table represent a linear or nonlinear function? Explain.

Question 1.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 36
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 36
The graph represents a straight line
We know that,
The straight line will be in the form of
y = mx + c or y = mx
Where,
m is the slope-intercept
c is the y-intercept that cuts through the y-axis
Hence, from the above,
We can conclude that the given graph is linear function

Question 2.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 37
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 37
From the given graph,
We can observe that the graph is not a straight line i.e., it is not linear
Hence, from the above,
We can conclude that the given graph is a non-linear function

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 38
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 38
From the given table,
We can observe that
There is a constant difference of 1 between the values of x and there is a constant difference of 2 between the values of y
The difference is constant for both the values of x and y
Hence, from the above,
We can conclude that the given table is a linear function

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 39
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 39
From the given table,
We can observe that
There is a constant difference of 1 between the values of x and there is a constant ratio of 2 between the values of y.
Since the operations are different between the values of x and y,
We can conclude that the given table is a non-linear function

Does the equation represent a linear or nonlinear function? Explain.

Question 5.
y = x + 9
Answer:
The given equation represents the linear function

Explanation:
The given function is:
y = x + 9
We know that,
The standard representation of the linear function is:
y = mx + c
When we compare the given equation with the standard representation of linear function,
We can conclude that the given equation is a linear function

Question 6.
y = \(\frac{3 x}{5}\)
Answer:
The given equation represents a linear function

Explanation:
The given function is:
y = \(\frac{3 x}{5}\)
5y = 3x
3x – 5y = 0
y = \(\frac{3}{5}\)x + 0
We know that,
The standard representation of a linear function is:
y = mx + c
When we compare the above function with the standard representation of a linear function,
We can conclude that the given function is a linear function

Question 7.
y = 5 – 2x²
Answer:
The given equation represents a non-linear function

Explanation:
The given function is:
y = 5 – 2x²
We know that,
The standard representation of a linear function is:
y = mx + c
When we compare the given function with the standard representation of a linear function,
We can conclude that the given function is a non-linear function

Question 8.
The linear function m = 50 – 9d represents the amount m (in dollars) of money you have after buying d DVDs.
(a) Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
The given linear function is:
m = 50 – 9d
Where,
m is the amount in dollars you have after buying d DVDs
m is the dependent variable
d is the independent variable
We know that,
The domain is defined for the independent variables
So,
Let
d = 0, 1, 2, 3, 4, 5……
Now,
m = 50 – 9 (0) = 50
m = 50 – 9 (1) = 41
m = 50 – 9 (2) = 32
m = 50 – 9 (3) = 23
m = 50 – 9 (4) = 14
m = 50 – 9 (5) = 5
m = 50 – 9 (6) = -4
Hence, from the above,
We can conclude that
The domain of the given linear function is: 0, 1, 2, 3, 4, and 5 [ Since from 6, -ve values are coming and the money will not be -ve ]
The domain is discrete [ Since a discrete graph is a series of unconnected points ]

(b) Graph the function using its domain.
Answer:
From part (a),
The domain of the given function is: 0, 1, 2, 3, 4, and 5
Hence,
The representation of the domain in the coordinate plane is:

Question 9.
Is the domain discrete or continuous? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 40
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 40
From the given table,
The ordered pairs are:
(1, 12), (2, 24), (3, 36)
Hence,
The representation of the ordered pairs in the coordinate plane is:

From the above graph,
We can say that the points are scattered or discrete and they are unconnected
Hence, from the above,
We can conclude that the domain for the given table is discrete

Question 10.
A 20-gallon bathtub is draining at a rate of 2.5 gallons per minute. The number g of gallons remaining is a function of the number m of minutes.
a. Does this situation represent a linear function? Explain.
Answer:
It is given that a 20-gallon bathtub is draining at a rate of 2.5 gallons per minute.
Where,
The number g of gallons remaining is a function of the number m of minutes.
So,
From the above,
We can say that the bathtub is draining at the constant rate
We know that,
From the property of linear function, the change will be constantly increasing or decreasing
So,
The representation of the linear function for this situation is:
g = 20 – 2.5x
Hence, from the above,
We can conclude that the given situation is a linear function

b. Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
From part (a),
We can conclude that the given situation is a linear function
So,
The domain of the function = \(\frac{Total volume of the bathtub}{The rate of draining}\)
= \(\frac{20}{2.5}\)
= \(\frac{200}{25}\)
= 8
Hence,
The domain of the function is: 0 ≤ x ≤ 8 [ Since the draining rate will not be -ve ]
From the representation of the domain,
We can conclude that the domain is continuous

c. Graph the function using its domain.
Answer:
From part (a),
The linear function is:
g = 20 – 2.5x
We know that,
The domain is: 0 ≤ x ≤ 8
So,
g = 20 – 2.5 (0) = 20
g = 20 – 2.5 (1) = 17.5
g = 20 – 2.5 (2) = 15
g = 20 – 2.5 (3) = 12.5
g = 20 – 2.5 (4) = 10
g = 20 – 2.5 (5) = 7.5
g = 20 – 2.5 (6) = 5
g = 20 – 2.5 (7) = 2.5
g = 20 – 2.5 (8) = 0
Hence,
The representation of the linear function using the domain in the coordinate plane is:

Write a real-life problem to fit the data shown in the graph. Is the domain of the function discrete or continuous? Explain.

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 41
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 41
The real-life problem that fits the given graph is:
An escalator moves upwards at a constant rate of 1step/minute.
So,
What is the rate escalator moves upwards after 8 minutes?
From the given graph,
We can say that the points are connected
Hence, from the above,
We can conclude that the domain of the function is continuous

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 42
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 42
The real-life problem that fits the given graph is:
If one company offers to pay you $450 per week and the other offers $10 per hour, and both ask you to work 40 hours per week, which company is offering the better rate of pay?
From the graph,
We can observe that the points are connected
Hence, from the above,
We can conclude that the domain is continuous

Linear Functions 3.2 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A linear equation in two variables is an equation that can be written in the form ________, where m and b are constants.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 1

Question 2.
VOCABULARY
Compare linear functions and nonlinear functions.
Answer:
Linear Function:
A Linear function is a relation between two variables that produces a straight line when graphed.
Example:
y = 2x + 3
Non-Linear Function:
A non-linear function is a function that does not form a line when graphed.
Example:
y = 6x³

Question 3.
VOCABULARY
Compare discrete domains and continuous domains.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 3

Question 4.
WRITING
How can you tell whether a graph shows a discrete domain or a continuous domain?
Answer:
A “Discrete domain” is a set of input values that consists of only certain numbers in an interval.
A “continuous domain” is a set of input values that consists of all numbers in an interval.
Sometimes, the set of points that represent the solutions of an equation are distinct, and other times the points are connected.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, determine whether the graph represents a linear or nonlinear function. Explain.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 43
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 44
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 44
From the above graph,
By using the vertical test, there is only one point going through each point and the given graph is a straight line
Hence, from the above,
We can conclude that the given graph is a linear function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 45
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 46
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 46
From the above graph,
By using the vertical test, each line passes through only one point and the given graph is a straight line
Hence, from the above,
We can conclude that the given graph is a linear function

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 47
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 48
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 48
From the above graph,
By using the vertical test, each line passes through each point and the given graph is not a straight line
Hence, from the above,
We can conclude that the given graph is a non-linear function

In Exercises 11–14, determine whether the table represents a linear or nonlinear function. Explain.

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 49
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 50
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 50
From the above table,
The difference between all the values of x is 2 which is a constant through all the values of x
The difference between all the values of y is not constant
Since the difference between all the values of y is not constant
The given table is a non-linear function

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 51
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 52
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 52
From the above table,
The difference between all the values of x is +1 throughout all the values of x
The difference between all the values of y is -15 throughout all the values of y
Since the differences are constant for all the values of x and y,
The given table is a linear function

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in determining whether the table or graph represents a linear function.

Question 15.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 53
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 15

Question 16.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 54
Answer:
By using the vertical test in the graph,
We can say that each line passes through one point and the given graph is a straight line
Hence, from the above,
We can conclude that the given graph is a linear function

In Exercises 17–24, determine whether the equation represents a linear or nonlinear function. Explain.

Question 17.
y = x2 + 13
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 17

Question 18.
y = 7 – 3x
Answer:
The given equation is a linear function

Explanation:
The given equation is:
y = 7 – 3x
Compare the given equation with the standard representation of the given linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is a linear function

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 55
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 19

Question 20.
y = 4x(8 – x)
Answer:
The given equation is not a linear function

Explanation:
The given equation is:
y = 4x (8 – x)
so,
y = 4x (8) – 4x (x)
y = 32x – 4x²
Compare the above equation with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is not a linear function

Question 21.
2 + \(\frac{1}{6}\) y = 3x + 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 21

Question 22.
y – x = 2x – \(\frac{2}{3}\)y
Answer:
The given equation is a linear fraction

Explanation:
The given equation is:
y – x = 2x – \(\frac{2}{3}\)y
So,
y + \(\frac{2}{3}\)y = 2x + x
\(\frac{3y}{3}\) + \(\frac{2y}{3}\) = 3x
\(\frac{5}{3}\)y = 3x
y = 3x × \(\frac{3}{5}\)
y = \(\frac{9}{5}\)x + 0
Compare the above equation with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is a linear function

Question 23.
18x – 2y = 26
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 23

Question 24.
2x + 3y = 9xy
Answer:
The given equation is not a linear fraction

Explanation:
The given equation is:
2x + 3y = 9xy
2x = 9xy – 3y
2x = y (9x – 3)
y = \(\frac{2}{9x – 3}\)x
Compare the above equation with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above,
We can conclude that the given equation is not a linear function

Question 25.
CLASSIFYING FUNCTIONS
Which of the following equations do not represent linear functions? Explain.
A. 12 = 2x2 + 4y2
B. y – x + 3 = x
C. x = 8
D. x = 9 – \(\frac{3}{4}\)y
E. y = \(\frac{5x}{11}\)
F = \(\sqrt{x}\) + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 25

Question 26.
USING STRUCTURE
Fill in the table so it represents a linear function.
Big Ideas Math Algebra 1 Solutions Chapter 3 Graphing Linear Functions 56
Answer:
The given table is:
Big Ideas Math Algebra 1 Solutions Chapter 3 Graphing Linear Functions 56
From the above table,
we can observe that the difference between all the values of x is 5 which is a constant
Now,
To find the difference between all the values of y which is a constant, we have to use the trial and error method.
Now,
If we add +1 to -1 and continue adding +1 to all the values of y, then
-1 + 1 = 0
0 + 1 = 1
1 + 1 = 2
2 + 1 = 3
But the last value is: 11
Now,
If we add +2 to -1 and continue adding +2 to all the values of y, then
-1 + 2 = 1
1 + 2 = 3
3 + 2 = 5
5 + 2 = 7
But the last value is: 11
Now,
If we add +3 to -1 and continue adding +3 to all the values of y, then
-1 + 3 = 2
2 + 3 = 5
5 + 3 = 8
8 + 3 = 11
The last value is also: 11
Hnece, from the above,
We can conclude that we have to add +3 to make all the values of y constant so that the given table represents a linear function
The completed table is:

In Exercises 27 and 28, find the domain of the function represented by the graph. Determine whether the domain is discrete or continuous. Explain.

Question 27.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 57
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 27

Question 28.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 58
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 58
We know that,
The domain is defined as the range of all the values of x
So,
From the above graph,
The domain is: 0, 1, 2, 3, 4, 5, 6, and 7
Hence, from the above,
We can conclude that the domain of the given graph is: 0, 1, 2, 3, 4, 5, 6, and 7

In Exercises 29–32, determine whether the domain is discrete or continuous. Explain.

Question 29.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 59
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 29

Question 30.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 60
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 60
From the given table,
The difference between the values of x is 1 which is constant throughout all of the values of x
The difference between the values of y is 3 which is constant throughout all of the values of y
Hence,
Since the difference is constant for both the values of x and y,
The domain of the given table is continuous

Question 31.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 61
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 31

Question 32.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 62
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 62
From the above table,
The difference between all the values of x is 1 which is a constant
The difference between all the values of y is 4 which is a constant
Hence,
Since the difference between all the values of x and y is constant,
The given function is a linear function
The domain of the given function is continuous

ERROR ANALYSIS
In Exercises 33 and 34, describe and correct the error in the statement about the domain. 33. xy214324682.5 is in the domain.

Question 33.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 63
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 33

Question 34.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 64
Answer:
We know that,
The domain is the range of all the values of x
Now,
From the given graph,
The domain of the given function is: 0, 1, 2, 3, 4, 5, and 6
From the given graph,
We can observe that the domain of the given graph is continuous because there are not any unconnected points in the graph

Question 35.
MODELING WITH MATHEMATICS
The linear function m = 55 – 8.5b represents the amount m (in dollars) of money that you have after buying b books.
a. Find the domain of the function. Is the domain discrete or continuous? Explain.
b. Graph the function using its domain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 65
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 35.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 35.2

Question 36.
MODELING WITH MATHEMATICS
The number y of calories burned after x hours of rock climbing is represented by the linear function y = 650x.
a. Find the domain of the function. Is the domain discrete or continuous? Explain.
b. Graph the function using its domain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 65.1
Answer:
a.
It is given that
The number y of calories burned after x hours of rock climbing is represented by the linear function
y = 650x.
It is given that x is the number of hours
Hence,
The domain of the given function is:
1 ≤ x ≤ 24
Since, the domain of the function is inequality,
The domain of the function is continuous

b.
The given function is:
y = 650x
From part (a),
The domain of the function is: 1 ≤ x ≤ 24
So,
y = 650 (1) = 650
y = 650 (2) = 1300
y = 650 (3) = 1950
.
.
.
.
y = 650 (12) = 7800
Hence,
The representation of the function with the domain in the coordinate plane is:

Question 37.
MODELING WITH MATHEMATICS
You are researching the speed of sound waves in dry air at 86°F. The table shows the distances d (in miles) sound waves travel in t seconds.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 66
a. Does this situation represent a linear function? Explain.
b. Find the domain of the function. Is the domain discrete or continuous? Explain.
c. Graph the function using its domain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 37

Question 38.
MODELING WITH MATHEMATICS
The function y = 30 + 5x represents the cost y (in dollars) of having your dog groomed and buying x extra services.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 67
a. Does this situation represent a linear function? Explain.
Answer:
The given function is:
y = 30 + 5x
We know that,
The standard representation of the linear function is:
y = mx + c
Compare the given function with the standard representation
Hence, from the above,
We can conclude that the given situation represents a linear function

b. Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
The given function is:
y = 30 + 5x
Where,
y is the amount in dollars
x is the cost of extra grooming services
From the above,
The maximum number of given extra grooming services is: 5
So,
We can use extra grooming service or not
Hence, from the above,
We can conclude that
The domain of the given function is: 0 ≤ x ≤ 5
Hence,
The domain of the given function is continuous

c. Graph the function using its domain.
Answer:
The given function is:
y = 30 + 5x
We know that,
The domain of the function is: 0 ≤ x ≤ 5
So,
y = 30 + 5(0) = 30
y = 30 + 5 (1) = 35
y = 30 + 5(2) = 40
y = 30 + 5 (3) = 45
y = 30 + 5 (4) = 50
y = 30 + 5 (5) = 55
Hence,
The representation of the function using its domain in the coordinate plane is:

WRITING In Exercises 39–42, write a real-life problem to fit the data shown in the graph. Determine whether the domain of the function is discrete or continuous. Explain.

Question 39.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 68
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 69
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 69
From the above graph,
The real-life situation is:
The temperature of a country in the winter season falls by 2°C
The domain of the graph is continuous as there are not any unconnected points

Question 41.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 70
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 41

Question 42.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 71
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 71
From the above graph,
The real-life situation is:
The number of ants in a colony increase by 2 times per day. So, the number of ants increases by how many times in 5 days?
The domain of the given graph is continuous since there is not any unconnected point in the graph

Question 43.
USING STRUCTURE
The table shows your earnings y (in dollars) for working x hours.
a. What is the missing y-value that makes the table represent a linear function?
b. What is your hourly pay rate?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 72
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 43

Question 44.
MAKING AN ARGUMENT
The linear function d = 50t represents the distance d (in miles) Car A is from a car rental store after t hours. The table shows the distances Car B is from the rental store.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 73
a. Does the table represent a linear or nonlinear function? Explain.
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 73
From the above table,
The values of x increases at a constant rate of 2 but the value of y increases by 120 1st time and by 130 2nd time
Hence,
We can observe that for the constant difference of the values of x, there is no constant difference of the values of y
Hence, from the above,
We can conclude that the given table is a linear function

b. Your friend claims Car B is moving at a faster rate. Is your friend correct? Explain.
Answer:
It is given that
The function represented by car A is:
d = 50t
The function represented by car B is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 73
From the above functions,
We can say that the distance traveled by car A increases at a constant rate whereas the distance traveled by car B increases at a random rate
Hence, from the above
We can conclude that the car B is moving at a faster rate when we observe the above table

MATHEMATICAL CONNECTIONS
In Exercises 45–48, tell whether the volume of the solid is a linear or nonlinear function of the missing dimension(s). Explain.

Question 45.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 45

Question 46.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.1
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.1
From the above figure,
We can observe that the given figure is a prism
We know that,
The volume of a prism = Area × height
= Length × Width × Height
So,
The volume of a prism (V) =3 × b × 4
V = 12b
Compare the above volume with y = mx + c
So,
V = 12b + 0
Hence, from the above,
We can conclude that the equation represents a Linear function

Question 47.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 47

Question 48.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.3
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 74.3
From the above figure,
We can observe that the given figure is a cone
We know that,
The volume of a cone = \(\frac{1}{3}\) πr²h
So,
The volume of a cone (V) = \(\frac{1}{3}\) × \(\frac{22}{7}\) × r² × 15
V = 770r²
Compare the above equation with
y = mx + c
But, the given equation is not in the form of y = mx + c
Hence, from the above,
We can conclude that the equation represents a non-linear function

Question 49.
REASONING
A water company fills two different-sized jugs. The first jug can hold x gallons of water. The second jug can hold y gallons of water. The company fills A jugs of the first size and B jugs of the second size. What does each expression represent? Does each expression represent a set of discrete or continuous values?
a. x + y
b. A + B
c. Ax
d. Ax + By
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 75
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 49

Question 50.
THOUGHT-PROVOKING
You go to a farmer’s market to buy tomatoes. Graph a function that represents the cost of buying tomatoes. Explain your reasoning.
Answer:
It is given that you go to a farmer’s market to buy tomatoes.
So,
To draw the graph that represents the cost of buying tomatoes,
The required relation is:
Cost of tomatoes ∝ Quantity or weight of tomatoes
We know that,
∝ represents the direct relation. In a graph, this relation can be represented in a straight line
Hence,
The representation of the relation of cost of buying tomatoes and weight of tomatoes is:

Question 51.
CLASSIFYING A FUNCTION
Is the function represented by the ordered pairs linear or nonlinear? Explain your reasoning.
(0, 2), (3, 14), (5, 22), (9, 38), (11, 46)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 51

Question 52.
HOW DO YOU SEE IT?
You and your friend go running. The graph shows the distances you and your friend run.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 76
a. Describe your run and your friend’s run. Who runs at a constant rate? How do you know? Why might a person not run at a constant rate?
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 76
From the graph,
The running represented by you is a straight line
The running represented by your friend is not a straight line
We know that,
A straight line has a constant rate
Hence, from the above,
We can conclude that you run at a constant rate and your friend does not run at a constant rate

b. Find the domain of each function. Describe the domains using the context of the problem.
Answer:
From part (a),
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 76
We know that,
The domain is defined as the range of the values of x
Hence, from the above,
We can conclude that
The domain of the function related to you is: 0 ≤ x ≤ 50
The domain of the function related to your friend is: 0 ≤ x ≤ 50

WRITING
In Exercises 53 and 54, describe a real-life situation for the constraints.

Question 53.
The function has at least one negative number in the domain. The domain is continuous.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 53

Question 54.
The function gives at least one negative number as an output. The domain is discrete.
Answer:
When you go on a world tour and for some days, you stayed in Antarctica,
The temperatures in Antarctica is at a negative temperature around the year and only for some months in the year, the temperature will be positive

Maintaining Mathematical Proficiency

Tell whether x and y show direct variation. Explain your reasoning.

Question 55.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 77
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 55

Question 56.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 78
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 78
From the graph,
We can observe that it is a straight line and passes through the origin
Hence, from the above,
We can conclude that x and y shows direct variation

Question 57.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 79
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 57

Evaluate the expression when x = 2.

Question 58.
6x + 8
Answer:
The given expression is: 6x + 8
When x = 2,
6x + 8 = 6 (2) + 8
= 12 + 8 = 20
Hence,
The value of the expression when x = 2 is: 20

Question 59.
10 – 2x + 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 59

Question 60.
4(x + 2 – 5x)
Answer:
The given expression is: 4 (x + 2 – 5x)
When x = 2,
4 (x + 2 – 5x) = 4 (2 + 2 – 5 (2) )
= 4 (4 – 10 )
= 4 (-6) = -24
Hence,
The value of the expression when x = 2 is: -24

Question 61.
\(\frac{x}{2}\) + 5x – 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.2 Question 61

Lesson 3.3 Function Notation

Essential Question

How can you use function notation to represent a function?
The notation f(x), called function notation, is another name for y. This notation is read as “the value of f at x” or “f of x.” The parentheses do not imply multiplication. You can use letters other than f to name a function. The letters g, h, j, and k are often used to name functions.

EXPLORATION 1
Matching Functions with Their Graphs
Work with a partner.
Match each function with its graph.
a. f (x) = 2x – 3
b. g(x) = -x + 2
c. h(x) = x2 – 1
d. j(x) = 2x2 – 3
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 80
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 81
Answer:
The given equations are:
a. f (x) = 2x – 3
b. g(x) = -x + 2
c. h(x) = x2 – 1
d. j(x) = 2x2 – 3
Now,
a.
The given equation is:
f(x) = 2x – 3
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph B) matches this equation
b.
The given equation is:
f(x) = 2x – 3
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph D) matches this equation
c.
The given equation is:
h(x) = x² – 1
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph A) matches this equation
d.
The given equation is:
j(x) = 2x² – 3
So,
The representation of the given equation in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches this equation

EXPLORATION 2
Evaluating a Function
Work with a partner.
Consider the function
f(x) = -x + 3.
Locate the points (x, f(x)) on the graph. Explain how you found each point.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 82
a. (-1, f(-1))
b. (0, f(0))
c. (1, f(1))
d. (2, f(2))
Answer:
The given function is:
f(x) = -x + 3
The graph for the given function is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 82
Now,
The simplified points are:
a.
(-1, f(-1)) = (-1, [-(-1) + 3]) = (-1, 4)
b.
(0, f(0)) = (0, [0 + 3]) = (0, 3)
c.
(1, f(1)) = (1, [-1 + 3]) = (1, 2)
d.
(2, f(2)) = (2, [-2 + 3]) = (2, 1)
So,
The simplified points are:
(-1, 4), (0, 3), (1, 2), (2, 1)
Hence,
The representation of the above points in the graph is:

Communicate Your Answer

Question 3.
How can you use function notation to represent a function? How are standard notation and function notation similar? How are they different?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 82.1
Answer:
“Function notation” is a simpler method of describing a function without a lengthy written explanation. The most frequently used function notation is f(x) which is read as “f” of “x”.
The standard notation and function notation are similar in the way of simplification
Difference betwwen Function notataion and standard notation:
The representation of function notation largely depends on the number of variables present in the function
Ex:
f(x,y) = 2xy + 3
f(x) = x + 3
The representation of the standard notation does not depend on the number of variables present in the equation.

3.3 Lesson

Monitoring Progress

Evaluate the function when x = −4, 0, and 3.

Question 1.
f(x) = 2x – 5
Answer:
The given function is:
f(x) = 2x – 5
Now,
When x = -4,
f(-4) = 2 (-4) – 5 = -8 – 5 = -13
When x = 0,
f(0) = 2 (0) – 5 = 0 – 5 = -5
When x = 3,
f(3) = 2 (3) – 5 = 6 – 5 = 1
Hence, from the above,
We can conclude that the values of f(x) when x = -4, 0, 3 are: -13, -5, and 1

Question 2.
g(x) = -x – 1
Answer:
The given function is:
g(x) = -x – 1
Now,
When x = -4,
g(-4) = -[-4] – 1 = 4 – 1 = 3
When x = 0,
g(0) = 0 – 1 = -1
When x = 3,
g(3) = -3 – 1 = -4
Hence, from the above,
We can conclude that the values of g(x) when x = -4, 0, 3 are: 3, -1, and -4

Question 3.
WHAT IF? In Example 2, let f(t) be the outside temperature (°F) t hours after 9 A.M. Explain the meaning of each statement.
a. f(4) = 75
b. f(m) = 70
c. f(2) = f(9)
d. f(6) > f(0)

Monitoring Progress

Find the value of x so that the function has the given value.

Question 4.
f(x) = 6x + 9; f(x) = 21
Answer:
The value of x is: 2

Explanation:
The give function is:
f(x) = 6x + 9 with f(x) = 21
So,
21 = 6x + 9
6x = 21 – 9
6x = 12
x = 12 / 6
x = 2
Hence, from the above,
We can conclude that the value of x in the given function is: 2

Question 5.
g(x) = \(-\frac{1}{2}\)x + 3; g(x) = -1
Answer:
The value of x is: 8

Explanation:
The given function is:
g(x) = \(-\frac{1}{2}\)x + 3 with g(x) = -1
So,
-1 = \(-\frac{1}{2}\)x + 3
\(-\frac{1}{2}\)x = -1 – 3
\(-\frac{1}{2}\)x = -4
–\(\frac{1}{2}\)x = -4
\(\frac{1}{2}\)x = 4
x = 2(4)
x = 8
Hence, from the above,
We can conclude that the value of x in the given function is: 8

Graph the linear function.

Question 6.
f(x) = 3x – 2
Answer:
The given function is:
f(x) = 3x – 2
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
f(-2) = 3 (-2) – 2 = -6 – 2 = -8
f(-1) = 3(-1) – 2 = -3 – 2 = -5
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 7.
g(x) = -x + 4
Answer:
The given function is:
g(x) = -x + 4
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
g(-2) = -[-2] + 4 = 2 + 4 = 6
g(-1) = -[-1] + 4 = 4 + 1 = 5
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 8.
h(x) = \(-\frac{3}{4}\)x – 1
Answer:
The given function is:
h(x) = \(-\frac{3}{4}\)x – 1
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
h(-2) = \(-\frac{3}{4}\) (-2) – 1
= \(\frac{3}{2}\) – 1
= \(\frac{1}{2}\)
h(-1) = \(-\frac{3}{4}\) (-1) – 1
= \(\frac{3}{4}\) – 1
= \(-\frac{1}{4}\)
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 9.
WHAT IF?
Let f(x) = 250 – 75x represent the second flight, where f(x) is the number of miles the helicopter is from its destination after x hours. Which flight takes less time? Explain.

Function Notation 3.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
When you write the function y = 2x + 10 as f(x) = 2x + 10, you are using ______________.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 1

Question 2.
REASONING
Your height can be represented by a function h, where the input is your age. What does h(14) represent?
Answer:
It is given that your height can be represented by a function h, where the input is your age
So,
h (14) means you are 14 years old

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, evaluate the function when x = –2, 0, and 5.

Question 3.
f(x) = x + 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 3

Question 4.
g(x) = 3x
Answer:
The given function is:
g (x) = 3x
When x = -2,
g (-2) = 3 (-2) = -6
When x = 0,
g (0) = 3 (0) = 0
When x = 5,
g (5) = 3 (5) = 15
Hence, from the above,
We can conclude that the values of g (x) when x = -2, 0, 5 is: -6, 0, and 15

Question 5.
h(x) = -2x + 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 5

Question 6.
r(x) = -x – 7
Answer:
The given function is:
r (x) = -x – 7
When x = -2,
r (-2) = -[-2] – 7
= 2 – 7 = -5
When x = 0,
r (0) = 0 – 7 = -7
When x = 5,
r (5) = -5 – 7 = -12
Hence, from the above,
We can conclude taht the values of r (x) when x = 0, -2, 5 is: -5, -7, and -12

Question 7.
p(x) = -3 + 4x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 7

Question 8.
b(x) = 18 – 0.5x
Answer:
The given function is:
b (x) = 18 – 0.5x
When x = -2,
b (-2) = 18 – 0.5 (-2)
= 18 + 1 = 19
When x = 0,
b (0) = 18 – 0.5 (0) = 18
When x = 5,
b (5) = 18 – 0.5 (5)
18 – 2.5 = 15.5
Hence, from the above,
We can conclude that the values of b (x) when x = 0, -2, 5 is: 19, 18, and 15.5

Question 9.
v(x) = 12 – 2x – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 9

Question 10.
n(x) = -1 – x + 4
Answer:
The given function is:
n (x) = -1 – x + 4
When x = -2,
n (-2) = -1 – [-2] + 4
= -1 + 2 + 4
= 6 – 1
= 5
When x = 0,
n (0) = -1 – 0 + 4
= -1 + 4 = 3
When x = 2,
n (2) = -1 – 2 + 4
= 4 – 3
= 1
Hence, from the above,
We can conclude that the values of n (x) when x = -2, 0, 5 is: 5, 3, and 1

Question 11.
INTERPRETING FUNCTION NOTATION
Let c(t) be the number of customers in a restaurant t hours after 8 A.M. Explain the meaning of each statement.
a. c(0) = 0
b. c(3) = c(8)
c. c(n) = 29
d. c(13) < c(12)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 11

Question 12.
INTERPRETING FUNCTION NOTATION
Let H(x) be the percent of U.S. households with Internet use x years after 1980. Explain the meaning of each statement.
a. H(23) = 55
b. H(4) = k
c. H(27) ≥ 61
d. H(17) + H(21) ≈ H(29)
Answer:
It is given that H (x) is the percent of U.S households with internet use x years after 1980
Now,
a.
The given function is:
H (23) = 55
The meaning of the above function is:
23 years after 1980, 55% of U.S households will be using the internet

b.
The given function is:
H(4) = k
The meaning of the above function is:
4 years after 1980, k% of U.S households will be using the internet

c.
The given function is:
H(27) ≥ 61
The meaning of the above function is:
27 years after 1980, more than 61% of U.S households will be using the internet

d.
The given function is:
H(17) + H(21) ≈ H(29)
The meaning of the above function is:
The percentage of U.S households using the internet after 29 years is equal to the sum of percentage of household using the internet after 17 years and 21 years

In Exercises 13–18, find the value of x so that the function has the given value.  

Question 13.
h(x) = -7x; h(x) = 63
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 13

Question 14.
t(x) = 3x; t(x) = 24
Answer:
The given function is:
t (x) = 3x with t (x) = 24
So,
24 = 3x
x = 24 / 3
x = 8
Hence, from the above,
We can conclude that the value of the given function is: 8

Question 15.
m(x) = 4x + 15; m(x) = 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 15

Question 16.
k(x) = 6x – 12; k(x) = 18
Answer:
The given function is:
k (x) = 6x – 12 with k (x) = 18
So,
6x – 12 = 18
6x = 12 + 18
6x = 30
x = 30 / 6
x = 5
Hence, from the above,
We can conclude that the value of the given function is: 5

Question 17.
q(x) = \(\frac{1}{2}\)x – 3; q(x) = -4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 17

Question 18.
j(x) = –\(\frac{4}{5}\)x + 7; j(x) = -5
Answer:
The given function is:
j (x) = —\(\frac{4}{5}\)x + 7 with j (x) = -5
So,
-5 = –\(\frac{4}{5}\)x + 7
-5 – 7 = –\(\frac{4}{5}\)x
–\(\frac{4}{5}\)x = -12
\(\frac{4}{5}\)x = 12
x = 12 × –\(\frac{5}{4}\)
x = 15
Hence, from the above,
We can conclude that the value of the given function is: 15

In Exercises 19 and 20, find the value of x so that f(x) = 7. 

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 83
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 19

Question 20.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 84
Answer:
The given function is:
f (x) = 7
So,
We know that,
y = f (x)
So,
From the above,
We can say that the value of y is: 7
Now,
To find the value of x, observe the location corresponding to the value of y
So,
The value of x is: -2
Hence,
The point of the graph where f (x) = 7 is: (-2, 7)

Question 21.
MODELING WITH MATHEMATICS
The function C(x) = 17.5x – 10 represents the cost (in dollars) of buying x tickets to the orchestra with a $10 coupon.
a. How much does it cost to buy five tickets?
b. How many tickets can you buy for $130?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 21

Question 22.
MODELING WITH MATHEMATICS
The function d(t) = 300,000t represents the distance (in kilometers) that light travels in t seconds.
a. How far does light travel in 15 seconds?
b. How long does it take light to travel 12 million kilometers?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 85
Answer:
The given function is:
d (t) = 300,000t
represents the distance in kilometers that travels in t seconds
a.
The distance traveled by light in 15 seconds is:
d (15) = 300,000 (15)
= 4,500,000 kilometers
Hence, from the above,
We can conclude that the distance traveled by light in 15 seconds is: 4,500,000 kilometers

b.
It is given that the total distance traveled by light is 12 million kilometers
So,
d (t) = 12 million kilometers
So,
12,000,000 = 300,000 (t)
t = 12,000,000 / 300,000
t = 40 seconds
Hence, from the above,
We can conclude that the time taken by light to travel 12 million kilometers is: 40 seconds

In Exercises 23–28, graph the linear function. 

Question 23.
p(x) = 4x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 23

Question 24.
h(x) = -5
Answer:
The given function is:
h (x) = -5
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 25.
d(x) = \(-\frac{1}{2} x\) – 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 25

Question 26.
w(x) = \(\frac{3}{5} x\) + 2
Answer:
The given function is:
w (x) = \(\frac{3}{5} x\) + 2
So,
w (x) = \(\frac{3}{5}\) x + 2
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 27.
g(x) = -4 + 7x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 27

Question 28.
f(x) = 3 – 6x
Answer:
The given function is:
f (x) = 3 – 6x
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 29.
PROBLEM-SOLVING
The graph shows the percent p(in decimal form) of battery power remaining in a laptop computer after t hours of use. A tablet computer initially has 75% of its battery power remaining and loses 12.5% per hour. Which computer’s battery will last longer? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 86
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 29.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 29.2

Question 30.
PROBLEM-SOLVING
The function C(x) = 25x + 50 represents the labor cost (in dollars) for Certified Remodeling to build a deck, where x is the number of hours of labor. The table shows sample labor costs from its main competitor, Master Remodeling. The deck is estimated to take 8 hours of labor. Which company would you hire? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 87
Answer:
It is given that the given function
C(x) = 25x + 50
represents the labor cost (in dollars) for certified Remodeling to build a deck
Where,
x is the number of hours of labor.
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 87
The given table shows the sample labor costs from its main competitor, Master Remodeling.
It is also given that the deck is estimated to take 8 hours of labor.
Now,
From the given table,
We can observe that the competitor ‘Master Remodeling’ completes the deck in 6 hours whereas for certified Remodeling, it will take 8 hours of labor to complete a deck thereby increases the cost. i.e., the avlue of C(x) also increases when compared to the Master Remodeling
Hence, from the above,
We can conclude that we choose “Master Remodeling” company

Question 31.
MAKING AN ARGUMENT
Let P(x) be the number of people in the U.S. who own a cell phone x years after 1990. Your friend says that P(x + 1) > P(x) for any x because x + 1 is always greater than x. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 31

Question 32.
THOUGHT-PROVOKING
Let B(t) be your bank account balance after t days. Describe a situation in which B(0) < B(4) < B(2).
Answer:
The given relation is:
B (0) < B (4) < B (2)
Let us consider
Sunday – Day 0
Monday -Day 1
Tuesday -Day 2
Wednesday – Day 3
Thursday – Day 4
Friday – Day 5
Saturday – Day 6
Now,
By using the above relation,
The situation we can assume is:
A man named A works at a company where he receives his salary every Tuesday (Day 2) of a normal week. He then spends some of his salary paying bills on Thursday (Day 4). On Sunday, he decides to spend all of his remaining salary on food, groceries, and transportation allowance for the following week

Question 33.
MATHEMATICAL CONNECTIONS
Rewrite each geometry formula using function notation. Evaluate each function when r = 5 feet. Then explain the meaning of the result.
a. Diameter, d = 2r
b. Area, A = πr2
c. Circumference, C = 2πr
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 88
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 33

Question 34.
HOW DO YOU SEE IT?
The function y = A(x) represents the attendance at a high school x weeks after a flu outbreak. The graph of the function is shown.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
a. What happens to the school’s attendance after the flu outbreak?
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
From the above graph,
We can observe that after the flu outbreak, the school’s attendance first decreased and then again steadily increased

b. Estimate A(13) and explain its meaning.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
Now,
We know that
The function notation and the standard notations are similar
So,
y = A (x)
So,
y = A (13)
So,
The value of x is: 13
To find the value of y or the given function, observe the graph for the value of y corresponding to the value of x i.e., 13
So,
From the graph,
y = 430 [Approximately]
Hence, from the above,
We can conclude that
The estimation of A(13) is:
A (13) = 430 {Approximately]

c. Use the graph to estimate the solution(s) of the equation A(x) = 400. Explain the meaning of the solution(s).
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
The given equation is:
A (x) = 400
From the above graph,
For the solution of the given equation i.e., the value of x,
Observe the location of 400 in the graph and its corresponding x-value
Hence,
From the graph,
We can observe that the value of x is: 1
Hence, from the above,
We can conclude that the solution of the given equation is: 1

d. What was the least attendance? When did that occur?
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
From the above graph,
We can observe that there is deep depreciation in the attendance i.e., least attendance of the students in a week
So,
By observing the graph,
We can say that
The least attendance of the students occurs in the 4th week
The least attendance of the students is: 350
Hence, from the above,
We can conclude that
The week that has the least attendance is: 4th week
The number of students of the least attendance is: 350students

e. How many students do you think are enrolled at this high school? Explain your reasoning.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 89
From the given graph,
We can observe that the y-axis represents the number of students enrolled in the high school
So,
The highest number on the y-axis represents the total number of students enrolled in the high school
Hence, from the above,
We can conclude that the total number of students in the high school is: 450 students

Question 35.
INTERPRETING FUNCTION NOTATION
Let f be a function. Use each statement to find the coordinates of a point on the graph of f.
a. f(5) is equal to 9.
b. A solution of the equation f(n) = -3 is 5.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 35

Question 36.
REASONING
Given a function f, tell whether the statement f(a + b) = f(a) + f(b) is true or false for all inputs a and b. If it is false, explain why.
Answer:
Let the given function is:
f or f(x) = mx + c
Now,
Let the values of a and b be integers
So,
Let,
a = 0 and b = 1
So,
f( a + b ) = f (1)
So,
f(1) = m + c
Now,
f(a) = f (0) = c
f (b) = f(1) = m + c
Hence, from the above,
We can conclude that
f (a + b) is not equal to f (a) + f(b)

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution. (Section 2.5)

Question 37.
-2 ≤ x – 11 ≤ 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 37

Question 38.
5a < -35 or a – 14 > 1
Answer:
The given inequality is:
5a < -35 or a – 14 > 1
So,
a < -35 / 5 or a > 1 + 14
a < -7 or a > 15
Hence,
The solutions of the given inequality are:
a < -7 or a > 15
The representation of the solution of the given inequality in the graph is:

Question 39.
-16 < 6k + 2 < 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 39

Question 40.
2d + 7 < -9 or 4d – 1 > -3
Answer:
The given inequality is:
2d + 7 < -9 or 4d – 1 > -3
2d < -9 – 7 or 4d > -3 + 1
2d < -16 or 4d > -2
d < -16 / 2 or d > -2 / 4
d < -8 or d > -1 /  2
Hence,
The solutions of the given inequality are:
d < -8 or d > -1 / 2
The representation of the solutions of the given inequality in the graph is:

Question 41.
5 ≤ 3y + 8 < 17
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.3 Question 41

Question 42.
4v + 9 ≤ 5 or -3v ≥ -6
Answer:
The given inequality is:
4v + 9 ≤5 or -3v ≥ -6
So,
4v ≤ 5 – 9 or 3v ≥ 6
4v ≤ -4 or v ≥ 6 / 3
v ≤ -4 / 4 or v ≥ 2
v ≤ -1 or v ≥ 2
Hence,
The solutions of the given inequality are:
v ≤ -1 or v ≥ 2
The representation of the solutions of the given inequality is:

Graphing Linear Functions Study Skills: Staying Focused During Class

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 90

Core Concepts
Section 3.1
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 91

Section 3.2
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 92

Section 3.3
Using FunctionNotation, p. 122

Mathematical Practices

Question 1.
How can you use technology to confirm your answers in Exercises 40–43 on page 110?
Answer:
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
By using the above properties,
We can find the domain and the range of the given Exercises
Hence,
We can confirm the answers in Exercises 40 – 43 on page 110

Question 2.
How can you use patterns to solve Exercise 43 on page 119?
Answer:
In Exercise 43 on page 119,
We can observe from the table that the difference between the values of x and y is constant
So,
By using the above property, we can find the constant difference between the values of x and y to complete the pattern

Question 3.
How can you make sense of the quantities in the function in Exercise 21 on page 125?
Answer:
In Exercise 21 on page 125,
Compare the given function with the standard representation of the linear function y = mx + c
So,
In the above function,
We know that,
m is the cost of x tickets in dollars
c is the constant
f (x) or y is the function notation of Cost function

Study Skills

Staying Focused during Class

As soon as class starts, quickly review your notes from the previous class and start thinking about math.
Repeat what you are writing in your head.
When a particular topic is difficult, ask for another example.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 94

Graphing Linear Functions 3.1 – 3.3 Quiz

Determine whether the relation is a function. Explain. (Section 3.1)

Question 1.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 95
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 95
We know that for a relationship to be a function,
Each input value has to match with each output value
Hence,
From the above table,
We can conclude that the given table is a function since each input value matches with each output value

Question 2.
(-10, 2), (-8, 3), (-6, 5), (-8, 8), (-10, 6)
Answer:
The given ordered pairs are:
(-10, 2), (-8, 3), (-6, 5), (-8, 8), (-10, 6)
We know that,
A relation is said to be a function if each input matches with only 1 output
So,
By observing the above-ordered pairs,
We can say that the input -8 comes 2 times
Hence, from the above,
We can conclude that the given ordered pairs do not represent a function

Find the domain and range of the function represented by the graph. (Section 3.1)

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 96
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 96
We know that,
The domain is the set of all x-values
The range is the set of all y-values
So,
The domain of the given graph is: {0, 1, 2, 3, 4}
The range of the given graph is: {1, -1, -3}

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 97
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 97

We know that,
The domain is the set of all x-values
The range is the set of all y-values
So,
The domain of the given graph is: {-1, -2, 1, 2}
The range of the given graph is: {-2, -1, 0, 1, 2}

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 98
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 98
We know that,
The domain is the set of all x-values
The range is the set of all y-values
So,
The domain of the given graph is: {-3, -2, -1, 1, 2, 3}
The range of the given graph is: {-1, 0, 1, 2, 3}

Determine whether the graph, table, or equation represents a linear or nonlinear function. Explain. (Section 3.2)

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 99
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 99
We know that,
A linear function must always represent a straight line irrespective of the straight line passes through the origin or passes through any other point
Hence, from the above,
We can conclude that the given graph is a linear function

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 100
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 100
The representation of the values of x and y in the table in the form of ordered pairs is:
(-5, 3), (0, 7), and (5, 10)
Represent the ordered pairs in the coordinate plane
The representation of the ordered pairs in the coordinate plane is:

From the above points, we can say it forms a straight line
A linear function must always represent a straight line irrespective of the straight line passes through the origin or passes through any other point
Hence, from the above,
We can conclude that the given table represents a linear function

Question 8.
y = x(2 – x)
Answer:
The given function is:
y = x (2 – x)
So,
y = 2 (x) – x (x)
y = 2x – x²
Compare the above function with the standard representation of the linear function
The standard representation of the linear function is:
y = mx + c
Hence, from the above comparison,
We can conclude that the given function is a non-linear function

Determine whether the domain is discrete or continuous. Explain. (Section 3.2)

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 101
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 101
From the above table,
The values of x have specific values of y
We know that,
The domain is defined as the set of all the values of x
Hence, from the above,
We can conclude that the domain of the given table is discrete

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 102
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 102
From the above table
We can observe that the values of x have specific values of y
We know that,
The domain is defined as the set of all the values of x
Hence, from the above,
We can conclude that the domain of the given table is discrete

Question 11.
For w(x) = -2x + 7, find the value of x for which w(x) = -3. (Section 3.3)
Answer:
The given function is:
w (x) = -2x + 7 with w (x)  = -3
So,
-3 = -2x + 7
-2x = -3 – 7
-2x = -10
2x = 10
x = 10 / 5
x = 2
Hence, from the above,
We can conclude that the value of the given function is: 2

Graph the linear function. ( Section 3.3)

Question 12.
g(x) = x + 3
Answer:
The given function is:
g (x) = x + 3
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 13.
p(x) = -3x – 1
Answer:
The given function is:
p (x) = -3x – 1
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) toplot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 14.
m(x) = \(\frac{2}{3}\)x
Answer:
The given function is:
m (x) = \(\frac{2}{3}\)x
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table for the given function is:

Hence,
The representation of the given function in the coordinate plane is:

Question 15.
The function m = 30 – 3r represents the amount m (in dollars) of money you have after renting r video games. (Section 3.1 and Section 3.2)
a. Identify the independent and dependent variables.
Answer:
The given function is:
m = 30 – 3r
We know that,
The independent variables are the values of x
The dependent variables are the values of y
Now,
Compare the given function with
y = mx + c
Hence,
The independent variable of the given function is: r
The dependent variable of the given function is: m

b. Find the domain and range of the function. Is the domain discrete or continuous? Explain.
Answer:
The given function is:
m = 30 – 3r
Now,
put the values -2, -1, 0, 1, 2 in the place of x and find the values of f(x) to plot a graph [ Remember you can take any value and any number of values]
So,
The completed table of the given function is:

We know that,
The domain is defined as the set of all the values of x
the range is defined as the set of all the values of y
Hence,
The domain of the given function is: {-2, -1, 0, 1, 2}
The range of the given function is: {36, 33, 30, 27, 24, 21}

c. Graph the function using its domain.
Answer:
The completed table for the given function from part (b) is:

Hence,
The representation of the given function using its domain in the coordinate plane is:

Question 16.
The function d(x) = 1375 – 110x represents the distance (in miles) a high-speed train is from its destination after x hours. (Section 3.3)
a. How far is the train from its destination after 8 hours?
Answer:
The given function is:
d (x) = 1375 – 110x
Where,
d (x) represents the distance (in miles)
x represents the time
Now,
It is given that we have to find the distance traveled by train after 8 hours i.e., the value of x is given
So,
d (x) = 1375 – 110 (8)
= 1375 – 880
= 495 miles
Hence, from the above,
We can conclude that the distance traveled by train after 8 hours is: 495 miles

b. How long does the train travel before reaching its destination?
Answer:
The given function is:
d (x) = 1375 – 110x
From part (a),
We find the value of d (x) as: 495 miles
So,
495 = 1375 – 110x
110x = 1375 – 495
110x = 880
x = 880 /110
x = 8 hours
Hence, from the above,
We can conclude that train travel for 8 hours before reaching its destination

Lesson 3.4 Graphing Linear Equations in Standard Form

Essential Question
How can you describe the graph of the equation Ax + By = C?
Answer:
When A and B are not both zero,
The graph of Ax + By = C is always a line.
Now,
Divide both sides by B
Because the form Ax + By = C can describe any line,
It is called the standard form of an equation for a line.

EXPLORATION 1
Using a Table to Plot Points
Work with a partner.
You sold a total of $16 worth of tickets to a fundraiser. You lost track of how many of each type of ticket you sold. Adult tickets are $4 each. Child tickets are $2 each.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 102.1
a. Let x represent the number of adult tickets. Let y represent the number of child tickets. Use the verbal model to write an equation that relates to x and y.
Answer:
It is given that you sold a total of $16 worth of tickets to a fundraiser. You lost track of how many of each type of ticket you sold. Adult tickets are $4 each. Child tickets are $2 each.
It is also given that
Let x be the number of adult tickets
Let y be the number of child tickets
So,
The total cost of tickets = (The cost of the child tickets + The cost of the adult tickets)
The total cost of tickets = (The number of children) × ( The cost of each child ticket ) + ( The numebr of adults ) × ( The cost of each adult ticket )
So,
16 = 2x + 4y
2 (x + 2y) = 16
x + 2y = 16 / 2
So,
x + 2y = 8
Hence, from the above,
We can conclude that the equation that relates x and y is:
x + 2y = 8

b. Copy and complete the table to show the different combinations of tickets you might have sold.
Answer:
The equation that represents the number of different tickets sold is:
x + 2y = 8
2y = 8 – x
y = \(\frac{8 – x}{2}\)
So,
The completed table that shows the different combinations of tickets you might have sold is:

c. Plot the points from the table. Describe the pattern formed by the points.
Answer:
The completed table that shows the different combinations of tickets you might have sold is from part (b) is:

The representation of the ordered pairs from the above table is:
(1, 4), (2, 3), (3, 3), (4, 2), (5, 2)
Hence,
The representation of the points from the table in the coordinate plane is:

Hence, from the graph,
We can observe that the pattern drawn by the points is a straight line

d. If you remember how many adult tickets you sold, can you determine how many child tickets you sold? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 104
Answer:
The given function from part (a) is in the form of
y = mx + c
Where,
y is the number of adult tickets
x is the number of child tickets
So,
If we know the number of adult tickets you sold, i.e., the value of y, then we can find the number of child tickets, i.e., the value of x by putting the value of y in the equation
y = mx + c
Hence, from the above,
We can conclude that if we know the number of adult tickets you sold, then we can find the number of child tickets you sold

EXPLORATION 2
Rewriting and Graphing an Equation
Work with a partner.
You sold a total of $48 worth of cheese. You forgot how many pounds of each type of cheese you sold. Swiss cheese costs $8 per pound. Cheddar cheese costs $6 per pound.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 105
a. Let x represent the number of pounds of Swiss cheese. Let y represent the number of pounds of cheddar cheese. Use the verbal model to write an equation that relates to x and y.
Answer:
it is given that you sold a total of $48 worth of cheese. You forgot how many pounds of each type of cheese you sold. Swiss cheese costs $8 per pound. Cheddar cheese costs $6 per pound.
It is also given that,
x represents the number of pounds of Swiss cheese
y represents the number of pounds of Cheddar cheese
Now,
The total cost = ( The cost of Swiss Cheese ) + ( The cost of Cheddar Cheese )
The total cost = ( The number of pounds of Swiss cheese ) × ( The cost of Swiss cheese ) + ( The number of pounds of Cheddar cheese ) × ( The cost of Cheddar cheese )
48 = 8x + 6y
8x + 6y = 48
2 (4x + 3y) = 48
4x + 3y = 48 / 4
4x + 3y = 12
Hence, from the above,
We can conclude that the equation that relates both x and y is:
4x + 3y = 12

b. Solve the equation for y. Then use a graphing calculator to graph the equation. Given the real-life context of the problem, find the domain and range of the function.
Answer:
The equation that relates to x and y from part (a) is:
4x + 3y = 12
3y = 12 – 4x
y = \(\frac{12 – 4x}{3}\)
So,
The representation of the equation in the coordinate plane is:

We know that,
The domain is the set of all the values of x in the graph
The range is the set of all the values of y in the graph
Now,
From the graph,
The domain of the equation is: { -4, -3, -2, -1, 0, 1, 2, 3 }
The range of the given equation is: {-1, -2, -3, -4, -5, -6, -7, -8, -9, 5, 6, 7, 8, 9, 10}

c. The x-intercept of a graph is the x-coordinate of a point where the graph crosses the x-axis. The y-intercept of a graph is the y-coordinate of a point where the graph crosses the y-axis. Use the graph to determine the x- and y-intercepts.
Answer:
The given equation is:
y = \(\frac{12 – 4x}{3}\)
It is given that the x-intercept of a graph is the x-coordinate of a point where the graph crosses the x-axis. The y-intercept of a graph is the y-coordinate of a point where the graph crosses the y-axis.
Now,
The graph from part (b) is:

Hence,
By observing the graph,
We can conclude that
The x-intercept is: 3
The y-intercept is: 4

d. How could you use the equation you found in part (a) to determine the x- and y-intercepts? Explain your reasoning.
Answer:
The given equation is:
y = \(\frac{12 – 4x}{3}\)
We know that,
We can obtain the x-intercept by making y term zero
We can obtain the y-intercept making x term zero
So,
The x-intercept is:
0 = \(\frac{12 – 4x}{3}\)
12 – 4x = 0
4x = 12
x = 12 / 4
x = 3
Hence,
The x-intercept is: 3
The y-intercept is:
y = \(\frac{12 – 4(0)}{3}\)
y = 12 / 3
y = 4
Hence,
The y-intercept is: 4

e. Explain the meaning of the intercepts in the context of the problem.
Answer:
An intercept of any function is a point where the graph of the function crosses, or intercepts, the x-axis or y-axis.  When the linear function is used to represent a real-world situation, the intercepts have significant meaning in the context of the problem.

Communicate Your Answer

Question 3.
How can you describe the graph of the equation Ax + By = C?
Answer:
When A and B are not both zero,
The graph of Ax + By = C is always a line.
Now,
Divide both sides by B
Because the form Ax + By = C can describe any line,
It is called the standard form of an equation for a line.

Question 4.
Write a real-life problem that is similar to those shown in Explorations 1 and 2.
Answer:
The real-life problem that is similar to those in Explorations  1 and 2 is:
The age of A is twice the age of B and the sum of the twice the age of A and the age of B is 24

3.4 Lesson

Monitoring Progress

Question 1.
y = -2.5
Answer:
The given equation is:
y = -2.5
Hence,
The representation of the given equation in the coordinate plane is:

Question 2.
x = 5
Answer:
The given equation is:
x = 5
Hence,
The representation of the given equation in the coordinate plane is:

Use intercepts to graph the linear equation. Label the points corresponding to the intercepts.

Question 3.
2x – y = 4
Answer:
The given equation is:
2x – y = 4
Rewrite the given equation in the standard form
We know that,
The standard form of the linear equation is:
y = mx + c
So,
y = 2x – 4
To find the value of x-intercept, put y = 0
2x – 4 = 0
2x = 4
x = 4 / 2
x = 2
To find the value of y-intercept, put x = 0
y = 2(0) – 4
y = -4
Hence,
The representation of the given equation corresponding to the intercepts in the coordinate plane is:

Question 4.
x + 3y = -9
Answer:
The given equation is:
x + 3y = -9
Rewrite the given equation in the standard form
We know that,
The standard form of the linear equation is:
y = mx + c
So,
3y = -9 – x
y = (-9 -x ) / 3
Now,
To find the value of x-intercept, put y = 0
-9 – x = 0
x = -9
To find the value of y-intercept, put x = 0
y = (-9 – 0) / 3
t = -9 / 3
y = -3
Hence,
The representation of the given equation corresponding to the intercepts in the coordinate plane is:

Question 5.
WHAT IF? You decide to rent tables from a different company. The situation can be modeled by the equation 4x + 6y = 180, where x is the number of small tables and y is the number of large tables. Graph the equation and interpret the intercepts.
Answer:
It is given that you decide to rent tables from a different company.
The given equation corresponding to the above situation is given as:
4x + 6y = 180
Where,
x is the number of small tables
y is the number of large tables
Now,
Rewrite the given equation in the standard form of the linear equation
We know that,
The standard form of the linear equation is:
y = mx + c
So,
6y = 180 – 4x
y = (180 – 4x) / 6
Now,
To find the value of the x-intercept, put y = 0
180 – 4x = 0
4x = 180
x = 180 / 4
x = 45
To find the value of the y-intercept, put x = 0
y = (180 – 4(0) ) / 6
y = 180 / 6
y = 30
Hence,
The representation of the given equation corresponding to the intercepts in the coordinate plane is:

Graphing Linear Equations in Standard Form 3.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
How are x-intercepts and y-intercepts alike? How are they different?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 1

Question 2.
WHICH ONE did DOESN’T BELONG?
Which point does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 105.1
Answer:
The given points are:
a. (0, -3)
b. (0, 0)
c. (4, -3)
d. (4, 0)
From the above points,
We can observe that
(0, -3) belongs to the y-axis
(0, 0) belongs to the origin
(4, -3) does not belong either to the x-axis or y-axis
(4, 0) belongs to the x-axis
Hence, from the above,
We can conclude that (4, -3) does not belong with the other three

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, graph the linear equation.

Question 3.
x = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 3

Question 4.
y = 2
Answer:
The given equation is:
y = 2
Hence,
The representation of the given equation in the coordinate plane is:

Question 5.
y = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 5

Question 6.
x = -1
Answer:
The given equation is:
x = -1
Hence,
The representation of the given equation in the coordinate plane is:

In Exercises 7–12, find the x- and y-intercepts of the graph of the linear equation.

Question 7.
2x + 3y = 12
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 7

Question 8.
3x + 6y = 24
Answer:
The given equation is:
3x + 6y = 24
Now,
To find the value of the x-intercept, put y = 0
So,
3x + 0 = 24
3x = 24
x = 24 / 3
x = 8
To find the value of the y-intercept, put x = 0
So,
0 + 6y = 24
6y = 24
y = 24 / 6
y = 4
Hence, from the above,
We can conclude that
The value of the x-intercept is: 8
The value of the y-intercept is: 4

Question 9.
-4x + 8y = -16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 9

Question 10.
-6x + 9y = -18
Answer:
The given equation is:
-6x + 9y = -18
Now,
To find the x-intercept, put y = 0
So,
-6x + 0 = -18
-6x = -18
6x = 18
x = 18 / 6
x = 3
To find the y-intercept, put x = 0
So,
0 + 9y = -18
9y = -18
y = -18 / 9
y = -2
Hence, from the above,
We can conclude that
The value of the x-intercept is: 3
The value of the y-intercept is: -2

Question 11.
3x – 6y = 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 11

Question 12.
-x + 8y = 4
Answer:
The given equation is:
-x + 8y = 4
Now,
To find the x-intercept, put y = 0
So,
-x + 0 = 4
-x = 4
x = -4
To find the y-intercept, put x = 0
So,
0 + 8y = 4
8y = 4
y = 4 / 8
y = 1 / 2
Hence, from the above,
We can conclude that
The value of the x-intercept is: -4
The value of the y-intercept is: 1/2

In Exercises 13–22, use intercepts to graph the linear equation. Label the points corresponding to the intercepts.

Question 13.
5x + 3y = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 13

Question 14.
4x + 6y = 12
Answer:
The given equation is:
4x + 6y = 12
Now,
To find the x-intercept, put y = 0
So,
4x + 0 = 12
4x = 12
x = 12 / 4
x = 3
To find the y-intercept, put x = 0
0 + 6y = 12
6y = 12
y = 12 / 6
y = 2
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 15.
-12x + 3y = 24
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 15

Question 16.
-2x + 6y = 18
Answer:
The given equation is:
-2x + 6y = 18
Now,
To find the x-intercept, put y = 0
So,
-2x + 0 = 18
-2x = 18
x = -18 / 2
x = -9
To find the y-intercept, put x = 0
So,
0 + 6y = 18
6y = 18
y = 18 / 6
y = 3
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 17.
-4x + 3y = -30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 17

Question 18.
-2x + 7y = -21
Answer:
The given equation is:
-2x + 7y = -21
Now,
To find the value of the x-intercept, put y = 0
-2x + 0 = -21
-2x = -21
2x = 21
x = 21/2
To find the value of the y-intercept, put x = 0
So,
0 + 7y = -21
7y = -21
y = -21 / 7
y = -3
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 19.
-x + 2y = 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 19

Question 20.
3x – y = -5
Answer:
The given equation is:
3x – y = -5
Now,
To find the x-intercept, put y = 0
So,
3x – 0 = -5
3x =  -5
x = -5/3
To find the y-intercept, put x = 0
So,
0 – y = -5
-y = -5
y = 5
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 21.
–\(\frac{5}{2}\)x + y = 10
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 21.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 21.2

Question 22.
–\(\frac{1}{2}\)x + y = -4
Answer:
The given equation is:
–\(\frac{1}{2}\)x + y = -4
Now,
To find the x-intercept, put y =0
So,
–\(\frac{1}{2}\)x + 0 = -4
–\(\frac{1}{2}\)x = -4
-x = -4(2)
-x = -8
x = 8
To find the y-intercept, put x = 0
So,
0 + y = -4
y = -4
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

Question 23.
MODELING WITH MATHEMATICS
A football team has an away game, and the bus breaks down. The coaches decide to drive the players to the game in cars and vans. Four players can ride in each car. Six players can ride in each van. There are 48 players on the team. The equation 4x + 6y = 48 models this situation, where x is the number of cars and y is the number of vans.
a. Graph the equation. Interpret the intercepts.
b. Find four possible solutions in the context of the problem.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 23.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 23.2
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 23.3

Question 24.
MODELING WITH MATHEMATICS
You are ordering shirts for the math club at your school. Short-sleeved shirts cost $10 each. Long-sleeved shirts cost $12 each. You have a budget of $300 for the shirts. The equation 10x + 12y = 300 models the total cost, where x is the number of short-sleeved shirts and y is the number of long-sleeved shirts.
a. Graph the equation. Interpret the intercepts.
Answer:
It is given that you are ordering shirts for the math club at your school. Short-sleeved shirts cost $10 each. Long-sleeved shirts cost $12 each. You have a budget of $300 for the shirts.
The above situation is represented by the equation
10x + 12y = 300
where,
x is the number of short-sleeved shirts
y is the number of long-sleeved shirts.
Now,
The given equation is:
10x + 12y = 300
To find the x-intercept, put y =0
So,
10x + 12 (0) = 300
10x = 300
x = 300 / 10
x = 30
To find the y-intercept, put x = 0
So,
10 (0) + 12y = 300
12y = 300
y = 300 / 12
y = 25
Hence,
The representation of the given equation along with the intercepts in the coordinate plane is:

b. Twelve students decide they want short-sleeved shirts. How many long-sleeved shirts can you order?
Answer:
The given equation is:
10x + 12y = 300
where,
x is the number of short-sleeved shirts
y is the number of long-sleeved shirts.
It is given that 12 students decided they want short-sleeved shirts
So,
The number of short-sleeved shirts = 12
So,
10 (12) + 12y = 300
120 + 12y = 300
12y = 300 – 120
12y = 180
y = 180 / 12
y = 15
Hence, from the above,
We can conclude that the number of long-sleeved shirts is: 15

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in finding the intercepts of the graph of the equation.

Question 25.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 106
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 25

Question 26.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 107
Answer:
The given equation is:
4x + 10y = 20
Now,
To find the value of the x-intercept, put y = 0
So,
4x + 10 (0) = 20
4x = 20
x = 20 / 4
x = 5
To find the value of the y-intercept, put x = 0
So,
4 (0) + 10y = 20
10y = 20
y = 20 / 10
y = 2
Hence, from the above,
We can conclude that
The x-intercept is: (5, 0)
The y-intercept is: (0, 2)

Question 27.
MAKING AN ARGUMENT
You overhear your friend explaining how to find intercepts to a classmate. Your friend says, “When you want to find the x-intercept, just substitute 0 for x and continue to solve the equation.” Is your friend’s explanation correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 27

Question 28.
ANALYZING RELATIONSHIPS
You lose track of how many 2-point baskets and 3-point baskets a team makes in a basketball game. The team misses all the 1-point baskets and still scores 54 points. The equation 2x + 3y = 54 models the total points scored, where x is the number of 2-point baskets made and y is the number of 3-point baskets made.
a. Find and interpret the intercepts.
Answer:
It is given that you lose track of how many 2-point baskets and 3-point baskets a team makes in a basketball game.
It is also given that the team misses all the 1-point baskets and still scores 54 points.
The above situation can be described by the equation:
2x + 3y = 54
where,
x is the number of 2-point baskets made
y is the number of 3-point baskets made.
So,
To find the value of the x-intercept, put y = 0
So,
2x = 54
x = 54 / 2
x = 27
To find the value of the y-intercept, put x = 0
3y = 54
y = 54 / 3
y = 18
Hence, from the above,
We can conclude that
The value of the x-intercept is: 27
The value of the y-intercept is: 18

b. Can the number of 3-point baskets made be odd? Explain your reasoning.
Answer:
From part (a),
The number of 3-point baskets can be represented as: y
So,
To make the 3-point baskets odd, the multiples of 3 must be odd
But, in this scenario, it is not possible
Hence, from the above,
We can conclude that the number of 3-point baskets made not be odd

c. Graph the equation. Find two more possible solutions in the context of the problem.
Answer:

The given equation is:
2x + 3y = 54
From the graph,
We can observe that there are many solutions like (3, 16), (6, 14), (10, 11) and so on
Now,
To satisfy the above equation, substitute the points we obtained from the graph
Hence, from the above,
We can conclude that the solutions to the given equation are: (3, 16) and (6, 14)

MULTIPLE REPRESENTATIONS
In Exercises 29–32, match the equation with its graph.

Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 108

Question 29.
5x + 3y = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 29

Question 30.
5x + 3y = -30
Answer:
The given equation is:
5x + 3y = -30
Now,
To find the value of the x-intercept, put y = 0
So,
5x = -30
x = -30 / 5
x = -6
To find the value of the y-intercept, put x = 0
So,
3y = -30
y = -30 / 3
y = -10
Hence, from the above,
We can conclude that the graph C) matches the given equation

Question 31.
5x – 3y = 30
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 31

Question 32.
5x – 3y = -30
Answer:
The given equation is:
5x – 3y = -30
Now,
To find the value of the x-intercept, put y = 0
So,
5x = -30
x = -30 / 5
x = -6
To find the value of the y-intercept, put x = 0
So,
-3y = -30
3y = 30
y = 30 / 3
y = 10
Hence, from the above,
We can conclude that the graph B) matches the given equation

Question 33.
MATHEMATICAL CONNECTIONS
Graph the equations x = 5, x = 2, y = -2, and y = 1. What enclosed shape do the lines form? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 33

Question 34.
HOW DO YOU SEE IT? You are organizing a class trip to an amusement park. The cost to enter the park is $30. The cost to enter with a meal plan is $45. You have a budget of $2700 for the trip. The equation 30x + 45y = 2700 models the total cost for the class to go on the trip, where x is the number of students who do not choose the meal plan and y is the number of students who do choose the meal plan.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 109
a. Interpret the intercepts of the graph.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 109
We know that,
The x-intercept is the point that cuts the x-axis and y must be 0
The y-intercept is the point that cuts the y-axis and x must be 0
Hence,
From the graph,
We can conclude that
The x-intercept is: (90, 0)
The y-intercept is: (0, 60)

b. Describe the domain and range in the context of the problem.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 109
We know that,
The domain is defined as the set of all the values of x
The range is defined as the set of all the values of y
Hence,
The domain of the given function is: {0, 10, 20, 30, 40, 50, 60, 70, 80, 90}
The range of the given function is: {0, 10, 20, 30, 40, 50, 60}

Question 35.
REASONING
Use the values to fill in the equation Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 110 so that the x-intercept of the graph is -10 and the y-intercept of the graph is 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 110.1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 35

Question 36.
THOUGHT-PROVOKING
Write an equation in the standard form of a line whose intercepts are integers. Explain how you know the intercepts are integers.
Answer:
We know that,
The equation in the standard form is:
y = mx + c
Now,
To find the value of the x-intercept, put y = 0
So,
mx = -c
x = –\(\frac{c}{m}\)
To find the value of the y-intercept, put x = 0
So,
y = 0 + c
y = c
Now, from the above
We can observe that to make the x and y-intercepts integers, the values of the intercepts must be the multiples of the variable c

Question 37.
WRITING
Are the equations of horizontal and vertical lines written in standard form? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 37

Question 38.
ABSTRACT REASONING
The x- and y-intercepts of the graph of the equation 3x + 5y = k are integers. Describe the values of k. Explain your reasoning.
Answer:
The given equation is:
3x + 5y = k
To find the value of the x-intercept, put y = 0
So,
3x = k
So,
The values of k so that k become integer is 0, 1, 2, and so on
To find the value of the y-intercept, put x = 0
So,
5y = k
So,
The values of k so that k becomes integer is 0, 1, 2, and so on
Hence, from the above,
We can conclude that the values of k so that the x and y-intercepts become integers are: 0, 1, 2, and so on

Maintaining Mathematical Proficiency

Simplify the expression.

Question 39.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 111
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 112
Answer:
The given expression is:
\(\frac{14 -18}{0 – 2}\)
= \(\frac{-4}{-2}\)
= \(\frac{4}{2}\)
= 2
Hence, from the above,
We can conclude that the value of the given expression is: 2

Question 41.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 113
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.4 Question 41

Question 42.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 114
Answer:
The given expression is:
\(\frac{12 – 17}{-5 – (-2)}\)
= \(\frac{12 – 17}{-5 + 2}\)
= \(\frac{-5}{-3}\)
= \(\frac{5}{3}\)
Hence, from the above,
We can conclude that the value of the given expression is: \(\frac{5}{3}\)

Lesson 3.5 Graphing Linear Equations in Slope-Intercept Form

EXPLORATION 1
Finding Slopes and y-Intercepts
Work with a partner.
Find the slope and y-intercept of each line.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 115
Answer:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 115
We know that,
The standard form of a linear equation is:
y = mx + c
Where,
m is the slope
c is the y-intercept
Now,
a.
The given equation is:
y = \(\frac{2}{3}\)x + 2
Now,
Compare the given equation with the standard form of the linear equation
By comparison,
We get,
m = \(\frac{2}{3}\) and c = 2

b.
The given equation is:
y = -2x – 1
Now,
Compare the given equation with the standard form of the linear equation
By comparison,
We get,
m = -2 and c = -1

EXPLORATION 2
Writing a Conjecture
Work with a partner.
Graph each equation. Then copy and complete the table. Use the completed table to write a conjecture about the relationship between the graph of y = mx + b and the values of m and b.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 116
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 117
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 117
Now,
The representation of the given equations in the coordinate plane is as follows:
Now,
a.
The given equation is:
y = –\(\frac{2}{3}\)x + 3
Hence,
The representation of the given equation in the coordinate plane is:

b.
The given equation is:
y = 2x – 2
Hence,
The representation of the given equation in the coordinate plane is:

c.
The given equation is:
y = -x + 1
Hence,
The representation of the given equation in the coordinate plane is:

d.
The given equation is:
y = x – 4
Hence,
The representation of the given equation in the coordinate plane is:

Now,
We know that,
The standard form of a linear equation is:
y = mx + b
Where,
m is the slope
b is the y-intercept
Hence,
The completed table for the above equations’ slopes and y-intercepts is:

Communicate Your Answer

Question 3.
How can you describe the graph of the equation y = mx + b?
a. How does the value of m affect the graph of the equation?
b. How does the value of b affect the graph of the equation?
c. Check your answers to parts (a) and (b) by choosing one equation from Exploration 2 and (1) varying only m and (2) varying only b.
Answer:
The graph of the equation
y = mx + b
is a straight line
Now,
a.
The value of m affects the steepness of the line. It also describes the direction (positive or negative). The value of m defines the constant rate of change of variables

b.
The value of b affects the line to where it should have the point of intersection with the y-axis

c.
The examples of how m and b varies are as follows:
y =-3x – 3
y = 2x + 8
y = -3x + 6
Hence,
The slopes and y-intercepts of the above equations are:
m = -3 and b = -3
m = 2 and b = 8
m = -3 and b = 6

3.5 Lesson

Monitoring Progress

Describe the slope of the line. Then find the slope.

Question 1.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 118
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 118
From the graph,
The given points are:
(-4, 3), (1, 1)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
The points are represented as (x, y)
So,
The first point is represented as (x1, y1)
The second point is represented as (x2, y2)
So,
(x1, y1) = (-4, 3) and (x2, y2) = (1, 1)
Hence,
m = \(\frac{1 – 3}{1 – [-4]}\)
m = \(\frac{-2}{1 + 4}\)
m = \(\frac{-2}{5}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{-2}{5}\)

Question 2.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 119
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 119
From the graph,
The given points are:
(3, 3), (-3, 1)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
The points are represented as (x, y)
So,
The first point is represented as (x1, y1)
The second point is represented as (x2, y2)
So,
(x1, y1) = (3, 3) and (x2, y2) = (-3, 1)
Hence,
m = \(\frac{1 – 3}{-3 – 3}\)
m = \(\frac{-2}{6}\)
m = \(\frac{-1}{3}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{-1}{3}\)

Question 3.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 120
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 120
From the graph,
The given points are:
(5, 4), (2, -3)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
The points are represented as (x, y)
So,
The first point is represented as (x1, y1)
The second point is represented as (x2, y2)
So,
(x1, y1) = (5, 4) and (x2, y2) = (2, -3)
Hence,
m = \(\frac{-3 – 4}{2 – 5}\)
m = \(\frac{-7}{-3}\)
m = \(\frac{7}{3}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{7}{3}\)

Monitoring Progress

The points represented by the table lie on a line. How can you find the slope of the line from the table? What is the slope of the line?

Question 4.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 121
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 121
From the above table,
The representations of the values of x and y in the form of ordered pairs are:
(2, 10), (4, 15), (6, 20), (8, 25)
Now,
To find the slope of a line,
Take any 2 ordered pairs and find the slope
Now,
Let,
(x1, y1) = (2, 10) and (x2, y2) = (4, 15)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
Hence,
m = \(\frac{15 – 10}{4 – 2}\)
m = \(\frac{5}{2}\)
Hence, from the above,
We can conclude that the slope of the given graph is: \(\frac{5}{2}\)

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 122
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 122
From the above table,
The representations of the values of x and y in the form of ordered pairs are:
(5, -12), (5, -9), (5, -6), (5, -3)
Now,
To find the slope of a line,
Take any 2 ordered pairs and find the slope
Now,
Let,
(x1, y1) = (5, -12) and (x2, y2) = (5, -9)
We know that,
The slope of the line when the two points are given is:
m = \(\frac{y2 – y1}{x2 – x1}\)
Hence,
m = \(\frac{-9 – [-12]}{5 – 5}\)
m = \(\frac{3}{0}\)
Hence, from the above,
We can conclude that the slope of the given graph is: undefined

Find the slope and the y-intercept of the graph of the linear equation.

Question 6.
y = -6x + 1
Answer:
The given equation is:
y = -6x + 1
We know that,
The standard representation of a linear equation is:
y  mx + c
Where,
m is the slope
c is the y-intercept
Now,
Compare the given equation with the standard representation of the linear equation
Hence,
The values of m and c are: -6 and 1

Question 7.
y = 8
Answer:
The given equation is:
y = 8
Rewrite the given equation as:
y = 0x + 8
We know that,
The standard representation of a linear equation is:
y  mx + c
Where,
m is the slope
c is the y-intercept
Now,
Compare the given equation with the standard representation of the linear equation
Hence,
The values of m and c are: 0 and 8

Question 8.
x + 4y = -10
Answer:
The given equation is:
x + 4y = -10
Rewrite the given equation as:
4y = -10 – x
y = \(\frac{-10 – x}{4}\)
y = \(\frac{-x}{4}\) – \(\frac{10}{4}\)
We know that,
The standard representation of a linear equation is:
y  mx + c
Where,
m is the slope
c is the y-intercept
Now,
Compare the given equation with the standard representation of the linear equation
Hence,
The values of m and ‘c’ are: –\(\frac{1}{4}\) and –\(\frac{10}{4}\)

Graph the linear equation. Identify the x-intercept.

Question 9.
y = 4x – 4
Answer:
The given equation is:
y = 4x – 4
Now,
To find the x-intercept, put y = 0
So,
0 = 4x – 4
4x = 4
x = 4 / 4
x = 1
Hence,
The representation of the given equation along with the x-intercept in the coordinate plane is:

Question 10.
3x + y = -3
Answer:
The given equation is:
3x + y = -3
Now,
To find the x-intercept, put y = 0
So,
3x + 0 = -3
3x = -3
x = -3 / 3
x = -1
Hence,
The representation of the given equation along with the x-intercept in the coordinate plane is:

Question 11.
x + 2y = 6
Answer:
The given equation is:
x + 2y = 6
Now,
To find the x-intercept, put y = 0
So,
x + 0 = 6
x = 6
Hence,
The representation of the given equation along with the x-intercept in the coordinate plane is:

Question 12.
A linear function h models a relationship in which the dependent variable decreases 2 units for every 5 units the independent variable increases. Graph h when h(0) = 4. Identify the slope, y-intercept, and x-intercept of the graph.
Answer:
It is given that a linear function h models a relationship in which the dependent variable decreases 2 units for every 5 units the independent variable increases.
We know that,
The independent variable is: x
the dependent variable is: y
So,
So,
x = -2 and y = +5
So,
We can say that the slope is represented as:
m = \(\frac{-2}{5}\)
It is also given that
h (0) = 2
So,
y = 2 at x = 0
Hence,
The y-intercept is: (0, 2)
Hence,
The representation of the equation in the standard form is:
y = mx + c
y = –\(\frac{2}{5}\)x + 2
Now,
The value of the x-intercept can be obtained by putting y = 0
So,
0 = –\(\frac{2}{5}\)x + 2
–\(\frac{2}{5}\)x = -2
\(\frac{2}{5}\)x = 2
x = \(\frac{5 × 2}{2}\)
x = 5
Hence, from the above,
We can conclude that
The slope is: –\(\frac{2}{5}\)
The x-intercept is: (5, 0)
The y-intercept is: (0, 2)

Question 13.
WHAT IF? The elevation of the submersible is modeled by h(t) = 500t – 10,000.
(a) Graph the function and identify its domain and range.
Answer:
The given function is:
h(t) = 500t – 10,000
So,
The representation of the given function in the coordinate plane is:

From the above graph,
We can say that the given equation is parallel to the y-axis
Hence,
There are no values for the domain since there are no values of x
The range of the given function is: -20,000 ≤ t ≤ 20,000

(b) Interpret the slope and the intercepts of the graph.
Answer:
The given function is:
h (t) = 500t – 10,000
Compare the given equation with the standard linear equation y = mx + c
So,
m = 500
c = -10,000
Now,
To find the x-intercept, put y = 0 or h (t) = 0
So,
500t – 10,000 = 0
500t = 10,000
t = 10,000 / 500
t = 20 or x = 20
To find the y-intercept, put x = 0 or t = 0
So,
h (t) = 500 (0) – 10,000
h (t) = -10,000
Hence, from the above,
We can conclude that
The slope is: 500
The x-intercept is: 20
The y-intercept is: -10,000

Graphing Linear Equations in Slope-Intercept Form 3.5 Exercises

Vocabulary and Core Concept

Question 1.
COMPLETE THE SENTENCE
The ________ of a nonvertical line passing through two points is the ratio of the rise to the run.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 1

Question 2.
VOCABULARY
What is a constant function? What is the slope of a constant function?
Answer:
A “Constant function” is a function where the output (y) is the same for every input (x) value.
The slope for a constant function will be 0

Question 3.
WRITING
What is the slope-intercept form of a linear equation? Explain why this form is called the slope-intercept form.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 3

Question 4.
WHICH ONE did DOESN’T BELONG? Which equation does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 122.1
Answer:
The given equations are:
a. y = -5x – 1
b. 2x – y = 8
c. y = x + 4
d. y = -3x + 13
Now,
Rewrite all the above equations in the form of
y = mx + c
So,
Now,
a.
The given equation is:
y = -5x – 1
Compare the above equation with
y = mx + c
So,
We will get,
m = -5 and c = -1

b.
The given equation is:
2x – y = 8
y = 2x – 8
Compare the above equation with
y = mx + c
So,
We will get
m = 8 and c = -8

c.
The given equation is:
y = x + 4
Compare the above equation with
y = mx + c
So,
We will get
m = 1 and c = 4

d.
The given equation is:
y = -3x + 13
Compare the above equation with
y = mx + c
So,
We will get
m = -3 and c = 3
Hence, from the above,
We can conclude that equation (b) does not belong with the other three since the slope of equation (b) is even whereas all the other slops are odd

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, describe the slope of the line. Then find the slope.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 123
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 124
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 124
From the graph,
We can observe that the slope falls from right to left
So,
The slope is positive
Now,
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(4, 3), (1, -1)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-1 – 3}{1 – 4}\)
m = \(\frac{-4}{-3}\)
m = \(\frac{4}{3}\)
Hence, from the above,
We can conclude that the slope of the line is: \(\frac{4}{3}\)

Question 7.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 125
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 7

Question 8.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 126
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 126
From the graph,
We can observe that the slope falls from left to right
So,
The slope is negative
Now,
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(0, 3), (5, -1)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-1 – 3}{5 – 0}\)
m = \(\frac{-4}{5}\)
m = –\(\frac{4}{5}\)
Hence, from the above,
We can conclude that the slope of the line is: –\(\frac{4}{5}\)

In Exercises 9–12, the points represented by the table lie on a line. Find the slope of the line.

Question 9.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 127
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 9

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 128
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 128
We know that,
To find the slope from a given table, we can take any x and y pair from the table
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(-1, -6), (2, -6)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-6 + 6}{2 + 1}\)
m = \(\frac{0}{3}\)
m = 0
Hence, from the above,
We can conclude that the slope of the line is: 0

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 129
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 11

Question 12.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 130
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 130
We know that,
To find the slope from a given table, we can take any x and y pair from the table
We have to represent the first pair as (x1, y1) and the second pair as (x2, y2)
Now,
The representation of the values of x and y in the form of ordered pairs to find a slope is:
(-4, 2), (-3, -5)
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{-5 – 2}{-3 – [-4]}\)
m = \(\frac{-7}{1}\)
m = -7
Hence, from the above,
We can conclude that the slope of the line is: -7

Question 13.
ANALYZING A GRAPH
The graph shows the distance y(in miles) that a bus travels in x hours. Find and interpret the slope of the line.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 131
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 13

Question 14.
ANALYZING A TABLE
The table shows the amount x(in hours) of time you spend at a theme park and the admission fee y (in dollars) to the park. The points represented by the table lie on a line. Find and interpret the slope of the line.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 132
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 132
It is given that the table shows the amount x(in hours) of time you spend at a theme park and the admission fee y (in dollars) to the park and the points represented by the table lie on a line.
Now,
The representations of the x and y values in the form of ordered pairs i.e., (x, y)
So,
Let (6, 54.99) and (7, 54.99) be the 2 ordered pairs
We know that,
m = \(\frac{y2 – y1}{x2 – x1}\)
m = \(\frac{54.99 – 54.99}{7 – 6}\)
m = \(\frac{0}{1}\)
m = 0
Hence, from the above,
We can conclude that the slope of the line is: 0

In Exercises 15–22, find the slope and the y-intercept of the graph of the linear equation.

Question 15.
y = -3x + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 15

Question 16.
y = 4x – 7
Answer:
The given equation is:
y = 4x – 7
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = 4 and c = -7
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 4
The y-intercept of the given equation is: -7

Question 17.
y = 6x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 17

Question 18.
y = -1
Answer:
The given equation is:
y = -1
Rewrite the given equation in the form of
y = mx + c
So,
y = 0x – 1
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = 0 and c = -1
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 0
The y-intercept of the given equation is: -1

Question 19.
-2x + y = 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 19

Question 20.
x + y = -6
Answer:
The given equation is:
x + y = -6
Rewrite the given equation in the form of the linear equation
y = mx + c
So,
y = -x – 6
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = -1 and c = -6
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: -1
The y-intercept of the given equation is: -6

Question 21.
-5x = 8 – y
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 21

Question 22.
0 = 1 – 2y + 14x
Answer:
The given equation is:
0 =1 – 2y + 14x
Rewrite the give equation in the form of
y = mx + c
So,
2y = 14x + 1
Divide by 2 into both sides
y = (14 / 2)x + (1 / 2)
y = 7x + \(\frac{1}{2}\)
Compare the given equation with the standard form of the linear equation
y = mx + c
By comparing, we will get
m = 7 and c = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 7
The y-intercept of the given equation is:\(\frac{1}{2}\)

ERROR ANALYSIS
In Exercises 23 and 24, describe and correct the error in finding the slope and the y-intercept of the graph of the equation.

Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 133
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 23

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 134
Answer:
The given equation is:
y = 3x – 6
Compare the above equation with the standard form of the linear equation
y = mx + c
So,
We will get
m = 3 and c = -6
Hence, from the above,
We can conclude that
The slope of the given equation (m) is: 3
The y-intercept of the given equation is: -6

In Exercises 25–32, graph the linear equation. Identify the x-intercept.

Question 25.
y = -x + 7
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 25

Question 26.
y = \(\frac{1}{2}\)x + 3
Answer:
The given linear equation is:
y = \(\frac{1}{2}\)x + 3
To find the x-intercepy, put y = 0
0 = \(\frac{1}{2}\)x + 3
–\(\frac{1}{2}\)x = 3
x = 3 (-2)
x = -6
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 27.
y = 2x
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 27

Question 28.
y = -x
Answer:
The given linear equation is:
y = -x
Rewrite the given equation in the form of y = mx + c
So,
y = -x + 0
To find the x-intercept, put y = 0
So,
0 = -x + 0
x = 0
Hence,
The representation of the given linear equation along with the x-intercept in the coordinate plane is:

Question 29.
3x + y = -1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 29

Question 30.
x + 4y = 8
Answer:
The given linear equation is:
x + 4y = 8
4y = -x + 8
Divide by 4 into both sides
y = –\(\frac{1}{4}\)x + (8 / 4)
y = –\(\frac{1}{4}\)x + 2
To find the x-intercept, put y = 0
0 = –\(\frac{1}{4}\)x + 2
\(\frac{1}{4}\)x = 2
x = 2(4)
x = 8
Hence,
The representation of the given linear equation along with the x-intercept in the coordinate plane is:

Question 31.
-y + 5x = 0
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 31

Question 32.
2x – y + 6 = 0
Answer:
The given linear equation is:
2x – y + 6 = 0
So,
y = 2x + 6
To find the x-intercept, put y = 0
So,
0 = 2x + 6
-2x = 6
x = -6 / 2
x = -3
Hence,
The representation of the given linear equation along with the x-intercept in the coordinate plane is:

In Exercises 33 and 34, graph the function with the given description. Identify the slope, y-intercept, and x-intercept of the graph.

Question 33.
A linear function f models a relationship in which the dependent variable decreases 4 units for every 2 units the independent variable increases. The value of the function at 0 is -2.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 33

Question 34.
A linear function h models a relationship in which the dependent variable increases 1 unit for every 5 units the independent variable decreases. The value of the function at 0 is 3.
Answer:
It is given that a linear function h models a relationship in which the dependent variable increases 1 unit for every 5 units the independent variable decreases.
We know that,
The independent variable is: x
The dependent variable is: y
So,
Slope (m) = y / x
m = 1 / (-5)
m = –\(\frac{1}{5}\)
It is given that the value of the function at 0 is 3
So,
The value of y at x = 0 is: 3
Hence,
The y-intercept of the given function is: 3
Hence,
The representation of the slope and the y-intercept in the standard form of the linear function s:
y = mx + c
So,
y = –\(\frac{1}{5}\)x + 3
To find the x-intercept, put y = 0
So,
0 = –\(\frac{1}{5}\)x + 3
\(\frac{1}{5}\)x = 3
x = 3(5)
x = 15
To find the y-intercept, put x = 0
So,
y = 3
Hence, from the above,
We can conclude that
The slope of the given function is: –\(\frac{1}{5}\)
The x-intercept is: 15
The y-intercept is: 3
The representation of the given function in the coordinate plane is:

Question 35.
GRAPHING FROM A VERBAL DESCRIPTION
A linear function r models the growth of your right index fingernail. The length of the fingernail increases 0.7 millimeters every week. Graph r when r(0) = 12. Identify the slope and interpret the y-intercept of the graph.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 35

Question 36.
GRAPHING FROM A VERBAL DESCRIPTION
A linear function m models the amount of milk sold by a farm per month. The amount decreases 500 gallons for every $1 increase in price. Graph m when m(0) = 3000. Identify the slope and interpret the x- and y-intercepts of the graph.
Answer:
It is given that the amount of milk decreases 500 gallons for every $1 increase
From the above,
We can say that,
The price is the independent variable
So,
x represents the price
The amount of milk is the dependent variable
So,
y represents the amount of milk
So,
Slope = y / x
Slope = -500 / 1
So,
Slope (m) = -500
It is given that,
m (0) = 3000
We know that,
m (0) is the functional representation of y
So,
y = 3000 when x = 0
Hence,
The y-intercept is: 3000
Now,
The representation of the slope and the y-intercept in the standard form of the linear equation is:
y = ,x + c
y = -500x + 3000
Now,
To find the x-intercept, put y = 0
So,
0 = -500x + 3000
500x = 3000
x = 3000 / 500
x = 6
To find the y-intercept, put x = 0
So,
y = -500 (0) + 3000
y = 3000
Hence, from the above,
We can conclude that
The slope of the given equation is: -500
The c-intercept or y-intercept of the given equation is: 3000
The x-intercept of the given equation is: 6

Question 37.
MODELING WITH MATHEMATICS
The function shown models the depth d (in inches) of snow on the ground during the first 9 hours of a snowstorm, where t is the time (in hours) after the snowstorm begins.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 135
a. Graph the function and identify its domain and range.
b. Interpret the slope and the d-intercept of the graph.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 37

Question 38.
MODELING WITH MATHEMATICS
The function c(x) = 0.5x + 70 represents the cost c (in dollars) of renting a truck from a moving company, where x is the number of miles you drive the truck.
a. Graph the function and identify its domain and range.
Answer:
The given function is:
c (x) = 0.5x + 70
Where,
represents the cost c (in dollars) of renting a truck from a moving company,
x is the number of miles you drive the truck.
Hence,
The representation of the given function in the coordinate plane is:

We know that,
The domain is the set of all the values of x that holds the given equation true
The range is defined as the set of all the values of y that holds the given equation true
Hence,
The domain of the given function is: -100 ≤ x ≤ 60
The range of the given function is: 20 ≤y ≤ 70

b. Interpret the slope and the c-intercept of the graph.
Answer:
The given function is:
c (x) = 0.5x + 70
Compare the above equation with the standard form of the linear function
y = mx + c
SO,
By comparing, we get
m = 0.5 and c = 70
Hence, from the above,
We can conclude that
The slope of the given function is: 0.5
The c-intercept is: 70

Question 39.
COMPARING FUNCTIONS
A linear function models the cost of renting a truck from a moving company. The table shows the cost y (in dollars) when you drive the truck x miles. Graph the function and compare the slope and the y-intercept of the graph with the slope and the c-intercept of the graph in Exercise 38.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 136
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 39

ERROR ANALYSIS
In Exercises 40 and 41, describe and correct the error in graphing the function.

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 137
Answer:
The given equation is:
y + 1 = 3x
Rewrite the above equation in the form of
y = mx + c
So,
The given equation can be rewritten as:
y = 3x – 1
Now,
Compare the above equation with the standard form of the linear equation
y = mx + c
So,
By comparison,
We get
m = 3 and c = -1
Hence,
The representation of the given equation in the coordinate plane is:

Question 41.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 138
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 41

Question 42.
MATHEMATICAL CONNECTIONS
Graph the four equations in the same coordinate plane.
3y = -x – 3
2y – 14 = 4x
4x – 3 – y = 0
x – 12 = -3y
a. What enclosed shape do you think the lines form? Explain.
Answer:
The given equations are:
a. 3y = -x – 3
b. 2y – 14 = 4x
c. 4x – 3 – y = 0
d. x – 12 = -3y
Now,
Rewrite the given equations in the form of
y = mx + c
So,
a.
The given equation is:
3y = -x – 3
Divide by 3 into both sides
y = –\(\frac{x}{3}\) – (3 / 3)
y = –\(\frac{1}{3}\)x – 1

b.
The given equation is:
2y – 14 = 4x
Divide by 2 into both sides
y – (14 / 2) = 2x
y – 7 = 2x
y = 2x + 7

c.
The given equation is:
4x – 3 – y = 0
y = 4x – 3

d.
The given equation is:
x – 12 = -3y
Divide by -3 into both sides
y = –\(\frac{x}[3]\) + (12 / 3)
y = –\(\frac{1}{3}\)x + 4
Hence,
The representations of the given four equations in the coordinate plane is:

Hence,
By observing the graph,
We can conclude that the enclosed lines in the graph form a rectangle

b. Write a conjecture about the equations of parallel lines.
Answer:
We can determine from their equations whether two lines are parallel by comparing their slopes.
If the slopes are the same and the y-intercepts are different, the lines are parallel.
If the slopes are different, the lines are not parallel.
Unlike parallel lines, perpendicular lines do intersect.

Question 43.
MATHEMATICAL CONNECTIONS The graph shows the relationship between the width y and the length x of a rectangle in inches. The perimeter of a second rectangle is 10 inches less than the perimeter of the first rectangle.
a. Graph the relationship between the width and length of the second rectangle.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 139
b. How does the graph in part (a) compare to the graph shown?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 43.2

Question 44.
MATHEMATICAL CONNECTIONS
The graph shows the relationship between the base length x and the side length (of the two equal sides) y of an isosceles triangle in meters. The perimeter of a second isosceles triangle is 8 meters more than the perimeter of the first triangle.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 140
a. Graph the relationship between the base length and the side length of the second triangle.
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 140
It is given that the graph shows the relationship between the base length x and the side length (of the two equal sides) y of an isosceles triangle in meters. The perimeter of a second isosceles triangle is 8 meters more than the perimeter of the first triangle.
Now,
From the graph,
The equation for the first Isosceles triangle is:
y = 6 – \(\frac{1}{2}\)x
Multiply by 2 into both sides
2y = 6 (2) – x
2y = 12 – x
2y + x = 12
The above equation represents the perimeter of the first Isosceles triangle
Where,
x is the base length
y is the side of the equal length like the third side
It is also given that the perimeter of a second isosceles triangle is 8 meters more than the perimeter of the first triangle.
So,
2y + x = 12 + 8
2y + x = 20
The above equation represents the perimeter of the second Isosceles triangle
Hence,
The representation of the perimeter of the second isosceles triangle in the coordinate plane is:

b. How does the graph in part (a) compare to the graph shown?
Answer:
The equation of the perimeter of the first isosceles triangle is:
y = 6 – \(\frac{1}{2}\)x
The equation of the perimeter of the second isosceles triangle is:
y = 10 – \(\frac{1}{2}\)x
Hence,
From comparing the above 2 equations with the standard form of the linear equation
y = mx + c
We can observe that the slopes are equal and only the y-intercepts are different
Hence, from the above,
We can conclude that the graphs of the perimeters of the first isosceles triangle and the second isosceles triangle are parallel

Question 45.
ANALYZING EQUATIONS
Determine which of the equations could be represented by each graph.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 141
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 45

Question 46.
MAKING AN ARGUMENT
Your friend says that you can write the equation of any line in slope-intercept form. Is your friend correct? Explain your reasoning.
Answer:
No, your friend is not correct

Explanation:
We can write the equation in slope-intercept form only when hat equation is in the linear form i.e., only x and y terms.
If there are exponential terms in an equation, then we can not write that equation in slope-intercept form
Hence, from the above,
We can conclude that your friend is not correct

Question 47.
WRITING
Write the definition of the slope of a line in two different ways.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 47

Question 48.
THOUGHT-PROVOKING
Your family goes on vacation to a beach 300 miles from your house. You reach your destination 6 hours after departing. Draw a graph that describes your trip. Explain what each part of your graph represents.
Answer:

Question 49.
ANALYZING A GRAPH
The graphs of the functions g(x) = 6x + a and h(x) = 2x + b, where a and b are constants, are shown. They intersect at the point (p, q).
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 142
a. Label the graphs of g and h.
b. What do a and b represent?
c. Starting at the point (p, q), trace the graph of g until you get to the point with the x-coordinate p + 2. Mark this point C. Do the same with the graph of h. Mark this point D. How much greater is the y-coordinate of point C than the y-coordinate of point D?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 49.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 49.2

Question 50.
HOW DO YOU SEE IT? Your commute to school by walking and by riding a bus. The graph represents your commute.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 143
a. Describe your commute in words.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 143
From the graph,
We can observe that your commute to school is different at different time intervals
Now,
From 0 to 10 seconds,
The distance you commuted gradually increases
From 10 to 14 seconds,
The distance you commuted is constant
From 14 to 18 seconds,
The distance you commuted abruptly increased

b. Calculate and interpret the slopes of the different parts of the graph.
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 143
From the graph,
We can observe that there are different commutes at different time intervals,
Now,
The x-axis represents the time
The y-axis represents the distance in miles
Now,
From 0 to 10 seconds,
The ordered pairs from the coordinate plane are:
(0, 0), (10, 0.5)
We know that,
The slope when 2 ordered pairs are given is represented as:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
From 0 to 10 seconds,
m = \(\frac{0.5 – 0}{10 – 0}\)
m = \(\frac{0.5}{10}\)
m = \(\frac{5}{100}\)
m = 0.05
From 10 to 14 seconds,
The ordered pairs from the coordinate plane are:
(10, 0.5), (14, 0.5)
We know that,
The slope when 2 ordered pairs are given is represented as:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
From 10 to 14 seconds,
m = \(\frac{0.5 – 0.5}{14 – 10}\)
m = \(\frac{0}{4}\)
m = 0
From 14 to 18 seconds,
The ordered pairs from the coordinate plane are:
(14, 0.5), (18, 2.5)
We know that,
The slope when 2 ordered pairs are given is represented as:
m = \(\frac{y2 – y1}{x2 – x1}\)
So,
From 14 to 18 seconds,
m = \(\frac{2.5 – 0.5}{18 – 14}\)
m = \(\frac{2}{4}\)
m = \(\frac{1}{2}\)
m = 0.5
Hence, fromthe above,
We can conclude that
The slope from 0 to 10 seconds is: 0.05
The slope from 10 to 14 seconds is: 0
The slope from 14 to 18 seconds is: 0.5

PROBLEM-SOLVING
In Exercises 51 and 52, find the value of k so that the graph of the equation has the given slope or y-intercept.

Question 51.
y = 4kx – 5; m = \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 51

Question 52.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 144
Answer:
The given equation is:
y = –\(\frac{1}{3}\)x + \(\frac{5}{6}\)k
Compare the above equation with the standard representation of the linear equation.
We know that,
The standard representation of the linear equation is:
y = mx + b
On comparison,
We get,
m = –\(\frac{1}{3}\) and b = \(\frac{5}{6}\)k
It is given that
The value of b is: -10
So,
\(\frac{5}{6}\)k = -10
5k = -10 × 6
k = \(\frac{-10 × 6}{5}\)
k = -12
Hence, from the above,
We can conclude that the value of k is: -12

Question 53.
ABSTRACT REASONING
To show that the slope of a line is constant, let (x1, y1) and (x2, y2) be any two points on the line y = mx + b. Use the equation of the line to express y1 in terms of x1 and y2 in terms of x2. Then use the slope formula to show that the slope between the points is m.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 53

Maintaining Mathematical Proficiency

Find the coordinates of the figure after the transformation.

Question 54.
Translate the rectangle with 4 units left.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 145
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 145
From the given graph,
The rectangle is covered up to 1 to 3 units at the x-axis and 0 to -4 units at the y-axis
So,
When we move the rectangle 4 units left, i.e., move the 1 and 3 units left for 4 units and move the 0 and -4 units left for 4 units
So,
The new rectangle in the graph is formed at:
At the x-axis:
1 – 4 = -3
3 – 4 = -1
At the y-axis:
0 – 4 = -4
-4 – 4 = -8
Hence,
The new rectangle is formed at (-1, -3) at the x-axis and at (-4, -8) at the y-axis
Hence,
The representation of the rectangle after the transformation in the coordinate plane is:

Question 55.
Dilate the triangle with respect to the origin using a scale factor of 2.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 146
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 55

Question 56.
Reflect the trapezoid in the y-axis.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 147
Answer:
Determine whether the equation represents a linear or nonlinear function. Explain.

Question 57.
y – 9 = \(\frac{2}{x}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 57

Question 58.
x = 3 + 15y
Answer:
The given function is:
x = 3 + 15y
Rewrite the above given function in the form of
y = mx + c
So,
15y = 3 – x
y = \(\frac{3 – x}{15}\)
y = \(\frac{3}{15}\) – \(\frac{1}{15}\)x
Hence,
The above function is in the form of
y = mx + c
Hence, from the above,
We can conclude that the given function is linear

Question 59.
\(\frac{x}{4}\) + \(\frac{y}{12}\) = 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.5 Question 59

Question 60.
y = 3x4 – 6
Answer:
The given function is:
y = 3x4 – 6
Rewrite the above given function in the form of
y = mx + c
But,
The above function is not in the form of
y = mx + c
Hence, from the above,
We can conclude that the given function is non-linear

Lesson 3.6 Transformations of Graphs of Linear Functions

Essential Question
How does the graph of the linear function f(x) = x compare to the graphs of g(x) = f(x) + c and h(x) = f(cx)? Comparing Graphs of Functions
Answer:
The given linear function is:
f (x) = x
The graph corresponding to the above function will be the vertical line
Now,
The given functions are:
g (x) = f (x) + c and h (x) = f (cx)
Now,
The graph of g (x) will be the vertical line translated i.e., addition or subtraction by c units
The graph of h (x) will be the graph of the function f(x) stretched or compressed by 1/c units

EXPLORATION 1
Comparing Graphs of Functions
Work with a partner.
The graph of f(x) = x is shown. Sketch the graph of each function, along with f, on the same set of coordinate axes. Use a graphing calculator to check your results. What can you conclude?
a. g(x) = x + 4
b. g(x) = x + 2
c. g(x) = x – 2
d. g(x) = x – 4
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 148
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 149
Answer:
The given functions are:
a. g(x) = x + 4
b. g(x) = x + 2
c. g(x) = x – 2
d. g(x) = x – 4
The other given function is:
f (x) = x
Now,
a.
The given function is:
g (x) = x + 4
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 4 units away from f(x) to the left side i.e., towards the positive y-axis

b.
The given function is:
g (x) = x + 2
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 2 units away from f(x) to the left side i.e., towards the positive y-axis

c.
The given function is:
g (x) = x – 2
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 2 units away from f(x) to the right side i.e., towards the negative y-axis

d,
The given function is:
g (x) = x – 4
Hence,
The representations of f(x) and g (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that g (x) translated 4 units away from f(x) to the right side i.e., towards the negative y-axis

EXPLORATION 2
Comparing Graphs of Functions
Work with a partner.
Sketch the graph of each function, along with f(x) = x, on the same set of coordinate axes. Use a graphing calculator to check your results. What can you conclude?
a. h(x) = \(\frac{1}{2}\)x
b. h(x) = 2x
c. h(x) = –\(\frac{1}{2}\)x
d. h(x) = -2x
Answer:
The given functions are:
a. h(x) = \(\frac{1}{2}\)x
b. h(x) = 2x
c. h(x) = –\(\frac{1}{2}\)x
d. h(x) = -2x
The other given function is:
f (x) = x
Now,
a.
The given function is:
h(x) = \(\frac{1}{2}\)x
Hence,
The representations of f (x) and h (x) in the same coordinate plane are:

Hence, from the above,
We can conclude that h (x) and f (x) passes through the origin and h (x) is steeper than f (x)

b.
The given function is:
h(x) = 2x
Hence,
The representation of f (x) and h(x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h(x) and f(x) passes through the origin and f (x) is steeper than h (x)

c.
The given function is:
h(x) = –\(\frac{1}{2}\)x
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) and f (x) are on the opposite axes

d.
The given function is:
h(x) = -2x
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are in the opposite axes

EXPLORATION 3
Matching Functions with Their Graphs
Work with a partner.
Match each function with its graph. Use a graphing calculator to check your results. Then use the results of Explorations 1 and 2 to compare the graph of k to the graph of f(x) = x.
a. k(x) = 2x – 4
b. k(x) = -2x + 2
c. k(x) = –\(\frac{1}{2}\)x + 4
d. k(x) = –\(\frac{1}{2}\)x — 2
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 150
Answer:
The given functions are:
a. k(x) = 2x – 4
b. k(x) = -2x + 2
c. k(x) = –\(\frac{1}{2}\)x + 4
d. k(x) = –\(\frac{1}{2}\)x — 2
The other given function is:
f (x) = x
Now,
a.
The given equation is:
k(x) = 2x – 4
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) translates 2 units from f (x) towards the positive x-axis

b.
The given function is:
k(x) = -2x + 2
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph A) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) and f (x) are on the opposite axes

c.
The given function is:
k(x) = –\(\frac{1}{2}\)x + 4
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph D) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) and f (x) are on opposite axes and k (x) only translates through only the positive and negative x-axes

d.
The given function is:
k(x) = –\(\frac{1}{2}\)x — 2
Hence,
The representation of k (x) and f (x) in the coordinate plane is:

Hence, from the above,
We can conclude that graph B) matches k (x)
By comparing f (x) and k (x),
We can say that k (x) and f(x) are on the opposite axes and k (x) translates through only the negative x-axis and the negative y-axis

Communicate Your Answer

Question 4.
How does the graph of the linear function f(x) = x compare to the graphs of g(x) = f(x) + c and h(x) = f(cx)?
Answer:
The given linear function is:
f (x) = x
The graph corresponding to the above function will be the vertical line
Now,
The given functions are:
g (x) = f (x) + c and h (x) = f (cx)
Now,
The graph of g (x) will be the vertical line translated i.e., addition or subtraction by c units
The graph of h (x) will be the graph of the function f(x) stretched or compressed by 1/c units

3.6 Lesson

Monitoring Progress
Using f, graph (a) g and (b) h. Describe the transformations from the graph of f to the graphs of g and h.

Question 1.
f(x) = 3x + 1; g(x) = f(x) – 2; h(x) = f(x – 2)
Answer:
The given functions are:
f (x) = 3x + 1; g (x) = f (x) – 2 and h (x) = f (x – 2)
Now,
It is given that,
g (x) = f (x) – 2
So,
g (x) = 3x + 1 – 2
g (x) = 3x – 1
It is given that,
h (x) = f (x – 2)
h (x) = 3 (x – 2) + 1
h (x) = 3 (x) + 3 (2) + 1
h (x) = 3x + 6 + 1
h (x) = 3x + 7
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) is translated 2 units away from f (x) toward the positive x-axis and h (x) is translated 5 units away from f (x) toward the positive y-axis

Question 2.
f(x) = -4x – 2; g(x) = -f(x); h(x) = f(-x)
Answer:
The given functions are:
f (x) = -4x – 2; g (x) = -f (x) and h (x) = f (-x)
Now,
It is given that,
g (x) = -f (x)
So,
g (x) = – (-4x – 2)
g (x) = 4x + 2
It is given that,
h (x) = f (-x)
h (x) = -4 (-x) – 2
h (x) = 4x – 2
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) is translated 2.5 units away from f (x) toward the positive y-axis and h (x) is translated 1 unit away from f (x) toward the positive x-axis

Using f, graph (a) g and (b) h. Describe the transformations from the graph of f to the graphs of g and h.

Question 3.
f(x) = 4x – 2; g(x) = f (\(\frac{1}{2}\)x ); h(x) = 2f(x)
Answer:
The given functions are:
f (x) = 4x – 2; g (x) = f (\(\frac{1}{2}\)x ) and h (x) = 2f (x)
Now,
It is given that,
g (x) = f (\(\frac{1}{2}\)x )
So,
g (x) = 4 ( f (\(\frac{1}{2}\)x ) – 2
g (x) = 2x – 2
It is given that,
h (x) = 2f (x)
h (x) = 2 (4x – 2)
h (x) = 2 (4x) – 2(2)
h (x) = 8x – 4
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) is translated 0.5 units away from f (x) toward the positive x-axis and h (x) is translated 0 units away from f (x) toward the origin

Question 4.
f(x) = -3x + 4; g(x) = f(2x); h(x) = \(\frac{1}{2}\)f(x)
Answer:
The given functions are:
f (x) = -3x + 4; g (x) = f (2x) and h (x) = \(\frac{1}{2}\)f(x)
Now,
It is given that,
g (x) = f (2x)
So,
g (x) = -3 (2x) + 4
g (x) = -6x + 4
It is given that,
h (x) = \(\frac{1}{2}\)f(x)
h (x) = \(\frac{1}{2}\) (3x + 4)
h (x) = \(\frac{3}{2}\)x + \(\frac{4}{2}\)
h (x) = \(\frac{3}{2}\)x + 2
Hence,
The representations of f (x), g (x), and h (x) in a coordinate plane is:

Hence, from the above,
We can conclude that
g (x) and f (x) are on the same axis and h (x) and f (x) are on the opposite axis

Question 5.
Graph f(x) = x and h(x) = \(\frac{1}{2}\)x – 2. Describe the transformations from the graph of f to the graph of h.
Answer:
The given functions are:
f (x) = x and h (x) = \(\frac{1}{2}\)x – 2
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) translated 4 units away from f (x) towards the positive x-axis

Transformations of Graphs of Linear Functions 3.6 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the relationship between f(x) = x and all other nonconstant linear functions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 1

Question 2.
VOCABULARY
Name four types of transformations. Give an example of each and describe how it affects the graph of a function.
Answer:
The four types of transformations that affect the graph of a function are:
a. Translation
Ex:
f (x) = x and g (x) = 2 f(x)
b. Rotation
Ex:
f (x) = 3x + 2 and g (x) = 3x – 2
c. Reflection
Ex:
f (x) =x and g (x) =-f (x)
d. Dilation
Ex:
f (x) = 3x + 6 and g (x) = f (2x)

Question 3.
WRITING
How does the value of a in the equation y = f(ax) affect the graph of y = f(x)? How does the value of a in the equation y = af(x) affect the graph of y = f(x)?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 151
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 3

Question 4.
REASONING
The functions f and g are linear functions. The graph of g is a vertical shrink of the graph of f. What can you say about the x-intercepts of the graphs of f and g? Is this always true? Explain.
Answer:
It is given that the functions f and g are linear functions
We know that,
f can be written as f (x)
g can be written as g (x)
It is also given that the graph of g is a vertical shrink of the graph of f.
So,
Since the graph of g shrinks, then the x-intercept of g will also shrink if we observe the functions of f and g
Hence, from the above,
We can conclude that the x-intercept of g will shrink and this is always true

In Exercises 5–10, use the graphs of f and g to describe the transformation from the graph of f to the graph of g.

Question 5.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 152
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 5

Question 6.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 153
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 153
From the above graph,
The given functions are:
f (x) = x – 3
g (x) = f (x + 4)
So,
g (x) = (x + 4) – 3
g (x) = x + 1
Hence,
When we observe f (x) and g (x), we can say that
g (x) shrinks by 2 units of f (x)

Question 7.
f(x) = \(\frac{1}{3}\)x + 3; g(x) = f(x) = -3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 7

Question 8.
f(x) = -3x + 4; g(x) = f(x) + 1
Answer:
The given functions are:
f (x) = -3x + 4
g (x) = f (x) + 1
So,
g (x) = -3x + 4 + 1
g (x) = -3x + 5
Hence,
The representation of f and g in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 1 unit away from f (x)

Question 9.
f(x) = -x – 2; g(x) = f(x + 5)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 9

Question 10.
f(x) = \(\frac{1}{2}\)x – 5; g(x) = f(x – 3)
Answer:
The given functions are:
f (x) = \(\frac{1}{2}\)x – 5
g (x) = f (x – 3)
So,
g (x) = \(\frac{1}{2}\) ( x – 3 ) – 5
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) is dilated from f (x)

Question 11.
MODELING WITH MATHEMATICS
You and a friend start biking from the same location. Your distance d (in miles) after t minutes is given by the function d(t) = \(\frac{1}{5}\)t. Your friend starts biking 5 minutes after you. Your friend’s distance f is given by the function f(t) = d(t – 5). Describe the transformation from the graph of d to the graph of f.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 154
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 11

Question 12.
MODELING WITH MATHEMATICS
The total cost C (in dollars) to cater an event with p people is given by the function C(p) = 18p + 50. The set-up fee increases by $25. The new total cost T is given by the function T(p) = C(p) + 25. Describe the transformation from the graph of C to the graph of T.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 155
Answer:
It is given that the total cost C (in dollars) to cater an event with p people is given by the function
C(p) = 18p + 50
It is also given that the set-up fee increases by $25.
So,
The new total cost T is given by the function
T(p) = C(p) + 25
So,
T (p) = 18p + 50 + 25
T (p) = 18p + 75
Hence,
The representation of T (p) and C (p) in the same coordinate plane is:

Hence, from the above,
We can conclude that T (p) translated 25 units away from C (p)

In Exercises 13–16, use the graphs of f and h to describe the transformation from the graph of f to the graph of h.

Question 13.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 156
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 13

Question 14.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 157
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 157
From the above graph,
The given functions are:
f (x) = -3x + 1
h (x) = f (-x)
So,
h (x) = -3 (-x) + 1
h (x) = 3x + 1
Hence,
The representations of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are perpendicular lines

Question 15.
f(x) = -5 – x; h(x) = f(-x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 15

Question 16.
f(x) = \(\frac{1}{4}\)x – 2; h(x) = -f(x)
Answer:
The given functions are:
f (x) = \(\frac{1}{4}\)x – 2
h (x) = – f(x)
So,
h (x) = – (\(\frac{1}{4}\)x – 2)
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) is a reflection of f (x) on the y-axis

In Exercises 17–22, use the graphs of f and r to describe the transformation from the graph of f to the graph of r.

Question 17.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 158
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 17

Question 18.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 159
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 159
From the above graph,
The given functions are:
f (x) = -x
r (x) = f (\(\frac{1}{4}\) x)
So,
r (x) = –\(\frac{1}{4}\)x
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that r (x) shrinks by \(\frac{1}{4}\) of f (x)

Question 19.
f(x) = -2x – 4; r(x) = f(\(\frac{1}{2}\)x )
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 19

Question 20.
f(x) = 3x + 5; r(x) = f (\(\frac{1}{3}\)x)
Answer:
The given functions are:
f (x) = 3x + 5
r (x) = f (\(\frac{1}{3}\)x)
So,
r (x) = 3 ( \(\frac{1}{3}\) x ) + 5
r (x) = x + 5
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that r (x) shrinks by 1 unit of f (x)

Question 21.
f(x) = \(\frac{2}{3}\)x + 1; r(x) = 3f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 21

Question 22.
f(x) = –\(\frac{1}{4}\)x – 2; r(x) = 4f(x)
Answer:
The given functions are:
f (x) = –\(\frac{1}{4}\)x – 2
r (x) = 4 f (x)
So,
r (x) = 4 [-\(\frac{1}{4}\)x – 2]
r (x) = -x – 8
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that r (x) translates 4 units away from  f (x)

In Exercises 23–28, use the graphs of f and h to describe the transformation from the graph of f to the graph of h.

Question 23.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 160
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 23

Question 24.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 161
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 161
From the above grah,
The given functions are:
f (x) = -2x – 6
h (x) = \(\frac{1}{3}\) f(x)
So,
h (x) = –\(\frac{1}{3}\) (2x + 6)
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) stretches by \(\frac{1}{3}\) of f (x)

Question 25.
f(x) = 3x – 12; h(x) = \(\frac{1}{6}\)f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 25

Question 26.
f(x) = -x + 1; h(x) = f(2x)
Answer:
The given functions are:
f (x) = -x + 1
h (x) = f (2x)
So,
h (x) = -(2x) + 1
h (x) = -2x + 1
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) stretches by 1 unit from f (x)

Question 27.
f(x) = -2x – 2; h(x) = f(5x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 27

Question 28.
f(x) = 4x + 8; h(x) = \(\frac{3}{4}\)f(x)
Answer:
The given functions are:
f (x) = 4x + 8
h (x) = \(\frac{3}{4}\) f (x)
So,
h (x) = \(\frac{3}{4}\) (4x + 8)
h (x) = \(\frac{3}{4}\) (4x) + \(\frac{3}{4}\) (8)
h (x) = 3x + 6
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) shrinks by 2 units from f (x)

In Exercises 29–34, use the graphs of f and g to describe the transformation from the graph of f to the graph of g.

Question 29.
f(x) = x – 2; g(x) = \(\frac{1}{4}\)f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 29

Question 30.
f(x) = -4x + 8; g(x) = -f(x)
Answer:
The given functions are:
f (x) = -4x + 8
g (x) = -f (x)
So,
g (x) = -4 (-x) + 8
g (x) = 4x + 8
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above graph,
We can conclude that g (x) is a reflection of f (x)

Question 31.
f(x) = -2x – 7; g(x) = f(x – 2)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 31

Question 32.
f(x) = 3x + 8; g(x) = f(\(\frac{2}{3}\)x)
Answer:
The given functions are:
f (x) = 3x + 8
g (x) = f (\(\frac{2}{3}\)x)
So,
g (x) = 3 ( \(\frac{2}{3}\) ) x + 8
g (x) = 2x + 8
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) shrink by 1 unit of the graph of f (x)

Question 33.
f(x) = x – 6; g(x) = 6f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 33

Question 34.
f(x) = -x; g(x) = f(x) -3
Answer:
The given functions are:
f (x) = -x
g (x) = f (x) – 3
So,
g (x) = -x – 3
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) is translated by 3 units away from f (x)

In Exercises 35–38, write a function g in terms of f so that the statement is true.

Question 35.
The graph of g is a horizontal translation 2 units right of the graph of f.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 35

Question 36.
The graph of g is a reflection in the y-axis of the graph of f.
Answer:
We know that,
f can be writen as f (x)
g can be written as g (x)
Now,
The given statement is:
The graph of g is a reflection of f(x) in the y-axis of the graph of f (x)
So,
Reflection in the y-axis means if f (x) is +ve, then the reflection of f (x) will be -ve and vice-versa
Hence,
The representation of the given statement in terms of f(x) is:
g (x) = f (-x)

Question 37.
The graph of g is a vertical stretch by a factor of 4 of the graph of f.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 37

Question 38.
The graph of g is a horizontal shrink by a factor of \(\frac{1}{5}\) of the graph of f.
Answer:
We know that,
f can be writen as f (x)
g can be written as g (x)
Now,
The given statement is:
The graph of g is a horizontal shrink by a factor of \(\frac{1}{5}\) of the graph
We know that,
“Shrink” is represented by ‘-‘
“Stretch” is represented by ‘+’
Hence,
The representation of the given statement in terms of f (x) is:
g (x) = – \(\frac{1}{5}\) f (x)

ERROR ANALYSIS In Exercises 39 and 40, describe and correct the error in graphing g.

Question 39.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 162
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 163
Answer:
The given functions are:
f (x) = -x + 3 and g (x) = f (-x)
So,
g (x) = -[-x] + 3
g (x) = x + 3
Hence,
The representations of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and g (x) are perpendicular lines from the graph

In Exercises 41–46, graph f and h. Describe the transformations from the graph of f to the graph of h.

Question 41.
f(x) = x; h(x) = \(\frac{1}{3}\)x + 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 41

Question 42.
f(x) = x; h(x) = 4x – 2
Answer:
The given functions are:
f (x) = x and h (x) = 4x – 2
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that the transformations are a vertical stretch by a factor of 4 followed by a vertical translation of 2 units down

Question 43.
f(x) = x; h(x) = -3x – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 43

Question 44.
f(x) = x; h(x) = –\(\frac{1}{2}\)x + 3
Answer:
The given functions are:
f (x) = x and h (x) = –\(\frac{1}{2}\)x + 3
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that the transformations are a vertical shrink by a factor of \(\frac{1}{2}\) followed by a vertical translation of 3 units up

Question 45.
f(x) = 2x; h(x) = 6x – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 45

Question 46.
f(x) = 3x; h(x) = -3x – 7
Answer:
The given functions are:
f (x) = 3x and h (x) = -3x – 7
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that the transformations are a vertical shrink by a factor of 3 followed by a vertical translation of 7 units down

Question 47.
MODELING WITH MATHEMATICS
The function t(x) = -4x +72 represents the temperature from 5 P.M. to 11 P.M., where x is the number of hours after 5 P.M. The function d(x) = 4x + 72 represents the temperature from 10 A.M. to 4 P.M., where x is the number of hours after 10 A.M. Describe the transformation from the graph of t to the graph of d.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 164
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 47

Question 48.
MODELING WITH MATHEMATICS
A school sells T-shirts to promote school spirit. The school’s profit is given by the function P(x) = 8x – 150, where x is the number of T-shirts sold. During the playoffs, the school increases the price of the T-shirts. The school’s profit during the play-offs is given by the function Q(x) = 16x – 200, where x is the number of T-shirts sold. Describe the transformations from the graph of P to the graph of Q.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 164.1
Answer:
It is given that a school sells T-shirts to promote school spirit. The school’s profit is given by the function
P(x) =8x- 150
where,
x is the number of T-shirts sold
It is also given that during the playoffs, the school increases the price of the T-shirts. The school’s profit during the play-offs is given by the function
Q(x) = 16x – 200
where,
x is the number of T-shirts sold.
Hence,
The representation of P (x) and Q(x) in the same coordinate plane is:

Hence, from the above graph,
We can conclude that
The x-axis shrink by a factor of 2
The y-axis shrink by 50

Question 49.
USING STRUCTURE
The graph of g(x) = a • f(x – b) + c is a transformation of the graph of the linear function f. Select the word or value that makes each statement true.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 165
a. The graph of g is a vertical ______ of the graph of f when a = 4, b = 0, and c = 0.
b. The graph of g is a horizontal translation ______ of the graph of f when a = 1, b = 2, and c = 0.
c. The graph of g is a vertical translation 1 unit up of the graph of f when a = 1, b = 0, and c = ____.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 49

Question 50.
USING STRUCTURE
The graph of h(x) = a • f(bx – c) + d is a transformation of the graph of the linear function f. Select the word or value that makes each statement true.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 166
a. The graph of h is a ______ shrink of the graph of f when a = \(\frac{1}{3}\), b = 1, c = 0, and d = 0.
b. The graph of h is a reflection in the ______ of the graph of f when a = 1, b = -1, c = 0, and d = 0.
c. The graph of h is a horizontal stretch of the graph of f by a factor of 5 when a = 1, b = _____, c = 0, and d = 0.
Answer:
The given original function is:
f (x) = x
The given choice of words to make the given statements true is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 166
The transformation of the function of f (x) is given as:
h (x) = a f (bx – c) + d
Where,
a, b, c, and d are constants
Now,
a.
The given values of constants are:
a = \(\frac{1}{3}\),
b = 1
c = 0
d = 0
So,
h (x) = \(\frac{1}{3}\)f (1 (x) – 0) + 0
h (x) = \(\frac{1}{3}\) f (x)
So,
h (x) = \(\frac{1}{3}\)x
Hence,
The representations of f (x) and h (x) in the coordinate plane is:

Hence, from the above,
We can conclude that the graph of h is a vertical shrink of the graph of f when a = \(\frac{1}{3}\), b = 1, c = 0, and d = 0.

b.
The given values of constants are:
a = 1
b = -1
c = 0
d = 0
So,
h (x) = 1f (-1 (x) – 0) + 0
h (x) =  f (-x)
So,
h (x) = -x
Hence,
The representations of f (x) and h (x) in the coordinate plane is:

Hence, from the above,
We can conclude that the graph of h is a reflection in the y-axis of the graph of f when a = 1, b = -1, c = 0, and d = 0

c.
The given values of constants are:
a = 1
b = p
c = 0
d = 0
So,
h (x) = 1f (p (x) – 0) + 0
h (x) =  f (px)
So,
h (x) = px
Hence,
The representations of f (x) and h (x) in the coordinate plane is:

Hence, from the above,
We can conclude that the graph of h is a horizontal stretch of the graph of f by a factor of 5 when a = 1, b = \(\frac{1}{5}\), c = 0, and d = 0

Question 51.
ANALYZING GRAPHS
Which of the graphs are related by only a translation? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 167
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 51

Question 52.
ANALYZING RELATIONSHIPS
A swimming pool is filled with water by a hose at a rate of 1020 gallons per hour. The amount v (in gallons) of water in the pool after t hours is given by the function v(t) = 1020t. How does the graph of v change in each situation?
a. A larger hose is found. Then the pool is filled at a rate of 1360 gallons per hour.
Answer:
It is given that a swimming pool is filled with water by a hose at a rate of 1020 gallons per hour. The amount v (in gallons) of water in the pool after t hours is given by the function
v(t) = 1020t
Now,
It is also given that a pool is filled at a rate of 1360 gallons per hour
So,
The overall rate change = 1340 / 1020
= 440 / 330
= 4 / 3
So,
The new rate of change is representad by the function
f (t) = (4 / 3) v (t)
Hence,
The representation of the graph of v using the functions v (t) = 1020t and v (t) = 1360t is:

b. Before filling up the pool with a hose, a water truck adds 2000 gallons of water to the pool.
Answer:
We know that,
From part (a),
The function for filling up the pool with a hose is given as:
v (t) = 1020t
It is given that before filling up the pool with a hose, a water truck adds 2000 gallons of water to the pool
So,
v (t) = 1020t + 2000
Hence,
The representations of the functions before filling the hose and  a water truck adds 2000 gallons of water are:

Question 53.
ANALYZING RELATIONSHIPS
You have $50 to spend on fabric for a blanket. The amount m (in dollars) of money you have after buying y yards of fabric is given by the function m(y) = -9.98y + 50. How does the graph of m change in each situation?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 168
a. You receive an additional $10 to spend on the fabric.
b. The fabric goes on sale, and each yard now costs $4.99.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 53

Question 54.
THOUGHT-PROVOKING
Write a function g whose graph passes through the point (4, 2) and is a transformation of the graph of f(x) = x.
Answer:
It is given that a function g whose graph passes through the point (4, 2) and is a transformation of the graph of f(x) = x
We know that,
The equation that passes through (h, k) and it is a transformation of another function is:
y = fx – h) + k
So,
From (4, 2)
As the translation between the values of x and y in the given point is 2,
We can make the value of ‘h’ as -2
We can make the value of k as -4
Hence,
The equation that passes through (4, 2) is:
y = (x + 2 ) – 4
y = x – 2
Hence, from the above,
We can conclude that the function that passes through (4, 2) is:
g (x) = x – 2

In Exercises 55–60, graph f and g. Write g in terms of f. Describe the transformation from the graph of f to the graph of g.

Question 55.
f(x) = 2x – 5; g(x) = 2x – 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 55

Question 56.
f(x) = 4x + 1; g(x) = -4x – 1
Answer:
The given equations are:
f (x) = 4x + 1
g (x) = -4x – 1
So,
g (x) = – (4x + 1)
g (x) = -f (x)
Hence,
The representations of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that f (x) and g (x) are on the opposite quadrants

Question 57.
f(x) = 3x + 9; g(x) = 3x + 15
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 57

Question 58.
f(x) = -x – 4 ; g(x) = x – 4
Answer:
The given equations are:
f (x) = -x – 4
g (x) = x – 4
Hence,
The representations of f (x) and g (x) in the coordinate plane are:

Hence, from the above,
We can conclude that f (x) and g (x) are perpendicular lines since only the slopes vary and the y-intercepts are the same

Question 59.
f(x) = x + 2; g(x) = \(\frac{2}{3}\)x + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 59

Question 60.
f(x) = x – 1; g(x) = 3x – 3
Answer:
The given equations are:
f (x) = x – 1
g (x) = 3x – 3
So,
g (x) = 3 (x – 1)
g (x) = 3 f (x)
Hence,
The representations of f (x) and g (x) in the coordinate plane are:

Hence, from the above,
We can conclude that g (x) translates 3 units away from f (x)

Question 61.
REASONING
The graph of f(x) = x + 5 is a vertical translation 5 units up of the graph of f(x) = x. How can you obtain the graph of f(x) = x + 5 from the graph of f(x) = x using a horizontal translation?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 61

Question 62.
HOW DO YOU SEE IT? Match each function with its graph. Explain your reasoning.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 169
a. a(x) = f(-x)
b. g(x) = f(x) – 4
c. h(x) = f(x) + 2
d. k(x) = f(3x)
Answer:

REASONING
In Exercises 63–66, find the value of r.

Question 63.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 170
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 63

Question 64.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 171
Answer:
The given equations are:
f (x) = -3x + 5
g (x) = f (rx)
So,
g (x) = -3 (rx) + 5
Where,
r is the transformational factor
From the graph,
The x-intercept of g(x) is: 5
Now,
To find he x-intercept, put y = 0
So,
0 = -3 (rx) + 5
3 (rx) = 5
3r (5) = 5
3r = 5 / 5
3r = 1
r = \(\frac{1}{3}\)
Hence, from the above,
WE can conclude that the value of r is: \(\frac{1}{3}\)

Question 65.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 172
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 65

Question 66.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 173
Answer:
The given equations are:
f (x) = \(\frac{1}{2}\)x + 8
g (x) = f (x) + r
So,
g (x) = \(\frac{1}{2}\)x + 8 + r
From the graph,
We can observe that f (x) and g (x) are both the parallel lines
So,
The slopes of f (x) and g (x) are the same and only the y-intercepts differ in f (x) and g (x)
From the graph,
We can observe that the y-intercept of g (x) is: 0
So,
g (x) = \(\frac{1}{2}\)x + 8 + 0
Hence, from the above,
We can conclude that the value of r is: 0

Question 67.
CRITICAL THINKING
When is the graph of y = f(x) + w the same as the graph of y = f(x + w) for linear functions? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 67

Maintaining Mathematical Proficiency

Solve the formula for the indicated variable.(Section 1.5)

Question 68.
Solve for h.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 174
Answer:
The given formula from the given figure is:
V = πr²h
Divide by πr² into both sides
So,
\(\frac{V}{πr²}\) = h
Hence, from the above,
We can conclude that the formula for ‘h’ is:
h = \(\frac{V}{πr²}\)

Question 69.
Solve for w.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 175
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 69

Solve the inequality. Graph the solution, if possible. (Section 2.6)

Question 70.
| x – 3 | ≤ 14
Answer:
The given absolute value inequality is:
| x + 3 | ≤ 14
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 ≤ 14 and x + 3 ≥ -14
x ≤ 14 – 3 and x ≥ -14 – 3
x ≤ 11 and x ≥ -17
Hence, from the above,
We can conclude that the solutions to the given absolute value inequality are:
x ≤11 and x ≥ -17
The representation of the solutions of the given inequality in the graph is:

Question 71.
| 2x + 4 | > 16
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 71

Question 72.
5 | x + 7 | < 25
Answer:
The given absolute value inequality is:
5 | x + 7 | < 25
So,
| x + 7 | < 25 / 5
| x + 7 | < 5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 7 < 5 and x + 7 > -5
x < 5 – 7 and x > -5 – 7
x < -2 and x > -12
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
x < -2 and x > -12
The representations of the solutions of the given absolute value inequality in the graph is:

Question 73.
-2 | x + 1 | ≥ 18
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.6 Question 73

Lesson 3.7 Graphing Absolute Value Functions

Essential Question
How do the values of a, h, and k affect the graph of the absolute value function g(x) = a | x – h | + k? The parent absolute value function is f(x) = | x | Parent absolute value function
The graph of f is V-shaped.
Answer:
The given absolute value function is:
g (x) = a | x – h | + k
The parent function for the given absolute value function is:
f (x) = | x |
It is also given that the graph of f is v-shaped
Now,
The given absolute value function has the constants a, x, h, and k
Now,
“x – h” represents the translation of the x-axis where h is the translation value
Ex:
x – h =x – 3
Where,
x is the original function and h is the translation factor of x
Now,
“k” represents the y-intercept of the given absolute value function
The value of k affects the graph of the given absolute value function to translate on the y-axis up and down or shrink and stretch

EXPLORATION 1
Identifying Graphs of Absolute Value Functions
Work with a partner.
Match each absolute value function with its graph. Then use a graphing calculator to verify your answers.
a. g(x) = | x – 2 |
b. g(x) = | x – 2 | + 2
c. g(x) = | x + 2 | + 2
d. g(x) = – | x + 2 | + 2
e. g(x) = 2 | x – 2 |
f. g(x) = – | x + 2 | + 2
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 176
Answer:
The given absolute value functions are:
a. g(x) = | x – 2 |
b. g(x) = | x – 2 | + 2
c. g(x) = | x + 2 | + 2
d. g(x) = – | x + 2 | + 2
e. g(x) = 2 | x – 2 |
f. g(x) = – | x + 2 | + 2
We know that,
| x | =x for x > 0
| x | = -x for x < 0
Now,
a.
The given absolute value function is:
g(x) = | x – 2 |
So,
g (x) = x – 2 or g (x) = – (x – 2)
g (x) = x – 2 or g (x) = 2 – x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, fro the above,
We can conclude that graph F) matches the given absolute value function
b.
The given absolute value function is:
g(x) = | x – 2 | + 2
So,
g (x) = x – 2 + 2 or g (x) = – (x – 2) + 2
g (x) = x or g (x) = 2 – x + 2
g (x) = x or g (x) = 4 – x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches the given absolute value function
c.
The given absolute value function is:
g(x) = | x + 2 | + 2
So,
g (x) = x + 2 + 2 or g (x) = – (x + 2 ) + 2
g (x) = x + 4 or  g (x) = -x – 2 + 2
g (x) = x + 4 or g (x) = -x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph C) matches the given absolute value function
d.
The given absolute value function is:
g(x) = – | x + 2 | + 2
So,
g (x) = – (x + 2 )  + 2 or g (x) = – [- (x + 2 ) ] + 2
g (x) = -x – 2 + 2 or g (x) = x + 2 + 2
g (x) = x or g (x) =x + 4
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph B) matches the given absolute value function
e.
The given absolute value function is:
g(x) = 2 | x – 2 |
So,
g (x) = 2 (x – 2) or g (x) = 2 ( – (x – 2) )
g (x) = 2x – 4 or g (x) = 2 (-x + 2 )
g (x)= 2x – 4 or g (x) = -2x + 4
Hence,
The representation of the given absolue value functions in the coordinate plane is:

Hence, from the above,
We can conclude that the graph A) matches the given absolute value function
f.
The given absolute value function is:
g(x) = – | x + 2 | + 2
So,
g (x) = – (x + 2) + 2 or g (x) = – [- (x – 2} ] + 7
g (x) = x + 4 or g (x) = 2 – x + 7
g (x) = x + 4 or g (x) = -x + 9
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that graph E) matches the given absolute value function

Communicate Your Answer

Question 2.
How do the values of a, h, and k affect the graph of the absolute value function g(x) = a | x – h | + k?
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 176.1
Answer:
The given absolute value function is:
g (x) = a | x – h | + k
From the above absolute value function,
We can say that
a, h and k are constants
Now,
“a” represents the constant to multiply g (x)
“h” represents the translation factor on the x-axis
“k” represents the y-intercept of the given function
The value of “h” and “k” affect the graph in such a way that the graph can move anywhere in the graph

Question 3.
Write the equation of the absolute value function whose graph is shown. Use a graphing calculator to verify your equation.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 177
Answer:
The given graphing calculator is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 177
From the above graph,
We can observe that the plot is in downwqrd direction i.e., the absolute value fuction is negative and the plot passes through 2 in the x-axis
Hence,
The absolute value function representing the given graph is:
f (x) = -| x + 1 | + 2

3.7 Lesson

Monitoring Progress

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 1.
h(x) = | x | – 1
Answer:
The given functions are:
f (x) = | x |
h (x) = | x | – 1
Hence,
The representation of f (x) and h (x) in  the same coordinate plane is:

Hence, from the above graph,
h (x) is translated vertically away from f (x) on the y-axis by 1unit
Now,
The domain of the given functions are: -10 ≤ x ≤ 10
The range of the given functions are: -1 ≤ y ≤ 10

Question 2.
n(x) = | x + 4 |
Answer:
The given functions are:
f (x) = | x |
h (x) = | x + 4 |
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above graph,
h (x) is translated from f (x) 4 units away on the x-axis
Now,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of h (x) is: -10 ≤ x ≤6
The range of f (x) is: 0 ≤ y ≤ 10
The range of h (x) is: 0 ≤y ≤ 6

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 3.
t(x) = -3| x |
Answer:
The given functions are:
f (x) = | x |
t (x) = -3 | x |
Hence,
The representation of f (x) and t (x) in the same coordinate plane is:

Hence, from the above graph,
t (x) is shrunk by 3units and rotated by 90° on the y-axis
Now,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of t (x) is: -3 ≤x ≤ 3
The range of f (x) is: 0 ≤ y ≤ 10
The range of t (x) is: -10 ≤ y ≤ 0

Question 4.
v(x) = \(\frac{1}{4}\)| x |
Answer:
The given functions are:
f (x) = | x |
v (x) = \(\frac{1}{4}\)| | x |
Hence,
The representation of f (x) and v (x) in the same coordinate plane is:

Hence, from the above graph,
t (x) translates \(\frac{1}{4}\)| units of f (x) on the y-axis
Now,
The domain of the given functions are: -10 ≤ x ≤ 10
The range of f (x) is: 0 ≤ y ≤ 10
The range of t (x) is: 0 ≤ y ≤ 2.5

Question 5.
Graph f(x) = | x – 1 | and g(x) = | \(\frac{1}{2}\)x – 1 |.
Compare the graph of g to the graph of f.
Answer:
The given functions are:
f (x) = | x – 1 |
g (x) = | \(\frac{1}{2}\)x – 1 |
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above graph,
g (x) translates 1 unit away from f (x) on the x-axis
Now,
The domain of f (x) is: -8 ≤ x ≤  10
The domain of g (x) is: -10 ≤  x ≤  10
The range of f (x) is: 0 ≤  y ≤ 10
The range of g (x) is: 0 ≤  y ≤  6

Question 6.
Graph f(x) = | x + 2 | + 2 and g(x) = | -4x + 2 | + 2.
Compare the graph of g to the graph of f.
Answer:
The given functions are:
f (x) = | x + 2 | + 2
g (x) = | -4x + 2 | + 2
Hence,
The representation of f (x) and g(x) in the same coordinate plane is:

Hence, from the above graph,
g (x) translates \(\frac{3}{2}\) units away from f (x) on the x-axis
Now,
The domain of f (x) is: -10 ≤ x ≤ 6
The domain of g (x) is:-1.5 ≤ x ≤ 2.5
The range of f (x) is: 0 ≤ y ≤ 10
The range of g (x) is: 2 ≤y ≤ 10

Question 7.
Let g(x) = | –\(\frac{1}{2}\)x + 2 | + 1.
(a) Describe the transformations from the graph of f(x) = | x | to the graph of g.
(b) Graph g.
Answer:
a.
The given functions are:
f (x) = | x |
g (x) = | –\(\frac{1}{2}\)x + 2 | + 1
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above graph,
We can observe that
g (x) translates 4 units away from f (x) on the positive x-axis
g (x) translates 2 units away from f (x) on the negative x-axis

b.
The function of g (x) is:
| –\(\frac{1}{2}\)x + 2 | + 1
We know that,
g can be written as g (x)
Hence,
The representation of g (x) in the coordinate plane is:

Graphing Absolute Value Functions 3.7 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The point (1, -4) is the _______ of the graph of f(x) = -3 | x – 1 | – 4.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 1

Question 2.
USING STRUCTURE
How do you know whether the graph of f(x) = a | x – h | + k is a vertical stretch or a vertical shrink of the graph of f(x) = | x | ?
Answer:
The given function is:
f (x) = a | x – h | + k
Where,
a is the vertical stretch or vertical shrink
h is the horizontal translation
k is the vertical shift
Now,
‘a’ will be the vertical stretch if the value of a is an integer
‘a’ will be the vertical shrink if the value of ‘a’ is a fraction
Hence, from the above,
We can conclude that the given function is a vertical stretch

Question 3.
WRITING
Describe three different types of transformations of the graph of an absolute value function.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 3

Question 4.
REASONING
The graph of which function has the same y-intercept as the graph of f(x) = | x – 2 | + 5? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 178
Answer:
The given absolute value functions are:
a. f (x) = | x – 2 | + 5
b. g (x) = | 3x – 2 | + 5
c. h (x) = 3 | x – 2 | + 5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
a.
The given absolute value function is:
f (x) = | x – 2 | + 5
So,
f (x) = x – 2 + 5 or f (x) = -(x – 2) + 5
f (x) = x + 3 or f (x) = -x + 2 + 5
f (x) = x + 3 or f (x) = -x + 7

b.
The given absolute value function is:
g (x) = | 3x – 2 | + 5
So,
g (x) = 3x – 2 + 5 or g (x) = – (3x – 2 ) + 5
g (x) = 3x + 3 or g (x) = -3x + 2 + 5
g (x) = 3x + 3 or g (x) = -3x + 7

c.
The given absolute value function is:
h (x) = 3 | x – 2 | + 5
So,
h (x) = 3 (x – 2) + 5 or h (x) = -3 (x – 2) + 5
h (x) = 3x – 6 + 5 or h (x) = -3x + 6 + 5
h (x) = 3x – 1 or h (x) = -3x + 11
Compare the solutions of the given absolute value functions with
f (x) = mx + c
Where,
m is the slope
c is the y-intercept
Hence, from the above,
We can conclude that a) and b ) have the same y-intercepts

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 5.
d(x) = | x | – 4
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 5

Question 6.
r(x) = | x | + 5
Answer:
The given functions are:
f (x) = | x |
r (x) = | x | + 5
Hence,
The representation of f (x) and r (x) in the coordinate plane is:

Hence, from the above,
We can conclude that r (x) is 5 units away from f (x)
Now,
The domain of the given absolute value function is: -5 ≤ x ≤ 5
The range of the given absolute value function is: 5 ≤y ≤10

Question 7.
m(x) = | x + 1 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 7

Question 8.
v(x) = | x – 3 |
Answer:
The given absolute value functions are:
f (x) = | x |
v (x) = | x – 3 |
Hence,
The representation of f(x) and v (x) in the coordinate plane is:

Hence, from the above,
We can conclude that v (x) translates 3 units away from f (x)
Now,
The domain of the given absolute value function is: -6 ≤ x ≤ 10
The range of the given absolute value function is: 6 ≤ y ≤ 10

Question 9.
p(x) = \(\frac{1}{3}\) | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 9

Question 10.
j(x) = 3 | x |
Answer:
The given absolute value functions are:
f (x) = | x |
j (x) = 3 | x |
Hence,
The representation of f (x) and j (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that j (x) vertically stretches 3 units away from f (x)
Now,
The domain of the given absolute value function is: -3.5 ≤ x ≤ 3.5
The range of the given absolute value function is: 0 ≤ y ≤ 10

Question 11.
a(x) = -5 | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 11

Question 12.
q(x) = – \(\frac{3}{2}\) | x |
Answer:
The given absolute value functions are:
f (x) = | x |
q (x) = –\(\frac{3}{2}\) | x |
Hence,
The representation of f (x) and q (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that q (x) shrinks by \(\frac{3}{2}\) units from f (x)
Now,
The domain of the given absolute value function is: -6.5 ≤ x ≤ 6.5
The range of the given absolute value function is: -10 ≤ y ≤ 0

In Exercises 13–16, graph the function. Compare the graph to the graph of f(x) = | x − 6 |.

Question 13.
h(x) = | x – 6 | + 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 13

Question 14.
n(x) = \(\frac{1}{2}\) | x – 6 |
Answer:
The given functions are:
f (x) = | x – 6 |
n (x) = \(\frac{1}{2}\) | x – 6 |
Hence,
The representation of f (x) and n (x) in the coordinate plane is:

Hence, from the above,
We can conclude that n (x) vertically stretches \(\frac{1}{2}\) units of f (x)

Question 15.
k(x) = -3 | x – 6 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 15

Question 16.
g(x) = | x – 1 |
Answer:
The given functions are:
f (x) = | x – 6 |
g (x) = | x – 1 |
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 4 units away from f (x) on the positive x-axis

In Exercises 17 and 18, graph the function. Compare the graph to the graph of f(x) = | x + 3 | − 2.

Question 17.
y(x) = | x + 4 | – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 17

Question 18.
b(x) = | x + 3 | + 3
Answer:
The given absolute value functions are:
f (x) = | x + 3 | – 2
b (x) = | x + 3 | + 3
Hence,
The representation of f (x) and b (x) in the coordinate plane is:

Hence, from the above,
We can conclude that f (x) and b (x) are parallel lines since the slopes are constant and there are different values of the y-intercepts

In Exercises 19–22, compare the graphs. Find the value of h, k, or a.

Question 19.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 179
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 19

Question 20.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 180
Answer:
The given graph is:
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 180
From the above graph,
The given functions are:
f (x) = | x |
t (x) = | x – h |
Where,
h is the horizontal translation
From the graph,
We can observe that the translation occurs at (1, 0) of f (x)
Hence, from the above,
We can conclude that the value of ‘h’ is: 1

Question 21.
Big Ideas Math Answer Key Algebra 1 Chapter 3 Graphing Linear Functions 181
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 21

Question 22.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 182
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 182
From the given graph,
The given graphs are:
f (x) = | x |
w (x) = a | x |
Where,
‘a’ is the vertical stretch
From the graph,
We can observe that the vertical stretch occurs at (0, 2)
Hence, from the above,
We can conclude that the value of a is: 2

In Exercises 23–26, write an equation that represents the given transformation(s) of the graph of g(x) = | x |.

Question 23.
vertical translation 7 units down
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 23

Question 24.
horizontal translation 10 units left
Answer:
The given statement is:
Horizontal translation 10 units left
Let the absolute value function be y = | x |
So,
“Horizontal translation” means the moving of value on the x-axis either to the left side or the right side i.e., either on the negative or positive x-axis
So,
The given statement in the absolute value function form is:
y = | x | + 10

Question 25.
vertical shrink by a factor of \(\frac{1}{4}\)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 25

Question 26.
vertical stretch by a factor of 3 and a reflection in the x-axis
Answer:
The given statement is:
Vertical stretch by a factor of 3 and a reflection in the x-axis
We know that,
The representation of the vertical stretch ‘a’ of the function is:
y = a.f (x)
The representation of the reflection of the function is:
y= f (-x)
Now,
Let the absolute value function be
y = x
Hence,
The representation of the given statement in the absolute value function form is:
Vertical stretch:
y = 3x
Reflection:
y = -3x

In Exercises 27–32, graph and compare the two functions.

Question 27.
f(x) = | x – 4 |; g(x) = | 3x – 4 |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 27

Question 28.
h(x) = | x + 5 |; t(x) = | 2x + 5 |
Answer:
The given functions are:
h (x) = | x + 5 |
t (x) = | 2x + 5 |
Hence,
The representation of h (x) and t (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that t (x) translates 6 units away from h (x) in the negative y-axis

Question 29.
p(x) = | x + 1 | – 2; q(x) = | \(\frac{1}{4}\)x + 1 | – 2
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 29

Question 30.
w(x) = | x – 3 | + 4; y(x) = | 5x – 3 | + 4
Answer:
The given absolute value functions are:
w (x) = | x – 3 | + 4
y (x) = | 5x – 3 | + 4
Hence,
The representation of w (x) and y (x) in the coordinate plane is:

Hence, from the above,
We can conclude that w (x) is 7 units away from y (x) in the positive y-axis

Question 31.
a(x) = | x + 2 | + 3; b(x) = | -4x + 2 | + 3
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 31

Question 32.
u(x) = | x – 1 | + 2; v(x) = | –\(\frac{1}{2}\)x – 1 | + 2
Answer:
The given absolute value functions are:
u (x) = | x – 1 | + 2
v (x) = | –\(\frac{1}{2}\)x – 1 | + 2
Hence,
The representation of u (x) and v (x) in the coordinate plane is:

Hence, from the above,
We can conclude that u (x) is 1 unit away from v (x) in the positive y-axis

In Exercises 33–40, describe the transformations from the graph of f(x) = | x | to the graph of the given function. Then graph the given function.

Question 33.
r(x) = | x + 2 | – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 33

Question 34.
c(x) = | x + 4 | + 4
Answer:
The given absolute value function is:
c (x) = | x + 4 | + 4
Hence,
The representation of c (x) in the coordinate plane is:

Hence, from the above,
We can conclude that c (x) translates 4 units away in the positive x-axis and 4 units vertically in the y-axis

Question 35.
d(x) = – | x – 3 | + 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 35

Question 36.
v(x) = -3| x + 1 | + 4
Answer:
The given absolute value function is:
v (x) = -3 | x + 1 | + 4
Hence,
The representation of v (x) in the coordinate plane is:

Hence, from the above,
We can conclude that v (x) translates 4 units vertically above and 2 units to the right side of the positive x-axis

Question 37.
m(x) = \(\frac{1}{2}\) | x + 4 | – 1
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 37

Question 38.
s(x) = | 2x – 2 | – 3
Answer:
The given absolute value function is:
s (x) = | 2x – 2 | – 3
Hence,
The representation of s (x) in the coordinate plane is:

Hence, from the above,
We can conclude that s (x) translates 4 units away to the right side of the positive x-axis

Question 39.
j(x) = | -x + 1 | – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 39

Question 40.
n(x) = | –\(\frac{1}{3}\)x + 1 | + 2
Answer:
The given absolute value function is:
n (x) = | –\(\frac{1}{3}\)x + 1 | + 2
Hence,
The representation of n (x) in the coordinate plane is:

Hence, from the above,
We can conclude that n (x) translates 4 units away from the origin

Question 41.
MODELING WITH MATHEMATICS
The number of pairs of shoes sold s (in thousands) increases and then decreases as described by the function s(t) = -2 | t – 15 | + 50, where t is the time (in weeks).
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 183
a. Graph the function.
b. What is the greatest number of pairs of shoes sold in 1 week?
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 41

Question 42.
MODELING WITH MATHEMATICS
On the pool table shown, you bank the five ball off the side represented by the x-axis. The path of the ball is described by the function p(x) = \(\frac{4}{3}\) | x – \(\frac{5}{4}\) |.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 184
a. At what point does the five-ball bank off the side?
Answer:
It is given that on the pool table shown, you bank the five ball off the side represented by the x-axis. The path of the ball is described by the function
p(x) = \(\frac{4}{3}\) | x – \(\frac{5}{4}\) |
Now,
To find the point where the five-ball bank off the side, draw the plot of the given absolute value function
So,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
The point where the five-ball offside is: (1, 0)

b. Do you make the shot? Explain your reasoning.
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 184
From the graph,
We can observe that the point is accurately going through the hole i.e., at (5, 5)
Hence, from the above,
We can conclude that you make the shot

Question 43.
USING TRANSFORMATIONS
The points A (-\(\frac{1}{2}\), 3) , B(1, 0), and C(-4, -2) lie on the graph of the absolute value function f. Find the coordinates of the points corresponding to A, B, and C on the graph of each function.
a. g(x) = f(x) – 5
b. h(x) = f(x – 3)
c. j(x) = -f(x)
d. k(x) = 4f(x)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 43.1
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 43.2
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 43.3

Question 44.
USING STRUCTURE
Explain how the graph of each function compares to the graph of y = | x | for positive and negative values of k, h, and a.
a.y = | x | + k
b. y = | x – h |
c. y = a| x |
d. y = | ax |
Answer:
The given absolute value functions are:
a.y = | x | + k
b. y = | x – h |
c. y = a| x |
d. y = | ax |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
It is given that
The parent function is:
y = | x |
Now,
a.
The given absolute value function is:
y = | x | + k
Hence,
The given absolute value function translates from the parent function k units away in the positive y-axis vertically
b.
The given absolute value function is:
y = | x – h |
Hence,
The given absolute value function translates from the parent function h units away in the positive x-axis horizontally
c.
The given absolute value function is:
y = a| x |
Hence,
The given absolute value function reflects from the parent function ‘a’ units away
d.
The given absolute value function is:
y = | ax |
Hence,
The given absolute value function dilates from the parent function with the value of ‘a’ units

ERROR ANALYSIS
In Exercises 45 and 46, describe and correct the error in graphing the function.

Question 45.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 185
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 45

Question 46.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 186
Answer:
The given absolute value function is:
y = -3 | x|
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = -3 (x) or y = -3 (-x)
So,
y = -3x or y = 3x
Hence,
The representation of the given absolute value function in the coordinate plane is:

Hence, from the above,
We can conclude that the given absolute value function in downward direction

MATHEMATICAL CONNECTIONS
In Exercises 47 and 48, write an absolute value function whose graph forms a square with the given graph.

Question 47.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 187
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 47

Question 48.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 188
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 188
From the graph,
The given absolute value function is:
y = | x – 3 | + 1
From the given absolute value function,
We can say that the x-axis translates 3 units away in the positive x-axis and the y-axis is a vertical stretch with the value of k as 1

Question 49.
WRITING
Compare the graphs of p(x) = | x – 6 | and q(x) = | x | – 6.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 49

Question 50.
HOW DO YOU SEE IT? The object of a computer game is to break bricks by deflecting a ball toward them using a paddle. The graph shows the current path of the ball and the location of the last brick.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 189
a. You can move the paddle up, down, left, and right. At what coordinates should you place the paddle to break the last brick? Assume the ball deflects at a right angle.
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 189
It is given that the paddle moves at the right angle using a ball
So,
From the graph,
The coordinates that are at a right angle to break the brick using a paddle is:
(3, 8) and (14, 8)

b. You move the paddle to the coordinates in part (a), and the ball is deflected. How can you write an absolute value function that describes the path of the ball?
Answer:
From part (a),
The coordinates that are used to break the brick using a paddle is:
(3, 8) and (14, 8)
We know that,
For absolute value function,
The parent function will always be:
f (x) = | x |
From part (a),
The y-axis is constant and the x-axis is translating 3 units away and 14 units away
We know that,
The value of the absolute value function with the translation of x-value and  y-value as k-value
So,
f (x) = | x – h | + k
Hence,
The absolute value function that describes the path of the ball is:
f (x) = | x – 3 | + 8
f (x) = | x – 14 | + 8

In Exercises 51–54, graph the function. Then rewrite the absolute value function as two linear functions, one that has the domain x < 0 and one that has the domain x ≥ 0.

Question 51.
y = | x |
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 51

Question 52.
y = | x | – 3
Answer:
The given absolute value function is:
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = x – 3 or y = -x – 3
Hence,
The representation of the given absolute value function in the coordinate plane is:

Question 53.
y = – | x | + 9
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 53

Question 54.
y = – 4 | x |
Answer:
The given absolute value function is:
y = -4 | x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = -4 (x) or y = -4 (-x)
y = -4x or y = 4x
Hence,
The representation of the given absolute value function in the coordinate plane is:

In Exercises 55–58, graph and compare the two functions.

Question 55.
f(x) = | x – 1 | + 2; g(x) = 4 | x – 1 | + 8
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 55

Question 56.
s(x) = | 2x – 5 | – 6; t(x) = \(\frac{1}{2}\) | 2x – 5 | – 3
Answer:
The given functions are:
s (x) = | 2x – 5 | – 6
t (x) = \(\frac{1}{2}\) | 2x – 5 | – 3
Hence,
The representation of s (x) and t (x) in the coordinate plane is:

Hence, from the above,
We can conclude that s (x) translates 60 units away from t (x)

Question 57.
v(x) = -2 | 3x + 1 | + 4; w(x) = 3 | 3x + 1 | – 6
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 57

Question 58.
c(x) = 4 | x + 3 | – 1; d(x) = –\(\frac{4}{3}\) | x + 3 | + \(\frac{1}{3}\)
Answer:
The given functions are:
c (x) = 4 | x + 3 | – 1
d (x) =-\(\frac{4}{3}\) | x + 3 | + \(\frac{1}{3}\)
Hence,
The representation of c (x) and d (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that d(x) and c (x) are opposite to each other

Question 59.
REASONING
Describe the transformations from the graph of g(x) = -2 | x + 1 | + 4 to the graph of h(x) =| x |. Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 59

Question 60.
THOUGHT-PROVOKING
Graph an absolute value function f that represents the route a wide receiver runs in a football game. Let the x-axis represent the distance (in yards) across the field horizontally. Let the y-axis represent the distance (in yards) down the field. Be sure to limit the domain so the route is realistic.
Answer:
It is given that
Let the x-axis represent the distance (in yards) across the field horizontally
Let the y-axis represent the distance (in yards) down the field.
Let,
The distance across the field horizontally is: 50 yards
The distance down the field is: 25 yards
Remember we can take any value in the x-axis and the y-axis
So,
(x, y) = (50, 25)
So,
The absolute value function f that represents the route a wide receiver runs in a football game is:
f (x) = | x – 50 | + 25
Hence,
The representation of the above absolute value function in the coordinate plane is:

Hence, from the above,
The domain of the given absolute value function is: x ≥ 0
The range of the given absolute value function is: y > 0

Question 61.
SOLVING BY GRAPHING
Graph y = 2 | x + 2 | – 6 and y = -2 in the same coordinate plane. Use the graph to solve the equation 2 | x + 2 | – 6 = -2. Check your solutions.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 61

Question 62.
MAKING AN ARGUMENT
Let p be a positive constant. Your friend says that because the graph of y =| x | + p is a positive vertical translation of the graph of y = | x |, the graph of y = | x + p | is a positive horizontal translation of the graph of y = | x |. Is your friend correct? Explain.
Answer:
It is given that your friend says that because the graph of y =| x | + p is a positive vertical translation of the graph of y = | x |, the graph of y = | x + p | is a positive horizontal translation of the graph of y = | x |.
Now,
The graph of
y = | x | + p is a positive vertical translation because the y-intercept in the given equation is positive
Now,
The graph of
y = | x + p | will be a horizontal translation with negative translation value
But, according to your friend,
y = | x + p | is a positive horizontal translation
Hence, from the above,
We can conclude that your friend is not correct

Question 63.
ABSTRACT REASONING
Write the vertex of the absolute value function f(x) = | ax – h | + k in terms of a, h, and k.
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 63

Maintaining Mathematical Proficiency

Solve the inequality. (Section 2.4)

Question 64.
8a – 7 ≤ 2(3a – 1)
Answer:
The given inequality is:
8a – 7 ≤ 2 (3a – 1)
So,
8a – 7 ≤ 2 (3a) – 2 (1)
8a – 7 ≤ 6a – 2
8a – 6a ≤ -2 + 7
2a ≤ 5
a ≤ \(\frac{5}{2}\)
Hence, from the above,
We can conclude that the solution to the given inequality is:
a ≤ \(\frac{5}{2}\)

Question 65.
-3(2p + 4) > -6p – 5
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 65

Question 66.
4(3h + 1.5) ≥ 6(2h – 2)
Answer:
The given inequality is:
4 (3h + 1.5) ≥ 6 (2h – 2)
So,
4 (3h) + 4 (1.5) ≥ 6 (2h) – 6 (2)
12h + 6 ≥ 12h – 12
12h – 12h + 6 ≥ -12
6 ≥ -12
Since the above inequality is true,
The given inequality has infinite solutions

Question 67.
-4(x + 6) < 2(2x – 9)
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 67

Find the slope of the line. (Section 3.5)

Question 68.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 190
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 190
From the above graph,
The points are: (0, 3), (-2, -2)
We know that,
The slope of the give line when (x1, y1), (x2, y2) are given is:
m= \(\frac{y2 – y1}{x2 – x1}\)
So,
(x1, y1) = (0, 3) and (x2, y2) = (-2, -2)
So,
m = \(\frac{-2 – 3}{-2 – 0}\)
m = \(\frac{-5}{-2}\)
m = \(\frac{5}{2}\)
Hence, from the above,
We can conclude that the slope of the given line is: \(\frac{5}{2}\)

Question 69
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 191
Answer:
Big Ideas Math Algebra 1 Answers Chapter 3 Graphing Linear Functions 3.7 Question 69

Question 70.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 192
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 192
From the above graph,
The given points are: (-3, 1), (1, -4)
We know that,
The slope of the give line when (x1, y1), (x2, y2) are given is:
m= \(\frac{y2 – y1}{x2 – x1}\)
So,
(x1, y1) = (-3, 1) and (x2, y2) = (1, -4)
So,
m = \(\frac{-4 – 1}{1 – [-3]}\)
m = \(\frac{-5}{1 + 3}\)
m = \(\frac{-5}{4}\)
Hence, from the above,
We can conclude that the slope of the given line is: –\(\frac{5}{4}\)

Graphing Linear Functions Performance Task: The Cost of a T-Shirt

3.4–3.7 What Did You Learn?

Core Vocabulary
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 193

Core Concepts
Section 3.4
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 194

Section 3.5
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 195

Section 3.6
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 196

Section 3.7
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 197

Mathematical Practices

Question 1.
Explain how you determined what units of measure to use for the horizontal and vertical axes in Exercise 37 on page 142.
Answer:
In Exercise 37 on page 142,
The given function is:
d (t)= (1/2)t + 6
We know that,
The horizontal axis represents the independent variable
The vertical axis represents the dependent variable
So,
The horizontal axis represents the time
The vertical axis represents the depth

Question 2.
Explain your plan for solving Exercise 48 on page 153.
Answer:
In Exercise 48 on page 153,
The profits of a school obtained by a school before playoffs and during the playoffs
So,
First, plot the 2 equations in a coordinate plane and compare the functions in the coordinate plane

Performance Task 

The Cost of a T-Shirt

You receive bids for making T-shirts for your class fundraiser from four companies. To present the pricing information, one company uses a table, one company uses a written description, one company uses an equation, and one company uses a graph. How will you compare the different representations and make the final choice? To explore the answers to this question and more, go to Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 198
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 199

Graphing Linear Functions Chapter Review

3.1 Functions (pp. 103 – 110)

Determine whether the relation is a function. Explain.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 200
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 200
We know that,
Every input has exactly one output.
x represents the input
y represents the output
Hence, from the table,
We can conclude that the given table is a function

Determine whether the relation is a function. Explain. 

Question 1.
(0, 1), (5, 6), (7, 9)
Answer:
The given ordered pairs are:
(0, 1), (5, 6), (7, 9)
We know that,
Every input has exactly one output
x represents the input
y represents the output
Hence, from the above,
We can conclude that the given relation is a function

Question 2.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 201
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 201
We know that,
Every input has exactly one output
x represents the input
y represents the output
Now,
From the graph,
The input ‘2’ has 2 outputs i.e., (2, 0), and (2, 2)
Hence, from the above,
We can conclude that the given graph is not a function

Question 3.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 202
Answer:
The given relation is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 202
We know that,
Every input has exactly one output
x represents the input
y represents the output
Hence, from the above,
We can conclude that the given relation is a function

Question 4.
The function y = 10x + 100 represents the amount y (in dollars) of money in your bank account after you babysit for x hours.
a. Identify the independent and dependent variables.
Answer:
The given function is:
y = 10x + 100
Where,
y represents the amount in dollars
x represents the number of hours
Now,
From the given function,
WE can say that
The independent variable of the given function is: x
The dependent variable of the given function is: y

b. You babysit for 4 hours. Find the domain and range of the function.
Answer:
The given function is:
y = 10x + 100
Where,
y represents the amount in dollars
x represents the number of hours
It is given that you babysit for 4 hours
So,
The given value of x is: 4
So,
The maximum value of y is:
y = 10 (4) + 100
y = 40 + 100
y = $140
We know that,
The amount and the time must not be the negative values
Hence,
The domain of the given function is: 0 ≤ x ≤ 4
The range of the give function is: 0 ≤ y ≤ 140

3.2 Linear Functions (pp. 111–120)

Does the table or equation represent a linear or nonlinear function? Explain.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 203
Answer:
As x increases by 4, y increases by different amounts. The rate of change is not constant.
So,
The function is nonlinear.

b. y = 3x – 4
Answer:
The equation is in the form y = mx + b.
So,
The equation represents a linear function.

Does the table or graph represent a linear or nonlinear function? Explain. 

Question 5.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 204
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 204
From the given table,
The difference between all the values of x is: 5
The difference between all the values of y is: -3
Since there is a constant difference present between the values of x and the values of y,
The given table is a linear function

Question 6.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 205
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 205
We know that,
The representation of a linear function in the graph is a “Straight line”
But,
We can say that the given graph is not a straight line from the graph
Hence, from the above,
We can conclude that the given graph is a non-linear function

Question 7.
The function y = 60 – 8x represents the amount y (in dollars) of money you have after buying x movie tickets.
(a) Find the domain of the function. Is the domain discrete or continuous? Explain.
Answer:
The given function is:
y = 60 – 8x
Where,
y represents the amount in dollars
x represents the number of movie tickets
We know that,
The domain is the set of all the values of x so that the given function will be satisfied
Since x represents the number of movie tickets, the value of x can’t be negative and the number of tickets will be infinity
Hence,
The domain of the given function is: 0 ≤ x ≤ ∞
Hence, from the above domain,
We can conclude that the domain is continuous

(b) Graph the function using its domain. Evaluate the function when x = -3, 0 and 5.
Answer:
The given function is:
y = 60 – 8x
Now,
The value of the given function when x = -3 is:
y = 60 – 8 (-3)
y = 60 + 24
y = 84
The value of the given function when x = 0 is:
y = 60 – 8 (0)
y = 60
The value of the given function when x = 5 is:
y = 60 – 8(5)
y = 60 – 40
y = 20
Now,
The representation of the given function along with its domain is:

Question 8.
f(x) = x + 8
Answer:
The given equation is:
f (x) = x + 8
We know that,
The standard representation of the function f (x) is: y
The standard representation of the linear equation is:
y = mx + c
Now,
y = x + 8
By comparing the given equation and the standard representation of the linear equation,
We can conclude that the given equation is a linear equation

Find the value of x, so that the function has the given value.

Question 9.
g(x) = 4 – 3x

Question 10.
k(x) = 7x; k(x) = 49
Answer:
The given function is:
k (x) = 7x with k (x) = 49
So,
49 = 7x
x = 49 / 7
x = 7
Hence, from the above,
We can conclude that the value of the given function is: 7

Question 11.
r(x) = -5x – 1; r(x) = 19
Answer:
The given function is:
r (x) = 5x – 1 with r (x) = 19
So,
19 = 5x – 1
5x = 19 + 1
5x = 20
x = 20 / 5
x = 4
Hence, from the above,
We can conclude that the value of the given function is: 4

Graph the linear function.

Question 12.
g(x) = -2x – 3
Answer:
The given function is:
g (x) = -2x – 3
Hence,
The representation of g (x) in the coordinate plane is:

Question 13.
h(x) = \(\frac{2}{3}\)x + 4
Answer:
The given function is:
h (x) = \(\frac{2}{3}\)x + 4
Hence,
The representation of h (x) in the coordinate plane is:

Question 14.
8x – 4y = 16
Answer:
The given function is:
8x – 4y = 16
4y = 8x – 16
y = \(\frac{8x  16}{4}\)
y = \(\frac{8x}{4}\) – \(\frac{16}{4}\)
y =2x – 4
Hence,
The representation of the given function in the coordinate plane is:

Question 15.
-12x – 3y = 36
Answer:
The given function is:
-12x – 3y = 36
Hence,
The representation of the given function in the coordinate plane is:

Question 16.
y = -5
Answer:
The given function is:
y = -5
Hence,
The representation of the given function in the coordinate plane is:

Question 17.
x = 6
Answer:
The given function is:
x = 6
Hence,
The representation of the given function in the coordinate plane is:

The points represented by the table lie on a line. Find the slope of the line. 

Question 18.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 206
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 206
We know that,
If the input and output values are given in the form of the table,
We can take any 2 pairs of the input and output values to find the slope
So,
The representation of ordered pairs to find the slope is:
(x1, y1) = (6, 9),
(x2, y2) = (11, 15)
Now,
We know that,
The slope of the line = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{15 – 9}{11 – 6}\)
= \(\frac{6}{5}\)
Hence, from the above,
We can conclude that the slope of the given line is: \(\frac{6}{5}\)

Question 19.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 207
Answer:

The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 207
We know that,
If the input and output values are given in the form of the table,
We can take any 2 pairs of the input and output values to find the slope
So,
The representation of ordered pairs to find the slope is:
(x1, y1) = (3, -5),
(x2, y2) = (3, -2)
Now,
We know that,
The slope of the line = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-2 – [-5]}{3 – 3}\)
= \(\frac{5 – 2}{0}\)
= Undefined or ∞
Hence, from the above,
We can conclude that the slope of the given line is: Undefined or ∞

Question 20.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 208
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 208
We know that,
If the input and output values are given in the form of the table,
We can take any 2 pairs of the input and output values to find the slope
So,
The representation of ordered pairs to find the slope is:
(x1, y1) = (-4, -1),
(x2, y2) = (-3, -1)
Now,
We know that,
The slope of the line = \(\frac{y2 – y1}{x2 – x1}\)
= \(\frac{-1 – [-1]}{-3 – [-4]}\)
= \(\frac{-1 + 1}{-3 + 4}\)
= 0
Hence, from the above,
We can conclude that the slope of the given line is: 0

Graph the linear equation. Identify the x-intercept. 

Question 21.
y = 2x + 4
Answer:
The given linear equation is:
y = 2x + 4
To find the x-intercept, put y = 0
So,
2x + 4 = 0
2x = -4
x = -4 / 2
x = -2
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 22.
-5x + y = -10
Answer:
The given linear equation is:
-5x + y = -10
To find the x-intercept, put y = 0
-5x + 0 = -10
-5x = -10
5x = 10
x = 10 / 5
x = 2
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 23.
x + 3y = 9
Answer:
The given linear equation is:
x + 3y = 9
To find the x-intercept, put y = 0
So,
x + 0 = 9
x = 9
Hence,
The representation of the given linear equation in the coordinate plane is:

Question 24.
A linear function h models a relationship in which the dependent variable decreases 2 units for every 3 units the independent variable increases. Graph h when h(0) = 2. Identify the slope, y-intercept, and x-intercept of the graph. Let f(x) = 3x + 4. Graph f and h. Describe the transformation from the graph of f to the graph of h.
Answer:
It is given that a linear function h models a relationship in which the dependent variable decreases 2 units for every 3 units the independent variable increases.
We know that,
x represents the independent variable
y represents the dependent variable
So,
The given x value is: 3
The given y value is: -2
So,
The rate of change (or) slope (m) = \(\frac{y}{x}\)
= \(\frac{-2}{3}\)
= –\(\frac{2}{3}\)
It is also given that h (0) = 2
That means the y-intercept of h (x) is 2 when x is 0
We know that,
The representation of the standad form of the linear equation is:
y = mx + c
So,
h (x) =-\(\frac{2}{3}\)x + 2
The other given function is:
f (x) = 3x + 4
Hence,
The representation of f (x0 and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) is 1.5 units away from f (x) on the x-axis

Let f(x) = 3x + 4.Graph f and h. Describe the transformation from the graph of f to the graph of h.

Question 25.
h(x) = f(x + 3)
ANswer:
The given functions are:
f (x) = 3x + 4
h (x) = f (x + 3)
So,
h (x) = 3 (x + 3) + 4
h (x) = 3 (x) + 3 (3) + 4
h (x) = 3x + 9 + 4
h (x) = 3x + 13
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are parallel lines

Question 26.
h(x) = f(x) + 1
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = f (x) + 1
So,
h (x) = 3x + 4 + 1
h (x) = 3x + 5
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that f (x) and h (x) are the paralel lines

Question 27.
h(x) = f(-x)
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = f (-x)
So,
h (x) = 3 (-x) + 4
h (x) = -3x + 4
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) and f (x) are inversions of each other

Question 28.
h(x) = -f(x)
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = -f (x)
So,
h (x) = – (3x + 4)
h (x) = -3x – 4
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) and f (x) are reflections to each other

Question 29.
h(x) = 3f(x)
Answer:
The given function srae:
f (x) = 3x + 4
h (x) = 3 f (x)
So,
h (x) = 3 (3x + 4)
Hence,
The representation of f (x) and  h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) is a vertical stretch of f (x)

Question 30.
h(x) = f(6x)
Answer:
The given functions are:
f (x) = 3x + 4
h (x) = f (6x)
So,
h (x) = 3 (6x) + 4
h (x) = 18x + 4
Hence,
The representation of f (x) and h (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that h (x) translates 2 units away from f (x) on the x-axis

Question 31.
Graph f(x) = x and g(x) = 5x + 1. Describe the transformations from the graph of f to the graph of g.
Answer:
The given functions are:
f (x) = x
g (x) = 5x + 1
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates away 8 units away from f (x)  on the x-axis

Graph the function. Compare the graph to the graph of f(x) = | x |. Describe the domain and range.

Question 32.
m(x) = | x | + 6
Answer:
The given absolute value functions are:
f (x) = | x |
m (x) = | x | + 6
Hence,
The representation of f (x) and m (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that m (x) translates 6 units up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of m (x) is: -4 ≤ x ≤ 4
The range of m (x) is: 0 ≤y ≤ 10
The range of f (x) is: 0 ≤ y ≤10
The range of m (x) is: 6 ≤y ≤10

Question 33.
p(x) = | x – 4 |
Answer:
The given absolute value functions are:
f (x) = | x |
p (x) = | x – 4 |
Hence,
The representation of f (x) and p (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that p (x) translates 1 unit up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of p (x) is: -6 ≤ x ≤ 10
The range of f (x) is: 0 ≤ y ≤10
The range of p (x) is: 6 ≤y ≤10

Question 34.
q(x) = 4 | x |
Answer:
The given absolute value functions are:
f (x) = | x |
q (x) = 4 | x |
Hence,
The representation of f (x) and q (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that q (x) translates 7 units up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of q (x) is: -2.5 ≤ x ≤ 2.5
The range of q (x) is: 0 ≤y ≤ 10
The range of f (x) is: 0 ≤ y ≤10

Question 35.
r(x) = –\(-\frac{1}{4}\)| x |
Answer:
The given absolute value functions are:
f (x) = | x |
r (x) = –\(-\frac{1}{4}\) | x |
So,
r (x) = \(\frac{1}{4}\) | x |
Hence,
The representation of f (x) and r (x) in the same coordinate plane is:

Hene, from the above,
We can conclude that r (x) translates  units up from the positive x-axis
Hence,
The domain of f (x) is: -10 ≤ x ≤ 10
The domain of r (x) is: -10 ≤ x ≤ 10
The range of r (x) is: 0 ≤y ≤ 2
The range of f (x) is: 0 ≤ y ≤10

Question 36.
Graph f(x) = | x – 2 | + 4 and g(x) = | 3x – 2 | + 4. Compare the graph of g to the graph of f.
Answer:
The given functions are:
f (x) = | x – 2 | + 4
g (x) = | 3x – 2 | + 4
Hence,
The representation of f (x) and g (x) in the same coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 5 units away from f (x) on the x-axis

Question 37.
Let g(x) = \(\frac{1}{3}\) | x – 1 | – 2.
(a) Describe the transformations from the graph of f(x) = | x | to the graph of g.
Answer:
The given functions are:
f (x) = | x |
g (x) = \(\frac{1}{3}\) | x – 1 | – 2
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that g (x) translates 2 units away from f (x) on the y-axis
(b) Graph g.
Answer:
It is given that
g (x) = \(\frac{1}{3}\) | x – 1 | – 2
Hence,
The representation of g (x) in the coordinate plane is:

Graphing Linear Functions Chapter Test

Determine whether the relation is a function. If the relation is a function, determine whether the function is linear or nonlinear. Explain. 

Question 1.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 209
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 209
We know that,
Every input has exactly one output
So,
The given table is a function
Now,
The constant between all the values of x is: 1
There is no constant difference between all the values of y
Since there is no constant rate of change,
We can conclude that the obtained function is a non-linear function

Question 2.
y = -2x + 3
Answer:
The given equation is:
y = -2x + 3
We know that,
The standard form of the linear equation is:
y = mx + c
For a relation o be a function, every input has exactly only one output
Hence, from the above,
We can conclude that the given equation is a function and it is a linear function

Question 3.
x = -2
Answer:
The given equation is:
x = -2
We know that,
A relation is said to be a function if every input has exactly one output
So,
For x = -2
The input is: 2
The output is: 0
So,
The given equation is a function
For a given function to be a linear function, it will be in the form
y = mx + c
Hence, from the above,
We can conclude that the given function is a non-linear function

Graph the equation and identify the intercept(s). If the equation is linear, find the slope of the line. 

Question 4.
2x – 3y = 6
Answer:
The given equation is:
2x – 3y = 6
3y = 2x – 6
y = \(\frac{2x – 6}{3}\)
y = \(\frac{2x}{3}\) – \(\frac{6}{3}\)
y = \(\frac{2}{3}\)x – 2
Hence,
The above equation is in the form of
y = mx + c
So,
The given equation is a linear equation
So,
m = \(\frac{2}{3}\)
The y-intercept is: -2
To find the x-intercept, put y = 0
\(\frac{2}{3}\)x – 2 = 0
x = 3
Hence, from the above,
We can conclude that
Slope (m) = \(\frac{2}{3}\)
The x-intercept is: 3
The y-intercept is: -2

Question 5.
y = 4.5
Answer:
The given equation is:
y = 4.5
The given equation is not in the form of
y = mx + c
Hence,
The given equation is non-linear

Question 6.
y = | x – 1 | – 2
Answer:
The given equation is:
y = | x – 1 | – 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
y = x – 1 – 2 or y = -(x – 1) – 2
y = x – 3 or y = -x + 1 – 2
y = x – 3 or y = -x – 1
The above 2 equations are in the form of
y = mx + c
Hence,
The slope is: 1 or -1
The y-intercept is: -3 or -1
To find the x-intercept, put y =0
So,
0 = x – 3 or 0 = -x – 1
x = 3 or x = -1
Hence, from the above,
We can conclude that
The slope is: 1 or -1
The x-intercept is: 3 or -1
The y-intercept is: -3 or -1

Find the domain and range of the function represented by the graph. Determine whether the domain is discrete or continuous. Explain. 

Question 7.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 210
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 210
From the graph,
We can observe that the graph is not a straight line
So,
We can say that the given graph is a non-linear function
Hence, from the above,
We can conclude that the domain is continuous by observing the graph

Question 8.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 211
Answer:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 211
From the graph,
We can observe that the points form a straight line
So,
We can say that the given graph is a linear function
Hence, from the above,
We can conclude that the domain is continuous since all the points are connected

Graph f and g. Describe the transformations from the graph of f to the graph of g. 

Question 9.
f(x) = x; g(x) = -x + 3
Answer:
The given functions are:
f (x) = x
g (x) = -x + 3
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that f (x) and g (x) are perpendicular lines with a slope of -1

Question 10.
f(x) = | x | ; g(x) = | 2x + 4 |
Answer:
The given functions are:
f (x) = | x |
g (x) = | 2x + 4 |
Hence,
The representation of f (x) and g (x) in the coordinate plane is:

Hence, from the above,
We can conclude that g (x) is 2 units away from f (x) on the x-axis

Question 11.
Function A represents the amount of money in a jar based on the number of quarters in the jar. Function B represents your distance from home over time. Compare the domains.
Answer:
It is given that
Function A represents the amount of money in a jar based on the number of quarters in the jar
Function B represents your distance from home over time
We know that,
The distance should be greater than or equal to 0
The amount is greater than or equal to a quarter of the amount of the money
Hence,
The domain of function A is: \(\frac{1}{4}\) ≤ x ≤ ∞
The domain of function B is: 0 ≤ x ≤ ∞

Question 12.
A mountain climber is scaling a 500-foot cliff. The graph shows the elevation of the climber over time.
a. Find and interpret the slope and the y-intercept of the graph.
b. Explain two ways to find f(3). Then find f(3) and interpret its meaning.
c. How long does it take the climber to reach the top of the cliff? Justify your answer.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 212
Answer:
a.
It is given that a mountain-climber is scaling a 500-foot cliff
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 212
From the graph,
The equation that shows the elevation of the climber over tie is:
f (x) = 125x + 50
We know that,
The standard representation of the output for the function output is: y
So,
y = 125x + 50
Compare the above equation with
y = mx + c
So,
m = 125 and the y-intercept is: 50
Hence, from the above,
We can conclude that
The slope of the given equation is: 125
The y-intercept of the given equation is: 50

b.
From part (a),
The given equation is:
f (x) = 125x + 50
So,
f (3) = 125 (3) + 50
f (3) = 375 + 50
f (3) = 425
Hence, from the above,
We can conclude that for 3 hours, the climber climbs 425 feet

c.
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 212
From the given graph,
The top of the hill is the maximum height i.e., the highest value on the y-axis
Hence, from the above,
We can conclude that it takes 4 hours to reach the top of the cliff

Question 13.
Without graphing, compare the slopes and the intercepts of the graphs of the functions f(x) = x + 1 and g(x) = f(2x).
Answer:
The given functons are:
f (x) = x + 1
g (x) = f (2x)
So,
g (x) = 2x + 1
Now,
Compare f (x) and g (x) with the standard linear equation
y = mx + c
So,
For f (x),
m = 1 and c = 1
Where,
c is the y-intercept
For g (x),
m = 2 and c = 1
Where,
c is the y-intercept

Question 14.
A rock band releases a new single. Weekly sales s (in thousands of dollars) increase and then decrease as described by the function s(t) = -2 | t – 20 | + 40, where t is the time (in weeks).
a. Identify the independent and dependent variables.
b. Graph s. Describe the transformations from the graph of f(x) = | x | to the graph of s.
Answer:
a.
It is given that a rock band releases a new single and weekly sales s (in thousands of dollars) increase and then decrease as described by the function
s(t) = -2 | t – 20 | + 40
where,
t is the time (in weeks)
Now,
The independent variable of the given function  is: t
The dependent variable of the given function is: s (t)

b.
The given absolute value functions are:
f (x) = | x |
s (t) = -2 | t – 20 | + 40
Hence,
The representation of f (x) and s (t) in the coordinate plane is:

Hence, from the above,
We can conclude that s (t) translates 5 units away from f (x) on the y-axis

Graphing Linear Functions Cumulative Assessment

Question 1.
You claim you can create a table of values that represents a linear function. Your friend claims he can create a table of values that represents a nonlinear function. Using the given numbers, what values can you use for x (the input) and y (the output) to support your claim? What values can your friend use?
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 213
Answer:
It is given that you claim you can create a table of values that represents a linear function. Your friend claims he can create a table of values that represents a nonlinear function.
Hence,
The values you and your friend use are:

Question 2.
A car rental company charges an initial fee of $42 and a daily fee of $12.
a. Use the numbers and symbols to write a function that represents this situation.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 214
b. The bill is $138. How many days did you rent the car?
Answer:
a.
It is given that a car rental company charges an initial fee of $42 and a daily fee of $12
So,
The function that represents the situation is:
f (x) = (42 + 12) × x
f (x) = 52x
Where,
x is the number of days that you rent the car

b.
It is given that the bill is: $138
So,
138 = 52x
x = 138 ÷ 52
x = 3
Hence, from the above,
We can conclude that you can rent the car for 3 days

Question 3.
Fill in values for a and b so that each statement is true for the inequality ax − b> 0.
a. When a = _____ and b = _____, x > \(\frac{b}{a}\).
b. When a = _____ and b = _____, x < \(\frac{b}{a}\).
Answer:
a.
The given inequality is:
ax – b > 0
ax > b
x > b / a
So,
The value of a is less than b and the value of b is greater than a
b.
The value of a is greater than b and the value of b is less than a

Question 4.
Fill in the inequality with <, ≤, >, or ≥ so that the solution of the inequality is represented by the graph.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 215
Answer:
The given number line is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 215
From the given number line,
The given inequality is:
-3 (x + 7) _____ -24
-3 (x) – 3 (7) _____ -24
-3x – 21 _____ -24
-3x ____ -24 + 21
-3x ____-3
3x ____ 3
x ____ 3 / 3
x ____ 1
From the number line
The marked line represented from 1 including 1 and continued till the right end of the number line
Hence,
x ≥ 1
Hence, from the above,
We can conclude that the symbol used for the given inequality is: ≥

Question 5.
Use the numbers to fill in the coefficients of ax + by = 40 so that when you graph the function, the x-intercept is -10 and the y-intercept is 8.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 216
Answer:
The given function is:
ax + by = 40
It is given that the x-intercept and the y-intercept is: -10 and 8
Now,
To find the x-intercept, put y = 0
So,
ax = 40
a (-10) = 40
a = -40 / 10
a = -4
To find the y-intercept, put x = 0
So,
by = 40
b (8) = 40
b = 40 / 8
b = 5
Hence, from the above,
We can conclude that the values of a and b are: -4 and 5

Question 6.
Solve each equation. Then classify each equation based on the solution. Explain your reasoning.
a. 2x – 9 = 5x – 33
Answer:
The given expression is:
2x – 9 = 5x – 33
2x – 5x = -33 + 9
-3x = -24
x = 24 / 3
x = 8
Hence, from the above,
We can conclude that the given expression has 1 solution

b. 5x – 6 = 10x + 10
Answer:
The given expression is:
5x – 6 = 10x + 10
5x – 10x = 10 + 6
-5x = 16
x = -16 / 5
Hence, from the above,
We can conclude that the given expression has only 1 solution

c. 2(8x – 3) = 4(4x + 7)
Answer:
The given expression is:
2 (8x – 3 ) = 4 (4x + 7)
2 (8x) – 2 (3) = 4 (4x) + 4(7)
16x – 6 = 16x + 28
Hence, from the above,
We can conclude that the given expression has no solution

d. -7x + 5 = 2(x – 10.1)
Answer:
The given expression is:
-7x + 5 = 2 (x – 10.1)
-7x + 5 = 2x – 20.2
-7x – 2x = -20.2 – 5
-9x = -25.2
x = 25.9 / 9
Hence, from the above,
We can conclude that the given expression has only 1 solution

e. 6(2x + 4) = 4(4x + 10)
Answer:
The given expression is:
6 (2x + 4) = 4 (4x + 10)
12x + 24 = 16x + 40
12x – 16x = 40 – 24
-4x = 16
x = -16 / 4
x = -4
Hence, from the above,
We can conclude that the given expression has only 1 solution

f. 8(3x + 4) = 2(12x + 16)
Answer:
The given expression is:
8 (3x + 4) = 2 (12x + 16)
24x + 32 = 24x + 32
Hence, from the above,
We can conclude that the given expression has no solution

Question 7.
The table shows the cost of bologna at a deli. Plot the points represented by the table in a coordinate plane. Decide whether you should connect the points with a line. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 217
Answer:
The given table is:
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 217
Hence,
The representation of the points in the coordinate plane is:

Hence, from the above,
We can say that we can connect the points in the graph

Question 8.
The graph of g is a horizontal translation right, then a vertical stretch, then a vertical translation down of the graph of f(x) = x. Use the numbers and symbols to create g.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 218
Answer:
The given function is:
f (x) = x
We know that,
f (x) can be re-written as f
g (x) can b ere-written as g
Now,
Let the operations performed on f (x) can be expressed in terms of g (x)
So,
Horizontal translation right:
g (x) = f (x) – 3 or g (x) = f (x) – 1 or g (x)  f (x) – (1/2)
Vertical stretch:
g (x) = 3 f (x)
Vertical translation down:
g (x) = f (x – 1)

Question 9.
What is the sum of the integer solutions of the compound inequality 2 | x – 5 | < 16?
A. 72
B. 75
C. 85
D. 88
Answer:
The given compound inequality is:
2 | x – 5 | < 16
So,
2 (x – 5) < 16 an d 2 (x – 5 ) > -16
2x – 10 < 16 and 2x – 10 > -16
2x < 26 and 2x >-6
x < 13 and x > -3
Hence,
The solution of the given compoud inequality is: -3 < x < 13
Now,
The sum of all the integers = -2 – 1 + 0 + 1 + 2 + 3 + 4 + 5+ 6 + 7 + 8 +9 + 10 + 11 + 12
= 75
Hence, from the above,
We can conclude that the sum of all the integers is: 75

Question 10.
Your bank offers a text alert service that notifies you when your checking account balance drops below a specific amount. You set it up so you are notified when your balance drops below $700. The balance is currently $3000. You only use your account for paying your rent (no other deposits or deductions occur). Your rent each month is $625.
a. Write an inequality that represents the number of months m you can pay your rent without receiving a text alert.
Answer:
It is given that your bank offers a text alert service that notifies you when your checking account balance drops below a specific amount. You set it up so you are notified when your balance drops below $700. The balance is currently $3000. You only use your account for paying your rent (no other deposits or deductions occur). Your rent each month is $625
Let m be the number of months
Hence,
The inequality that represents the number of months you can pay rent without receiving a text is:
3000 – 625x > 700

b. What is the maximum number of months you can pay your rent without receiving a text alert?
Answer:
From part (a),
The inequality that represents the number of months m without receiving a text alert is:
3000 – 625x > 700
3000 – 700 > 625x
2300 > 625x
2300 / 625 > x
3.68 > x
x < 4 months [ Since the number of months will not be in decimals]
Hence, from the above,
We can conclude that the maximum number of months you can pay your rent without receiving a text alert is: 4 months

c. Suppose you start paying rent in June. Select all the months you can pay your rent without making a deposit.
Big Ideas Math Answers Algebra 1 Chapter 3 Graphing Linear Functions 219
Answer:
From part (b),
The maximum number of months is:
x < 4 months
It is given that you start paying rent in June
So,
All the months you can pay your rent without making a deposit is:
June, July, August

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities

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Big Ideas Math Book Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities

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Solving Linear Inequalities Maintaining Mathematical Proficiency

Graph the number.

Question 1.
6
Answer:
The number line representing 6 is:

Question 2.
| 2 |
Answer:
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
The number line representing | 2 | is:

Question 3.
| -1 |
Answer:
We know that,
| -x | = x for x > 0
| -x | = -x for x < 0
So,
The number line representing | -1 | is:

Question 4.
2 + | -2 |
Answer:
The given absolute value expression is:
2 + | -2 |
We know that,
| -x | =x for x > 0
| -x | = -x for x < 0
So,
2 + | -2 | = 2 + 2                                                 2 + | – 2 | = 2 – 2
= 4                                                                       = 0
Hence,
The number line representing the values of 2 + | – 2 | is:

Question 5.
1 – | -4 |
Answer:
The given absolute value expression is:
1 – | – 4 |
We know that,
| -x | = x for x > 0
| – x | = -x for x < 0
So,
1 – | – 4 | = 1 – 4                                      1 – | – 4 | = 1 – ( -4 )
= -3                                                          = 5
Hence,
The number line representing the values of 1 – | – 4 | is:

Question 6.
-5 + | 3 |
Answer:
The given absolute value expression is:
-5 + | 3 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
-5 + | 3 | = -5 + 3                                 -5 + | 3 | = -5 -3
= -2                                                        = -8
Hence,
The number line representing the values of -5 + | 3 | is:

Complete the statement with <, >, or =.

Question 7.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 1
Answer:
We know that,
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Hence, from the above,
We can conclude that
2 < 9

Question 8.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 2
Answer:
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Hence, from the above,
We can conclude that
-6 < 5

Question 9.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 3
Answer:
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
-12 < -4

Question 10.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 4
Answer:
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
-7 > -13

Question 11.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 5
Answer:
We know that,
| -x | = x for x > 0
| -x | = -x for x < 0
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
8 = 8 ( or ) -8 = -8

Question 12.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 6
Answer:
We know that,
| -x | = x for x > 0
| -x | = -x for x < 0
The order of the numbers in the ascending order is:
-9, -8, -7, -6, -5, -4 ,-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
We know that,
The greater number in the positive integers is the lesser number in the negative integers
Hence, from the above,
We can conclude that
-10 < 18    ( or ) -10 > -18

Question 13.
ABSTRACT REASONING
A number a is to the left of a number b on the number line. How do the numbers -a and -b compare?
Answer:
It is given that a number a is to the left of a number b on the number line.
So,
The representation of a and b on the number line is:

Hence, from the above number line
We can conclude that
-a > -b

Solving Linear Inequalities Mathematical Practices

Monitoring Progress

Use a graphing calculator to solve the inequality.

Question 1.
2x + 3 < x – 1
Answer:
The given inequality is:
2x + 3 < x – 1
We know that,
When we convert the mathematical symbols from LHS to RHS, the sign of that mathematical symbol change. i.e.,
+ will be converted into – and vice – versa
× will be converted into ÷ and vice-versa
So,
2x – x < -1 – 3
x < – 4
Hence,
The representation of the solved inequality in the number line is:

Question 2.
-x – 1 > -2x + 2
Answer:
The given inequality is:
-x – 1 > -2x + 2
We know that,
When we convert the mathematical symbols from LHS to RHS, the sign of that mathematical symbol change. i.e.,
+ will be converted into – and vice – versa
× will be converted into ÷ and vice-versa
So,
-x + 2x > 2 + 1
x > 3
Hence,
The representation of the solved inequality in the number line is:

Question 3.
\(\frac{1}{2}\)x + 1 > \(\frac{3}{2}\)x + 3
Answer:
The given inequality is:
\(\frac{1}{2}\)x + 1 > \(\frac{3}{2}\)x + 3
We know that,
When we convert the mathematical symbols from LHS to RHS, the sign of that mathematical symbol change. i.e.,
+ will be converted into – and vice – versa
× will be converted into ÷ and vice-versa
So,
\(\frac{1}{2}\)x – \(\frac{3}{2}\)x > 3 – 1
\(\frac{1 – 3}{2}\)x > 2
\(\frac{-2}{2}\)x > 2
\(\frac{-1}{1}\)x > 2
-x > 2
Multiply with –  both sides
So,
x > -2
Hence,
The representation of the solved inequality in the number line is:

Lesson 2.1 Writing and Graphing Inequalities

Essential Question

How can you use an inequality to describe a real-life statement?
Answer:
Inequality is just a relationship between two amounts, in which the amounts are not the same.
Example:
She completed her work before anyone else. i.e., her finish time was less than everyone else

EXPLORATION 1
Writing and Graphing Inequalities

Work with a partner. Write an inequality for each statement. Then sketch the graph of the numbers that make each inequality true.
a. Statement The temperature t in Sweden is at least -10°C.
Inequality Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 7
Graph Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 8
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 8
It s given that the temperature t in Sweden is at least -10°C
The meaning of the above statement is that the temperature of Sweden is -10°C or greater than -10°C
Hence,
The representation of the temperature of Sweden in the graph is:

b. Statement The elevation e of Alabama is at most 2407 feet.
Inequality Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 9
Graph Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 10
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 10
It is given that the elevation ‘ e ‘ of Alabama is at most 2,407 feet
From the given graph,
We can say that the gap between each elevation is 1,000 feet
So,
The elevation of Alabama will come between 2000 and 3000 in the number line
Hence,
The representation of the elevation of Alabama in the given graph is:

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 11
Answer:
EXPLORATION 2
Writing Inequalities
Work with a partner. Write an inequality for each graph. Then, in other words, describe all the values of x that make each inequality true.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 12
Answer:
The given number lines are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 12
a)
From the given number line,
The value of x marked from 1 to 4
Hence,
The inequality representing the given number line is:
x ≥ 1
b)
From the given number line,
The value of x marked after 1 till the end of the number line
Hence,
The inequality representing the given number line is:
x > 1
c)
From the given number line,
The value of x marked from 1 till the end of the left side of the number line.
Hence,
The inequality representing the given number line is:
x ≤ 1
d)
From the given number line,
The value of x marked before 1 till the end of the left side of the number line
Hence,
The inequality representing the given number lie is:
x < 1

Communicate Your Answer

Question 3.
How can you use an inequality to describe a real-life statement?
Answer:
Inequality is just a relationship between two amounts, in which the amounts are not the same.
Example:
A ball in the net is worth two in the bush. i.e., the value of a single ball in the net is greater than the value of a single ball in the bush.

Question 4.
Write a real-life statement that involves each inequality.
a. x < 3.5
b. x ≤ 6
c. x > -2
d. x ≥ 10
Answer:
a)
The given inequality is:
x < 3.5
The real-life situation that involves the given inequality is:
The number of birds that are less than the height of 3.5 inches
b)
The given inequality is:
x ≤ 6
The real- life situation that involves the given inequality is:
The number of people that are buying the mobiles from a mobile store
c)
The given inequality is:
x > -2
The real-life situation that involves the given inequality is:
The number of integers that are greater than -2
d)
The given inequality is:
x ≥ 10
The real-life situation that involves the given inequality is:
The number of chocolates that are distributed greater than or equal to 10 children

2.1 Lesson

Monitoring Progress

Write the sentence as an inequality.

Question 1.
A number b is fewer than 30.4.
Answer:
The given sentence is:
A number b is fewer than 30.4
Hence,
The representation of the given sentence in the form of inequality is:
b < 30.4

Question 2.
–\(\frac{7}{10}\) is at least twice a number k minus 4.
Answer:
The given sentence is:
–\(\frac{7}{10}\) is at least twice a number k minus 4.
Hence,
The representation of the given sentence in the form of inequality is:
–\(\frac{7}{10}\) = 2k – 4

Tell whether −6 is a solution to the inequality.

Question 3.
c + 4 < -1
Answer:
-6 is not a solution to the given inequality

Explanation:
The given equation is:
c + 4 = -1
So,
c = -4 – 1
c = -5
Hence, from the above,
We can conclude that -6 is not a solution to the given inequality

Question 4.
10 ≤ 3 – m
Answer:
-6 is not a solution to the given inequality

Explanation:
The given inequality is:
10 ≤ 3 – m
So,
-m ≤ 10 – 3
-m ≤ 7
Multiply by – on both sides
m ≤ -7
Hence, from the above,
We can conclude that -6 is not a solution to the given inequality

Question 5.
21 ÷ x ≥ -3.5
Answer:
-6 is a solution to the given inequality

Explanation:
The given inequality is:
21 ÷ x ≥ -3.5
So,
-21 ÷ 3.5 ≥ x
-210 ÷ 35 ≥ x
-6 ≥ x
x ≤ -6
Hence, from the above,
We can conclude that -6 is a solution to the given inequality

Question 6.
4x – 25 > -2
Answer:
-6 is not a solution to the given inequality

Explanation:
The given inequality is:
4x – 25 > -2
So,
4x > -2 + 25
4x > 23
x > 23 ÷ 4
x > 5.75
Hence, from the above,
We can conclude that -6 is not a solution to the given inequality

Graph the inequality.

Question 7.
b > -8
Answer:
The given inequality is:
b > -8
Hence,
The representation of the given inequality in the number line is:

Question 8.
1.4 ≥ g
Answer:
The given inequality is:
1.4 ≥ g
So,
g ≤ 1.4
Hence,
The representation of the given inequality in the number line is:

Question 9.
r < \(\frac{1}{2}\)
Answer:
The given inequality is:
r < \(\frac{1}{2}\)
So,
r < 0.5
Hence,
The representation of the given inequality in the number line is:

Question 10.
v ≥ \(\sqrt{36}\)
Answer:
The given inequality is:
v > \(\sqrt{36}\)
So,
v > 6
Hence,
The representation of the given inequality in the number line is:

Question 11.
Write an inequality that represents the graph.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 13
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 13
From the given graph,
We can say that the marked line is from -6 and continued after -6 till the last number on the right side of the number line.
Hence,
The representation of the given number line in the form of inequality is:
x ≥ -6

Writing and Graphing Inequalities 2.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
A mathematical sentence using the symbols <, >, ≤, or ≥ is called a(n)_______.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 1

Question 2.
VOCABULARY
Is 5 in the solution set of x + 3 > 8? Explain.
Answer:
5 is not in the solution of the given inequality
x + 3 > 8

Explanation:
The given inequality is:
x + 3 > 8
So,
x > 8 – 3
x > 5
Hence, from the above,
We can conclude that 5 is not in the solution of the given inequality

Question 3.
ATTENDING TO PRECISION
Describe how to graph an inequality.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 3

Question 4.
DIFFERENT WORDS, SAME QUESTION
Which is different? Write “both” inequalities.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 14
Answer:
The given inequalities in the worded form are:
a) w is greater than or equal to -7
b) w is not less than -7
c) w is no more than -7
d) w is at least -7
Now,
a)
The given worded form in the form of inequality is:
w ≥ -7
b)
The given worded form in the form of inequality is:
w > -7
c)
The given worded form in the form of inequality is:
w ≥ -7
d)
The given worded form in the form of inequality is:
w ≥ -7

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, write the sentence as an inequality.

Question 5.
A number x is greater than 3.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 5

Question 6.
A number n plus 7 is less than or equal to 9.
Answer:
The given worded form is:
A number n plus 7 is less than or equal to 9
Hence,
The representation of the given worded form in the form of inequality is:
n + 7 ≤ 9

Question 7.
Fifteen is no more than a number t divided by 5.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 7

Question 8.
Three times a number w is less than 18.
Answer:
The given worded form is:
Three times a number w is less than 18
Hence,
The representation of the given worded form in the form of inequality is:
3w < 18

Question 9.
One-half of a number y is more than 22.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 9

Question 10.
Three is less than the sum of a number s and 4.
Answer:
The given worded form is:
Three is less than the sum of a number s and 4
Hence,
The representation of the given worded form in the form of inequality is:
3 < s + 4

Question 11.
Thirteen is at least the difference between a number v and 1.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 11

Question 12.
Four is no less than the quotient of a number x and 2.
Answer:
The given worded form is:
Four is no less than the quotient of a number x and 2
Hence,
The representation of the given worded form in the form of inequality is:
4 > x ÷ 2

Question 13.
MODELING WITH MATHEMATICS
On a fishing trip, you catch two fish. The weight of the first fish is shown. The second fish weighs at least 0.5 pound more than the first fish. Write an inequality that represents the possible weights of the second fish.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 15
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 13

Question 14.
MODELING WITH MATHEMATICS
There are 430 people in a wave pool. Write an inequality that represents how many more people can enter the pool.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 16
Answer:
The inequality that represents the number of more people that can enter the pool is:
430 + x = 600
Where,
x is the number of more people that can enter the pool

Explanation:
It is given that there are 430 people in a wave pool
It is also given that the maximum capacity in a pool is: 600
Let,
x be the number of more people that can enter the pool
Hence,
The inequality that represents the number of more people that can enter the pool is:
430 + x = 600

In Exercises 15–24, tell whether the value is a solution to the inequality.

Question 15.
r + 4 > 8; r = 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 15

Question 16.
5 – x < 8; x = -3
Answer:
-3 is not a solution of the given inequality

Explanation:
The given inequality is:
5 – x < 8
It is given that x = -3
So,
5 – ( -3 ) < 8
5 + 3 < 8
8 < 8
Hence, from the above,
We can conclude that x = -3 is not a solution to the given inequality

Question 17.
3s ≤ 19; s = -6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 17

Question 18.
17 ≥ 2y ; y = 7
Answer:
y = 7 is a solution of the given inequality

Explanation:
The given inequality is:
17 ≥ 2y
It is given that y = 7
So,
17 ≥ 2 ( 7 )
17 ≥ 14
Hence, from the above,
We can conclude that y = 7 is a solution of the given inequality

Question 19.
-1 > –\(\frac{x}{2}\); x = 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 19

Question 20.
–\(\frac{4}{z}\) ≥ 3; z = 2
Answer:
z = 2 is not a solution of the given inequality

Explanation;
The given inequality is:
–\(\frac{4}{z}\) ≥ 3
It is given that z = 2
So,
–\(\frac{4}{2}\) ≥ 3
-2 ≥ 3
Hence, from the above,
We can conclude that z = 2 is not a solution of the given inequality

Question 21.
14 ≥ -2n + 4; n = -5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 21

Question 22.
-5 ÷ (2s) < -1; s = 10
Answer:
s = 10 is a solution of the given inequality

Explaantion:
The given inequality is:
-5 ÷ ( 2s ) < -1
It is given that s = 10
So,
-5 ÷ 2 ( 10 ) < -1
-5 ÷ 20 < -1
–\(\frac{1}{4}\) < -1
\(\frac{1}{4}\) < 1
1 < 4
Hence, from the above,
We can conclude that s = 10 is a solution of the given inequality

Question 23.
20 ≤ \(\frac{10}{2z}\) + 20; z = 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 23

Question 24.
\(\frac{3m}{6}\) – 2 > 3; m = 8
Answer:
m = 8 is not a solution of the given inequality

Explanation:
The given inequality is:
\(\frac{3m}{6}\) – 2 > 3
It is given that m = 8
So,
\(\frac{3 × 8}{6}\) > 3 + 2
4  3 + 2
4 > 5
Hence, from the above,
We can conclude that m = 8 is not a solution of the given inequality

Question 25.
MODELING WITH MATHEMATICS
The tallest person who ever lived was approximately 8 feet 11 inches tall.
a. Write an inequality that represents the heights of every other person who has ever lived.
b. Is 9 feet a solution of the inequality? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 25

Question 26.
DRAWING CONCLUSIONS
The winner of a weight-lifting competition bench-pressed 400 pounds. The other competitors all bench-pressed at least 23 pounds less.
a. Write an inequality that represents the weights that the other competitors bench-pressed.
Answer:
It is given that the winner of a weight-lifting competition bench pressed 400 pounds whereas the other competitors all bench-pressed at least 23 pounds less.
Let,
x be the number of pounds that all the other competitors’ bench-pressed
So,
The inequality that represents the weights that the other competitors bench-pressed is:
x  + 23 ≤ 400

b. Was one of the other competitors able to bench-press 379 pounds? Explain.
Answer:
No, one of the other competitors won’t be able to bench-press 379 pounds

Explanation:
From part (a),
The inequality that represents the weights that the other competitors bench-pressed is:
x + 23 ≤ 400
x ≤ 400 – 23
x ≤ 377 pounds
Hence, from the above,
We can conclude that one of the other competitors won’t be able to bench-press 379 pounds

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in determining whether 8 is in the solution set of the inequality.

Question 27.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 17
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 18
Answer:
8 is a solution set of the given inequality

Explanation:
The given inequality is:
\(\frac{1}{2}\)x + 2 ≤ 6
It is given that x = 8
So,
\(\frac{1}{2}\) × 8 ≤ 6 – 2
\(\frac{8}{2}\) ≤ 4
4 ≤ 4
Hence, from the above,
We can conclude that 8 is a solution set of the given inequality

In Exercises 29–36, graph the inequality.

Question 29.
x ≥ 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 29

Question 30.
z ≤ 5
Answer:
The given inequality is:
z ≤ 5
Hence,
The representation of the given inequality in the number line is:

Question 31.
-1 > t
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 31

Question 32.
-2 < w
Answer:
The given inequality is:
-2 < w
So,
w > -2
Hence,
The representation of the given inequality in the number line is:

Question 33.
v ≤ -4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 33

Question 34.
s < 1
Answer:
The given inequality is:
s < 1
Hence,
The representation of the given inequality in the number line is:

Question 35.
\(\frac{1}{4}\) < p
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 35

Question 36.
r ≥ -| 5 |
Answer:
The given inequality is:
r ≥ – | 5 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
r ≥ -5   ( or ) r ≥ 5
Hence,
The representation of the given inequalities in the number line is:

In Exercises 37–40, write and graph an inequality for the given solution set.

Question 37.
{x | x < 7}
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 37

Question 38.
{n | n ≥ -2}
Answer:
The given inequality is:
{n | n ≥ -2}
The given inequality can be rewritten as:
n ≥ -2
Hence,
The representation of the given inequality in the number line is:

Question 39.
{z | 1.3 ≤ z}
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 39

Question 40.
{w | 5.2 > w}
Answer:
The given inequality is:
{w | 5.2 > w}
The given inequality can be rewritten as:
5.2 > w
So,
w < 5.2
From the above value,
We can say that 5.2 lies between 5 and 6
Hence,
The representation of the given inequality in the number line is:

In Exercises 41–44, write an inequality that represents the graph.

Question 41.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 19
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 41

Question 42.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 21
Answer:
The given number line is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 21
From the above number line,
We can say that the marked line started from -2 and ended at the last value on the right side of the number line.
Hence,
The representation of the inequality for the given number line is:
x ≥ -2

Question 43.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 22
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 43

Question 44.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 23
Answer:
The given number line is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 23
From the above number line,
We can say that the marked line started from -1 excluding – and continued till the end of the number line in the left side of the number line
Hence,
The representation of the inequality for the given number line is:
x < -1

Question 45.
ANALYZING RELATIONSHIPS
The water temperature of a swimming pool must be no less than 76°F. The temperature is currently 74°F. Which graph correctly shows how much the temperature needs to increase? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 24
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 45

Question 46.
MODELING WITH MATHEMATICS
According to state law for vehicles traveling on state roads, the maximum total weight of a vehicle and its contents depends on the number of axles on the vehicle. For each type of vehicle, write and graph an inequality that represents the possible total weights w (in pounds) of the vehicle and its contents.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 25
Answer:
It is given that the maximum total weight of a vehicle and its contents depends on the number of axles on the vehicle.
Now,
Let,
w be the total weight of the vehicle and its contents
For the 2 axles vehicle, if the maximum weight is 40,000 lb and w is the possible total weight of the vehicle and its contents, then
w < 40,000
The representation of the given inequality in the number line is:

For the 3 axles vehicle, if the maximum weight is 60,000 lb and w is the possible total weight of the vehicle and its contents, then
w < 60,000
The representation of the given inequality in the number line is:

For the 4 axles vehicle, if the maximum weight is 80,000 lb and w is the possible total weight of the vehicle and its contents, then
w < 80,000
The representation of the given inequality in the number line is:

Question 47.
PROBLEM-SOLVING
The Xianren Bridge is located in Guangxi Province, China. This arch is the world’s longest natural arch, with a length of 400 feet. Write and graph an inequality that represents the lengths ℓ (in inches) of all other natural arches.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 26
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 47

Question 48.
THOUGHT-PROVOKING
A student works no more than 25 hours each week at a part-time job. Write an inequality that represents how many hours the student can work each day.
Answer:
It is given that a student works no more than 25 hours each week at a part-time job
Let the number of days that students work each week be x
So,
The inequality that represents the number of hours the students work each week is:
x < 25
We know that,
1 week = 7 days
So,
7x < 25
Hence,
The inequality that represents the number of hours the students work each day is:
x < \(\frac{25}{7}\)
x < 3.5
Hence,
The representation of the inequality in the number line is:

Question 49.
WRITING
Describe a real-life situation modeled by the inequality 23 + x ≤ 31.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 49

Question 50.
HOW DO YOU SEE IT?
The graph represents the known melting points of all metallic elements (in degrees Celsius). Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27
a. Write an inequality represented by the graph.
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27
From the above graph,
We can say that the marked line started from -38.87 and continued till the end of the right line of the number line
Hence,
The inequality that represents the given number line is:
x ≥ -38.87

b. Is it possible for a metallic element to have a melting point of -38.87°C? Explain.
Answer:
Yes, it is possible for a metallic element to have a melting point of -38.87°C.

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27
From the above graph,
The starting point is -38.87
Hence, from the above,
We can conclude that it is possible for a metallic element to have a melting point of -38.87°C

Question 51.
DRAWING CONCLUSIONS
A one-way ride on a subway costs $0.90. A monthly pass costs $24. Write an inequality that represents how many one-way rides you can buy before it is cheaper to buy the monthly pass. Is it cheaper to pay the one-way fare for 25 rides? Explain.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 27.1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 51

Question 52.
MAKING AN ARGUMENT
The inequality x ≤ 1324 represents the weights (in pounds) of all mako sharks ever caught using a rod and reel. Your friend says this means no one using a rod and reel has ever caught a mako shark that weighs 1324 pounds. Your cousin says this means someone using a rod and reel has caught a mako shark that weighs 1324 pounds. Who is correct? Explain your reasoning.
Answer:
Your cousin is correct

Explanation:
It is given that the inequality x ≤ 1324 represents the weights (in pounds) of all mako sharks ever caught using a rod and reel. Your friend says this means no one using a rod and reel has ever caught a mako shark that weighs 1324 pounds. Your cousin says this means someone using a rod and reel has caught a mako shark that weighs 1324 pounds.
Now,
The given inequality is:
x ≤ 1324
The meaning of inequality is the value of x is less than or equal to 1324 pounds
Now,
Your friend says this means no one using a rod and reel has ever caught a mako shark that weighs 1324 pounds
But, from the given inequality,
The statement of your friend is wrong
Your cousin says this means someone using a rod and reel has caught a mako shark that weighs 1324 pounds
But from the given inequality,
The statement of your cousin is correct.
Hence, from the above,
We can conclude that your cousin is correct

Question 53.
CRITICAL THINKING
Describe a real-life situation that can be modeled by more than one inequality.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 53

Question 54.
MODELING WITH MATHEMATICS
In 1997, Superman’s cape from the 1978 movie Superman was sold at an auction. The winning bid was $17,000. Write and graph an inequality that represents the amounts all the losing bids.
Answer:
It is given that in 1997, Superman’s cape from the 1978 movie Superman was sold at an auction. The winning bid was $17,000.
So,
The bid must not be less than $17,000, otherwise, the bid will lose
Hence,
The inequality that represents the amounts of all losing bids is:
x < 17,000
The representation of the inequality in the number line is:

MATHEMATICAL CONNECTIONS
In Exercises 55–58, write an inequality that represents the missing dimension x.

Question 55.
The area is less than 42 square meters.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 28
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 55

Question 56.
The area is greater than or equal to 8 square feet.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 29
Answer:
The inequality that represents the value of x is:
x ≥ \(\frac{8}{5}\)

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 29
From the above figure,
We can say that the given figure is a Right-angled triangle
We know that,
The area of the triangle = \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × x × 10
= 5x
It is given that the area of the triangle is greater than or equal to 8 square feet
So,
5x ≥ 8 square feet
x ≥ \(\frac{8}{5}\)
Hence, from the above,
We can conclude that the inequality that represents x is:
x ≥ \(\frac{8}{5}\)

Question 57.
The area is less than 18 square centimeters.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 30
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 57

Question 58.
The area is greater than 12 square inches.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 31
Answer:
The inequality that represents the value of x is:
x > 6

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 31
From the above figure,
We can say that the given figure is the rectangle.
We know that,
The area of the rectangle = Length × Width
= 2 × x
It is given that the area of the rectangle is greater than 12 square inches
So,
2x > 12
x > 12 / 2
x > 6
Hence, from the above,
We can conclude that the inequality that represents the value of x is:
x > 6

Question 59.
WRITING
A runner finishes a 200-meter dash in 35 seconds. Let r represent any speed (in meters per second) faster than the runner’s speed.
a. Write an inequality that represents r. Then graph the inequality.
b. Every point on the graph represents a speed faster than the runner’s speed. Do you think every point could represent the speed of a runner? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 59

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.(Section 1.1)

Question 60.
x + 2 = 3
Answer:
The value of x is: 1

Explanation:
The given equation is:
x + 2 = 3
So,
x = 3 – 2
x = 1
Hence, from the above
We can conclude that the value of x is: 1

Question 61.
y – 9 = 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 61

Question 62.
6 = 4 + y
Answer:
The value of y is: 2

Explanation:
The given equation is:
6 = 4 + y
So,
6 – 4 = y
2 = y
y = 2
Hence, from the above,
We can conclude that the value of y is: 2

Question 63.
-12 = y – 11
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 63

Solve the literal equation for x.(Section 1.5)

Question 64.
v = x • y • z
Answer:
The value of x is: \(\frac{v}{yz}\)

Explanation:
The given equation is:
v = x ⋅ y ⋅ z
x = \(\frac{v}{yz}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{v}{yz}\)

Question 65.
s = 2r + 3x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 65

Question 66.
w = 5 + 3(x – 1)
Answer:
The value of x is: \(\frac{w – 2}{3}\)

Explanation:
The given equation is:
w = 5 + 3 ( x – 1 )
So,
w = 5 + 3 ( x ) – 3 ( 1 )
w = 5 + 3x – 3
w = 3x + 2
3x = w – 2
x = \(\frac{w – 2}{3}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{w – 2}{3}\)

Question 67.
n = \(\frac{2x + 1}{2}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.1 Question 67

Lesson 2.2 Solving Inequalities Using Addition or Subtraction

Essential Question
How can you use addition or subtraction to solve an inequality?
EXPLORATION 1
Quarterback Passing Efficiency
Work with a partner.
The National Collegiate Athletic Association (NCAA) uses the following formula to rank the passing efficiencies P of quarterbacks.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 32
Answer:
The formula used to rank the passing efficiencies P of the quarterbacks is:

By comparing the coefficients, we get
Y = 8.4
C = 100
T = 330
N = 200
A = 1
Now,
a) T < C
= 330 < 100
Hence, from the above,
We can conclude that the given inequality is not true
b) C + N ≤ A
= ( 100 + 200 ) ≤ 1
= 300 ≤ 1
Hence, from the above,
We can conclude that the given inequality is not true
c) N < A
= 200 < 1
Hence, from the above,
We can conclude that the given inequality is not true
d) A – C ≥ M
We know that,
M = C – N
So,
A-C ≥ C – N
( 1 – 100 ) ≥ ( 100 – 200 )
= -99 ≥ -100
= 99 ≥ 100
Hence, from the above,
We can conclude that the given inequality is not true

EXPLORATION 2
Finding Solutions of Inequalities
Work with a partner.
Use the passing efficiency formula to create a passing record that makes each inequality true. Record your results in the table. Then describe the values of P that make each inequality true.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 33
a. P < 0
b. P + 100 ≥ 250
c. P – 250 > -80
Answer:
Using the data from Exploration 1,
The completed table is:

Now,
The formula for passing efficiency is:

a) P < 0
To make P < 0,
The value of A or any of the values of the numerator must be less than 0
b) P + 100 ≥ 250
P ≥ 250 – 100
P ≥ 150
Hence, from the above,
We can conclude that the value of P must be greater than or equal to 150
c ) P – 250 > -80
P > -80 + 250
P > 170
Hence, from the above,
We can conclude that the value of P must be greater than 170

Communicate Your Answer

Question 3.
How can you use addition or subtraction to solve an inequality?
Answer:
If you want to add or subtract from one side of the equation, you must perform the same operation to the other side of the equation. When solving inequalities by adding or subtracting, our goal is to have the variable on its own

Question 4.
Solve each inequality.
a. x + 3 < 4 b. x – 3 ≥ 5 c. 4 > x – 2
d. -2 ≤ x + 1
Answer:
The given inequalities are:
a) x + 3 < 4
b) x – 3 ≥ 5
c) 4 > x – 2
d) -2 ≤ x + 1
Now,
a)
The given inequality is:
x + 3 < 4
So,
x < 4 – 3
x < 1
Hence, from the above,
We can conclude that x < 1
b)
The given inequality is:
x – 3 ≥ 5
x ≥ 5 + 3
x ≥ 8
Hence, from the above,
We can conclude that x ≥ 8
c)
The given inequality is:
4 > x – 2
So,
4 + 2 > x
6 > x
x < 6
Hence, from the above,
We can conclude that x < 6
d)
The given inequality is:
-2 ≤ x + 1
So,
-2 – 1 ≤ x
-3 ≤ x
x ≥ -3
Hence, from the above,
We can conclude that x ≥ -3

2.2 Lesson

Monitoring Progress

Solve the inequality. Graph the solution.

Question 1.
b – 2 > -9
Answer:
The given inequality is:
b – 2 > -9
So,
b > -9 + 2
b > -7
Hence, from the above,
We can conclude that the value of b is greater than -7
The representation of the inequality in the number line is:

Question 2.
m – 3 ≤ 5
Answer:
The given inequality is:
m – 3 ≤ 5
So,
m ≤ 5 + 3
m ≤ 8
Hence, from the above,
We can conclude that the value of m is less than or equal to 8
The representation of the inequality in the number line is:

Question 3.
\(\frac{1}{4}\) > y – \(\frac{1}{4}\)
Answer:
The given inequality is:
\(\frac{1}{4}\) > y – \(\frac{1}{4}\)
So,
\(\frac{1}{4}\) + \(\frac{1}{4}\) > y
\(\frac{1 + 1}{4}\) > y
\(\frac{2}{4}\) > y
\(\frac{1}{2}\) > y
y < \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of y is less than \(\frac{1}{2}\)
The representation of the inequality in the number line is:

Solve the inequality. Graph the solution.

Question 4.
k + 5 ≤ -3
Answer:
The given inequality is:
k + 5 ≤ -3
So,
k ≤ -3 – 5
k ≤ -8
Hence, from the above,
We can conclude that the value of k is less than or equal to -8
The representation of the inequality in the number line is:

Question 5.
\(\frac{5}{6}\) ≤ z + \(\frac{1}{6}\)
Answer:
The given inequality is:
\(\frac{5}{6}\) ≤ z + \(\frac{1}{6}\)
So,
\(\frac{5}{6}\) – \(\frac{1}{6}\) ≤ z
\(\frac{5 – 1}{6}\) ≤ z
\(\frac{4}{6}\) ≤ z
\(\frac{2}{3}\) ≤ z
z ≥ \(\frac{2}{3}\)
z ≥ 0.6
z ≥ 1 ( Approx. )
Hence, from the above,
We can conclude that the value of z is approximately greater than or equal to 1
The representation of the inequality in the number line is:

Question 6.
p + 0.7 > -2.3
Answer:
The given inequality is:
p + 0.7 > -2.3
So,
p > -2.3 – 0.7
p > -3
Hence, from the above,
We can conclude that the value of p is greater than -3
The representation of the inequality in the number line is:

Monitoring Progress

Question 7.
The microwave oven uses only 1000 watts of electricity. Does this allow you to have both the microwave oven and the toaster plugged into the circuit at the same time? Explain your reasoning.
Answer:
Yes, this allows you to have both the microwave oven and the toaster plugged into the circuit at the same time

Explanation:
It is given that the microwave oven uses only 1000 watts of electricity.
We know that,
The toaster consumes less electricity than the microwave oven
Hence, from the above,
We can conclude that 1000 watts of electricity allow you to have both the microwave oven and the toaster plugged into the circuit at the same time

Solving Inequalities Using Addition or Subtraction 2.2 Exercises

In Exercises 3−6, tell which number you would add to or subtract from each side of the inequality to solve it.

Question 1.
VOCABULARY
why is the inequality x ≤ 6 equivalent to the inequality x – 5 ≤ 6 – 5 ?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 1

Question 2.
WRITING
Compare solving equations using addition with solving inequalities using addition.
Answer:
Solving equations and inequalities using addition is very similar. You have to add the same quantity on every side and the sign between the two sides does not change.
It remains the equal sign  “=”  in the case of equations and the inequality sign in the case of inequalities

Question 3.
k + 11 < -3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 3

Question 4.
v – 2 > 14
Answer:
The value of v is: 16

Explanation:
The given equation is:
v – 2 > 14
So,
v  – 2 + 2 > 14 + 2
v > 16
Hence, from the above,
We can conclude that we have to add 2 on both sides

Question 5.
-1 ≥ b – 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 5

Question 6.
-6 ≤ 17 + p
Answer:
The value of p is: p ≥ -23

Explanation:
The given inequality is:
-6 ≤ 17 + p
So,
-6 + 6 ≤ 17 + 6 + p
0 ≤ 23 + p
-23 ≤ p
p ≥ -23
Hence, from the above,
We can conclude that we have to add 6 on both sides

In Exercises 7−20, solve the inequality. Graph the solution.

Question 7.
x – 4 < -5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 7

Question 8.
1 ≤ s – 8
Answer:
The given inequality is:
1 ≤ s – 8
1 + 8 ≤ s – 8 + 8
9 ≤ s
s ≥ 9
Hence,
The solution to the given inequality is:
s ≥ 9
The representation of the inequality in the number line is:

Question 9.
6 ≥ m – 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 9

Question 10.
c – 12 > -4
Answer:
The given inequality is:
c – 12 > -4
So,
c – 12 + 4 > -4 + 4
c – 8 > 0
c > 0 + 8
c > 8
Hence, from the above,
We can conclude that the solution to the given inequality is:
c > 8
The representation of the inequality in the number line is:

Question 11.
r + 4 < 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 11

Question 12.
-8 ≤ 8 + y
Answer:
The given inequality is:
-8 ≤ 8 + y
So,
-8 + 8 ≤ 8 + 8 + y
0 ≤ 16 + y
-16 ≤ y
y ≥ -16
Hence, from the above,
We can conclude that the solution to the given inequality is:
y ≥ -16
The representation of the inequality in the number line is:

Question 13.
9 + w > 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 13

Question 14.
15 ≥ q + 3
Answer:
The given inequality is:
15 ≥ q + 3
So,
15 – 15 ≥ q + 3 – 15
0 ≥ q – 12
12 ≥ q
q ≤ 12
Hence, from the above,
We can conclude that the solution to the given inequality is:
q ≤ 12
The representation of the inequality in the number line is:

Question 15.
h – (-2) ≥ 10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 15

Question 16.
-6 > t – (-13)
Answer:
The given inequality is:
-6 > t – ( -13 )
So,
-6 + 6 > t + 13 + 6
0 > t + 19
-19 > t
t < -19
Hence, from the above,
We can conclude that the solution to the given inequality is:
t < -19
The representation of the inequality in the number line is:

Question 17.
j + 9 – 3 < 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 17

Question 18.
1 – 12 + y ≥ -5
Answer:
The given inequality is:
1 – 12 + y ≥ -5
So,
-11 + y ≥ -5
-11 + 5 + y ≥ -5 + 5
6 + y ≥ 0
y ≥ -6
Hence, from the above,
We can conclude that the solution to the given inequality is:
y ≥ -6
The representation of the inequality in the number line is:

Question 19.
10 ≥ 3p – 2p – 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 19

Question 20.
18 – 5z + 6z > 3 + 6
Answer:
The given inequality is:
18 – 5z + 6z > 3 + 6
So,
18 + z > 9
18 – 9 + z > 9 – 9
z + 9 > 0
z > -9
Hence, from the above,
We can conclude that the solution to the given inequality is:
z > -9
The representation of the inequality in the number line is:

In Exercises 21−24, write the sentence as an inequality. Then solve the inequality.

Question 21.
A number plus 8 is greater than 11.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 21

Question 22.
A number minus 3 is at least -5.
Answer:
The given worded form is:
A number minus 3 is at least -5
Let the number be x
So,
The representation of the given worded form in the form of inequality is:
x – 3 ≥ -5
So,
x – 3 + 5 ≥ -5 + 5
x + 2 ≥ 0
x ≥ -2
Hence, the solution to the given worded form inequality is:
x ≥ -2

Question 23.
The difference of a number and 9 is fewer than 4.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 23

Question 24.
Six is less than or equal to the sum of a number and 15.
Answer:
The given worded form is:
Six is less than or equal to the sum of a number and 15
Let the number be x
So,
The representation of the given worded form in the form of inequality is:
6 ≤ x + 15
6 – 6 ≤ x + 15 – 6
0 ≤ x + 9
-9 ≤ x
x ≥ -9
Hence, from the above,
We can conclude that the solution to the given worded form of the inequality is:
x ≥ -9

Question 25.
MODELING WITH MATHEMATICS
You are riding a train. Your carry-on bag can weigh no more than 50 pounds. Your bag weighs 38 pounds.
a. Write and solve an inequality that represents how much weight you can add to your bag.
b. Can you add both a 9-pound laptop and a 5-pound pair of boots to your bag without going over the weight limit? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 25

Question 26.
MODELING WITH MATHEMATICS
You order the hardcover book shown from a website that offers free shipping on orders of $25 or more. Write and solve an inequality that represents how much more you must spend to get free shipping.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 34
Answer:
The inequality that represents the more money you must spend to get free shipping is:

Explanation:
It is given that you order the hardcover book shown from a website that offers free shipping on orders of $25 or more.
From the figure,
The cost of a hardcover book is: $19.76
Let
The amount more money you must spend to get free shipping to be x
So,
x + 19.76 ≥ 25
x + 19.76 – 25 ≥ 25 – 25
x – 5.24 ≥ 0
x ≥ $5.24
Hence, from the above,
We can conclude that the amount more money you must spend to get free shipping is: $5.24
ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in solving the inequality or graphing the solution.

Question 27.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 35
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 36
Answer:
The given inequality is:
-10 + x ≥ -9
-10 + 9 + x ≥ -9 + 9
-1 + x ≥ 0
x ≥ 1
Hence, from the above,
We can conclude that the solution to the given inequality is:
x ≥ 1
The representation of the inequality in the number line is:

Question 29.
PROBLEM-SOLVING
An NHL hockey player has 59 goals so far in a season. What are the possible numbers of additional goals the player can score to match or break the NHL record of 92 goals in a season?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 29

Question 30.
MAKING AN ARGUMENT
In an aerial ski competition, you perform two acrobatic ski jumps. The scores on the two jumps are then added together.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 37

a. Describe the score that you must earn on your second jump to beat your competitor.
Answer:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 37
Let the score of your second acrobatic ski jump be x
Now,
From the above table,
We can say that your score must be greater than your competitor’s score to beat your competitor
Hence,
The inequality that represents the score you must earn on your second jump to beat your competitor is:
x > 119.8

b. Your coach says that you will beat your competitor if you score 118.4 points. A teammate says that you only need 117.5 points. Who is correct? Explain.
Answer:
Your coach and your teammate both are correct

Explanation:
The given table is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 37
From the above table,
In the 1st acrobatic ski jump,
Your competitor’s score is: 117.1
Your score is: 119.5
If you want to beat your competitor, then you have to score more than 117.1
So,
x > 117.1
It is given that your coach says that you will beat your competitor if you score 118.4 points. A teammate says that you only need 117.5 points
So,
According to your coach,
You have to score 118.4 points to beat your competitor
So,
118.4 > 117.1
According to a teammate,
You have to score 117.5 points to beat your competitor
So,
117.5 > 117.1
Hence, from the above,
We can conclude that your coach and your teammate both are correct

Question 31.
REASONING
Which of the following inequalities are equivalent to the inequality x – b < 3, where b is a constant? Justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 38
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 31

MATHEMATICAL CONNECTIONS
In Exercises 32 and 33, write and solve an inequality to find the possible values of x.

Question 32.
Perimeter < 51.3 inches
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 39
Answer:
The value of x is greater than 21.6 inches

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 39
From the above figure,
We can say that the figure is the triangle
We know that,
The perimeter is the sum of all of the sides
So,
The perimeter of the triangle = 15.5 + 14.2 + x
It is given that
Perimeter < 51.3 inches
So,
51.3 < 29.7 + x
51.3 – 29.7 < x
21.6 < x
x > 21.6
Hence, from the above,
We can conclude that the value of x is greater than 21.6 inches

Question 33.
Perimeter ≤ 18.7 feet
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 39.1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 33

Question 34.
THOUGHT-PROVOKING
Write an inequality that has the solution shown in the graph. Describe a real-life situation that can be modeled by inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 40
Answer:
The given number line is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 40
From the above number line,
We can observe that the marked line is starting from 16 and continued till the left end of the number line
So,
The inequality that represents the given number line is:
x ≤ 16
The real-life situation that can be modeled by inequality is:
A lift carrying no more than 16 people

Question 35.
WRITING
Is it possible to check all the numbers in the solution set of an inequality? When you solve the inequality x – 11 ≥ -3, which numbers can you check to verify your solution? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 35

Question 36.
HOW DO YOU SEE IT?
The diagram represents the numbers of students in a school with brown eyes, brown hair, or both.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 41
Determine whether each inequality must be true. Explain your reasoning.
a. H ≥ E
b. H + 10 ≥ E
c. H ≥ X
d. H + 10 ≥ X
e. H > X
f. H + 10 > X
Answer:
The given diagram is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 41
From the above diagram,
We can observe that,
The spaces occupied by H and E are equal
The spaces occupied by H and E are greater than X
So,
We can say that
H = E; H > X; E > X
Now,
a. H ≥ E
From the above diagram,
The given inequality is not true
b. H + 10 ≥ E
From the above diagram,
The given inequality is not true
c. H ≥ X
From the above diagram,
The given inequality is not true
d. H + 10 ≥ X
From the above diagram,
The given inequality is not true
e. H > X
From the above diagram,
The given inequality is true
f. H + 10 > X
From the above diagram,
The given inequality is true

Question 37.
REASONING
Write and graph an inequality that represents the numbers that are not solutions to each inequality.
a. x + 8 < 14
b. x – 12 ≥ 5.7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 37

Question 38.
PROBLEM-SOLVING
Use the inequalities c – 3 ≥ d, b + 4 < a + 1, and a – 2 ≤ d – 7 to order a, b, c, and d from least to greatest.
Answer:
The order of a, b, c, and d from least to greatest is:
b, a, c, and d

Explanation:
The given inequalities are:
A) c – 3 ≥ d
B) b + 4 < a + 1
C) a – 2 ≤ d – 7
Now,
A)
The given inequality is:
c – 3 ≥ d
c ≥ d + 3
B)
The given inequality is:
b + 4 < a + 1
b < a + 1 – 4
b < a – 3
b + 3 < a
a > b + 3
C)
The given inequality is:
a – 2 ≤ d – 7
a ≤ d – 7 + 2
a ≤ d – 5
a + 5 ≤ d
d ≥ a + 5
Hence, from the above,
We can conclude that the order of a, b, c, d from least to greatest is:
b, a, c, and d

Maintaining Mathematical Proficiency

Find the product or quotient.

Question 39.
7 • (-9)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 39

Question 40.
-11 • (-12)
Answer:
The given expression is:
-11 ⋅ ( -12 )
= 11 ⋅ 12 [ Since we know that – × – = + ]
= 132
Hence, from the above,
We can conclude that the product of the given expression is: 132

Question 41.
-27 ÷ (-3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 41

Question 42.
20 ÷ (-5)
Answer:
The given expression is:
20 ÷ ( -5 )
= -20 ÷ 5 [ Since we know that + ÷ – = – ]
= -4
Hence, from the above,
We can conclude that the product of the given expression is: -4

Solve the equation. Check your solution.(Section 1.1)

Question 43.
6x = 24
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 43

Question 44.
-3y = -18
Answer:
The given equation is:
-3y = -18
y = -18 ÷ ( -3 )
y = 6 [ Sice we know that – ÷ – = + ]
Hence, fro the above,
We can conclude that the value of y in the given equation is: 6

Question 45.
\(\frac{s}{-8}\) = 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.2 Question 45

Question 46.
\(\frac{n}{4}\) = -7.3
Answer:
The given equation is:
\(\frac{n}{4}\) = -7.3
n = -7.3 × 4
n = -29.2
Hence, from the above,
We can conclude that the value of n in the given equation is: -29.2

Lesson 2.3 Solving Inequalities Using Multiplication or Division

Essential Question

How can you use division to solve inequality?

EXPLORATION 1
Writing a Rule
Work with a partner.
a. Copy and complete the table. Decide which graph represents the solution of the inequality 6 < 3x. Write the solution to the inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 42
Answer:
a)
The completed table is:

The given number lines are:

From the above two number lines,
We can say that the expression 6 < 3x will be true if the value of x is greater than 2
From the 2nd number line,
We can observe that the value of x is greater than 2
b)
Rule:
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.The given inequality is:
6 < 3x
3x > 6
x > 6 ÷ 3
x > 2
So,
From the table,
x > 2 means x = 3, 4, 5
So,
The value of x is: 3, 4, 5
Now,
i. 2x < 4
Answer:
The given inequality is:
2x < 4
x < 4 / 2
x < 2
Now,
From the above table,
We can observe that x < 2 will be held true if the value of x is -1, 0, 1

ii. 3 ≥ 3x
Answer:
The given inequality is:
3 ≥ 3x
1 ≥ x
x ≤ 1
Now,
From the above table,
We can observe that x ≤ 1 will be held true if the value of the x is -1, 0, 1

iii. 2x < 8
Answer:
The given inequality is:
2x < 8
x < 8 / 2
x < 4
Now,
From the above table,
We can observe that x < 4 will be held true if the value of x is -1, 0, 1, 2, 3

iv. 6 ≥ 3x
Answer:
The given inequality is:
6 ≥ 3x
So,
6 / 3 ≥ x
2 ≥ x
x ≤ 2
Now,
from the above table,
We can observe that x ≤ 2 will be held true if the value of x is -1, 0, 1, 2

EXPLORATION 2
Writing a Rule
Work with a partner.
a. Copy and complete the table. Decide which graph represents the solution of the inequality 6 < -3x. Write the solution of the inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 43
Answer:
The complete table is:

From the given number lines,
The first graph indicates the inequality
x < -2
The second graph indicates the inequality
x > -2
Hence, from the table and the number line,
The inequality that represents the table is:
x < -2

b. Use a table to solve each inequality. Then write a rule that describes how to use division to solve the inequalities.
Rule:
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.
i. -2x < 4
Answer:
The given inequality is:
-2x < 4
-x < 4 / 2
-x < 2
x < -2
Hence,
From the above table,
We can observe that the values of x for the inequality x < -2 is -5, -4, -3

ii. 3 ≥ -3x
Answer:
The given inequality is:
3 ≥ -3x
1 ≥ -x
-1 ≥ x
x ≤ -1
Hence,
From the above table,
We can observe that the values of x for the inequality x ≤ -1 is -5, -4, -3, -2, -1

iii. -2x < 8
Answer:
The given inequality is:
-2x < 8
x < -8 /2
x < -4
Hence,
From the above table,
We can observe that the values of x for the inequality x < -4 is  -5

iv.6 ≥ -3x
Answer:
The given inequality is:
6 ≥ -3x
6 / ( -3 ) ≥ x
-2 ≥ x
x ≤ -2
Hence,
From the above table,
We can observe that the values of x for the inequality x ≤ -2 is -5, -4, -3, -2

Communicate Your Answer

Question 3.
How can you use division to solve inequality?
Answer:
You can use division to solve the inequality by using the Division Property of Inequality
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.

Question 4.
Use the rules you wrote in Explorations 1(b) and 2(b) to solve each inequality.
a. 7x < -21
b. 12 ≤ 4x
c. 10 < -5x d. -3x ≤ 0
Answer:
Rule:
Division Property of Inequality:
If you divide one side of an inequality by a number, you can divide the other side of the inequality by the same number.
By using the above rule, solve the given inequalities
Now,
a. 7x < -21
Answer:
The given inequality is:
7x < -21
x < -21 /
x < -3
Hence, from the above,
We can conclude that the solution to the given inequality is x < -3

b. 12 ≤ 4x
Answer:
The given inequality is:
12 ≤ 4x
12 / 4 ≤ x
3 ≤ x
x ≥ 3
Hence, from the above,
We can conclude that the solution to the given inequality is x ≥ 3

c. 10 < -5x
Answer:
The given inequality is:
10 < -5x
10 / -5 < x
2 < x
x > 2
Hence, from the above,
We can conclude that the solution to the given inequality is x > 2

d. -3x ≤ 0
Answer:
The given inequality is:
-3x ≤ 0
x ≤ 0 / -3
x ≤ 0
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 0

2.3 Lesson 

Monitoring Progress Solve the inequality. Graph the solution. 

Question 1.
\(\frac{n}{7}\) ≥ -1
Answer:
The given inequality is:
\(\frac{n}{7}\) ≥ -1
n ≥ -1 ( 7 )
n ≥ -7
Hence, from the above,
We can conclude that the solution o the inequality is n ≥ -7
The representation of the inequality in the number line is:

Question 2.
-6.4 ≥ \(\frac{1}{5}\)w
Answer:
The given inequality is:
-6.4 ≥ \(\frac{1}{5}\)w
-6.4 ( 5 ) ≥ w
-32 ≥ w
w ≤ -32
Hence, from the above,
We can conclude that the solution to the given inequality is w ≤ -32
The representation of the inequality in the number line is:

Question 3.
4b ≥ 36
Answer:
The given inequality is:
4b ≥ 36
b ≥ 36 / 4
b ≥ 9
Hence, from the above,
We can conclude that the solution to the given inequality is b ≥ 9
The representation of the inequality in the number line is:

Question 4.
-18 > 1.5q
Answer:
The given inequality is:
-18 > 1.5q
-18 / 1.5 > q
-180 / 15 > q
-12 > q
q < -12
Hence, from the above,
We can conclude that the solution to the given inequality is q < -12
The representation of the inequality in the number line is:

Monitoring Progress

Solve the inequality. Graph the solution.

Question 5.
\(\frac{p}{-4}\) < 7
Answer:
The given inequality is:
\(\frac{p}{-4}\) < 7
–\(\frac{p}{4}\) < 7
-p < 7 ( 4 )
-p < 28
p < -28
Hence, from the above,
We can conclude that the solution to the given inequality is p < -28
The representation of the inequality in the number line is:

Question 6.
\(\frac{x}{-5}\) ≤ -5
Answer:
The given inequality is:
\(\frac{x}{-5}\) ≤ -5
–\(\frac{x}{5}\) ≤ -5
\(\frac{x}{5}\) ≤ 5
x ≤ 5 ( 5 )
x ≤ 25
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 25
The representation of the inequality in the number line is:

Question 7.
-1 ≥ –\(\frac{1}{10}\)z
Answer:
The given inequality is:
-1 ≥ –\(\frac{1}{10}\)z
1 ≥ \(\frac{1}{10}\)z
1 ( 10 ) ≥ z
10 ≥ z
z ≤ 10
Hence, from the above,
We can conclude that the solution to the given inequality is z ≤ 10
The representation of the inequality in the number line is:

Question 8.
-9m > 63
Answer:
The given inequality is:
-9m > 63
m > -63 / 9
m > -7
Hence, from the above
We can conclude that the solution to the given inequality is m > -7
The representation of the inequality in the number line is:

Question 9.
-2r ≥ -22
Answer:
The given inequality is:
-2r ≥ -22
2r ≥ 22
r ≥ 22 / 2
r ≥ 11
Hence, from the above
We can conclude that the solution to the given inequality is r ≥ 11
The representation of the inequality in the number line is:

Question 10.
-0.4y ≥ -12
Answer:
The given inequality is:
-0.4y ≥ -12
0.4y ≥ 12
y ≥ 12 / 0.4
y ≥ 120 / 4
y ≥ 30
Hence, from the above
We can conclude that the solution to the given inequality is y ≥ 30
The representation of the inequality in the number line is:

Question 11.
You have at most $3.65 to make copies. Each copy costs $0.25. Write and solve an inequality that represents the number of copies you can make.
Answer:
The inequality that represents the number of copies you can make is:
x + 0.25 ≤ 3.65

Explanation:
It is given that you have at most $3.65 i.e., you have a maximum of only $3.65 to make copies and it is also given that each copy costs $0.25.
Let,
The additional cost to make the copies be $x
So,
The total cost = x + 0.25
But,
The total cost won’t be greater than 3.65
Hence,
The inequality that represents the number of copies you can make is:
x + 0.25 ≤ 3.65

Question 12.
The maximum speed limit for a school bus is 55 miles per hour. Write and solve an inequality that represents the number of hours it takes to travel 165 miles in a school bus.
Answer:
The inequality that represents the number of hours it takes to travel 165 miles in a school bus is:
x ≥ 3

Explanation:
It is given that the maximum speed limit for a school bus is 55 miles per hour.
Let,
x be the number of hours
So,
The inequality that represents the number of hours it takes to travel 165 miles in a school bus is:
55x ≥ 165
x ≥ 165 / 55
x ≥ 3
Hence, from the above,
We can conclude that the inequality that represents the number of hours it takes to travel 165 miles in a school bus is:
x ≥ 3

Solving Inequalities Using Multiplication or Division 2.3 Exercises

In Exercises 3–10, solve the inequality. Graph the solution.

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how solving 2x < -8 is different from solving -2x < 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 1

Question 2.
OPEN-ENDED
Write an inequality that is solved using the Division property of Inequality where the inequality symbol needs to be reversed.
Answer:
The required inequality is:
-6 < 3x
Using the Division Property of Inequality,
-6 / 3 < ( 3 / 3 )x
-2 < x
x > -2
Hence, from the above,
We can conclude that the inequality is solved using the Division Property of Inequality wheer the inequality symbol needs to be reversed

Question 3.
4x < 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 3

Question 4.
3y ≤ -95.
Answer:
The given inequality is:
3y ≤ -95
y ≤ -95 / 3
y ≤ -31.6
y ≤ -32 [ Approximate value ]
Hence, from the above,
We can conclude that the solution to the given inequality is y ≤ -32
The representation of the inequality in the number line is:

Question 5.
-20 ≤ 10n
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 5

Question 6.
35 < 7t
Answer:
The given inequality is:
35 < 7t
35 / 7 < t
5 < t
t > 5
Hence, from the above,
We can conclude that the solution to the given inequality is t > 5
The representation of the inequality in the number line is:

Question 7.
\(\frac{x}{2}\) > -2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 7

Question 8.
\(\frac{a}{4}\) < 10.2
Answer:
The given inequality is:
\(\frac{a}{4}\) < 10.2
a < 10.2 ( 4 )
a < 40.8
a < 40 [ Approximate value ]
Hence, from the above,
We can conclude that the solution to the given inequality is a < 40
The representation of the inequality in the number line is:

Question 9.
20 ≥ \(\frac{4}{5}\)w
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 9

Question 10.
-16 ≤ \(\frac{8}{3}\)t
Answer:
The given inequality is:
-16 ≤ \(\frac{8}{3}\)t
-16 ( 3) ≤ 8t
-48 ≤ 8t
-48 / 8 ≤ t
-6 ≤ t
t ≥ -6
Hence, from the above,
We can conclude that the solution to the given inequality is t ≥ -6
The representation of the inequality in the number line is:

In Exercises 11–18, solve the inequality. Graph the solution.

Question 11.
-6t < 12
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 11

Question 12.
-9y > 9
Answer:
The given inequality is:
-9y > 9
y > -9 / 9
y > -1
Hence, from the above,
We can conclude that the solution to the given inequality is y > -1
The representation of the inequality in the number line is:

Question 13.
-10 ≥ -2z
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 13

Question 14.
-15 ≤ -3c
Answer:
The given inequality is:
-15 ≤ -3c
15 ≤ 3c
15 / 3 ≤ c
5 ≤ c
c ≥ 5
Hence, from the above,
We can conclude that the solution to the given inequality is c ≥ 5
The representation of the inequality in the number line is:

Question 15.
\(\frac{n}{-3}\) ≥ 1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 15

Question 16.
\(\frac{w}{-5}\) ≤ 16
Answer:
The given inequality is:
\(\frac{w}{-5}\) ≤ 16
w ≤ -16 ( 5 )
w ≤ -80
Hence, from the above,
We can conclude that the solution to the given inequality is w ≤ -80
The representation of the inequality in the number line is:

Question 17.
-8 < –\(\frac{1}{4}\)m
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 17

Question 18.
-6y > –\(\frac{2}{3}\)
Answer:
The given inequality is:
-6 > –\(\frac{2}{3}\)y
6  > \(\frac{2}{3}\)y
6 ( 3 ) > 2y
18 > 2y
18 / 2 > y
9 > y
y < 9
Hence, from the above,
We can conclude that the solution to the given inequality is y < 9
The representation of the inequality in the number line is:

Question 19.
MODELING WITH MATHEMATICS
You have $12 to buy five goldfish for your new fish tank. Write and solve an inequality that represents the prices you can pay per fish.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 19

Question 20.
MODELING WITH MATHEMATICS
A weather forecaster predicts that the temperature in Antarctica will decrease by 8°F each hour for the next 6 hours. Write and solve an inequality to determine how many hours it will take for the temperature to drop at least 36°F.
Answer:
The inequality to determine the number of hours it will take for the temperature to drop at least 36° F is:
x ≥ 4.5

Explanation:
It is given that a weather forecaster predicts that the temperature in Antarctica will decrease by 8°F each hour for the next 6 hours.
Now,
Let the number of hours that will take to drop the temperature be x
So,
8x ≥ 36
x ≥ 36 / 8
x ≥ 4.5
Hence, from the above,
We can conclude that the inequality to determine the number of hours it will take for the temperature to drop at least 36° F is:
x ≥ 4.5 hours

USING TOOLS
In Exercises 21–26, solve the inequality. Use a graphing calculator to verify your answer.

Question 21.
36 < 3y
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 21

Question 22.
17v ≥ 51
Answer:
The given inequality is:
17v ≥ 51
v ≥ 51 / 17
v ≥ 3
Hence, from the above,
We can conclude that the solution to the given inequality is v ≥ 3
The representation of the inequality in the number line is:

Question 23.
2 ≤ –\(\frac{2}{9}\)x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 23

Question 24.
4 > \(\frac{n}{-4}\)
Answer:
The given inequality is:
4 > \(\frac{n}{-4}\)
4 ( -4 ) > n
-16 > n
n < -16
Hence, from the above,
We can conclude that the solution to the given inequality is n < -16
The representation of the inequality in the number line is:

Question 25.
2x > \(\frac{3}{4}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 25

Question 26.
1.1y < 4.4
Answer:
The given inequality is:
1.1y < 4.4
\(\frac{11}{10}\)y > \(\frac{44}{10}\)
y > \(\frac{10 × 44}{11 × 10}\)
y > 4
Hence, from the above,
We can conclude that the solution to the given inequality is y > 4
The representation of the inequality in the number line is:

ERROR ANALYSIS
In Exercises 27 and 28, describe and correct the error in solving the inequality.

Question 27.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 45
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 27

Question 28.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 46
Answer:
The given inequality is:
-4y ≤ -32
4y ≤ 32
y ≤ 32 /4
y ≤ 8
Hence, from the above,
We can conclude that the solution to the given inequality is:
y ≤ 8

Question 29.
ATTENDING TO PRECISION
You have $700 to buy a new carpet for your bedroom. Write and solve an inequality that represents the costs per square foot that you can pay for the new carpet. Specify the units of measure in each step.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 47
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 29

Question 30.
HOW DO YOU SEE IT?
Let m > 0. Match each inequality with its graph. Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 48
Answer:
The given inequalities are:
a. \(\frac{x}{m}\) < -1
b. \(\frac{x}{m}\) > 1
c. \(\frac{x}{m}\) < 1
d. –\(\frac{x}{m}\) < 1
It is given that m > 0 i.e., m is a positive number
Let,
The value of x is: 1
Now,
a.
The given inequality is:
\(\frac{x}{m}\) < -1
\(\frac{1}{m}\) < -1
1 < -m
-1 < m
m > -1
Hence,,
From the number lines, D) represents the required number line  for this inequality
b.
The given inequality is:
\(\frac{x}{m}\) > 1
x > m
1 > m
m < 1
Hence,
From the number lines, B) represents the required number line for this inequality
c.
The given inequality is:
\(\frac{x}{m}\) < 1
x < m
1 < m
m > 1
Hence,
From the number lines, A) represents the required number line for this inequality
d.
The given inequality is:
–\(\frac{x}{m}\) < 1
-x < -m
x < m
1 < m
m > 1
Hence,
From the number lines, A) represents the required number line for this inequality

Question 31.
MAKING AN ARGUMENT
You run for 2 hours at a speed no faster than 6.3 miles per hour.
a. Write and solve an inequality that represents the possible numbers of miles you run.
b. A marathon is approximately 26.2 miles. Your friend says that if you continue to run at this speed, you will not be able to complete a marathon in less than 4 hours. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 31

Question 32.
THOUGHT-PROVOKING
The inequality
\(\frac{x}{4}\) ≤ 5 has a solution of x = p. Write a second inequality that also has a solution of x = p.
Answer:
The second inequality that also has a solution x = p is:
x ≤ 20

Explanation:
The given inequality is:
\(\frac{x}{4}\) ≤ 5
It is given that the given inequality has a solution of x = p
So,
The second inequality that also has a second solution of x = p is:
\(\frac{x}{4}\) ≤ 5
x ≤ 5 ( 4 )
x ≤ 20
Hence, from the above,
We can conclude that the second inequality that also has a solution of x = p is:
x ≤ 20

Question 33.
PROBLEM-SOLVING
The U.S. Mint pays $0.02 to produce every penny. How many pennies are produced when the U.S. Mint pays more than $6 million in production costs?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 33

Question 34.
REASONING
Are x ≤ \(\frac{2}{3}\) and -3x ≤ -2 equivalent? Explain your reasoning.
Answer:
Yes,
x ≤ \(\frac{2}{3}\) and -3x ≤ -2 are equivalent

Explanation:
The given inequalities are -3x ≤ -2 and x ≤ \(\frac{2}{3}\)
Now,
x ≤ \(\frac{2}{3}\)
3x ≤ 2
Now,
-3x ≤ -2
Multiply with ‘-‘ on both sides
We know that,
– × – = +
So,
3x ≤ 2
Hence, from the above,
We can conclude that the given two inequalities are equivalent

Question 35.
ANALYZING RELATIONSHIPS
Consider the number line shown.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 49
a. Write an inequality relating A and B.
b. Write an inequality relating -A and -B.
c. Use the results from parts (a) and (b) to explain why the direction of the inequality symbol must be reversed when multiplying or dividing each side of an inequality by the same negative number.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 35

Question 36.
REASONING
Why might solving the inequality \(\frac{4}{x}\) ≥ 2 by multiplying each side by x lead to an error? (Hint: Consider x > 0 and x < 0.)
Answer:
The given inequality is:
\(\frac{4}{x}\) ≥ 2
4 ≥ 2x
4 / 2 ≥ x
2 ≥ x
x ≤ 2
It is given that to consider x > 0 and x < 0 i.e., x as a positive number and a negative number
Now,
Multiply the given inequality by x on both sides and consider x as positive
4x ≥ 2x²
4x / 2 ≥ x²
2x / x ≥ x
2 ≥ x
x ≤ 2
Now,
Multiply the given inequality by x on both sides and consider x as negative
-4x ≥ -2x²
4x / 2 ≥ x²
2x ≥ x²
2x / x ≥ x
2 ≥ x
x ≤ 2
Hence, from the above,
We can conclude that there will be no error even when we multiply the given inequality with x on both sides

Question 37.
MATHEMATICAL CONNECTIONS
The radius of a circle is represented by the formula r = \(\frac{C}{2π}\). Write and solve an inequality that represents the possible circumferences C of the circle.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 50
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 37

Question 38.
CRITICAL THINKING
A water-skiing instructor recommends that a boat pulling a beginning skier has a speed less than 18 miles per hour. Write and solve an inequality that represents the possible distances d (in miles) that a beginner can travel in 45 minutes of practice time.
Answer:
The inequality that represents the possible distances that a beginner can travel in 45 minutes of practice time is:
d < 13.5 miles

Explanation:
It is given that a water-skiing instructor recommends that a boat pulling a beginning skier has a speed less than 18 miles per hour
The given speed is in terms of miles per hour
So,
The time should also be in hours
But the given time is in minutes
So,
We know that,
60 minutes = 1 hour
So,
45 minutes = \(\frac{45}{60}\) hours
= \(\frac{3}{4}\) hours
It is also given that the distance is d
We know that,
Speed = \(\frac{Distance}{Time}\)
So,
Distance = Speed × Time
Hence,
The inequality that represents the possible distances that a beginner can travel in 45 minutes of practice time is:
d < 18 ( \(\frac{3}{4}\) )
d < \(\frac{18 × 3}{4}\)
d < \(\frac{54}{4}\)
d < \(\frac{27}{2}\)
d < 13.5 miles
Hence, from the above,
We can conclude that the inequality that represents the possible distances that a beginner can travel in 45 minutes of practice time is:
d < 13.5 miles

Question 39.
CRITICAL THINKING
A local zoo employs 36 people to take care of the animals each day. At most, 24 of the employees work full time. Write and solve an inequality that represents the fraction of employees who work part-time. Graph the solution. Maintaining Mathematical Proficiency Solve the equation. Check your solution. (Section 1.2 and Section 1.3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 39

Question 40.
5x + 3 = 13
Answer:
The given equaltion is:
5x + 3 = 13
5x = 13 – 3
5x = 10
x = 10 / 2
x = 5
Hence, from the above,
We can conclude that the solution to the given equaltion is x = 5

Question 41.
\(\frac{1}{2}\)y – 8 = -10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 41

Question 42.
-3n + 2 = 2n – 3
Answer:
The given equation is:
-3n + 2 = 2n – 3
-3n – 2n = -3 – 2
-5n = -5
5n = 5
n = 5 / 5
n = 1
Hence, from the above,
We can conclud ethat the solution to the given equation is n = 1

Question 43.
\(\frac{1}{2}\)z + 4 = \(\frac{5}{2}\)z – 8 Tell which number is greater.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 43

Question 44.
0.8, 85%
Answer:
The relation between the given numbers are:
0.8 < 85%

Explanation:
The given numbers are: 0.8 and 85%
We know that,
x% = \(\frac{x}{100}\)
We can rewrite 0.8 as 0.80 since both 0.8 and 0.80 are the same
So,
85% = \(\frac{85}{100}\)
0.8 = \(\frac{80}{100}\)
By comparison, we can get
0.8 is less than 85%
Hence, from the above,
We can conclude that
0.8 < 85%

Question 45.
\(\frac{16}{30}\), 50%
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 45

Question 46.
120%, 0.12
Answer:
The relation between the given numbers are:
120% > 0.12

Explanation:
The given numbers are: 120% and 0.12
We know that,
x% = \(\frac{x}{100}\)
Now,
120% = \(\frac{120}{100}\)
0.12 = \(\frac{12}{100}\)
By comparison, we can get
120% is greater than 0.12
Hence, from the above,
We can conclude that
120% > 0.12

Question 47.
60%, \(\frac{2}{3}\)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.3 Question 47

Lesson 2.4 Solving Multi-step Inequalities

Essential Question
How can you solve a multi-step inequality?

EXPLORATION 1
Solving a Multi-Step Inequality
Work with a partner.

• Use what you already know about solving equations and inequalities to solve each multi-step inequality. Justify each step.
• Match each inequality with its graph. Use a graphing calculator to check your answer.
a. 2x + 3 ≤ x + 5
b. -2x + 3 > x + 9
c. 27 ≥ 5x + 4x
d. -8x + 2x – 16 < -5x + 7x
e. 3(x – 3) – 5x > -3x – 6
f. -5x – 6x ≤ 8 – 8x – x
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 51
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 51.1
Answer:
The given inequalities are:
a. 2x + 3 ≤ x + 5
b. -2x + 3 > x + 9
c. 27 ≥ 5x + 4x
d. -8x + 2x – 16 < -5x + 7x
e. 3(x – 3) – 5x > -3x – 6
f. -5x – 6x ≤ 8 – 8x – x
The given graphing calculators are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 51.1
Now,
From the graphing calculators, we can observe that the graph is divided into 4 parts.
The first part indicates +x
The second part indicates -x
Now,
a.
The given inequality is:
2x + 3 ≤ x + 5
So,
2x – x ≤ 5 – 3
x ≤ 2
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 2
The graph B) matches the solution of the given inequality

b.
The given inequality is:
-2x + 3 > x + 9
So,
-2x – x > 9  -3
-3x > 6
x > -6 / 3
x > -2
Hence, from the above,
We can conclude that the solution to the given inequality is x > -2
The graph C) matches the solution of the given inequality

c.
The given inequality is:
27 ≥ 5x + 4x
So,
27 ≥ 9x
27 / 9 ≥ x
3 ≥ x
x ≤ 3
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 3
The graph E) matches the solution of the given inequality

d.
The given inequality is:
-8x + 2x – 16 < -5x + 7x
So,
-6x – 16 < 2x
-6x – 2x < 16
-8x < 16
x < -16 / 8
x < -2
Hence, from the above,
We can conclude that the solution to the given inequality is x < -2
Graph A) matches the solution of the given inequality

e.
The given inequality is:
3(x – 3) – 5x > -3x – 6
So,
3 ( x ) – 3 ( 3 ) – 5x > -3x – 6
3x – 9 – 5x > -3x – 6
-2x – 9 > -3x – 6
-2x + 3x > -6 + 9
x > 3
Hence, from the above,
We can conclude that the solution to the given inequality is x > 3
The graph D) matches the solution of the given inequality

f.
The given inequality is:
-5x – 6x ≥ 8 – 8x – x
So,
-11x ≥ 8 – 9x
-11x + 9x ≥ 8
-2x ≥ 8
x ≥ -8 / 2
x ≥ -4
Hence, from the above,
We can conclude that the solution to the given inequality is x ≥ -4
The graph F) matches the solution of the given inequality

Question 2.
How can you solve a multi-step inequality?
Answer:
The general procedure for solving multi-step inequality is as follows:
a) Clear parenthesis i.e., Brackets on both sides of the inequality and collect like terms
b) Addor subtract terms so the variable is on one side and the constant is on another side of the inequality sign

Question 3.
Write two different multi-step inequalities whose solutions are represented by the graph.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 52
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 52
From the given graph,
We can observe that the marking started from -1 but by excluding -1 and continued till the left end of the graph
So,
The inequality represented by the graph is given as:
x < -1

2.4 Lesson

Monitoring Progress
Solve the inequality. Graph the solution.

Question 1.
4b – 1 < 7
Answer:
The given inequality is:
4b – 1 < 7
So,
4b < 7 + 1
4b < 8
b < 8 / 4
b < 2
Hence, from the above,
We can conclude that the solution to the given inequality is b < 2
The representation of the inequality in the number line is:

Question 2.
8 – 9c ≥ -28
Answer:
The given inequality is:
8 – 9c ≥ -28
-9c ≥ -28 – 8
-9c ≥ -36
9c ≥ 36
c ≥ 36 / 9
c ≥ 4
Hence, from the above,
We can conclude that the solution to the given inequality is c ≥ 4
The representation of the inequality in the number line is:

Question 3.
\(\frac{n}{-2}\) + 11 > 12
Answer:
The given inequality is:
\(\frac{n}{-2}\) + 11 > 12
\(\frac{n}{-2}\) > 12 – 11
\(\frac{n}{-2}\) > 1
n > 1 (-2)
n > -2
Hence, from the above,
We can conclude that the solution to the given inequality is n > -2
The representation of the inequality in the number line is:

Question 4.
6 ≥ 5 – \(\frac{v}{3}\)
Answer:
The given inequality is:
6 ≥ 5 – \(\frac{v}{3}\)
6 – 5 ≥ – \(\frac{v}{3}\)
-1 ≥ \(\frac{v}{3}\)
-3 ≥ v
v ≤ -3
Hence, from the above,
We can conclude that the solution to the given inequality is v ≤ -3
The representation of the inequality in the number line is:

Solve the inequality.

Question 5.
5x – 12 ≤ 3x – 4
Answer:
The given inequality is:
5x – 12 ≤ 3x – 4
5x – 3x ≤ 12 – 4
2x ≤ 8
x ≤ 8 / 2
x ≤ 4
Hence, from the above,
We can conclude that the solution to the given inequality is x ≤ 4

Question 6.
2(k – 5) < 2k + 5
Answer:
The given inequality is:
2 ( k – 5 ) < 2k + 5
So,
2 ( k ) – 2 ( 5 ) < 2k + 5
2k – 10 < 2k + 5
2k – 2k < 5 + 10
-10 < 5
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 7.
-4(3n – 1) > -12n + 5.2
Answer:
The given inequality is:
-4 ( 3n – 1 ) > -12n + 5.2
So,
-4 ( 3n ) – 4 ( -1 ) > -12n + 5.2
-12n + 4 > -12n + 5.2
-12n + 12n + 4 > 5.2
4 > 5.2
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 8.
3(2a – 1) ≥ 10a – 11
Answer:
The given inequality is:
3 ( 2a – 1 ) ≥ 10a – 11
So,
3 ( 2a ) – 3 ( 1 ) ≥ 10a – 11
6a – 3 ≥ 10a – 11
6a – 10a ≥ -11 + 3
-4a ≥ -8
4a ≥ 8
a ≥ 8 / 4
a ≥ 2
Hence, from the above,
We can conclude that the solution to the given inequality is a ≥ 2

Question 9.
WHAT IF?
You need a mean score of at least 85 points to advance to the next round. What scores in the fifth game will allow you to advance?
Answer:

Solving Multi-step Inequalities 2.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Compare solving multi-step inequalities and solving multi-step equations.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 1

Question 2.
WRITING
Without solving, how can you tell that the inequality 4x + 8 ≤ 4x – 3 has no solution?
Answer:
The given inequality is:
4x + 8  ≤ 4x – 3
Rearrange the variables ad constant terms
So,
4x – 4x ≤ -3 – 8
0 + 8 ≤ -3
8 ≤ -3
Hence, from the above,
We can observe that there is no variable x
Hence,
We can conclude that there is no solution for the given inequality

In Exercises 3–6, match the inequality with its graph.

Question 3.
7b – 4 ≤ 10
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 3

Question 4.
4p + 4 ≥ 12
Answer:
The given inequality is:
4p + 4 ≥ 12
So,
4p ≥ 12 – 4
4p ≥ 8
p ≥ 8 / 4
p ≥ 2
Hence, from the above,
We can conclude that the solution to the given inequality is p ≥ 2
The graph represented for the solution is A) [ Se the graphs from Exercise 6 ]

Question 5.
-6g + 2 ≥ 20
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 5

Question 6.
3(2 – f) ≤ 15

Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 53
Answer:
The given inequality is:
3 (2 – f) ≤  15
So,
3 (2) – 3 (f) ≤  15
6 – 3f ≤  15
-3f ≤  15 – 6
-3f ≤  9
f ≤  -9 / 3
f ≤  -3
Hence, from the above,
We can conclude that the solution to the given inequality is f ≤  -3
The graph C) represents the solution of the given inequality

In Exercises 7–16, solve the inequality. Graph the solution.

Question 7.
2x – 3 > 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 7

Question 8.
5y + 9 ≤ 4
Answer:
The given inequality is:
5y + 9 ≤ 4
So,
5y ≤  4  -9
5y ≤  -5
y ≤  -5 / 5
y ≤  -1
Hence, from the above,
We can conclude that the solution to the given inequality is y ≤  -1
The representation of the solution in the graph is:

Question 9.
-9 ≤ 7 – 8v
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 9

Question 10.
2 > -3t – 10
Answer:
The given inequality is:
2 > -3t – 10
So,
2 + 10 > -3t
12 > -3t
-12 / 3 > t
-4 > t
t < -4
Hence, from the above,
We can conclude that the solution to the given inequality is t < -4
The representation of the solution of the inequality in the graph is:

Question 11.
\(\frac{w}{2}\) + 4 > 5
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 11

Question 12.
1 + \(\frac{m}{3}\) ≤ 6
Answer:
The given inequality is:
1 + \(\frac{m}{3}\) ≤ 6
So,
\(\frac{m}{3}\) ≤ 6 – 1
\(\frac{m}{3}\) ≤ 5
m ≤ 5 (3)
m ≤ 15
Hence, from the above,
We can conclude that the solution to the given inequality is m ≤ 15
The representation of the solution of the inequality in the graph is:

Question 13.
\(\frac{p}{-8}\) + 9 > 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 13

Question 14.
3 + \(\frac{r}{-4}\) ≤ 6
Answer:
The given inequality is:
3 + \(\frac{r}{-4}\) ≤ 6
\(\frac{r}{-4}\) ≤ 6 – 3
\(\frac{r}{-4}\) ≤ 3
r ≤ 3 (-4)
r ≤ -12
Hence, from the above,
We can conclude that the solution to the given inequality is r ≤ -12
The representation of the solution of the given inequality in the graph is:

Question 15.
6 ≥ -6(a + 2)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 15

Question 16.
18 ≤ 3(b – 4)
Answer:
The given inequality is:
18 ≤ 3 (b – 4)
So,
18 ≤ 3 (b) – 3 (4)
18 ≤ 3b – 12
18 + 12 ≤ 3b
30 ≤ 3b
30 / 3 ≤ b
10 ≤ b
b ≥ 10
Hence, from the above,
We can conclude that the solution to the given inequality is b ≥ 10
The representation of the solution of the given inequality in the graph is:

In Exercises 17–28, solve the inequality.

Question 17.
4 – 2m > 7 – 3m
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 17

Question 18.
8n+ 2 ≤ 8n – 9
Answer:
The given nequality is:
8n + 2 ≤ 8n – 9
So,
8n – 8n + 2 ≤ -9
2 ≤ 9
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 19.
-2d – 2 < 3d + 8
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 19

Question 20.
8 + 10f > 14 – 2f
Answer:
The given inequality is:
8 + 10f > 14 – 2f
10f + 2f > 14 – 8
12f > 6
f > 6 / 12
f > \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the solution to the given inequality is f > \(\frac{1}{2}\)

Question 21.
8g – 5g – 4 ≤ -3 + 3g
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 21

Question 22.
3w – 5 > 2w + w – 7
Answer:
The given inequality is:
3w – 5 > 2w + w – 7
So,
3w – 5 > 3w – 7
3w – 5 – 3w > -7
-5 > -7
5 > 7
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 23.
6(ℓ + 3) < 3(2ℓ + 6)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 23

Question 24.
2(5c – 7) ≥ 10(c – 3)
Answer:
The given inequality is:
2 (5c – 7) ≥ 10 (c – 3)
So,
2 (5c) – 2 (7) ≥ 10 (c) – 10 (3)
10c – 14 ≥ 10c – 30
10c – 10c – 14 ≥ -30
-14 ≥ -30
14 ≥ 30
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 25.
4 (\(\frac{1}{2}\)t – 2 )> 2(t – 3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 25

Question 26.
15 (\(\frac{1}{3}\)b + 3 ) ≤ 6(b + 9)
Answer:
The given inequality is:
15 ( (\(\frac{1}{3}\)b + 3 ) ) ≤ 6 (b + 9)
So,
15 ( (\(\frac{1}{3}\)b ) ) + 15 (3) ≤ 6 (b) + 6 (9)
5b + 45 ≤ 6b + 54
5b – 6b ≤ 54 – 45
-b ≤ 9
b ≤ -9
Hence, from the above,
We can conclude that the solution to the given inequality is b ≤ -9

Question 27.
9j – 6 + 6j ≥ 3(5j – 2)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 27

Question 28.
6h – 6 + 2h < 2(4h – 3)
Answer:
The given inequality is:
6h – 6 + 2h < 2 (4h – 3)
So,
8h – 6 < 2 (4h) – 2 (3)
8h – 6 < 8h – 6
8h – 8h < -6 + 6
0 < 0
Hence, from the above,
We can conclude that there is no solution for the given inequality

ERROR ANALYSIS
In Exercises 29 and 30, describe and correct the error in solving the inequality.

Question 29.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 54
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 29

Question 30.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 55
Answer:
The given inequality is:
-2 (1 – x) ≤ 2x – 7
-2 (1) + 2 (x) ≤ 2x – 7
-2 + 2x ≤ 2x – 7
-2 + 2x – 2x ≤ -7
-2 ≤ -7
2 ≤  7
Hence, from the above,
We can conclude that the given inequality has no solution

Question 31.
MODELING WITH MATHEMATICS
Write and solve an inequality that represents how many $20 bills you can withdraw from the account without going below the minimum balance.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 56
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 31

Question 32.
MODELING WITH MATHEMATICS
A woodworker wants to earn at least $25 an hour making and selling cabinets. He pays $125 for materials. Write and solve an inequality that represents how many hours the woodworker can spend building the cabinet.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 57
Answer:
The inequality that represents the number of hours the woodworker can spend build the cabinet is:
x ≥ 5

Explanation:
It is given that a woodworker wants to earn at least $25 an hour making and selling cabinets. He pays $125 for materials.
Let x be the number of hours taken by the woodworker to build the cabinet
So,
The inequality that represents the number of hours the woodworker can spend build the cabinet is:
25 (x) ≥ 125
x ≥ 125 / 25
x ≥ 5
Hence, from the above,
We can conclude that the inequality that represents the number of hours the woodworker can spnd build the cabinet is x ≥ 5

Question 33.
MATHEMATICAL CONNECTIONS
The area of the rectangle is greater than 60 square feet. Write and solve an inequality to find the possible values of x.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 58
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 33

Question 34.
MAKING AN ARGUMENT
Forest Park Campgrounds charges a $100 membership fee plus $35 per night. Woodland Campgrounds charges a $20 membership fee plus $55 per night. Your friend says that if you plan to camp for four or more nights, then you should choose Woodland Campgrounds. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
It is given that the Forest Park Campgrounds charges a $100 membership fee plus $35 per night and the Woodland Campgrounds charges a $20 membership fee plus $55 per night.
It is also given that your friend says that if you plan to camp for four or more nights, then you should choose Woodland Campgrounds.
So,
Let x be the number of nights you spend
Now,
The charge of the Forest Park Campgrounds for x nights = 100 + 35x
The charge of the Woodland Campgrounds = 20 + 55x
Now,
It is given that for four or more nights i.e., x ≥ 4, you will choose Woodland Campgrounds
So,
We have to verify the above statement
So,
20 + 55x ≥ 100 + 35x
55x – 35x ≥ 100 – 20
20x ≥ 80
x ≥ 80 / 20
x ≥ 4
Hence, from the above,
We can conclude that your friend is correct.

Question 35.
PROBLEM-SOLVING
The height of one story of a building is about 10 feet. The bottom of the ladder on the truck must be at least 24 feet away from the building. How many stories can the ladder reach? Justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 59
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 35.1
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 35.2

Question 36.
HOW DO YOU SEE IT?
The graph shows your budget and the total cost of x gallons of gasoline and a car wash. You want to determine the possible amounts (in gallons) of gasoline you can buy within your budget.
a. What is your budget?
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 60
Answer:
Your budget is: $40

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 60
From the graph,
We can observe that x-axis represents the amount of gasoline ( gallons ) and the y-axis represents the budget
From the graph,
The line is represented parallel to the x-axis i.e., Budget
So,
From the graph,
The given equation is:
y = 40
Hence, from the above,
We can conclude that the budget is: $40

b. How much does a gallon of gasoline cost? How much does a car wash cost?
Answer:
The cost of a gallon of gasoline is: $3.55
The cost of a car wash is: $8

Explanation:
It is given that the graph shows your budget and the total cost of x gallons of gasoline and a car wash.
From the graph,
We know that,
The x-axis represents the amount of gasoline ( gallons )
So,
The equation represented by the x-axis is:
y = 3.55x + 8
So,
From the equation,
We know that,
The coefficient of x represents the cost of x gallons of gasoline and the constant term represents the cost of a car wash
Hence, from the above,
We can conclude that
The cost of a gallon of gasoline is: $3.55
The cost of a car wash is: $8

c. Write an inequality that represents the possible amounts of gasoline you can buy.
Answer:
The inequality that represents the possible amounts of gasoline you can buy is:
x ≤ 9 gallons

Explanation:
From the graph,
We know that,
The x-axis represents the amounts of gasoline you can buy
We know that,
y = 3.55x + 8
We know that,
The budget will be less than or equal to $40
So,
3.55x + 8 ≤ 40
3.55x ≤ 40 – 8
3.55x ≤ 32
x ≤ 32 / 3.55
x ≤ 9 [ Approx. ]
Hence, from the above,
We can conclude that the inequality that represents the possible amounts of gasoline you can buy is:
x ≤ 9

d. Use the graph to estimate the solution of your inequality in part (c).
Answer:
From part (c),
The required inequality is:
3.55x + 8 ≤ 40
From the graph,
We can observe the x-axis and y-axis intersected at one particular point.
The particular point is the estimated solution of the inequality
So,
The particular point = (x , y ) = ( 9 , 40 )
Hence, from the above,
We can conclude that the estimated solution of your inequality in part (c) is ( 9, 40 )

Question 37.
PROBLEM-SOLVING
For what r values of r will the area of the shaded region be greater than or equal to 9(π – 2)?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 37

Question 38.
THOUGHT-PROVOKING
A runner’s times (in minutes) in the four races he has completed are 25.5, 24.3, 24.8, and 23.5. The runner plans to run at least one more race and wants to have an average time of less than 24 minutes. Write and solve an inequality to show how the runner can achieve his goal.
Answer:
The inequality that shows the runner can achieve his goal is:
x < 21.9 minutes

Explanation:
It is given that a runner’s times (in minutes) in the four races he has completed are 25.5, 24.3, 24.8, and 23.5.
It is also given that the runner plans to run at least one more race and wants to have an average time of less than 24 minutes.
Now,
Let x be the time in the fifth race [ Since it is given that the runner plans to run at least one more race ]
We know that,
Average = \(\frac{The sum of all the items}{The number of items}\)
Now,
Average = \(\frac{25.5 + 24.3 + 24.8 + 23.5 + x}{5}\)
Average = \(\frac{98.1 + x}{5}\)
It is given that the average time will be less than 24 minutes
So,
\(\frac{98.1 + x}{5}\) < 24
98.1 + x < 24 (5)
98.1 + x < 120
x < 120 – 98.1
x < 21.9 minutes
Hence, from the above,
We can conclude that the inequality that shows the runner can achieve his goal is x < 21.9 minutes

REASONING
In Exercises 39 and 40, find the value of a for which the solution of the inequality is all real numbers.

Question 39.
a(x + 3) < 5x + 15 – x
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 39

Question 40.
3x + 8 + 2ax ≥ 3ax – 4a
Answer:
The value of a is: 3

Explanation:
The given inequality is:
3x + 8 + 2ax ≥ 3ax – 4a
So,
3x + 2ax – 3ax ≥ -4a – 8
3x – ax ≥ -4a – 8
It is given that the solution of the given inequality is all real numbers
So,
x = 0
So,
To make x = 0,
We can observe from the inequality that ‘a’ must be equal to 3
Hence, from the above,
We can conclude that the value of a is: 3

Maintaining Mathematical Proficiency
Write the sentence as an inequality. (Section 2.1)

Question 41.
Six times a number y is less than or equal to 10.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 41

Question 42.
A number p plus 7 is greater than 24.
Answer:
The given worded form is:
A number p plus 7 is greater than 24
Hence,
The representation of the worded form in the form of inequality is:
p + 7 > 24

Question 43.
The quotient of a number r and 7 is no more than 18.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.4 Question 43

Solving Linear Inequalities Study Skills: Analyzing Your Errors

2.1–2.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 61

Core Concepts
Section 2.1
Representing Linear Inequalities, p. 57

Section 2.2
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 62

Section 2.3
Multiplication and Division Properties of Inequality (c > 0), p. 68
Multiplication and Division Properties of Inequality (c < 0), p. 69

Section 2.4
Solving Multi-Step Inequalities, p. 74
Special Solutions of Linear Inequalities, p. 75

Mathematical Practices

Question 1.
Explain the meaning of the inequality symbol in your answer to Exercise 47 on page 59. How did you know which symbol to use?
Answer:
In Exercise 47 on page 59,
The inequality symbol we used is: ≤
The meaning of ≤ is ” Less than or equal to ”
In Exercise 47,
It is given that the Xianren bridge arch is the longest natural arch with a length of 400 feet i.e., there is no arch longer than the Xianren bridge arch and the remaining natural arches are shorter than the Xianren arch
Hence,
The lengths of all the arches including the Xianren arch will be represented by the inequality symbol “≤”

Question 2.
In Exercise 30 on page 66, why is it important to check the reasonableness of your answer in part (a) before answering part (b)?
Answer:
In part (a), it is given that you have to beat your competitor with your score.
So,
Your score must be greater than your competitor
Then only you can solve part (b).

Question 3.
Explain how considering the units involved in Exercise 29 on page 71 helped you answer the question.
Answer:
In Exercise 29 on page 71,
The mat that is given is in the form of the square since its length and width both are the same
So,
We know that,
Area of the square = Side²
Let the cost per square foot be x
So,
The inequality formed will be like:
( Area of the square ) ⋅ x ≤ $700

Study Skills

Analyzing Your Errors

Application Errors
What Happens: You can do numerical problems, but you struggle with problems that have context.
How to Avoid This Error: Do not just mimic the steps of solving an application problem. Explain out loud what the question is asking and why you are doing each step. After solving the problem, ask yourself, “Does my solution make sense?”
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 63

Solving Linear Inequalities Maintaining 2.1 – 2.4 Quiz

Write the sentence as an inequality. (Section 2.1)

Question 1.
A number z minus 6 is greater than or equal to 11.
Answer:
The given worded form is:
A number z minus 6 is greater than or equal to 11
Hence,
The representation of the worded form in the form of inequality is:
z – 6 ≥ 11

Question 2.
Twelve is no more than the sum of -1.5 times a number w and 4.
Answer:
The given worded form is:
Twelve is no more than the sum of -1.5 times a number w and 4
Hence,
The representation of the worded form in the form of inequality is:
12 ≤ -1.5w + 4

Write an inequality that represents the graph.(Section 2.1)

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 64
Answer:
The inequality that represents the given graph is:
x < 0

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 64
From the above number line,
We can observe that the marked line started from 0 that excludes 0 and continued till the left end of the number line
Hence,
The inequality that represents the given graph is:
x < 0

Question 4.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 65
Answer:
The inequality that represents the given graph is:
x ≥ 8

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 65
From the above graph,
The marked line started from 8 that includes 8 and continued till the right end of the number line.
Hence,
The inequality that represents the given graph is:
x ≥ 8

Solve the inequality. Graph the solution. (Section 2.2 and Section 2.3)

Question 5.
9 + q ≤ 15
Answer:
The solution to the given inequality is:
q ≤ 6

Explanation:
The given inequality is:
9 + q ≤ 15
So,
q ≤ 15 – 9
q ≤ 6
Hence, from the above,
We can conclude that the solution to the given inequality is q ≤ 6
The representation of the solution of the inequality in the graph is:

Question 6.
z – (-7) < 5
Answer:
The solution to the given inequality is:
z < -2

Explanation:
The given inequality is:
z – (-7) < 5
So,
z + 7  < 5
z < 5 – 7
z < -2
Hence, from the above,
We can conclude that the solution to the given inequality is z < -2
The representation of the solution of the inequality in the graph is:

Question 7.
-3 < y – 4
Answer:
The solution to the given inequality is:
y > 1

Explanation:
The given inequality is:
-3 < y – 4
So,
-3 + 4 < y
1 < y
y > 1
Hence, from the above,
We can conclude that the solution to the given inequality is y > 1
The representation of the solution of inequality in the graph is:

Question 8.
3p ≥ 18
Answer:
The solution to the given inequality is:
p ≥ 6

Explanation:
The given inequality is:
3p ≥ 18
p ≥ 18 / 3
p ≥ 6
Hence, from the above,
We can conclude that the solution to the given inequality is p ≥ 6
The representation of the solution of the inequality in the graph is:

Question 9.
6 > \(\frac{w}{-2}\)
Answer:
The solution to the given inequality is:
w < -12

Explanation:
The given inequality is:
6 > \(\frac{w}{-2}\)
So,
6 (-2) > w
-12 > w
w < -12
Hence, from the above,
We can conclude that the solution to the given inequality is w < -12
The representation of the solution of the inequality in the graph is:

Question 10.
-20x > 5
Answer:
The solution to the given inequality is:
x > – \(\frac{1}{4}\)
So,
x > -0.25

Explanation:
The given inequality is:
-20x > 5
So,
x > -5 / 20
x > – \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the solution to the given inequality is x > –\(\frac{1}{4}\)
The representation of the solution of the inequality in the graph is:

Solve the inequality. (Section 2.4)

Question 11.
3y – 7 ≥ 17
Answer:
The solution to the given inequality is:
y ≥ 8

Explanation:
The given inequality is:
3y – 7 ≥ 17
So,
3y ≥ 17 + 7
3y ≥ 24
y ≥ 24 / 3
y ≥ 8
Hence, from the above,
We can conclude that the solution to the given inequality is y ≥ 8

Question 12.
8(3g – 2) ≤ 12(2g + 1)
Answer:
There is no soution for the given inequality

Explanation:
The given inequality is:
8 (3g – 2) ≤ 12 (2g + 1)
So,
8 (3g) – 8 (2) ≤ 12 (2g) + 12 (1)
24g – 16 ≤ 24g + 12
24g – 24g – 16 ≤ 12
-16 ≤ 12
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 13.
6(2x – 1) ≥ 3(4x + 1)
Answer:
There is no solution for the given inequality

Explanation:
The given inequality is:
6 (2x – 1)≥ 3 (4x + 1)
So,
6 (2x) – 6 (1) ≥ 3 (4x) + 3 (1)
12x – 6 ≤ 12x + 3
12x – 6 – 12x ≤ 3
-6 ≤ 3
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 14.
Three requirements for a lifeguard training course are shown. (Section 2.1)
a. Write and graph three inequalities that represent the requirements.
b. You can swim 250 feet, tread water for 6 minutes, and swim 35 feet underwater without taking a breath. Do you satisfy the requirements of the course? Explain.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 66
Answer:
a.
The given requirements are:
1. Swim at least 100 yards
2. Tread water for at least 5 minutes
3. Swim 10 yards or more underwater without taking a breath
Now,
1.
The inequality that represents the first requirement is:
x ≥ 100 yards
2.
The inequality that represents the second requirement is:
x ≥ 5 minutes
3.
The inequality that represents the third requirement is:
x ≥ 10 yards

b.
It is given that you can swim 250 feet, tread water for 6 minutes, and swim 35 feet underwater without taking a breath.
From part (a),
The inequalities for the three requirements are:
x ≥ 100 yards
x ≥ 5 minutes
x ≥ 10 yards
Hence, from the above,
We can conclude that you satisfied the requirements of the course

Question 15.
The maximum volume of an American white pelican’s bill is about 700 cubic inches. A pelican scoops up 100 cubic inches of water. Write and solve an inequality that represents the additional volumes the pelican’s bill can contain. (Section 2.2)
Answer:
The inequality that represents the additional volumes of the pelican’s bill can contain is:
x  ≤ 600

Explanation:
It is given that the maximum volume of an American white pelican’s bill is about700 cubic inches. A pelican scoops up 100 cubic inches of water.
Let x be the additional volumes the pelican’s bill can contain
It is also given that the maximum volume is 700 cubic inches and a pelican’s bill scoops up 100 cubic inches of water
Hence,
The inequality that represents the additional volumes of the pelican’s bill can contain is:
x + 100 ≤ 700
x ≤ 700 – 100
x ≤ 600

Question 16.
You save $15 per week to purchase one of the bikes shown. (Section 2.3and Section 2.4)
a. Write and solve an inequality to find the numbers of weeks you need to save to purchase a bike.
b. Your parents give you $65 to help you buy the new bike. How does this affect your answer in part (a)? Use an inequality to justify your answer.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 67
Answer:
a.
It is given that you save $15 per week to purchase one of the bikes
From the fig,
It is given that the starting price of the bike is $120
Now,
Let x be the number of weeks
Hence,
The inequality that representing the number of weeks you need to save to purchase a bike is:
15 ⋅ x ≥ 120
15x ≥ 120
x ≥ 120 / 15
x ≥ 8
Hence, from the above,
We can conclude that the number of weeks you need to save to purchase a bike is 8 weeks

b.
It is given that your parents give you $65 to help you buy the new bike.
We know that,
The starting price of the bike is: $120
So,
The remaining amount that needs to purchase a bike = 120 – 65
= $55

Lesson 2.5 Solving Compound Inequalities

Essential Question
How can you use inequalities to describe intervals on the real number line?
EXPLORATION 1
Describing Intervals on the Real Number
Work with a partner.
In parts (a)–(d), use two inequalities to describe the interval.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 68
e. Do you use “and” or “or” to connect the two inequalities in parts (a)–(d)? Explain.
Answer:
(a) – (d):
The given graphs are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 68
From the given graphs,
We can observe that the marked lines are from a certain number to a certain number
Now,
a.
In the given graph,
The marked line starting from -3 excluding -3 and continued till -6 including -6
So,
The inequality that represents the given interval is:
x ≥ -6 and x < -3
Hence,
The compound inequality that represents the given interval is:
-6 ≤ x < -3

b.
In the given graph,
The marked line starting from -5 excluding -5 and continued till 4 including 4
So,
The inequality that represents the given interval is:
x > -5 and x < 4
Hence,
The compound inequality that represents the given interval is:
-5 < x < 4

c.
In the given graph,
The marked line starting from -4 including -4 and continued till 5 including 5
So,
The inequality that represents the given interval is:
x ≥ -4 and x ≤ 5
Hence,
The compound inequality that represents the given interval is:
-4 ≤ x ≤ 5

d.
In the given graph,
The marked line starting from -3 excluding -3 and continued till 6 including 6
So,
The inequality that represents the given interval is:
x > -3 and x < 6
Hence,
The compound inequality that represents the given interval is:
-3 < x < 6

e.
We use “and” to connect the two inequalities in parts (a) – (d)
Now,
From (a) – (d),
We can observe that,
For Half-open interval,
We use ≥ and <
For closed interval,
We use ≥ and ≤
For open interval,
We use > and <

EXPLORATION 2
Describing Two Infinite Intervals
Work with a partner. In parts (a)–(d), use two inequalities to describe the interval.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 69
e. Do you use “and” or “or” to connect the two inequalities in parts (a)–(d)? Explain.
Answer:
The given graphs are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 69
(a) – (d):
a.
In the given graph,
The first marked line starting from -6 including -6 and continued till the end of the left side of the graph
The second marked line starting from 3 excluding 3 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x ≤ -6 and x > 3
Hence,
The compound inequality that represents the given interval is:
-6 ≤ x > 3

b.
In the given graph,
The first marked line starting from -5 excluding -5 and continued till the end of the left side of the graph
The second marked line starting from 4 including 4 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x < -5 and x ≥ 4
Hence,
The compound inequality that represents the given interval is:
-5 < x ≥ 4

c.
In the given graph,
The first marked line starting from -4 including -4 and continued till the end of the left side of the graph
The second marked line starting from 5 including 5 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x ≤ -4 and x ≥ 5
Hence,
The compound inequality that represents the given interval is:
-4 ≤ x ≥ 5

d.
In the given graph,
The first marked line starting from -3 excluding -3 and continued till the end of the left side of the graph
The second marked line starting from 6 excluding 6 and continued till the end of the right side of the graph
So,
The inequality that represents the given intervals are:
x < -3 and x > 6
Hence,
The compound inequality that represents the given interval is:
-3 < x > 6

e.
We use “and” to connect the two inequalities in parts (a) – (d)
Now,
From (a) – (d),
We can observe that,
For Half-open interval,
We use ≥ and <
For closed interval,
We use ≥ and ≤
For open interval,
We use > and <

Communicate Your Answer

Question 3.
How can you use inequalities to describe intervals on the real number line?
Answer:
An inequality that includes the boundary point indicated by the “or equal” part of the symbols ≤ and ≥ with a closed dot on the number line. The symbol (∞) indicates the interval is unbounded to the right. Express ordering relationships using the symbol < for “less than” and > for “greater than.”

2.5 Lesson

Monitoring Progress

Write the sentence as an inequality. Graph the inequality.

Question 1.
A number d is more than 0 and less than 10.
Answer:
The given worded form is:
A number d is more than 0 and less than 10
Hence,
The representation of the given worded form in the form of inequality is:
d > 0 and d < 10
The representation of the inequalities in the form of compound inequality is:
0 < d < 10
The representation of the compound inequality in the graph is:

Question 2.
A number a is fewer than -6 or no less than -3.
Answer:
The given worded form is:
A number a is fewer than -6 or no less than -3
Hence,
The representation of the given worded form in the form of inequality is:
a < -6 or a > -3
The representation of the inequalities in the graph is:

Solve the inequality. Graph the solution.

Question 3.
5 ≤ m + 4 < 10
Answer:
The given inequality is:
5 ≤ m + 4 < 10
Subtract by 4 on both sides
5 – 4 ≤ m + 4 – 4 < 10 – 4
1 ≤ m < 6
Hence, from the above,
We can conclude that the compound inequality for the given inequality is: 1 ≤ m < 6
The representation of the compound inequality in the graph is:

Question 4.
-3 < 2k – 5 < 7
Answer:
The given inequality is:
-3 < 2k – 5 < 7
Add 5 on both sides
-3 + 5 < 2k – 5 + 5 < 7 + 5
2 < 2k < 12
Divide by 2 on both sides
1 < k < 6
Hence, from the avove,
We can conclude that the compound inequality for the given inequality is: 1 < k < 6
The representation of the compound inequality in the graph is:

Question 5.
4c + 3 ≤ -5 or c – 8 > -1
Answer:
The given inequality is:
4c + 3 ≤ -5 or c – 8 > -1
So,
4c ≤ -5 – 3 or c > -1 + 8
4c ≤ -8 or c > 7
c ≤ -8 / 4 or c > 7
c ≤ -2 or c > 7
Hence, from the above,
We can conclude that the solutions to the given inequality are: c ≤ -2 or c > 7
The representations of the solutions in the graph are:

Question 6.
2p + 1 < -7 or 3 – 2p ≤ -1
Answer:
The given inequality is:
2p + 1 < -7 or 3 – 2p ≤ -1
So,
2p < -7 – 1 or -2p ≤ -1 – 3
2p < -8 or -2p ≤ -4
p < -8 / 2 or p ≤ -4 / (-2)
p < -4 or p ≤ 2
Hence, from the above,
We can conclude that the solutions to the given inequality are: p < -4 or p ≤ 2
The representations of the solutions in the graph are:

Question 7.
Write and solve a compound inequality that represents the temperature rating (in degrees Fahrenheit) of the winter boots.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 70
Answer:
The range of the temperature rating ( In °C ) of the winter boots is: -40°C to 15°C
We know that,
°F = ( °C × \(\frac{9}{5}\) ) + 32
Now,
To convert -40°C into °F,
°F = ( -40°C × \(\frac{9}{5}\) ) + 32
°F = -72 + 32
°F = – 40°F
Now,
To convert 15°C into °F,
°F = ( 15 × \(\frac{9}{5}\) ) + 32
°F = 27 + 32
°F = 59°F
Let t be the temperature in °F
So,
The representation of the temperature ranges in the form of inequalities is:
t > -40°F and t < 59°F
Hence,
The representation of the temperature ranges in the form of compound inequality is:
-40°F < t < 59°F

Solving Compound Inequalities 2.5 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING
Compare the graph of -6 ≤ x ≤ -4 with the graph of x ≤ -6 or x ≥ -4.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 1

Question 2.
WHICH ONE do DOESN’T BELONG?
Which compound inequality does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 70.1
Answer:
The given inequalities are:
a. a > 4 or a < -3
b. a< -2 or a > 8
c. a > 7 or a < -5
d. a < 6 or a > -9
Now,
Represent all the inequalities in the graph
So,
a.
The given inequality is:
a > 4 or a < -3
The representation of the inequalities in the graph is:

b.
The given inequality is:
a < -2 or a > 8
The representation of the inequalities in the graph is:

c.
The given inequality is:
a > 7 or a < -5
The representation of the inequalities in the graph is:

d.
The given inequality is:
a < 6  or a > -9
The representation of the inequalities in the graph is:

From the above,
We can conclude that inequality c) does not belong with the other three inequalities

In Exercises 3–6, write a compound inequality that is represented by the graph.

Question 3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 71
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 3

Question 4.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 72
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 72
From the above graph,
We can observe that the marked line started from 7 excluding 7 and continued till 14 excluding 14
Hence,
The representation of the inequalities from the given graph is:
x > 7 and x < 14
The representation of the inequalities in the form of compound inequality is:
7 < x < 14

Question 5.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 73
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 5

Question 6.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 74
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 74
From the above graph,
We can observe that
The first marked line started from 4 including 4 and continued till the end of the left end of the graph
The second marked line started from 6 excluding 6 and continued till the end of the right end of the graph
Hence,
The representation of the given graph in the form of inequality is:
x ≤ 4 or x > 6

In Exercises 7–10, write the sentence as an inequality. Graph the inequality.

Question 7.
A number p is less than 6 and greater than 2.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 7

Question 8.
A number n is less than or equal to -7 or greater than 12.
Answer:
The given worded form is:
A number n is less than or equal to -7 or greater than 12
Hence,
The representation of the given worded form in the form of inequality is:
n ≤ -7 or n > 12
The representation of the inequalities in the graph is:

Question 9.
A number m is more than -7\(\frac{2}{3}\) or at most -10.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 9

Question 10.
A number r is no less than -1.5 and fewer than 9.5.
Answer:
The given worded form is:
A number r is no less than -1.5 and fewer than 9.5
Hence,
The representation of the given worded form in the form of inequality is:
r > -1.5 and r < 9.5
Hence,
The representation of the given worded form in the form of compound inequality is:
-1.5 < r < 9.5
The representation of the compound inequality in the graph is:

Question 11.
MODELING WITH MATHEMATICS
Slitsnails are large mollusks that live in deep waters. They have been found in the range of elevations shown. Write and graph a compound inequality that represents this range.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 75
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 11

Question 12.
MODELING WITH MATHEMATICS
The life zones on Mount Rainier, a mountain in Washington, can be approximately classified by elevation, as follows.
Low-elevation forest: above 1700 feet to 2500 feet mid-elevation forest: above 2500 feet to 4000 feet Subalpine: above 4000 feet to 6500 feet
Alpine: above 6500 feet to the summit
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 76
Write a compound inequality that represents the elevation range for each type of plant life.
a. trees in the low-elevation forest zone
b. flowers in the subalpine and alpine zones
Answer:
It is given that the elevations of the life zones on Mount Rainier are as follows:
Low-elevation forest: above 1700 feet to 2500 feet
Mid-elevation forest: above 2500 feet to 4000 feet
Subalpine: above 4000 feet to 6500 feet
Alpine: above 6500 feet to the summit
Let the elevation range for each type of plant life be x
So,
a.
The inequality that represents the elevation range of trees in the low-elevation forest zone is:
x > 1700 feet and x < 2500 feet
Hence,
The representation of the elevation range of trees in the low-elevation forest zone in the form of compound inequality is:
1700 feet < x < 2500 feet
b.
The inequalities that represent the elevation range of trees in the subalpine and alpine zones is:
Subalpine zone: x > 4000 feet and x < 6500 feet
Alpine zone: x > 6500 feet
Hence,
The representation of the elevation range of trees in the Subalpine zone and Alpine zone in the form of compound inequality is:
Subalpine zone: 4000 feet < x < 6500 feet
Alpine zone: x > 6500 feet

In Exercises 13–20, solve the inequality. Graph the solution.

Question 13.
6 < x + 5 ≤ 11
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 13

Question 14.
24 > -3r ≥ -9
Answer:
The given inequality is:
24 > -3r ≥ -9
Divide by 3 on both sides
( 24 / 3 ) > ( -3r / 3 ) ≥ ( -9 / 3 )
8 > -r ≥ -3
Multiply with ‘-‘ on both sides
-8 < r ≤ 3 [ Because when we multiply with ‘-‘, we have to change the inequality signs also ]
Hence,
The representation of the solution of the given inequality in the graph is:

Question 15.
v + 8 < 3 or -8v < -40
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 15

Question 16.
-14 > w + 3 or 3w ≥ -27
Answer:
The given inequality is:
-14 > w + 3 or 3w ≥ -27
So,
-14 – 3 > w or w ≥ -27 / 3
-17 > w or w ≥ -9
w < -17 or w ≥ -9
Hence,
The representation of the solutions of the given inequality in the graph is:

Question 17.
2r + 3 < 7 or -r + 9 ≤ 2
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 17

Question 18.
-6 < 3n + 9 < 21
Answer:
The given inequality is:
-6 < 3n + 9 < 21
Subtract with 9 on both sides
-6 – 9 < 3n + 9 – 9 < 21 – 9
-15 < 3n < 12
Divide by 3 on both sides
( -15 / 3 ) < ( 3n / 3 ) < ( 12 / 3 )
-5 < n < 4
Hence,
The representation of the solution of the inequality in the graph is:

Question 19.
-12 < \(\frac{1}{2}\)(4x + 16) < 18
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 19

Question 20.
35 < 7(2 – b) or \(\frac{1}{3}\)(15b – 12) ≥ 21
Answer:
The given inequality is:
35 < 7 (2 – b) or \(\frac{1}{3}\) ( 15b – 12 ) ≥ 21
So,
35 < 7 (2) – 7 (b) or 15b – 12 ≥ 21 (3)
35 < 14 – 7b or 15b – 12 ≥ 63
7b < 14 – 35 or 15b ≥ 63 + 12
7b < -21 or 15b ≥ 75
b < -21 / 7 or b ≥ 75 / 15
b < -3 or b ≥ 5
Hence,
The representation of the solutions of the given inequality in the graph is:

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in solving the inequality or graphing the solution.

Question 21.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 77
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 21

Question 22.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 78
Answer:
The given inequality is:
x – 2 > 3 or x + 8 < -2
So,
x > 3 + 2 or x < -2 – 8
x > 5 or x < -10
Hence,
The representation of the solutions of the inequality in the graph is:

Question 23.
MODELING WITH MATHEMATICS
Write and solve a compound inequality that represents the possible temperatures (in degrees Fahrenheit) of the interior of the iceberg.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 79
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 23

Question 24.
PROBLEM-SOLVING
A ski shop sells skis with lengths ranging from 150 centimeters to 220 centimeters. The shop says the length of the skis should be about 1.16 times a skier’s height (in centimeters). Write and solve a compound inequality that represents the heights of skiers the shop does not provide skis for.
Answer:
The compound inequality that represents the heights of skiers the shop does not provide skis for is:
129.3 centimeters > h > 189.7 centimeters

Explanation:
It is given that a ski shop sells skis with lengths ranging from 150 centimeters to 220 centimeters.
It is also given that the shop says the length of the skis should be about 1.16 times a skier’s height (in centimeters).
Now,
Let x be the possible length of the skis
Let h be the possible height of the skiers for whom the shop does not provide skis for
So,
The inequality that represents the lengths of the skis is:
x < 150 centimeters or x > 220 centimeters
So,
The compound inequality that represents the length of the skis is:
150 centimeters > x > 220 centimeters
But,
x = 1.16h
So,
150 centimeters > 1.16h > 220 centimeters
(150 / 1.16) centimeters > h > (220 / 1.16) centimeters
129.3 centimeters > h > 189.7 centimeters
Hence, from the above,
We can conclude that the compound inequality that represents the heights of skiers the shop does not provide skis for is:
129.3 centimeters > h > 189.7 centimeters

In Exercises 25–30, solve the inequality. Graph the solution, if possible.

Question 25.
22 < -3c + 4 < 14
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 25

Question 26.
2m – 1 ≥ 5 or 5m > -25
Answer:
The given inequality is:
2m – 1≥ 5 or 5m > -25
So,
2m ≥ 5 + 1 or m > -25 / 5
2m ≥ 6 or m > -5
m ≥ 6 / 2 or m > -5
m ≥ 3 or m > -5
Hence, from the above
We can conclude that the solutions to the given inequality are:
m ≥ 3 or m > -5
The compound inequality of the solutions of the given inequality is:
m > -5
The representation of the solutions of the given inequality in the graph is:

Question 27.
-y + 3 ≤ 8 and y + 2 > 9
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 27

Question 28.
x – 8 ≤ 4 or 2x + 3 > 9
Answer:
The given inequality is:
x – 8 ≤ 4 or 2x + 3 > 9
So,
x ≤ 4 + 8 or 2x >  9 – 3
x ≤ 12 or 2x > 6
x ≤ 12 or x > 6 / 2
x ≤ 12 or x > 3
Hence, from the above,
We can conclude that the solutions to the given inequality are:
x ≤ 12 or x > 3
The compound inequality of the solutions of the given inequality is:
3 < x ≤ 12
The representation of the solutions of the inequalities in the graph is:

Question 29.
2n + 19 ≤ 10 + n or -3n + 3 < -2n + 33
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 29

Question 30.
3x – 18 < 4x – 23 and x – 16 < -22
Answer:
The given inequality is:
3x – 18 < 4x – 23 and x – 16 < -22
So,
3x – 4x < -23 + 18 and x < -22 + 16
-x < -5 and x < – 6
x < 5 and x < -6
Hence, from the above,
We can conclude that the solutions to the given inequality are:
x < 5 and x < -6
The compound inequality that represents the solutions of the given inequality is:
x < 5
The representation of the solutions of the given inequality in the graph is:

Question 31.
REASONING
Fill in the compound inequality Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 110 and 5(x + 2) ≥ 2(x + 8) with <, ≤, >, or ≥ so that the solution is only one value.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 31

Question 32.
THOUGHT-PROVOKING
Write a real-life story that can be modeled by the graph.
Answer:
The real-life situation that can be modeled by the graph is:
Suppose we have 100 children in a school. Out of the 100 children, there are 60 boys and the rest are girls.
The attendance of all the boys in a particular month is 50% and all of the girls is 70%. Mark the percentage attendance that is between the attendance of all the boys and girls?
The graph representing the above real-life situation is:

Question 33.
MAKING AN ARGUMENT
The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Use the triangle shown to write and solve three inequalities. Your friend claims the value of x can be 1. Is your friend correct? Explain.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 111
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 33

Question 34.
HOW DO YOU SEE IT?
The graph shows the annual profits of a company from 2006 to 2013.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 80
a. Write a compound inequality that represents the annual profits from 2006 to 2013.
Answer:
The given graph is:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 80
From the graph,
We can observe that the annual profits from 2006 to 2013
So,
The minimum value in the graph is: $65 million
The highest value in the graph is: $100 million
Let x be the annual profit
Hence,
The inequalities that represent the annual profits from 2006 to 2013 are:
x ≥ $65 million and x ≤ $100 million
So,
The compound inequality that represents the annual profits from 2006 to 2013 is:
$65 million ≤ x ≤ $100 million

b. You can use the formula P = R – C to find the profit P, where R is the revenue and C is the cost. From 2006 to 2013, the company’s annual cost was about $125 million. Is it possible the company had annual revenue of $160 million from 2006 to 2013? Explain.
Answer:
Yes, it is possible that the company had annual revenue of $160 million from 2006 to 2013

Explanation:
From part (a),
The total profit from 2006 to 2013 is:
P = 65 + 70 + 85 + 50 + 65 + 70 + 90
= $575 million
It is given that you can use the formula P = R – C to find the profit P, where R is the revenue and C is the cost.
It is also given that from 2006 to 2013, the company’s annual cost was about $125 million.
So,
P = R – C
R = P + C
R = 575 + 125
= $700 million
Hence, from the above,
We can conclude that it is possible that the company had annual revenue of $160 million from 2006 to 2013

Maintaining Mathematical Proficiency

Solve the equation. Graph the solutions, if possible.

Question 35.
| \(\frac{d}{9}\) | = 6
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 35

Question 36.
7 | 5p – 7 | = -21
Answer:
The given absolute value equation is:
7 | 5p – 7 | = 21
So,
| 5p – 7 | = -21 / 7
| 5p – 7 | = -3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
5p – 7 = -3 or 5p – 7 = – (-3)
5p = -3 + 7 or 5p = 3 + 7
5p = 4 or 5p = 10
p = \(\frac{4}{5}\) or p = 2
p = 0.8 or p = 2
p = 1 [Approx. value] or p = 2
Hence, from the above,
We can conclude that the solutions to the given absolute value equation are:
p = 1 or p = 2
The representation of the solutions of the given absolute value equation in the graph is:

Question 37.
| r + 2 | = | 3r – 4 |
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 37.1
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 37.2

Question 38.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 81
Answer:
The given absolute value equation is:
| \(\frac{1}{2}\)w – 6 | = | w + 7 |
So,
\(\frac{1}{2}\)w – 6 = w + 7
\(\frac{1}{2}\)w – w = 7 + 6
–\(\frac{1}{2}\)w = 13
w = 13 (-2)
w = -26
Hence, from the above,
We can conclude that the solution to the given absolute value equation is:
w = -26
The representation of the solution of the given absolute value equation in the graph is:

Find and interpret the mean absolute deviation of the data.

Question 39.
1, 1, 2, 5, 6, 8, 10, 12, 12, 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.5 Question 39

Question 40.
24, 26, 28, 28, 30, 30, 32, 32, 34, 36
Answer:
The given numbers are:
24, 26, 28, 28, 30, 30, 32, 32, 34, 36
We know that,
Mean = \(\frac{The sum of the given numbers}{The total number of numbers}\)
= \(\frac{24 + 26+ 28 + 28 + 30 + 30 + 32 + 32 + 34 + 36}{10}\)
= \(\frac{300}{10}\)
= 30
Now,
We know that,
Absolute deviation = ( The mean ) – ( Given number )
So,
Absolute deviations are:
30 – 24, 30 – 26, 30 – 28, 30 – 28, 30 – 30, 30 – 30, 30 – 32, 30 – 32, 30 – 34, 30 – 36
= 6, 4, 2, 2, 0, 0, -2, -2, -6, -4
So,
Absolute Mean deviation = \(\frac{Sum of Absolute deviations}{The numebr of absolute deviations}\)
= \(\frac{6 + 4 + 2 + 2 + 0 + 0 – 2 – 2 – 6 – 4}{10}\)
= \(\frac{0}{10}\)
= 0
Hence, from the above,
We can conclude that there is no deviation

Lesson 2.6 Solving Absolute Value Inequalities

Essential Question
How can you solve an absolute value inequality? Solving an Absolute Value Inequality Algebraically
EXPLORATION 1
Solving an Absolute Value Inequality Algebraically
Work with a partner.

Consider the absolute value inequality | x + 2 | ≤ 3.
a. Describe the values of x + 2 that make the inequality true. Use your description to write two linear inequalities that represent the solutions of the absolute value inequality.
b. Use the linear inequalities you wrote in part (a) to find the solutions of the absolute value inequality.
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 82
c. How can you use linear inequalities to solve an absolute value inequality?
Answer:
The given absolute value equation is:
| x + 2 | ≤ 3
a.
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 2 ≤ 3 or x + 2 ≤ -3
x ≤ 3 – 2 or x ≤ -3 – 2
x ≤ 1 or x ≤ -5
Hence, from the above,
We can conclude that the solutions to the given absolute value equation is:
x ≤ 1 or x ≤ -5

b.
The given absolute value equation is:
| x + 2 | ≤ 3
From part (a),
The solutions of the given absolute value equation are:
x ≤ 1 or x ≤ -5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
For x = 1,
| 1 + 2 | ≤ 3
| 3 | ≤ 3
3 ≤ 3
For x = -5,
| -5 + 2 | ≤ 3
| -3 | ≤ 3
3 ≤ 3

c.
The following are the steps to solve the absolute value inequality:
A) Isolate the absolute value expression on the left side of the inequality
B) If the number on the other side of the inequality sign is negative, then your equation either has no solution or all real numbers as solutions
C) Remove the absolute value bars by setting up a compound inequality
D) Solve the inequalities

EXPLORATION 2
Solving an Absolute Value Inequality Graphically
Work with a partner.
Consider the absolute value inequality
| x + 2 | ≤ 3.
a. On a real number line, locate the point for which | x + 2 | ≤ 3
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 83
Answer:
The given absolute value inequality is:
| x + 2 | ≤ 3
From Exploration 1,
The solutions of the absolute value inequality | x + 2 | ≤ 3 are:
x ≤ 1 and x ≤ -5
The compound inequality of the solution of the given absolute value inequality is:
x ≤ 1 ( Since the absolute value inequality can never be negative )
Now,
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 83
Hence,
The representation of the solutions of the absolute value inequality is:

b. Locate the points that are within 3 units from the point you found in part (a). What do you notice about these points?
Answer:
From part (a),
The compound inequality that represents the solution of the given absolute value inequality is:
x ≤ 1
The points that are within 3 units from the point x ≤ 1, i.e., the given point + 3
So,
1 + 3 = 4
Hence,
The point we have to locate in the graph is:
x ≤ 4
Hence,
The representation of the compound inequality of the given absolute value inequality in the graph is:

c. How can you use a number line to solve an absolute value inequality?
Answer:
You begin the marking of the points on the number line by making it into separate equations and then solving them separately. An absolute value equation has no solution if the absolute value expression equals a negative number since an absolute value can never be negative. You can write an absolute value inequality as a compound value inequality

EXPLORATION 3
Solving an Absolute Value Inequality Numerically
Work with a partner.

Solving an Absolute Value Inequality Numerically
Work with a partner. Consider the absolute value inequality | x + 2 | ≤ 3.
a. Use a spreadsheet, as shown, to solve the absolute value inequality.
b. Compare the solutions you found using the spreadsheet with those you found in Explorations 1 and 2. What do you notice?
c. How can you use a spreadsheet to solve an absolute value inequality?
Answer:

Communicate Your Answer

Question 4.
How can you solve an absolute value inequality?
Answer:
Isolate the absolute value expression on the left side of the inequality. If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions. Use the sign of each side of your inequality to decide which of these cases holds.

Question 5.
What do you like or dislike about the algebraic, graphical, and numerical methods for solving an absolute value inequality? Give reasons for your answers.
Answer:
We have to use algebraic, graphical, and numerical methods for solving an absolute value inequality depending on the situation
The algebraic and graphical methods for solving an absolute value inequality are useful when the size of the inequality is small i.e., have only 1 variable

2.6 Lesson

Monitoring Progress

Solve the inequality. Graph the solution, if possible

Question 1.
| x | ≤ 3.5
Answer:
The given absolute value inequality is:
| x | ≤ 3.5
We know that,
| x | =x for x > 0
| x | = -x for x < 0
So,
x ≤ 3.5
x ≤ 4 [Approximate value}
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
x ≤ 4 [Approximate value]
The representation of the solution of the given absolute value inequality in the graph is:

Question 2.
| k – 3 | < -1
Answer:
The given absolute value inequality is:
| k – 3 | < -1
We know that,
If the number on the other side of the absolute value expression is negative, then the given absolute value equation has no solution or has real numbers as a solution
Hence, from the above,
We can conclude that the given absolute value inequality has no solution

Question 3.
| 2w – 1 | < 11
Answer:
The given absolute value inequality is:
| 2w – 1 | < 11
We know that,
| x | =x for x > 0
| x | = -x for x < 0
So,
2w – 1 < 11
2w < 11 + 1
2w < 12
w < 12 /  2
w < 6
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
w < 6
The representation of the solution of the given absolute value inequality in the graph is:

Question 4.
| x + 3 | > 8
Answer:
The given absolute value inquality is:
| x + 3 | > 8
We know that,
| x | = xfor x > 0
| x | = -x for x < 0
So,
x + 3 > 8
x > 8 – 3
x > 5
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
x > 5
The representation of the solution of the given absolute value inequality in the graph is:

Question 5.
| n + 2 | – 3 ≥ -6
Answer:
The given absolute value inequality is:
| n + 2 | – 3 ≥ -6
So,
| n + 2 | ≥ -6 + 3
| n + 2 | ≥ -3
We know that,
If the number on the other side of the absolute value expression is negative, then the given absolute value equation has no solution or has real numbers as a solution
Hence, from the above,
We can conclude that the given absolute value inequality has no solution

Question 6.
3 | d + 1 | – 7 ≥ -1
Answer:
The given absolute value inequality is:
3 | d + 1 | – 7 ≥ -1
So,
3 | d + 1 | ≥ -1 + 7
3 | d + 1 | ≥ 6
| d + 1 | ≥ 6 / 3
| d + 1 | ≥ 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Hence,
d + 1 ≥ 2
d ≥ 2 – 1
d ≥ 1
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is:
d ≥ 1
The representation of the solution of the given inequality in the graph is:

Question 7.
WHAT IF?
You are willing to pay the mean price with an absolute deviation of at most $75. How many of the computer prices meet your condition?
Answer:

Solving Absolute Value Inequalities 2.6 Exercises

Question 1.
REASONING
Can you determine the solution of | 4x – 2 | ≥ -6 without solving it? Explain
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 1

Question 2.
WRITING
Describe how solving | w – 9 | ≤ 2 is different from solving | w – 9 | ≥ 2.
Answer:
The given absolute value inequalities are:
| w – 9 | ≤ 2 and | w – 9 | ≥ 2
Now,
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
w – 9 ≤ 2 and w – 9 ≥ 2
w ≤ 2 + 9 and w ≥ 2 + 9
w ≤ 11 and w ≥ 11
w – 9 ≤ -2 and w – 9 ≥ -2
w ≤ -2 + 9 and w ≥ -2 + 9
w ≤ 7 and w ≥ 7
Hence, from the above,
We can conclude that the given absolute value inequalities are different by observing the above values for the 2 absolute value inequalities

Monitoring Progress and Modeling with Mathematics 

In Exercises 3–18, solve the inequality. Graph the solution, if possible.

Question 3.
| x | < 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 3

Question 4.
| y | ≥ 4.
Answer:
The given absolute value inequality is:
| y | ≥ 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Hence,
The solutions for the given absolute value inequality is:
y ≥ 4 or y ≤ -4
The representation of the compound inequality solution of the given absolute value inequality is:

Question 5.
| d + 9 | > 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 5

Question 6.
| h – 5 | ≤ 10
Answer:
The given absolute value inequality is:
| h – 5 | ≤ 10
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
h – 5 ≤ 10
h ≤ 10 + 5 and h ≥ -10 + 5
h ≤ 15 and h ≥ -5
Hence,
The compound inequality of the solutions of the given absolute value inequality are:
-5 ≤  h ≤  15
The representation of the compound inequality in the graph is:

Question 7.
| 2s – 7 | ≥ -1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 7

Question 8.
| 4c + 5 | > 7
Answer:
The given absolute value inequality is:
| 4c + 5 | > 7
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4c + 5 > 7 or 4c + 5 < -7
4c > 7 – 5 or 4c < -7 – 5
4c > 2 or 4c < -12
c > \(\frac{2}{4}\) or c < –\(\frac{4}{12}\)
c > \(\frac{1}{2}\) or c < –\(\frac{1}{3}\)
Hence,
The solutions of the given absolute value equation are:
c > \(\frac{1}{2}\) or c < –\(\frac{1}{3}\)
c > 1 or c < 0 [ Approximate values of the inequality ]
The representation of the solutions of the given absolute value inequality in the graph is:

Question 9.
| 5p + 2 | < -4
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 9

Question 10.
| 9 – 4n | < 5
Answer:
The given absolute value inequality is:
| 9 – 4n | < 5
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
9 – 4n < 5 and 9 – 4n > -5
9 – 5 < 4n and 9 + 5 > 4n
4 < 4n and 14 > 4n
4n > 4 and 4n < 14
n > 4/ 4 and n < 14 / 4
n > 1 and n < 7 / 2
n > 1 and n < 3.5
Hence,
The solutions of the given absolute value inequality are:
n > 1 and n < 3.5
The compound solution of the solutions of the given absolute value inequality is:
1 < n < 3.5
The representation of the compound inequality of the solutions of the given inequality in the graph is:

Question 11.
| 6t – 7 | – 8 ≥ 3
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 11

Question 12.
| 3j – 1 | + 6 > 0
Answer:
The given absolute value inequality is:
| 3j – 1 | + 6 > 0
So,
| 3j – 1 | > 0 – 6
| 3j – 1 | > -6
We know that,
By definition of absolute values, the absolute value of an expression must be greater than or equal to 0
So,
The expression | 3j – 1 |  will always be greater than -6
Hence,
For the given absolute value inequality, all the real numbers are solutions

Question 13.
3 | 14 – m | > 18
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 13

Question 14.
-4 | 6b – 8 | ≤ 12
Answer:
The given absolute value inequality is:
-4 | 6b – 8 | ≤ 12
So,
| 6b – 8 | ≤ -3
We know that,
By definition of absolute values, the absolute value of an expression must be greater than or equal to 0.
So,
The expression | 6b – 8 | must not be less than -3
Hence,
For the given absolute value inequality, there is no solution

Question 15.
2 | 3w + 8 | ≥ 13
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 15

Question 16.
3 | 2 – 4u | + 5 < -13
Answer:
The given absolute value inequality is:
3 | 2 – 4u | + 5 < -13
So,
3 | 2 – 4u | < -13 – 5
3 | 2 – 4u | < -18
| 2 – 4u | < -6
We know that,
By definition of absolute values, the absolute value of an expression must be greater than or equal to 0.
So,
The expression | 2 – 4u | must not be less than -6
Hence,
For the given absolute value inequality, there is no solution

Question 17.
6 | -f + 3 | + 7 > 7
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 17

Question 18.
\(\frac{2}{3}\) |4v + 6 | – 2 ≤ 10
Answer:
The given absolute value inequality is:
\(\frac{2}{3}\) | 4v + 6 | – 2 ≤ 10
\(\frac{2}{3}\) | 4v + 6 | ≤ 10 + 2
\(\frac{2}{3}\) | 4v + 6 | ≤ 12
| 4v + 6 | ≤ 12 × \(\frac{3}{2}\)
| 4v + 6 | ≤ 18
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4v + 6 ≤ 18 and 4v + 6 ≥ -18
4v ≤ 18 – 6 and 4v ≥ -18 – 6
4v ≤ 12 and 4v ≥ -24
v ≤ 12 / 4 and v ≥ -24 / 4
v ≤ 3 and v ≥ -6
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
v ≤ 3 and v ≥ -6
The representation of the compound inequality of the solutions of the given absolute value inequality is:
-6 ≤ v ≤ 3
The representation of the compound inequality in the graph is:

Question 19.
MODELING WITH MATHEMATICS
The rules for an essay contest say that entries can have 500 words with an absolute deviation of at most 30 words. Write and solve an absolute value inequality that represents the acceptable numbers of words.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 19

Question 20.
MODELING WITH MATHEMATICS
The normal body temperature of a camel is 37°C. This temperature varies by up to 3°C throughout the day. Write and solve an absolute value inequality that represents the range of normal body temperatures (in degrees Celsius) of a camel throughout the day.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 84
Answer:
The absolute value inequality that represents the range of normal body temperatures of a camel throughout the day is:
34°C ≤ t ≤ 40°C

Explanation:
It is given that the normal body temperature of a camel is 37°C and this temperature varies by up to 3°C throughout the day.
Let t be the temperature in °C
So,
The absolute value inequality that represents the range of normal body temperatures of a camel throughout the day is:
37 – 3 ≤ t ≤ 37 + 3
34°C ≤ t ≤ 40°C

ERROR ANALYSIS
In Exercises 21 and 22, describe and correct the error in solving the absolute value inequality.

Question 21.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 85
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 21

Question 22.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 86
Answer:
The given absolute value inequality is:
| x + 4 | > 13
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 4 > 13 or x + 4 < -13
x > 13 – 4 or x < -13 – 4
x > 9 or x < -17
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
x > 9 or x < -17

In Exercises 23–26, write the sentence as an absolute value inequality. Then solve the inequality.

Question 23.
A number is less than 6 units from 0.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 23

Question 24.
A number is more than 9 units from 3.
Answer:
The given worded form is:
A number is more than 9 units from 3
Let n be the number
Hence,
The representation of the given worded form in the form of absolute value inequality is:
| n – 3 | > 9
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
n – 3 > 9 or n – 3 < -9
n > 9 + 3 or n < -9 + 3
n > 12 or n < -6
Hence, from the above,
We can conclude that the solutions o the given absolute value inequality is:
n > 12 or n < -6

Question 25.
Half of a number is at most 5 units from 14.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 25

Question 26.
Twice a number is no less than 10 units from -1.
Answer:
The given worded form is:
Twice a number is no less than 10 units from -1
Let x be the number
So,
The representation of the given worded form in the form of absolute value inequality is:
| 2x – (-1) | > 10
So,
| 2x + 1 | > 10
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Hence,
2x + 1 > 10 or 2x + 1 < -10
2x > 10 – 1 or 2x < -10 – 1
2x > 9 or 2x < -11
x > \(\frac{9}{2}\) or x < –\(\frac{11}{2}\)
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality is:
x > \(\frac{9}{2}\) or x < –\(\frac{11}{2}\)

Question 27.
PROBLEM SOLVING
An auto parts manufacturer throws out gaskets with weights that are not within 0.06 pound of the mean weight of the batch. The weights (in pounds) of the gaskets in a batch are 0.58, 0.63, 0.65, 0.53, and 0.61. Which gasket(s) should be thrown out?
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 27

Question 28.
PROBLEM-SOLVING
Six students measure the acceleration (in meters per second ) of an object in free fall. The measured values are shown. The students want to state that the absolute deviation of each measured value x from the mean is at most d. Find the value of d.
10.56, 9.52, 9.73, 9.80, 9.78, 10.91
Answer:
The value of d is: 0.86

Explanation:
The given numbers are:
10.56, 9.52, 9.73, 9.80, 9.78, 10.91
We know that,
Mean = \(\frac{Sum of the given observations}{Number of observations}\)
= \(\frac{10.56 + 9.52 + 9.73 + 9.80 + 9.78 + 10.91}{6}\)
= \(\frac{60.3}{6}\)
= 10.05
Now,
We know that,
Absolute deviation = Mean – ( Given observation)
It is given that that the measured value is x
Hence,
The representation of the absolute deviation in the form of absolute value inequality is:
| x – 10.05 | ≤ d
So,
| 10.56 – 10.05 | ≤ d
| 9.52 – 10.05 | ≤ d
| 9.73 – 10.05 | ≤ d
| 9.80 – 10.05 | ≤ d
| 9.78 – 10.05 | ≤ d
| 10.91 – 10.05 | ≤ d
So,
d ≥ 0.51
d ≥ 0.53
d ≥ 0.32
d ≥ 0.25
d ≥ 0.27
d ≥ 0.86
Hence, from the above,
We can conclude that the value of d will be the highest deviation from the mean value
The value of d is: 0.86

MATHEMATICAL CONNECTIONS
In Exercises 29 and 30, write an absolute value inequality that represents the situation. Then solve the inequality.

Question 29.
The difference between the areas of the figures is less than 2.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 86.1
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 29

Question 30.
The difference between the perimeters of the figures is less than or equal to 3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 87
Answer:
The given figures are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 87
From the above,
The given figures are: Rectangle and square
We know that,
The perimeter of a rectangle = 2 ( Length + Width )
The perimeter of a square = 4 ( Side )
So,
The perimeter of a rectangle = 2 ( 3 + x + 1 ) = 2 ( x + 4 )
The perimeter of a square = 4 (x) = 4x
It is given that the difference of the perimeters of the given figures is less than or equal to 3
So,
| 2 ( x + 4 ) – 4x | ≤ 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 (x + 4) – 4x ≤3 and 2 ( x + 4 ) – 4x ≥ -3
2x + 8 – 4x ≤ 3 and 2x + 8 – 4x ≥ -3
8 – 2x ≤ 3 and 8- 2x ≥ -3
-2x ≤ 3 – 8 and -2x ≥ -3 – 8
-2x ≤ -5 and -2x ≥ -11
2x ≤ 5 and 2x ≥ 11
x ≤ \(\frac{5}{2}\) and x ≥ \(\frac{11}{2}\)
Hence, from the above,
We can conclude that the compound inequality for the solutions of the given absolute value inequality is:
\(\frac{11}{2}\) ≤ x ≤ \(\frac{5}{2}\)

REASONING
In Exercises 31–34, tell whether the statement is true or false. If it is false, explain why.

question 31.
If a is a solution of | x + 3 | ≤ 8, then a is also a solution of x + 3 ≥ -8.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 31

Question 32.
If a is a solution of | x + 3 | > 8, then a is also a solution of x + 3 > 8.
Answer:
The given statement is true

Explanation:
The given statement is:
If a is a solution of | x + 3 | > 8, then a is also a solution of x + 3 > 8
Now,
The given inequality is:
| x + 3 | > 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 > 8 or x + 3 < -8
We know that,
The value of the absolute value expression must be greater than or equal to 0
So,
x + 3 < -8 has no solution
So,
a is a solution of x + 3 > 8
Hence, from the above,
We can conclude that the given statement is true

Question 33.
If a is a solution of | x + 3 | ≥ 8, then a is also a solution of x + 3 ≥ -8.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 33

Question 34.
If a is a solution of x + 3 ≤  -8, then a is also a solution of | x + 3 | ≥ 8.
Answer:
The given statement is true

Explanation:
The given statement is:
If a is a solution of x + 3 ≤ -8, then a is a solution of | x + 3 | ≥ 8
Now,
The given absolute value inequality is:
| x + 3 | ≥ 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 ≥ 8 or x + 3 ≤ -8
If a is a solution of x + 3  ≥ 8, then a is also a solution of x + 3 ≤ -8
Hence, from the above,
We can conclude that the given statement is true

Question 35.
MAKING AN ARGUMENT
One of your classmates claims that the solution of | n | > 0 is all real numbers. Is your classmate correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 35

Question 36.
THOUGHT-PROVOKING
Draw and label a geometric figure so that the perimeter P of the figure is a solution to the inequality | P – 60 | ≤ 12.
Answer:
The given absolute value inequality is:
| p – 60 | ≤ 12
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
p – 60 ≤ 12 and p – 60 ≥ -12
p ≤ 12 + 60 and p ≥ -12 + 60
p ≤ 72 and p ≥ 48
Hence,
The representation of the compound inequality of the perimeter of the figure is:
48 ≤ p ≤ 72
Now, find the value of 4p i.e., the perimeter of the square
So,
192 ≤ 4p ≤ 288
Hence, from the above,
We can conclude that the perimeter is represented for the perimeter of the square

Question 37.
REASONING
What is the solution of the inequality | ax + b | < c, where c < 0? What is the solution of the inequality | ax + b | > c, where c < 0? Explain.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 37

Question 38.
HOW DO YOU SEE IT?
Write an absolute value inequality for each graph.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 88
How did you decide which inequality symbol to use for each inequality?
Answer:
The given graphs are:
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 88
a.
From the given graph,
The 1st marked line starts from -1 including -1 and continued till the left end of the number line
The 2nd marked line starts from 5 including from 5 and continued till the right end of the number line
Hence,
The representation of the marked lines in the form of absolute value inequality is:
| x + 1 | ≤ 3 and | x – 5 | ≥ 1

b.
From the given graph,
The marked line starts from -1 excluding -1 and continued till 5 on the right end of the number line
Hence,
The representation of the marked line in the form of absolute value inequality is:
| x + 1 | > 6
c.
From the given graph,
The marked line starts from -1 including -1 and continued till 5 on the right end of the number line
Hence,
The representation of the marked line in the form of absolute value inequality is:
| x + 1 | ≥ 6
d.
Fro the given graph,
The 1st marked line starts from -1 excluding -1 and continued till the left end of the number line
The 2nd marked line starts from 5 excluding 5 and continued till the right end of the number line
Hence,
The representation of the marked lines in the form of absolute value inequality is:
| x + 1 | < 3 and | x – 5 | > 1
If the marked point is indicated on the number line and if that marked line continued till the right end of the number line, then we will use ≥
If the marked point is not indicated on the number line and if that marked line continued till the right end of the number line, then we will use >
If the marked point is indicated on the number line and if that marked line continued till the left end of the number line, then we will use ≤
If the marked point is not indicated on the number line and if that marked line continued till the left end of the number line, then we will use <

Question 39.
WRITING
Explain why the solution set of the inequality | x | < 5 is the intersection of two sets, while the solution set of the inequality | x | > 5 is the union of two sets.
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 39

Question 40.
PROBLEM-SOLVING
Solve the compound inequality below. Describe your steps.
| x – 3 | < 4 and | x + 2 | > 8
Answer:
The given inequalities are:
| x + 3 | < 4 and | x + 2  | > 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x + 3 < 4 or x + 3 > -4 and x + 2 > 8 or x + 2 < -8
x < 4 – 3 or x > -4 – 3 and x > 8 – 2 or x < -8 – 2
x < 1 or x > -7 and x > 6 or x < -10
Hence,
The compound inequality of the solutions of the given absolute value inequalities are:
-7 < x < 1 and -10 > x > 6

Maintaining Mathematical Proficiency

Plot the ordered pair in a coordinate plane. Describe the location of the point.

Question 41.
A(1, 3)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 41

Question 42.
B(0, -3)
Answer:
In the graph,
X-axis:
The right side of the x-axis will be: +ve
The left side of the x-axis will be: -ve
Y-axis:
The top side of the y-axis will be: +ve
The downside of the y-axis will be: -ve
We know that,
The graph is divided into 4 quadrants. They are:
1st Quadrant: x: +ve and y: -ve
2nd Quadrant: x: -ve and y: +ve
3rd Quadrant: x: -ve and y: -ve
4th Quadrant: x: +ve and y: -ve
Hence,
B(0, -3) will be in the 4th Quadrant
The representation of point B in the graph is:

Question 43.
C(-4, -2)
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 43

Question 44.
D(-1, 2)
Answer:
In the graph,
X-axis:
The right side of the x-axis will be: +ve
The left side of the x-axis will be: -ve
Y-axis:
The top side of the y-axis will be: +ve
The downside of the y-axis will be: -ve
We know that,
The graph is divided into 4 quadrants. They are:
1st Quadrant: x: +ve and y: -ve
2nd Quadrant: x: -ve and y: +ve
3rd Quadrant: x: -ve and y: -ve
4th Quadrant: x: +ve and y: -ve
Hence,
D(-1, 2) will be in the 2nd Quadrant
The representation of point D in the graph is:

Copy and complete the table.

Question 45.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 89
Answer:
Big Ideas Math Algebra 1 Answer Key Chapter 2 Solving Linear Inequalities 2.6 Question 45

Question 46.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 90
Answer:
The completed table is:

Solving Linear Inequalities Performance Task: Grading Calculations

2.5–2.6 What Did You Learn?

Core Vocabulary
compound inequality, p. 82
absolute value inequality, p. 88
absolute deviation, p. 90

Core Concepts
Section 2.5
Writing and Graphing Compound Inequalities, p. 82
Solving Compound Inequalities, p. 83

Section 2.6
Solving Absolute Value Inequalities, p. 88

Mathematical Practices

Question 1.
How can you use a diagram to help you solve Exercise 12 on page 85?
Answer:
In Exercise 12 on page 85,
The given diagram represents the maximum elevation of Mount Rainier
In the diagram,
The given maximum elevation of Mount Rainier is: 14,410 ft
So,
This 14,410 ft is divided into different types of elevation in Exercise 12

Question 2.
In Exercises 13 and 14 on page 85, how can you use structure to break down the compound inequality into two inequalities?
Answer:
In Exercises 13 and 14 on page 85,
First, perform the necessary mathematical operations so that only the variable must be present
So,
After performing the necessary mathematical operations, the structure of the compound inequality will be like
a < x < b or a > x > b
The breakdown of the compound inequality a < x < b will be like x >a and x < b
The breakdown of the compound inequality a > x > b will be like x < a and x > b
We can put the different types of inequality symbols in the breakdown of inequalities

Question 3.
Describe the given information and the overall goal of Exercise 27 on page 91.
Answer:
The given information is:
An auto parts manufacturer throws out gaskets with weights that are not within 0.06 pounds of the mean weight of the batch. The weights (in pounds) of the gaskets in a batch are 0.58, 0.63, 0.65, 0.53, and 0.61
The overall goal of Exercise 27:
Find the mean of the given weights and find the absolute deviation of the gaskets
The absolute deviation of the gaskets that are less than 0.0 pounds of the man weight should be thrown out

Question 4.
For false statements in Exercises 31–34 on page 92, use examples to show the statements are false.
Answer:
In Exercises 31 – 34 on page 92,
Exercises 33 and 34 are false
The Exercise 33 is: | x + 3 | ≥ 8
The Exercise 34 is: | x + 3 | ≥ 8
So,
In Exercise 33,
a is a solution for x + 3 ≥ 8 and x + 3 ≤ -8
In Exercise 34,
a is a solution of x + 3 ≥ 8 and x + 3 ≤ -8

Performance Task

Grading Calculations

You are not doing as well as you had hoped in one of your classes. So, you want to figure out the minimum grade you need on the final exam to receive the semester grade that you want. Is it still possible to get an A? How would you explain your calculations to a classmate?
To explore the answers to this question and more, go to Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 91
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 91.1

Solving Linear Inequalities Chapter Review

2.1 Writing and Graphing Inequalities (pp. 53–60)
a. A number x plus 36 is no more than 40. Write this sentence as an inequality.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 91.2
b. Graph w > −3.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 92

Write the sentence as an inequality.

Question 1.
A number d minus 2 is less than -1.
Answer:
The given worded form is:
A number d minus 2 is less than -1
Hence,
The representation of the given worded form in the form of inequality is:
d – 2 < -1

Question 2.
Ten is at least the product of a number h and 5.
Answer:
The given worded form is:
Ten is at least the product of a number h and 5
Hence,
The representation of the given worded form in the form of inequality is:
10 ≥ h(5)
10 ≥ 5h
10 / 5 ≥ h
2 ≥ h
h ≤ 2
Hence, from the above,
We can conclude that the representation of the given worded form in the form of inequality is:
h ≤ 2

Graph the inequality.

Question 3.
x > 4
Answer:
The given inequality is:
x > 4
Hence,
The representation of the given inequality in the graph is:

Question 4.
y ≤ 2
Answer:
The given inequality is:
y ≤ 2
Hence,
The representation of the given inequality in the graph is:

Question 5.
-1 ≥ z
Answer:
The given inequality is:
-1 ≥ z
z ≤ -1
Hence,
The representation of the given inequality in the graph is:

2.2 Solving Inequalities Using Addition or Subtraction (pp. 61–66)

Solve x + 2.5 ≤ −6. Graph the solution.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 93

Solve the inequality. Graph the solution.

Question 6.
p + 4 < 10
Answer:
The given inequality is:
p + 4 < 10
So,
p < 10 – 4
p < 6
Hence, from the above,
We can conclude that the solution to the given inequality is p < 6
The representation of the solution of the given inequality in the graph is:

Question 7.
r – 4 < -6
Answer:
The given inequality is:
r – 4 < -6
So,
r < -6 + 4
r < -2
Hence, from the above,
We can conclude that the solution to the given inequality is r < -2
The representation of the solution of the given inequality in the graph is:

Question 8.
2.1 ≥ m – 6.7
Answer:
The given inequality is:
2.1 ≥ m – 6.7
So,
2.1 + 6.7 ≥ m
8.8 ≥ m
9 ≥ m [Approximate value]
m ≤ 9
Hence, from the above,
We can conclude that the solution to the given inequality is m ≤ 9
The representation of the solution of the given inequality in the graph is:

2.3 Solving Inequalities. Using Multiplication or Division

Solve \(\frac{n}{-10}\) > 5. Graph the solution.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 94

Solve the inequality. Graph the solution.

Question 9.
3x > -21
Answer:
The given inequality is:
3x > -21
So,
x > -21 / 3
x > -7
Hence, from the above,
We can conclude that the solution to the given inequality is x > -7
The representation of the solution of the given inequality in the graph is:

Question 10.
-4 ≤ \(\frac{g}{5}\)
Answer:
The given inequality is:
-4 ≤ \(\frac{g}{5}\)
-4 (5) ≤ g
-20 ≤ g
g ≥ -20
Hence, from the above,
We can conclude that the solution to the given inequality is g ≥ -20
The representation of the solution of the given inequality in the graph is:

Question 11.
–\(\frac{3}{4}\)n ≤ 3
Answer:
The given inequality is:
–\(\frac{3}{4}\)n ≤ 3
n ≤ 3 × –\(\frac{4}{3}\)
n ≤ —\(\frac{3 × 4}{3}\)
n ≤ -4
Hence, from the above,
We can conclude that the solution to the given inequality is n ≤ -4
The representation of the solution of the given inequality in the graph is:

Question 12.
\(\frac{s}{-8}\) ≥ 11
Answer:
The given inequality is:
\(\frac{s}{-8}\) ≥ 11
s ≥ 11 (-8)
s ≥ -88
Hence, from the above,
We can conclude that the solution to the given inequality is s ≥ -88
The representation of the solution of the given inequality in the graph is:

Question 13.
36 < 2q
Answer:
The given inequality is:
36 < 2q
So,
36 / 2 < q
18 < q
q > 18
Hence, from the above,
We can conclude that the solution to the given inequality is q > 18
The representation of the solution of the given inequality in the graph is:

Question 14.
-1.2k > 6
Answer:
The given inequality is:
-1.2k > 6
So,
k > 6 / -1.2
k > -60 / 12
k > -5
Hence, from the above,
We can conclude that the solution to the given inequality is k > -5
The representation of the solution of the given inequality in the graph is:

2.4 Solving Multi-step Inequalities

Solve 22 + 3y ≥ 4. Graph the solution.
Big Ideas Math Algebra 1 Answers Chapter 2 Solving Linear Inequalities 95

Solve the inequality. Graph the solution, if possible.

Question 15.
3x – 4 > 11
Answer:
The given inequality is:
3x – 4 > 11
So,
3x > 11 + 4
3x > 15
x > 15 / 3
x > 5
Hence, from the above,
We can conclude that the solution to the given inequality is x > 5
The representation of the solution of the given inequality in the graph is:

Question 16.
-4 < \(\frac{b}{2}\) + 9
Answer:
The given inequality is:
-4 < \(\frac{b}{2}\) + 9
-4 – 9 < \(\frac{b}{2}\)
-13 < \(\frac{b}{2}\)
-13 (2) < b
-26 < b
b > -26
Hence, from the above,
We can conclude that the solution of the given inequality is b > -26
The representation of the solution of the given inequality in the graph is:

Question 17.
7 – 3n ≤ n + 3
Answer:
The given inequality is:
7 – 3n ≤ n + 3
-3n – n ≤ 3 – 7
-4n ≤ -4
4n ≤ 4
n ≤ 4 / 4
n ≤ 1
Hence, from the above,
We can conclude that the solution to the given inequality is n ≤ 1
The representation of the solution of the given inequality in the graph is:

Question 18.
2(-4s + 2) ≥ -5s – 10
Answer:
The given inequality is:
2 (-4s +2) ≥ -5s – 10
So,
2 (-4s) + 2 (2) ≥ -5s – 10
-8s + 4 ≥ -5s – 10
-8s + 5s ≥ -10 –  4
-3s ≥ -14
3s ≥ 14
s ≥ 14 / 3
s ≥ 5 [ Approximate value }
Hence, from the above,
We can conclude that the solution of the given inequality is s ≥ 5
The representation of the solution of the given inequality in the graph is:

Question 19.
6(2t + 9) ≤ 12t – 1
Answer:
The given inequality is:
6(2t + 9) ≤ 12t – 1
So,
6 (2t) + 6 (9) ≤ 12t – 1
12t + 54 ≤ 12t – 1
1t – 12t + 54 ≤ -1
54 ≤ -1
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 20.
3r – 8 > 3(r – 6)
Answer:
The given inequality is:
3r – 8 > 3 ( r – 6 )
So,
3r – 8 > 3 (r) – 3 (6)
3r – 8 > 3r – 18
3r – 8 – 3r > -18
-8 > -18
8 > 18
Hence, from the above,
We can conclude that there is no solution to the given inequality

2.5 Solving Compound Inequalities

Solve −1 ≤ −2d + 7 ≤ 9. Graph the solution.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 96

Question 21.
A number x is more than -6 and at most 8. Write this sentence as an inequality. Graph the inequality.
Answer:
The given worded form is:
A number x is more than -6 and at most 8
The representation of the given worded form in the form of inequality is:
x > -6 and x ≤ 8
Hence,
The representation of the solutions of the given worded form in the form of compound inequality is:
-6 < x ≤ 8
The representation of the compound inequality in the graph is:

Solve the inequality. Graph the solution.

Question 22.
19 ≥ 3z + 1 ≥ -5
Answer:
The given inequality is:
19 ≥ 3z + 1 ≥ -5
Subtract with -1 on both sides
So,
19 – 1 ≥ 3z + 1 – 1 ≥ -5 – 1
18 ≥ 3z ≥ -6
Divide by 3 on both sides
So,
(18 / 3) ≥ (3z / 3) ≥ (-6 / 3)
6 ≥ z ≥ -2
-2 ≤ z ≤ 6
Hence, from the above,
We can conclude that the compound inequality solution to the given inequality is:
-2 ≤ z ≤ 6
The representation of the compound inequality in the graph is:

Question 23.
\(\frac{r}{4}\) < -5 or -2r – 7 ≤ 3
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 97
Answer:
The given inequality is:
\(\frac{r}{4}\) < -5 or -2r – 7 ≤ 3
r < -5 (4) or -2r ≤ 3 + 7
r < -20 or -2r ≤ 10
r < -20 or r ≤ -5
Hence, from the above,
We can conclude that the solutions to the given inequality are:
r < -20 or r ≤ -5
The representations of the solutions of the given inequality in the graph is:

2.6 Solving Absolute Value Inequalities

Solve the inequality. Graph the solution, if possible.

Question 24.
| m | ≥ 10
Answer:
The given absolute value inequality is:
| m | ≥ 10
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
m ≥ 10 or m ≤ -10
We know that,
The value of absolute value expression must be greater than or equal to 0
So,
m ≥ 10
Hence, from the above,
We can conclude that the solution of the given absolute value inequality is m ≥ 10
The representation of the solution of the given absolute value inequality in the graph is:

Question 25.
| k – 9 | < -4
Answer:
The given absolute value inequality is:
| k – 9 | < 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
k – 9 < 4 and k – 9 > -4
k < 4 + 9 and k > -4 + 9
k < 13 and k > 5
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
k < 13 and k > 5
The representation of the compound inequality of the solutions of the given absolute inequality is:
5 < k < 13
The representation of the compound inequality in the graph is:

Question 26.
4 | f – 6 | ≤ 12
Answer:
The given absolute value inequality is:
4 | f – 6 | ≤ 12
| f – 6 | ≤ 3
We know that,
| x | = x for x > 0
| X | = -x for x < 0
So,
f – 6 ≤ 3 and f – 6 ≥ -3
f ≤ 3 + 6 and f ≥ -3 + 6
f ≤ 9 and f ≥ 3
Hence, from the above,
We can conclude that the solutions of the given absolute value inequality are:
f ≤ 9 and f ≥ 3
The representation of the compound inequality of the solutions of the given absolute value inequality is:
3 ≤ f ≤ 9
The representation of the compound inequality in the graph is:

Question 27.
5 | b + 8 | – 7 > 13
Answer:
The given absolute value inequality is:
5 | b + 8 | – 7 > 13
5 | b + 8 | > 13 + 7
5 | b + 8 | > 20
| b + 8 | > 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
b + 8  > 4 or b + 8 < -4
b > 4 + 8 or b < -4 – 8
b > 12 or b < -12
We know that,
The value of absolute value expression must be greater than or equal to 0
So,
b > 12
Hence, from the above,
We can conclude that the solution to the given absolute value inequality is b > 12
The representation of the solution of the given absolute value inequality in the graph is:

Question 28.
| -3g – 2 | + 1 < 6
Answer:
The given absolute value inequality is:
| -3g – 2 | + 1 < 6
– | 3g + 2 | < 6 – 1
– | 3g + 2 | < 5
| 3g + 2 | < -5
We know that,
The value of absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that there is no solution to the given absolute value inequality

Question 29.
| 9 – 2j | + 10 ≥ 2
Answer:
The given absolute value inequality is:
| 9 – 2j | + 10 ≥  2
| 9 – 2j | ≥ 2 – 10
| 9 – 2j | ≥  -8
We know that,
The value of absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that there is no solution to the given absolute value inequality

Question 30.
A safety regulation states that the height of a guardrail should be 106 centimeters with an absolute deviation of no more than 7 centimeters. Write and solve an absolute value inequality that represents the acceptable heights of a guardrail.
Answer:
The absolute value inequality that represents the acceptable heights of a guardrail is:
| x – 106 | ≤ 7 centimeters

Explanation:
It is given that a safety regulation states that the height of a guardrail should be 106 centimeters with an absolute deviation of no more than 7 centimeters.
Now,
We know that,
Absolute deviation = ( Mean ) – ( The given observation )
Let the Mean be x
So,
Absolute deviation = x – 106
Hence,
The absolute value inequality that represents the acceptable heights of a guardrail is:
| x – 106 | ≤ 7
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 106 ≤ 7 and x – 106 ≥ -7
x ≤ 7 + 106 and x ≥ -7 + 106
x ≤ 113 and x ≥ 99
Hence, from the above,
We can conclude that the absolute value inequality that represents the acceptable heights of a guardrail is:
| x – 106 | ≤ 7

Solving Linear Inequalities Chapter Test

Write the sentence as an inequality.

Question 1.
The sum of a number y and 9 is at least -1.
Answer:
The given worded form is:
The sum of a number y and 9 is at least -1
Hence,
The representation of the given worded form in the form of inequality is:
y + 9 ≥ -1

Question 2.
A number r is more than 0 or less than or equal to -8.
Answer:
The given worded form is:
A number r is more than 0 or less than or equal to -8
Hence,
The representation of the given worded form in the form of inequality is:
r > 0 or r ≤ -8

Question 3.
A number k is less than 3 units from 10.
Answer:
The given worded form is:
A number k is less than 3 units from 10
Hence,
The representation of the given worded form in the form of inequality is:
k – 10 < 3

Solve the inequality. Graph the solution, if possible.

Question 4.
\(\frac{x}{2}\) – 5 ≥ -9
Answer:
The given inequality is:
\(\frac{x}{2}\) – 5 ≥ -9
\(\frac{x}{2}\) ≥ -9 + 5
\(\frac{x}{2}\) ≥ -4
x ≥ -4 (2)
x ≥ -8
Hence, from the above,
We can conclude that the solution to the given inequality is x ≥ -8
The representation of the solution of the given inequality in the graph is:

Question 5.
-4s < 6s + 1
Answer:
The given inequality is:
-4s < 6s + 1
So,
-4s – 6s < 1
-10s < 1
10s < -1
s < –\(\frac{1}{10}\)
s < -0.1
Hence, from the above,
We can conclude that the solution to the given inequality is s < -0.1
The representation of the solution of the given inequality in the graph is:

Question 6.
4p + 3 ≥ 2(2p + 1)
Answer:
The given inequality is:
4p + 3 ≥ 2 (2p + 1)
So,
4p + 3 ≥ 2 (2p) + 2 (1)
4p + 3 ≥ 4p + 2
4p – 4p + 3 ≥ 2
3 ≥ 2
Hence, from the above,
We can conclude that there is no solution for the given inequality

Question 7.
-7 < 2c – 1 < 10
Answer:
The given inequality is:
-7 < 2c – 1 < 10
Add 1 on both sides
So,
-7 + 1 < 2c – 1 + 1 < 10 + 1
-6 < 2c < 11
Divide by 3 on both sides
(-6 / 2) < (2c / 2) < (11 / 2)
-3 < c < 5.5
-3 < c < 6 [Approximate inequality]
Hence, from the above,
We can conclude that the solution to the given inequality is -3 < c < 6
The representation of the solution of the given inequality in the graph is:

Question 8.
-2 ≤ 4 – 3a ≤ 13
Answer:
The given inequality is:
-2 ≤ 4 – 3a ≤ 13
Subtract 4 on both sides
-2 – 4 ≤ 4 – 4 – 3a ≤ 13 – 4
-6 ≤ -3a ≤ 9
Divide by – on both sides
6 ≥ 3a ≥ -9
Divide by 3 on both sides
(6 / 3) ≥ (3a / 3) ≥ (-9 / 3)
2 ≥ a ≥ -3
-3 ≤ a ≤ 2
Hence, from the above,
We can conclude that the solution to the given inequality is -3 ≤ a ≤ 2
The representation of the solution of the inequality in the graph is:

Question 9.
-5 < 2 – h or 6h + 5 > 71
Answer:
The given inequality is:
-5 < 2 – h or 6h + 5 > 71
-5 – 2 < h or 6h > 71 – 5
-7 < -h or 6h > 66
7 < h or h > 66 / 6
h > 7 or h > 11
Hence, from the above,
We can conclude that the solution to the given inequality is h > 7 [Since h > 11 is before  h > 7]
The representation of the solution of the given inequality in the graph is:

Question 10.
| 2q + 8 | > 4
Answer:
The given absolute value inequality is:
| 2q + 8 | > 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2q + 8 > 4 or 2q + 8 < -4
2q > 4 – 8 or 2q < -4 – 8
2q > -4 or 2q < -12
We know that,
The value of the absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that the given absolute value inequality has no solution

Question 11.
-2 | y – 3 | – 5 ≥ -4
Answer:
The given absolute value inequality is:
-2 | y – 3 | – 5 ≥ -4
So,
-2 | y – 3 | ≥ -4 + 5
-2 | y – 3 | ≥1
| y – 3 | ≥ –\(\frac{1}{2}\)
We know that,
The value of the absolute value expression must be greater than or equal to 0
Hence, from the above,
We can conclude that the given absolute inequality has no solution

Question 12.
4 | -3b + 5 | – 9 < 7
Answer:
The given absolute value inequality is:
4 | -3b + 5 | – 9 < 7
So,
4 | -3b + 5 | < 7 + 9
4 | -3b + 5 | < 16
| -3b + 5 | < 16 / 4
| -3b + 5 | < 4
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
-3b + 5 < 4 and -3b + 5 > -4
-3b < 4 – 5 and -3b > -4 – 5
-3b < -1 and -3b > -9
3b < 1 and b > 3
b < \(\frac{1}{3}\) and b > 3
b < 0 and b > 3 [ Approximate value of inequality ]
Hence, from the above,
We can conclude that the solutions of the given inequality are b < 0 and b > 3
The representation of the solutions of the given inequality in the graph is:

Question 13.
You start a small baking business, and you want to earn a profit of at least $250 in the first month. The expenses in the first month are $155. What are the possible revenues that you need to earn to meet the profit goal?
Answer:
The possible revenues that you need to earn to meet the profit goal is:
R ≥ $405

Explanation:
It is given that you start a small baking business, and you want to earn a profit of at least $250 in the first month. The expenses in the first month are $155.
Now,
Let
P is the profit
R is the Revenue
E is the expenses
We know that,
P = R – E
It is given that P ≥ $250
So,
R – E ≥250
R – 155 ≥ 250
R ≥ 250 + 155
R ≥ 405
Hence, from the above,
We can conclude that the possible revenues that you need to earn to meet the profit goal is:
R ≥ $405

Question 14.
A manufacturer of bicycle parts requires that a bicycle chain have a width of 0.3 inches with an absolute deviation of at most 0.0003 inches. Write and solve an absolute value inequality that represents the acceptable widths.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 96.1
Answer:
The absolute value inequality that represents the acceptable widths is:
0.2997 inches ≤ w ≤0.3003 inches

Explanation:
It is given that a manufacturer of bicycle parts requires that a bicycle chain have a width of 0.3 inches with an absolute deviation of at most 0.0003 inches.
We know that,
Absolute deviation = Mean – Observation
Let the Mean be x
So,
x – 0.3 ≤ 0.0003
Hence,
The absolute value inequality that represents the acceptable widths is:
| x – 0.3 | ≤ 0.0003
-0.0003 ≤ x – 0.3 ≤ 0.0003
Add with 0.3 on both sides
-0.0003 + 0.3 ≤ x – 0.3 + 0.3 ≤ 0.0003 + 0.3
0.2997 ≤ x ≤ 0.3003
Hence, from the above,
We can conclude that the absolute value inequality that represents the acceptable widths is:
0.2997 inches ≤ w ≤0.3003 inches

Question 15.
Let a, b, c, and d be constants. Describe the possible solution sets of the inequality ax + b < cx + d.
Answer:

Write and graph a compound inequality that represents the numbers that are not solutions to the inequality represented by the graph shown. Explain your reasoning.

Question 16.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 97.1
Answer:
The given graph is:
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 97.1
From the graph,
The 1st marked line starts from -3 including -3 and continued till the left end of the number line
The 2nd marked line starts from 2 excluding 2 and continued till the right end of the number line
Hence,
The representation of the inequalities from the graph is:
x ≤ -3 and x > 2

Question 17.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 98
Answer:
The given graph is:
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 98
From the graph,
The marked line starts from -4 including -4 and continued till -1 including -1 to the right end of the number line
Hence,
The representation of the inequality from the graph is:
x ≥ -4 and x ≤ -1
Hence,
The representation of the inequalities in the form of compound inequality is:
–4 ≤ x ≤ -1

Question 18.
A state imposes a sales tax on items of clothing that cost more than $175. The tax applies only to the difference in the price of the item and $175.
a. Use the receipt shown to find the tax rate (as a percent).
b. A shopper has $430 to spend on a winter coat. Write and solve an inequality to find the prices p of coats that the shopper can afford. Assume that p ≥ 175.
c. Another state imposes a 5% sales tax on the entire price of an item of clothing. For which prices would paying the 5% tax to be cheaper than paying the tax described above? Write and solve an inequality to find your answer and list three prices that are solutions.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 99
Answer:
a.
From the given receipt,
The price is: $295
The total price is: $302.50
It is given that
Tax = Product price – $175
Now,
Let R be the tax rate
So,
R (product price – 175) = 7.5
R ( 295 – 175 ) = 7.5
R ( 120 ) = 7.5
%R = \(\frac{7.5}{120}\) × 100
%R = \(\frac{75}{1200}\) × 100
%R = 0.0625 × 100
%R = 6.25%
Hence, from the above,
We can conclude that the tax rate using the receipt is: 6.25%

b.
It is given that a shopper has $430 to spend on a winter coat
Let p be the number of coats
It is also given that to assume p ≥ 175
We know that,
The total price = The given price + Tax
So,
The representation of the inequality that represents the price of coats is:
p + R ( p – 175 ) ≤ 430
p + 0.0625 (p – 175)  ≤ 430
1.0625p – 10.9375 ≤ 430
1.0625p ≤ 430 + 10.9375
1.0625p ≤ 440.9375
p ≤ 440.9375 / 1.0625
p ≤ 415
Hence, from the above,
We can conclude that the prices of the coats that the shopper can afford are:
p ≤ $415

c.
It is given that another state imposes a 5% sales tax on the entire price of an item of clothing.
Now,
Let t be the tax rate that another state-imposed
Let x be the price
So,
(1+0.05) × t ≤ t + 0.0625 (t – 175)
1.05t ≤ 1.0625t – 10.9375
10.9375 ≤ 1.0625t – 1.05t
10.9375 ≤ 0.0125t
10.9375 / 0.0125 ≤ t
875 ≤ t
t ≥ 875
hence, from the above,
We can conclude that $875 is cheaper

Solving Linear Inequalities Maintaining Cumulative Assessment

Question 1.
The expected attendance at a school event is 65 people. The actual attendance can vary by up to 30 people. Which equation can you use to find the minimum and maximum attendances?
A. | x – 65 | = 30
B. | x + 65 | = 30
C. | x – 30 | = 65
D. | x + 30 | = 65
Answer:
It is given that the expected attendance at a school event is 65 people. The actual attendance can vary by up to 30 people.
So,
The absolute value equation that represents the maximum and minimum attendances are:
| x – Expected maximum attendance | ≤ 30
Hence,
| x – 65 | ≤ 30
| x – 65 | = 30
Hence, from the above,
We can conclude that option A represents the minimum and maximum attendances at a school event

Question 2.
Fill in values for a and b so that each statement is true for the inequality ax + 4 ≤ 3x + b.
a. When a = 5 and b = _____, x ≤ -3.
b. When a = _____ and b = _____, the solution of the inequality is all real numbers.
c. When a = _____ and b = _____, the inequality has no solution.
Answer:
The given inequality is:
ax + 4  ≤ 3x + b
a.
When a = 5,
5x + 4 ≤ 3x + b
5x – 3x + 4 ≤ b
2x + 4 ≤ b
2x ≤ b – 4
x ≤ ( b – 4 ) / 2
So,
The solution x ≤ -3 will be possible only if b = -2

b.
The given inequality is:
ax + 4 ≤ 3x + b
Compare the like coefficients
Hence,
The values of a and b so that the solution of an inequality is all real numbers are:
a = 3, b ≥ 4

c.
The given inequality is:
ax + 4 ≤ 3x + b
Hence,
The values of a and b so that the inequality has no solution are:
a = 3 and b < 4

Question 3.
Place each inequality into one of the two categories.
Big Ideas Math Algebra 1 Solutions Chapter 2 Solving Linear Inequalities 101
Answer:
The inequalities that have at least one integer solution must have x variable
The inequalities that have no integer solution must not have x variable
So,
The given inequalities are:
a. 5x – 6 + x ≥ 2x – 8
b. x – 8 + 4x ≤ 3 (x – 3) + 2x
c. 2 (3x + 8) > 3 (2x + 6)
d. 9x – 3 < 12 or 6x + 2 > -10
e. 17 < 4x + 5 < 21
f. 5 (x – 1) ≤ 5x – 3
Hence, from the above,
We can conclude that
The inequalities that have at least one integer solution are: a, d, e
The inequalities that have no integer solution are: b, c, f

Question 4.
Admission to a play costs $25. A season pass costs $180.
a. Write an inequality that represents the number x of plays you must attend for the season pass to be a better deal.
b. Select the number of plays for which the season pass is not a better deal.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 104
Answer:
a.
It is given that admission to a play costs $25 and a season pass costs $180
Let the number of plays be x
So,
The total number of pays = 25x
Hence,
The inequality that represents the number x of plays you must attend for the season pass to be a  better deal is:
25x > 180
x > 180 / 25
x > 7.2
Hence, from the above,
We can conclude that the inequality that represents the number x of plays you must attend for the season pass to be a better deal is:
x > 7.2

b.
From part (a),
The inequality that represents the season pas to be a better deal is:
x > 7.2
So,
The inequality that represents the season pass not to be a better deal is:
x < 7.2
Hence, from the given numbers,
The number of plays that the season pass is not a better deal is: 0, 1, 2, 3, 4, 5, 6, and 7

Question 5.
Select the values of a that make the solution of the equation 3(2x – 4) = 4(ax – 2) positive.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 105
Answer:
The given equation is:
3 (2x – 4) = 4 (ax – 2)
So,
3 (2x) – 3 (4) = 4 (ax) – 4 (2)
6x – 12 = 4ax – 8
6x – 12 + 8 = 4ax
6x – 4 = 4ax
6x – 4ax = 4
x (6 – 4a) = 4
Now,
The given numbers are: -2, -1, 0, 1, 2, 3, 4, 5
So,
x (6 – 4(-2) ) = 4
x ( 6 – 4 ( -1) ) = 4
x ( 6 – 4(0) ) = 4
x ( 6 – 4(1) ) = 4
x ( 6 – 4 (2) ) = 4
x ( 6 – 4 (3) ) = 4
x ( 6 – 4 (4) ) = 4
x ( 6 – 4 (5) ) = 4
Hence, from the above,
We can conclude that the values of a so that the solution is positive are: -2, -1, 0, 1

Question 6.
Fill in the compound inequality with <, ≤, , ≥, or > so the solution is shown in the graph.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 107
Answer:
From the graph,
The representation of the inequality is:
-2 ≤ x < 3
Now,
The given inequality is:
4x – 18 ____ -x – 3 and -3x – 9 _____ -3
4x + x ____ -3 + 18 and -3x ___ -3 + 9
5x ____ 15 and -3x ___ 6
x ____ 15 / 5 and x ____ -3 / 6
x ____ 3 and x ____ -2
Hence,
The blanks that can fill the inequality are:
x < 3 and x ≥ -2

Question 7.
You have a $250 gift card to use at a sporting goods store.
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 108
a. Write an inequality that represents the possible numbers x of pairs of socks you can buy when you buy 2 pairs of sneakers. Can you buy 8 pairs of socks? Explain.
b. Describe what the inequality 60 + 80x ≤ 250 represents in this context.
Answer:
a.
It is given that
The cost of a pair of sneakers is: $80
The cost of a pair of socks is: $12
It is given that there are 2 pairs of sneakers and x pairs of socks
Hence,
The inequality that represents the possible number of x pairs of socks you can buy when you buy 2 pairs of sneakers is:
2 (80) + 12x ≤ 250
160 + 12x ≤ 250
12x ≤ 250 – 160
12x ≤ 90
x ≤ 90 / 12
x ≤ 7.5
Hence, from the above,
We can conclude that you can buy only 7 pairs of socks

b.
The given inequality is:
60 + 8x ≤ 250
12 (5) + 8x ≤ 250
Hence, from the above,
We can conclude that you can buy 5 pairs of sneakers in this context

Question 8.
Consider the equation shown, where a, b, c, and d are integers.
ax + b = cx + d
Student A claims the equation will always have one solution. Student B claims the equation will always have no solution. Use the numbers shown to answer parts (a)–(c).
Big Ideas Math Answer Key Algebra 1 Chapter 2 Solving Linear Inequalities 109
a. Select values for a, b, c, and d to create an equation that supports Student A’s claim.
Answer:
The given equation is:
ax + b = cx + d
Student A claims the equation will have always one solution
Hence,
From the given numbers,
The values of a, b, c, and d are:
a = -1, b = 2, c = -1, d = 4
So,
The value of x is:
-x + 2 = x + 4
-x – x = 4 – 2
-2x = 2
x = -2 /2
x = -1
Hence, from the above,
We can conclude that according to Student A’s claim,
The values of a, b, c, and d are:
a = -1, b = 2, c = 1, d = 4

b. Select values for a, b, c, and d to create an equation that supports Student B’s claim.
Answer:
The given equation is:
ax + b = cx + d
Student B claims that the equation will have no solution
So,
The value of x must be equal to 0
So,
a must be equal to c
Hence,
The values of a, b, c, and d are: a = c and the values of b and d will be any number from the given numbers

c. Select values for a, b, c, and d to create an equation that shows both Student A and Student B are incorrect.
Answer:
The given equation is:
ax + b = cx + d
For the claims of both Student A and Student B to be false,
The given equation must have the real numbers as a solution
So,
To have the real numbers as a solution, the values of a and b will be arbitrary from the given numbers

Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations

Big Ideas Math Book Algebra 1 Answer Key Chapter 1 Solving Linear Equations

Solving Linear Equations Maintaining Mathematical Proficiency

Add or subtract.

Question 1.
-5 + (-2)
Answer:
-5 + (-2 ) = -7

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-5 + ( -2 ) = -5 – 2
= – ( 5 + 2 )
= -7
Hence, from the above,
We can conclude that,
-5 + ( -2 ) = -7

Question 2.
0 + (-13)
Answer:
0 + -13 = -13

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
0 + ( -13 ) = 0 – 13
= -13
Hence, from the above,
We can conclude that,
0 + ( -13 ) = -13

Question 3.
-6 + 14
Answer:
-6 + 14 = 8

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-6 + 14 = +14 – 6
= +8
= 8
Hence, from the above,
We can conclude that
-6 +14 = 8

Question 4.
19 – (-13)
Answer:
19 – ( -13 ) = 32

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
19 – ( -13 ) = 10 + 9 + 10 + 3
= 20 + 12
= 32
Hence, from the above,
We can conclude that,
19 – (-13 ) = 32

Question 5.
-1 – 6
Answer:
-1 -6 = -7

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-1 – 6 = – ( 1 + 6 )
= -7
Hence, from the above,
We can conclude that
-1 -6 = -7

Question 6.
– 5 – (-7)
Answer:
-5 – ( -7 ) = 2

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-5 – ( -7 ) = -5 + 7
= 7 – 5
= 2
Hence, from the above,
We can conclude that
-5 – ( -7 ) = 2

Question 7.
17 + 5
Answer:
17 + 5 = 22

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
17 + 5 = 15 + 2 + 5
= 22
Hence, from the above,
We can conclude that
17 + 5 = 22

Question 8.
8 + (-3)
Answer:
8 + ( -3 ) = 5

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
8 + ( -3 ) = 8 – 3
= 5
Hence, from the above,
We can conclude that
8  + ( -3 ) = 5

Question 9.
11 – 15
Answer:
11 – 15 = -4

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
11 – 15 = -15 + 11
= – ( 15 – 11 )
= -4
Hence, from the above,
We can conclude that,
11 – 15 = -4

Multiply or divide.

Question 10.
-3(8)
Answer:
-3(8) = -3 × 8
= -24

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-3(8) = -3 × +8
= -24
Hence, from the above,
We can conclude that
-3 ( 8 ) = -24

Question 11.
-7 • (-9)
Answer:
-7 . ( -9 ) = -7 × +9
= -63

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-7 . ( 9 ) = -7 × +9
= -63
Hence, from the above,
We can conclude that
-7 . ( 9 ) = -63

Question 12.
4 • (-7)
Answer:
4 . ( -7 ) = 4 × ( -7 )
= -28

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
4. (-7 ) = +4 × -7
= -28
Hence, from the above,
We can conclude that
4 . ( -7 ) = -28

Question 13.
-24 ÷ (-6)
Answer:
-24 ÷ ( -6 ) = 4

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-24 ÷ ( -6 ) = + ( 24 ÷ 6 )
= 4
Hence, from the above,
We can conclude that
-24 ÷ ( -6 ) = 4

Question 14.
-16 ÷ 2
Answer:
-16 ÷ 2 = -8

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-16 ÷ 2 = -8
Hence, from the above,
We can conclude that
-16 ÷ 2 = -8

Question 15.
12 ÷ (-3)
Answer:
12 ÷ ( -3 ) = -4

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
12 ÷ ( -3 )
= 12 ÷ -3
= -4
Hence, from the above,
We can conclude that
12 ÷ ( -3 ) = -4

Question 16.
6 • 8
Answer:
6 . 8 = 6 × 8
= 48

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
6 . 8 = 6 × 8
= 48
Hence, from the above,
We can conclude that
6 . 8 = 48

Question 17.
36 ÷ 6
Answer:
36 ÷ 6 = 6

Explanation:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
36 ÷ 6 = ( 30 + 6 ) ÷ 6
= ( 30 ÷ 6 ) + ( 6 ÷ 6 )
= 5 + 1
= 6
Hence, from the above,
We can conclude that
36 ÷ 6 = 6

Question 18.
-3(-4)
Answer:
-3 ( -4 ) = -3 × -4
= -12

Explanation:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
We know that,
The result of any mathematical operation follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
So,
-3 ( -4 ) = -3 × -4
= +12
Hence, from the above,
We can conclude that
-3 ( – 4 ) = 12

Question 19.
ABSTRACT REASONING
Summarize the rules for
(a) adding integers,
(b) subtracting integers,
(c) multiplying integers, and
(d) dividing integers.
Give an example of each.
Answer:
a) Adding integers:
We know that,
The result of any mathematical operation i.e., Addition or subtraction follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
Example:
Let take the two numbers -2 and 3
So,
The addition of -2 and 3 is:
-2 + 3 = +1
= 1 ( Since the big number has a positive sign )

b) Subtracting integers:
We know that,
The result of any mathematical operation i.e., Addition or subtraction follows the sign of the big number and if the two numbers have the same sign, then the result of that also operation also follows the same sign and if there is no sign before the number, then we can take that number as positive number i.e., +a
Example:
Let take the two numbers -3 and +8
So,
The subtraction of -3 and +8 is:
-3 – ( +8 ) = -3 – 8
= – ( 3 + 8 )
= -11 ( Since both the numbers have a negative sign )

c) Multiplying integers:
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
Example:
Let take the two numbers +8 and -3
So,
The multiplication of +8 and -3 is:
+8 ( -3 ) = 8 × -3
= -24 ( Since + × – = – )

d) Dividing integers:
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
Example:
Let take the two numbers -12 and -2
So,
The division of -12 and -2 is:
-12 ÷ ( -2 ) = -12 ÷ -2
= ( -10 + -2 ) ÷ -2
= ( -10 ÷ -2 ) + ( -2 ÷ -2 )
= 5 + 1
= 6
Hence, from the above,
We can conclude all the rules for the four basic mathematical operations.

Solving Linear Equations Monitoring Progress

Solve the problem and specify the units of measure.

Question 1.
The population of the United States was about 280 million in 2000 and about 310 million in 2010. What was the annual rate of change in population from 2000 to 2010?
Answer:
The annual rate of change in population from 2000 to 2010 is: 30 million

Explanation:
It is given that the population of the United States was about 280 million in 2000 and about 310 million in 2010.
So,
The annual rate of change in population from 2000 to 2010 = ( The population of United States in 2010 ) – ( The population of United States in 2000 )
= 310 – 280
= 30 million
Hence, from the above,
We can conclude that the annual rate of change in population from 2000 to 2010 is: 30 million

Question 2.
You drive 240 miles and use 8 gallons of gasoline. What was your car’s gas mileage (in miles per gallon)?
Answer:
Your car’s gas mileage ( in miles per gallon ) is: 30

Explanation:
It is given that you drive 240 miles and use 8 gallons of gasoline.
So,
The mileage of your car = ( The total number of miles driven by your car ) ÷ ( The number of gallons of gasoline used by your car )
= 240 ÷ 8
= ( 160 + 80 ) ÷ 8
= ( 160 ÷ 8 ) + ( 80 ÷ 8 )
= 20 + 10
= 30 miles
Hence, from the above,
We can conclude that the mileage of your car is: 30 miles per gallon

Question 3.
A bathtub is in the shape of a rectangular prism. Its dimensions are 5 feet by 3 feet by 18 inches. The bathtub is three-fourths full of water and drains at a rate of 1 cubic foot per minute. About how long does it take for all the water to drain?
Answer:
The total time taken for the water to drain is: 2,430 minutes

Explanation:
It is given that a bathtub is in the shape of a rectangular prism. Its dimensions are 5 feet by 3 feet by 18 inches. The bathtub is three-fourths full of water and drains at a rate of 1 cubic foot per minute.
So,
The volume of the rectangular prism = The dimensions of the rectangular prism
= 5 × 3 × 18 × 12
= 3,240 cubic feet
Now,
The volume of the bathtub which is three-fourths full of water = \(\) {3}{4}[\latex] × 3240
= 2,430 cubic feet
It is also given that that the bathtub drains at a rate of 1 cubic foot per minute.
So,
The time taken to drain 2,430 cubic feet of water in minutes = 2,430 × 1
= 2,430 minutes
Hence, from the above,
We can conclude that the time taken for the water to drain from the bathtub at a rate of 1 cubic foot per minute is: 2,430 minutes

Lesson 1.1 Solving Simple Equations

Essential Question
How can you use simple equations to solve real-life problems?

Exploration 1
Measuring Angles

Work with a partner. Use a protractor to measure the angles of each quadrilateral. Copy and complete the table to organize your results. (The notation m∠A denotes the measure of angle A.) How precise are your measurements?

Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 2

EXPLORATION 2
Making a Conjecture

Work with a partner. Use the completed table in Exploration 1 to write a conjecture about the sum of the angle measures of a quadrilateral. Draw three quadrilaterals that are different from those in Exploration 1 and use them to justify your conjecture.
Answer:
The completed table is:

From the above table,
We can say that the sum of all the angle in any quadrilateral is: 360 degrees
So,
From the above table,
The angles in Quadrilateral a is: 60 degrees, 125 degrees, 120 degrees, and 55 degrees
We know that,
The quadrilateral will have n angles based on the shape.
The shape which has more than 3 sides is called a Quadrilateral.
Ex: Rectangle, Square, Pentagon, Hexagon, etc.

From the above quadrilaterals,
We can say that all the sides in each quadrilateral are equal.
So,
The angles in each quadrilateral are also equal.
So,
In a rectangle, there are 4 sides
So,
By measuring, we can observe that each angle of a rectangle is: 90 degrees
Hence,
The sum of all angles in a rectangle = 90 + 90 + 90 + 90 = 360 degrees
In a pentagon, there are 5 sides
By measuring, we can observe that each rectangle of a pentagon
Hence,
The sum of all the angles in a pentagon = = 72+ 72 + 72 + 72 + 72 = 360 degrees
In a Hexagon, there are 6 sides
So,
By measuring, we can observe that each angle of a hexagon is: 60 degrees
Hence,
The sum of all angle is a Hexagon = 60 + 60 + 60 + 60 + 60 + 60 = 360 degrees
Hence, from the above-drawn quadrilaterals,
We can conclude that our conjecture is proven.

EXPLORATION 3
Applying Your Conjecture

Work with a partner. Use the conjecture you wrote in Exploration 2 to write an equation for each quadrilateral. Then solve the equation to find the value of x. Use a protractor to check the reasonableness of your answer.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 3
Answer:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 3
From Exploration 2, the proven conjecture is:
The sum of all angles in a quadrilateral is: 360 degrees
Now,
In Quadrilateral a.,
By using the above-proven conjecture,
85 + 80 + 100 + x = 360 degrees
265 + x = 360 degrees
x = 360 – 265
= 95 degrees
So,
The angle x is: 95 degrees
In Quadrilateral b.,
By using the above-proven conjecture,
72 + 78 + 60 + x = 360 degrees
210 + x = 360 degrees
x = 360 – 210
= 150 degrees
So,
The angle of x is: 150 degrees
In Quadrilateral c.,
By using the above-proven conjecture,
90 + 90  +30 + x = 360 degrees
210 + x = 360 degrees
x = 360 – 210
= 150 degrees
So,
The angle of x is: 150 degrees

Communicate Your Answer

Question 4.
How can you use simple equations to solve real-life problems?

Question 5.
Draw your own quadrilateral and cut it out. Tear off the four corners of the quadrilateral and rearrange them to affirm the conjecture you wrote in Exploration 2. Explain how this affirms the conjecture.
Answer:
Your Quadrilateral is:

From the above Quadrilateral,
We can observe that the tear-off corners of the quadrilateral are joined and it becomes the triangle.
So,
In the above Quadrilateral, there are two quadrilaterals.
We know that,
The sum of all angles in a triangle is: 180 degrees
So,
The sum of all angles in the two triangles = 180 + 180 = 360 degrees
These two triangles form a quadrilateral.
So,
The sum of all angles in a quadrilateral is: 360 degrees
Hence, from the above,
We can conclude that we can prove Exploration 2’s conjecture by your own example.

1.1 Lesson

Monitoring Progress

Solve the equation. Justify each step. Check your solution.

Question 1.
n + 3 = -7
Answer:
The value of n is: -10

Explanation:
The given equation is:
n + 3 = -7
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
n + 3 = -7
n = -7 – (+3 )
n = -7 – 3
= -10
Hence from the above,
We can conclude that the value of n is: -10

Question 2.
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
Answer:
The value of g is: –\(\frac{1}{3}\)

Explanation:
The given equation is:
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
g – \(\frac{1}{3}\) = –\(\frac{2}{3}\)
g = –\(\frac{2}{3}\) + \(\frac{1}{3}\)
g = \(\frac{-2 + 1}{3}\)
g = \(\frac{-1}{3}\)
g = –\(\frac{1}{3}\)
Hence, fromthe above,
We can conclude that the value of g is: –\(\frac{1}{3}\)

Question 3.
-6.5 = p + 3.9
Answer:
The value of p is: -10.4

Explanation:
The given equation is:
-6.5 = p + 3.9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-6.5 = p + 3.9
p = -6.5 – 3.9
= – ( 6.5 + 3.9 )
= – 10.4
Hence, from the above,
We can conclude that the value of p is: -10.4

Monitoring Progress

Solve the equation. Justify each step. Check your solution.

Question 4.
\(\frac{y}{3}\) = -6
Answer:
The value of y is: -18

Explanation:
The given equation is:
\(\frac{y}{3}\) = -6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{y}{3}\) = -6
\(\frac{y}{1}\) × \(\frac{1}{3}\) = -6
\(\frac{y}{1}\) = -6 ÷ \(\frac{1}{3}\)
y = -6 × -3
y = -18
Hence, from the above,
We can conclude that the value of y is: -18

Question 5.
9π = πx
Answer:
The value of x is: 9

Explanation:
The given equation is:
9π = πx
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
9π = πx
9 × π = π × x
x = ( 9 × π ) ÷ π
x = 9
Hence, from the above,
We can conclude that the value of x is: 9

Question 6.
0.05w = 1.4
Answer:
The value of w is: 28

Explanation:
The given equation is:
0.05w = 1.4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
0.05w = 1.4
0.05 × w = 1.4
\(\frac{5}{100}\) × w = \(\frac{14}{10}\)
w = \(\frac{14}{10}\) ÷ \(\frac{5}{100}\)
w = \(\frac{14}{10}\) × \(\frac{100}{5}\)
w = \(\frac{14 × 100}{10 × 5}\)
w = \(\frac{28}{1}\)
w = 28
Hence, from the above,
We can conclude that the value of w is: 28

Monitoring Progress

Question 7.
Suppose Usain Bolt ran 400 meters at the same average speed that he ran the 200 meters. How long would it take him to run 400 meters? Round your answer to the nearest hundredth of a second.
Answer:
The time it took for him to run 400 meters is: 0.50 seconds

Explanation:
It is given that Usain Bolt ran 400 meters at the same average speed that he ran the 200 meters.
We know that,
Speed = Distance ÷ Time
But, it is given that the average speed is the same.
Hence,
Speed = Constant
So,
Since speed is constant, distance is directly proportional to time.
So,
The time taken by Usain Bolt to run 400 meters = 200 ÷ 400
= ( 2 × 100 ) ÷ ( 4 × 100 )
= 10 ÷ 20
= 0.50 seconds ( 0.5 and 0.50 are the same values Only for the representation purpose, we will add ‘0’ after 5 )
Hence from the above,
We can conclude that the time is taken by Usain Bolt to run 400 meters when rounded-off to the nearest hundredth is: 0.50 seconds

Monitoring Progress

Question 8.
You thought the balance in your checking account was $68. When your bank statement arrives, you realize that you forgot to record a check. The bank statement lists your balance as $26. Write and solve an equation to find the amount of the check that you forgot to record.
Answer:
The amount of the check that you forgot to record is: $42

Explanation:
It is given that you thought the balance in your checking account was $68 and when your bank statement arrives, you realize that you forgot to record a check and the bank statement lists your balance as $26.
Now,
Let the amount you forgot to record be: x
So,
The total balance in your checking account = ( The listed balance ) + ( The amount that you forgot to record a check )
68 = 26 + x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
68 = 26 + x
x = 68 – 26
x = $42
Hence, from the above,
We can conclude that the amount that forgot to record is: $42

Solving Simple Equations 1.1 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 5–14, solve the equation. Justify each step. Check your solution.

Question 1.
VOCABULARY Which of the operations +, -, ×, and ÷ are inverses of each other?
Answer:
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Hence, from the above,
We can conclude that,
+ is inverse of –  and vice-versa
× is inverse of ÷ and vice-versa

Question 2.
VOCABULARY Are the equations -2x = 10 and -5x = 25 equivalent? Explain.
Answer:
The equations -2x = 10 and -5x = 25 are equivalent

Explanation:
The given equations are:
-2x = 10 and -5x = 25
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
So,
From -2x = 10,
x = 10 ÷ ( -2 )
x = -10 ÷ 2
x = -5
From -5x = 25,
x = 25 ÷ ( -5 )
x = -25 ÷ 5
x = -5
Hence, from the above,
We can conclude that the equations -2x = 10 and -5x = 25 are equivalent

Question 3.
WRITING Which property of equality would you use to solve the equation 14x = 56? Explain.
Answer:
The given equation is:
14x = 56
So,
It can be re-written as
14 × x = 56
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = 56 ÷ 14
x = 4
Hence, from the above,
We can conclude that the value of x is: 4

Question 4.
WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 501
Answer:
The equation C) does not belong to the other three

Explanation:
Let the given equations be named as A), B), C), and D)
So,
The given equations are:
A) 8 = x ÷ 2
B) 3 = x ÷ 4
C) x – 6 = 5
D) x ÷ 3 = 9
So,
From the above equations,
The equations A, B), and D) are dividing the numbers whereas equation C) subtracting the numbers
Hence, from the above,
We can conclude that,
The equation C) does not belong to the other three.

Question 5.
x + 5 = 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q5

Question 6.
m + 9 = 2
Answer:
The value of m is: -5

Explanation:

Question 7.
y – 4 = 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q7

Question 8.
s – 2 = 1
Answer:
The value of s is: 3

Explanation;

Question 9.
w + 3 = -4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q9

Question 10.
n – 6 = -7
Answer:
The value of n is: -1

Explanation:

Question 11.
-14 = p – 11

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q11

Question 12.
0 = 4 + q
Answer:
The value of q is: -4

Explanation;

Question 13.
r + (-8) = 10

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q13

Question 14.
t – (-5) = 9
Answer:
The value of t is: 4

Explanation;

Question 15.
MODELING WITH MATHEMATICS
A discounted amusement park ticket costs $12.95 less than the original price p. Write and solve an equation to find the original price.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 4

Answer:
The equation for the original price is:
p = x + $12.95

Explanation:
It is given that a discounted amusement park ticket costs $12.95 less than the original price p.
So,
Let the discounted amusement park ticket be: x
The given original price is: p
So,
The discounted amusement park ticket cost = p – $12.95
x = p – 12.95
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
p = x + $12.95
Hence, from the above,
We can conclude that the equation for the original price is:
p = x + $12.95

Question 16.
MODELING WITH MATHEMATICS
You and a friend are playing a board game. Your final score x is 12 points less than your friend’s final score. Write and solve an equation to find your final score.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 5

Answer:
Your final score is:
x = ( The score of your friend ) – 12

Explanation:
It is given that you and a friend are playing a board game. Your final score x is 12 points less than your friend’s final score.
So,
The scores table is shown below:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 5
From the above table,
The final score of your friend is: 195
Let the final score of yours is: x
So,
x = ( The final score of your friend ) – 12
= 195 – 12
= 183 points
Hence, from the above,
We can conclude that the final score of yours is: 183 points

USING TOOLS
The sum of the angle measures of a quadrilateral is 360°. In Exercises 17–20, write and solve an equation to find the value of x. Use a protractor to check the reasonableness of your answer.

Question 17.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q17

Question 18.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 7

Answer:
The value of x is: 85 degrees

Explanation:
We know that,
The sum of angles in a quadrilateral is: 360 degrees
So,
150 + 48 + 77 + x = 360
275 + x = 360
x = 360 – 275
x = 85 degrees
Hence, from the above,
We can conclude that the value of x is: 85 degrees

Question 19.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q19

Question 20.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 9

Answer:
The value of x is: 100 degrees

Explanation:
We know that,
The sum of all angles in a quadrilateral is: 360 degrees
So,
115 + 85 + 60 + x = 360
260 + x = 360
x = 360 – 260
x = 100 degrees
Hence, from the above,
We can conclude that the value of x is: 100 degrees

In Exercises 21–30, solve the equation. Justify each step. Check your solution.

Question 21.
5g = 20

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q21

Question 22.
4q = 52
Answer:
The value of g is: 13

Explanation:
The given equation is:
4g = 52
4 × g = 52
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
g = 52 ÷ 4
= ( 44 + 8 ) ÷ 4
= ( 44 ÷ 4 ) + ( 8 ÷ 4 )
= 11 + 2
= 13
Hence, from the above,
We can conclude that the value of g is: 13

Question 23.
p ÷ 5 = 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q23

Question 24.
y ÷ 7 = 1
Answer:
The value of y is: 7

Explanation:
The given equation is:
y ÷ 7 = 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
y = 1 × 7
y = 7
Hence, from the above,
We can conclude that the value of y is: 7

Question 25.
-8r = 64

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q25

Question 26.
x ÷(-2) = 8
Answer:
The value of x is: -16

Explanation:
The given equation is:
x ÷ ( -2 ) = 8
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ÷ ( -2 ) = 8
x = 8 × ( -2 )
x = -16
Hence, from the above,
We can conclude that the value of x is: -16

Question 27.
\(\frac{x}{6}\) = 8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q27

Question 28.
\(\frac{w}{-3}\) = 6
Answer:
The value of w is: -18

Explanation:
The given equation is:
\(\frac{w}{-3}\) = 6
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{w}{-3}\) = 6
w = 6 × ( -3 )
w = -18
Hence, from the above,
We can conclude that the value of w is: -18

Question 29.
-54 = 9s

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q29

Question 30.
-7 = \(\frac{t}{7}\)
Answer:
The value of t is: -49

Explanation:
The given equation is:
-7 = \(\frac{t}{7}\)
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-7 = \(\frac{t}{7}\)
t = -7 × 7
t = -49
Hence, from the above
We can conclude that the value of t is: -49

In Exercises 31– 38, solve the equation. Check your solution.

Question 31.
\(\frac{3}{2}\) + t = \(\frac{1}{2}\)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q31

Question 32.
b – \(\frac{3}{16}\) = \(\frac{5}{16}\)
Answer:
The value of b is: \(\frac{1}{2}\)

Explanation:
The given equation is:
b – \(\frac{3}{16}\) = \(\frac{5}{16}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
b = \(\frac{5}{16}\) + \(\frac{3}{16}\)
Since the denominators of both the numerators are equal, add the numerators making the denominator common
So,
b = \(\frac{5 + 3}{16}\)
b = \(\frac{8}{16}\)
b = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of b is: \(\frac{1}{2}\)

Question 33.
\(\frac{3}{7}\)m = 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q33

Question 34.
–\(\frac{2}{5}\)y = 4
Answer:
The value of y is: 10

Explanation:
The given equation is:
–\(\frac{2}{5}\)y = 4
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
–\(\frac{2}{5}\)y = 4
–\(\frac{2}{5}\) × y = 4
y = 4 ÷ –\(\frac{2}{5}\)
y = 4 × –\(\frac{5}{2}\)
y = -4 × –\(\frac{5}{2}\)
y = –\(\frac{4}{1}\) × –\(\frac{5}{2}\)
y = –\(\frac{4 × 5}{1 × 2}\)
y = 10
Hence, from the above,
We can conclude that the value of y is: 10

Question 35.
5.2 = a – 0.4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q35

Question 36.
f + 3π = 7π
Answer:
The value of f is: 4π

Explanation:
The given equation is:
f + 3π = 7π
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
f + 3π = 7π
f = 7π – 3π
f = π ( 7 – 3 )
f = π ( 4 )
f = 4π
Hence, from the above,
We can conclude that the value of f is: 4π

Question 37.
– 108π = 6πj

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q37

Question 38.
x ÷ (-2) = 1.4
Answer:
The value of x is: –\(\frac{14}{5}\)

Explanation:
The given equation is:
x ÷ ( -2 ) = 1.4
We know that,
A) – ÷ – = +
B) + ÷ – = –
C) + ÷ + = +
D) – ÷ + = –
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ÷ ( -2 ) = 1.4
x ÷ ( -2 ) = \(\frac{14}{10}\)
x ÷ ( -2 ) = \(\frac{7}{5}\)
x = \(\frac{7}{5}\) × ( -2 )
x = – \(\frac{7}{5}\) × \(\frac{2}{1}\)
x = –\(\frac{14}{5}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{14}{5}\)

ERROR ANALYSIS
In Exercises 39 and 40, describe and correct the error in solving the equation.

Question 39.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 10

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q39

Question 40.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 11

Answer:
A negative 3 should have been multiplied to each side.
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
So,
-(\(\frac{m}{3}\) ) =-4
-3  ( \(\frac{m}{3}\) ) = -4 ( -3 )
3 ( \(\frac{m}{3}\) ) = -4 ( -3 )
3 ( \(\frac{m}{3}\) ) = 12
\(\frac{m}{3}\)  × \(\frac{3}{1}\) = 12
m = 12
Hence, from the above,
We can conclude that the value of m is: 12

Question 41.
ANALYZING RELATIONSHIPS
A baker orders 162 eggs. Each carton contains 18 eggs. Which equation can you use to find the number x of cartons? Explain your reasoning and solve the equation.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 12

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q41

MODELING WITH MATHEMATICS
In Exercises 42– 44, write and solve an equation to answer the question.

Question 42.
The temperature at 5 P.M. is 20°F. The temperature at 10 P.M. is -5°F. How many degrees did the temperature fall?
Answer:
The fall in temperature is: 25 degrees Fahrenheit

Explanation:
It is given that the temperature at 5 P.M. is 20°F and the temperature at 10 P.M. is -5°F.
So,
The fall in temperature = ( The temperature at 5 P.M ) – ( The temperature at 10 P.M )
We know that,
A) – × – = +
B) + × – = –
C) + × + = +
D) – × + = –
So,
The fall in temperature = 20 – ( -5 )
= 20 + 5
= 25 degrees Fahrenheit
Hence, from the above,
We can conclude that the fall in temperature is: 25 degrees Fahrenheit

Question 43.
The length of an American flag is 1.9 times its width. What is the width of the flag?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q43

Question 44.
The balance of an investment account is $308 more than the balance 4 years ago. The current balance of the account is $4708. What was the balance 4 years ago?
Answer:
The balance 4 years ago is: $4,400

Explanation:
It is given that the balance of an investment account is $308 more than the balance 4 years ago. The current balance of the account is $4708.
So,
The current balance of the account = ( The balance of an investment account 4 years ago ) + $308
Let the balance of an investment account four years ago be x.
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
$4,708 = x + $308
x = 4,708 – 308
x = $4,400
Hence, from the above,
We can conclude that the balance of an investment account four years ago is: $4,400

Question 45.
REASONING
Identify the property of equality that makes Equation 1 and Equation 2 equivalent.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 14

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q45

PROBLEM-SOLVING
Question 46.

Tatami mats are used as a floor covering in Japan. One possible layout uses four identical rectangular mats and one square mat, as shown. The area of the square mat is half the area of one of the rectangular mats.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15
a. Write and solve an equation to find the area of one rectangular mat.
Answer:
The area of one rectangular mat is: 18 ft²

Explanation:
It is given that the tatami mats are used as a floor covering in Japan and for that, one layout of tatami mats requires the four identical rectangular mats and the one square mat
So,
The total area of the tatami mats = ( The area of the four rectangular mats ) + ( The area of the one square mat )
The give tatami mat layout is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15
From the layout,
We can observe that
The total area of the layout is: 81 ft²
So,
The total area of tatami mats = 81 ft²
( The area of the four rectangular mats ) + ( The area of the one square mat ) = 81 ft²
It is also given that
The area of a square mat is half of one of the rectangular mats
So,
Area of the square mat = ( Area of the rectangular mat ) ÷ 2
So,
( The area of the four rectangular mats ) + \(\frac{Area of the rectangular mat}{2}\)  = 81 ft²
4 ( The area of the rectangular mat ) + \(\frac{Area of the rectangular mat}{2}\)  = 81 ft²
\(\frac{8}{2}\) ( The area of the rectangular mat ) + \(\frac{1}{2}\) ( The area of the rectangular mat ) = 81 ft²
( The area of the rectangular mat ) [ \(\frac{8}{2}\) + \(\frac{1}{2}\) ] = 81 ft²
\(\frac{9}{2}\) ( The area of the rectangular mat ) = 81 ft²
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
The area of the rectangular mat = 81 ÷ \(\frac{9}{2}\)
= 81 × \(\frac{2}{9}\)
= \(\frac{81}{1}\) × \(\frac{2}{9}\)
= \(\frac{81 × 2}{9 × 1}\)
= 18 ft²
Hence, from the above,
We can conclude that the area of one rectangular mat is: 18 ft²

b. The length of a rectangular mat is twice the width. Use Guess, Check, and Revise to find the dimensions of one rectangular mat.
Answer:
The dimensions of the rectangular mat are:
Length: 6 ft
Width: 3 ft

Explanation:
From the above problem,
The area of the rectangular mat = 18 ft²
It is given that the length of a rectangular mat is twice the width.
We know that the area of the rectangle = ( Length ) × ( Width )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
It is also given that the length of a rectangular mat is twice the width
So,
Length of a rectangular mat = 2 × Width
Now,
The area of the rectangular mat = Length × Width
18 = 2 × Width × Width
Width × Width = 18 ÷ 2
Width × Width = 9
From guessing,
We can say that
Width of the rectangular mat = 3 ft
Now,
The length of the rectangular mat = 2 × 3
= 6 ft
Hence, from the above,
We can conclude that the dimensions of the rectangular mat are:
Length: 6 ft
Width: 3 ft

Question 47.
PROBLEM-SOLVING
You spend $30.40 on 4 CDs. Each CD costs the same amount and is on sale for 80% of the original price.
a. Write and solve an equation to find how much you spend on each CD.
b. The next day, the CDs are no longer on sale. You have $25. Will you be able to buy 3 more CDs? Explain your reasoning.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 15.1

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q47

Question 48.
ANALYZING RELATIONSHIPS
As c increases, does the value of x increase, decrease, or stay the same for each equation? Assume c is positive.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 16

Answer:
Let assume the values of c be: 0,1,2,3
So,
The completed table by taking the values of c is:

By taking the values of c positive i.e., 0, 1, 2, 3
We can observe that as the value of c increases, the values of x sometimes increasing and sometimes stays the same but not decreasing.

Question 49.
USING STRUCTURE
Use the values -2, 5, 9, and 10 to complete each statement about the equation ax = b – 5.
a. When a = ___ and b = ___, x is a positive integer.
b. When a = ___ and b = ___, x is a negative integer.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q51

Question 50.
HOW DO YOU SEE IT?
The circle graph shows the percents of different animals sold at a local pet store in 1 year.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 17
a. What percent is represented by the entire circle?
Answer:
The percent represented by the entire circle is = 69 % + x %

Explanation:
It is given that the circle represents the percent of different animals sold at a local store for 1 year
Now,
The given circle is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 17
We know that,
In terms of percentages, any circle represents 100%
So,
The equation representing the circle is:
The percentage of different animals in the circle = 48 + 5 + 9 + 7 +x
The percentage of different animals in the circle= 69% + x%

b. How does the equation 7 + 9 + 5 + 48 + x = 100 relate to the circle graph? How can you use this equation to find the percent of cats sold?
Answer:
The percent of cats sold is: 31%

Explanation:
We know that,
In terms of percentages, any circle represents 100%
So,
The total percent of animals = The percent of animals that are represented by the circle
100% = 69% + x%
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x% = 100% – 69%
x% = 31%
Hence, from the above,
We can conclude that the percent of cats is: 31%

Question 51.
REASONING
One-sixth of the girls and two-sevenths of the boys in a school marching band are in the percussion section. The percussion section has 6 girls and 10 boys. How many students are in the marching band? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q51

Question 52.
THOUGHT-PROVOKING
Write a real-life problem that can be modeled by an equation equivalent to the equation 5x = 30. Then solve the equation and write the answer in the context of your real-life problem.

Answer:
Let suppose there is some number of boys. The number of girls is five times of the boys and the total number of girls is 30. Find the number of boys?
Ans:
Let,
The number of boys is x.
It is given that the number of girls is five times of boys.
So,
The number of girls = 5x
It is also given that
The number of girls = 30
So,
5x = 30
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 × x = 30
x = 30 ÷ 5
x = 6
Hence, from the above,
We can conclude that the number of boys is: 6

MATHEMATICAL CONNECTIONS
In Exercises 53–56, find the height h or the area of the base B of the solid.

Question 53.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q53

Question 54.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 19

Answer:
The height of the cuboid is: 9 cm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 19
We know that,
The volume of a cuboid = L × B × H
We know that,
The cuboid is made from a rectangle
We know that,
The area of a rectangle = L × H
So,
The volume of a cuboid = A × B
From the above figure,
The given volume is: 1323 cm³
The given Area is: 147 cm²
So,
1323 = 147 × H
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
H = 1323 ÷ 147
H = 9
Hence, from the above,
We can conclude that the height of the cuboid is: 9 cm

Question 55.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 20

Answer:

Question 56.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 21

Answer:
The height of the prism is: \(\frac{5}{6}\) ft

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 21
We know that,
The volume of the prism =  Area × Height
From the above figure,
The volume of the prism = 35 ft³
The area of the prism = 30 ft²
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
35 = 30 × H
H = 35 ÷ 30
H = \(\frac{5}{6}\) ft
Hence, from the above,
We can conclude that the height of the prism is: \(\frac{5}{6}\) ft

Question 57.
MAKING AN ARGUMENT
In baseball, a player’s batting average is calculated by dividing the number of hits by the number of at-bats. The table shows Player A’s batting average and the number of at-bats for three regular seasons.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 22
a. How many hits did Player A have in the 2011 regular season? Round your answer to the nearest whole number.
b. Player B had 33 fewer hits in the 2011 season than Player A but had a greater batting average. Your friend concludes that Player B had more at-bats in the 2011 season than Player A. Is your friend correct? Explain. Maintaining

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q57

Maintaining Mathematical Proficiency

Use the Distributive Property to simplify the expression.

Question 58.
8(y + 3)
Answer:
8 ( y + 3 ) = 8y + 24

Explanation:
The given expression is: 8 ( y + 3 )
We know that,
By using the Distributive Property,
a ( b + c ) = ( a × b ) + ( a × c )
So,
By using the above Property,
8 ( y + 3 ) = ( 8 × y ) + ( 8 × 3 )
= 8y + 24
Hence, from the above,
We can conclude that
8 ( y + 3 ) = 8y + 24

Question 59.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 23

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q59

Question 60.
5(m + 3 + n)
Answer:
5 ( m + 3 + n ) = 5m + 5n + 15

Explanation:
The given expression is: 5 ( m + 3 + n )
By using the Distributive Property,
a ( b + c ) = ( a × b ) + ( a × c )
So,
By using the above Property,
5 ( m + 3 + n ) = ( 5 × m ) + ( 5 × 3 ) + ( 5 × n )
= 5m + 15 + 5n
Hence, from the above,
We can conclude that,
5 ( m + 3 + n ) = 5m + 15 + 5n

Question 61.
4(2p + 4q + 6)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q61

Copy and complete the statement. Round to the nearest hundredth, if necessary.

Question 62.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 24

Answer:
The missing number is: \(\frac{1}{12}\)

Explanation:
Let the missing number be: x
So,
The given equation is:
\(\frac{5L}{min}\) = \(\frac{x L}{h}\)
We know that,
1 hour = 60 minutes
So,
1 min = \(\frac{1}{60}\) hour
So,
\(\frac{5 L}{min}\) = \(\frac{5 L × 1}{60h}\)
\(\frac{5 L}{min}\) = \(\frac{1 L }{12h}\)
So,
x = \(\frac{1}{12}\)
Hence, from the above,
We can conclude that,
The missing number is: \(\frac{1}{12}\)

Question 63.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 25

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q63

Question 64.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 26

Answer:
The missing number is: \(\frac{1}{12}\)

Explanation:
Let the missing number be: x
So,
The given equation is:
\(\frac{7 gal}{min}\) = \(\frac{x qin}{sec}\)
We know that,
1 min = 60 seconds
1 quintal = 100 kg
1 gallon = 3.78 kg = 4 kg
So,
1 gallon = 0.04 quintal
1 sec = \(\frac{1}{60}\) min
So,
\(\frac{7 gal}{min}\) = \(\frac{x qin × 1}{60min}\)
\(\frac{7 gal}{min}\) = \(\frac{1 L }{12h}\)
So,
x = \(\frac{1}{12}\)
Hence, from the above,
We can conclude that,
The missing number is: \(\frac{1}{12}\)

Question 65.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 27

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.1-Q65

Lesson 1.2 Solving Multi-step Equations

Essential Question

How can you use multi-step equations to solve real-life problems?

EXPLORATION 1
Solving for the Angle Measures of a Polygon

Work with a partner. The sum S of the angle measures of a polygon with n sides can be found using the formula S = 180(n – 2). Write and solve an equation to find each value of x. Justify the steps in your solution. Then find the angle measures of each polygon. How can you check the reasonableness of your answers?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 27.1
Answer:
The given polygons are:

It is given that,
The sum S of the angle measures of a polygon with n sides can be found using the formula S = 180(n – 2).
a)
The number of sides (n ) = 3
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 3 – 2 )
= 180 ( 1 )
= 180
Now,
The given sides of a polygon are: 30, 9x, (30 + x )
So,
30 + 9x + 30 + x = 180
60 + 10x = 180
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
10x = 180 – 60
10x = 120
x = 120 ÷ 10
x = 12
Hence, from the above,
The angle measures of the given polygon are:
30, 9 × 12, 30 + 12
= 30, 108, 45 degrees
b)
The number of sides (n ) = 3
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 3 – 2 )
= 180 ( 1 )
= 180
Now,
The given sides of a polygon are: 30, 9x, (30 + x )
So,
50 + x + 10 + 20 + x = 180
80 + 2x = 180
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2x = 180 – 80
2x = 100
x = 100 ÷ 2
x = 50
Hence, from the above,
The angle measures of the given polygon are:
50, 50 + 10, 50 + 20
= 50, 60, 70 degrees
c)
The number of sides (n ) = 4
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 4 – 2 )
= 180 ( 2 )
= 360
Now,
The given sides of a polygon are: 50, x, ( 2x + 20), ( 2x + 30 )
So,
50 + x + 2x + 20 + 2x + 30 = 360
100 + 5x = 360
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5x = 360 – 100
5x = 260
x = 260 ÷ 5
x = 52
Hence, from the above,
The angle measures of the given polygon are:
50,52, 2 (52) + 20, 2(52) + 30
= 50, 52, 124, 134 degrees
d)
The number of sides (n ) = 4
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 4 – 2 )
= 180 ( 2 )
= 360
Now,
The given sides of a polygon are: x, x + 42, x + 35, x – 17
So,
x + x + 42 + x + 35 + x – 17 = 360
60 + 4x = 360
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4x = 360 – 60
4x = 300
x = 300 ÷ 4
x = 75
Hence, from the above,
The angle measures of the given polygon are:
75,  75 + 42,  75 + 35 , 75 – 17
= 75, 117, 110, 58 degrees
e)
The number of sides (n ) = 5
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 5 – 2 )
= 180 ( 3 )
= 540
Now,
The given sides of a polygon are: (4x + 15), (5x + 10), (8x + 8), (3x + 5), (5x + 2)
So,
(4x + 15)+ (5x + 10)+ (8x + 8)+ (3x + 5)+ (5x + 2) = 540
40 + 25x = 540
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
25x = 540 – 40
25x = 500
x = 500 ÷ 25
x = 20
Hence, from the above,
The angle measures of the given polygon are:
(4. 20 + 15)+ (5. 20 + 10)+ (8.20 + 8)+ (3. 20 + 5)+ (5. 20 + 2)
= 95, 110, 168, 65, 102 degrees
f)
The number of sides (n ) = 5
So,
The sum of angles ( S ) = 180 ( n – 2 )
= 180 ( 5 – 2 )
= 180 ( 3 )
= 540
Now,
The given sides of a polygon are: (2x + 25), (3x + 16), (2x + 8), (4x – 18), (3x – 7)
So,
(2x + 25) + (3x + 16) + (2x + 8) + (4x – 18) + (3x – 7) = 540
24 + 14x = 540
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
14x = 540 – 24
14x = 516
x = 540 ÷ 14
x = 38.5
x = 39
Hence, from the above,
The angle measures of the given polygon are:
(2. 39 + 25), (3.39 + 16), (2.39 + 8), (4.39 – 18), (3.39 – 7)
= 103, 133, 86, 138, 110 degrees

EXPLORATION 2
Work with a partner.

a. Draw an irregular polygon.
Answer:

b. Measure the angles of the polygon. Record the measurements on a separate sheet of paper.
Answer:

c. Choose a value for x. Then, using this value, work backward to assign a variable expression to each angle measure, as in Exploration 1.
d. Trade polygons with your partner.
e. Solve an equation to find the angle measures of the polygon your partner drew. Do your answers seem reasonable? Explain.

Communicate Your Answer

Question 3.
How can you use multi-step equations to solve real-life problems?

Question 4.
In Exploration 1, you were given the formula for the sum S of the angle measures of a polygon with n sides. Explain why this formula works.
Answer:
We know that,
The sum of the angles in a triangle is: 180 degrees
The triangle is also a quadrilateral
So,
A quadrilateral can be formed by the minimum of the three lines
So,
The minimum sum of all the angles in a quadrilateral is: 180 degrees
Now,
Let suppose we form a quadrilateral with 4 sides.
So,
The sum of all the angles in a quadrilateral = 360 degrees = 180 degrees × 2
= 180 degrees ( 4 sides -2 )
Let suppose we form a quadrilateral with 5 sides
So,
The sum of all the angles in a quadrilateral = 540 degrees = 180 degrees × 3
= 180 degrees ( 5 -2 )
Hence, in general,
We can conclude that the sum of all the angles with n sides in a quadrilateral = 180 degrees ( n-2 )

Question 5.
The sum of the angle measures of a polygon is 1080º. How many sides does the polygon have? Explain how you found your answer.
Answer:
The number of sides the polygon with 1080° have: 6

Explanation:
It is given that the sum of all angle measures of a polygon is: 1080°
We know that,
The sum of angle measures with n sides in a polygon = 180° ( n – 2 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
1080° =180° ( n – 2 )
n – 2 = 1080 ÷ 180
n – 2 = 6
n = 6 + 2
n = 8
Hence, from the above,
We can conclude that the number of sides of the polygon with sum of the angles 1080° is: 6

1.2 Lesson

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
-2n + 3 = 9
Answer:
The value of n is: -3

Explanation:
The given equation is:
-2n + 3 = 9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2n + 3 = 9
-2n = 9 – (+3 )
n = 6 ÷ ( -2 )
= -3
Hence from the above,
We can conclude that the value of n is: -3

Question 2.
-21 = \(\frac{1}{2}\) – 11

Question 3.
-2x – 10x + 12 = 18
Answer:
The value of x is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
-2x – 10x + 12 = 18
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-( 2x + 10x ) = 18 – 12
-12x = 6
x = 6 ÷ ( -12 )
x = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{1}{2}\)

Monitoring Progress

Solve the equation. Check your solution.

Question 4.
3(x + 1) + 6 = -9
Answer:
The value of x is: -6

Explanation:
The given equation is:
3 ( x + 1 ) + 6 = -9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3 ( x + 1 ) = -9 – (+6 )
By using the Distributive property,
3 ( x + 1 ) = 3x + 3
So,
3x + 3 = -15
3x = -15 – ( +3 )
3x = -18
x = -18 ÷ 3
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Question 5.
15 = 5 + 4(2d – 3)
Answer:
The value of d is:\(\frac{11}{4}\)

Explanation:
The given equation is:
15 = 5 + 4 ( 2d – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
4 ( 2d – 3 ) = 15 -5
4 ( 2d – 3 ) = 10
By using the Distributive property,
4 ( 2d – 3 ) = 4 (2d ) -4 (3 )
= 8d – 12
So,
8d – 12 = 10
8d = 10 + 12
8d = 22
d = 22 ÷ 8
d = \(\frac{11}{4}\)
Hence, from the above,
We can conclude that the value of d is: \(\frac{11}{4}\)

Question 6.
13 = -2(y – 4) + 3y
Answer:
The value of y is: 5

Explanation:
The given equation is:
13 = -2 ( y – 4 ) + 3y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
By using the Distributive Property,
-2 ( y – 4 ) = -2y + 8
So,
13 = -2y + 8 + 3y
13 = y + 8
y = 13 – 8
y = 5
Hence, from the above,
We can conclude that the value of y is: 5

Question 7.
2x(5 – 3) – 3x = 5
Answer:
The value of x is: 5

Explanation:
The given equation is:
2x ( 5 – 3 ) – 3x = 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2x ( 2 ) – 3x = 5
4x – 3x = 5
x = 5
Hence, from the above,
We can conclude that the value of y is: 5

Question 8.
-4(2m + 5) – 3m = 35
Answer:
The value of m is: -5

Explanation:
The given equation is:
-4 ( 2m + 5 ) – 3m = 35
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property,
-4 ( 2m + 5 ) = -4 (2m ) + 5 ( -4 )
= -8m -20
So,
-8m -20 -3m = 35
-11m – 20 = 35
-11m = 35 + 20
-11m = 55
m = 55 ÷ ( -11 )
m = -5
Hence, from the above,
We can conclude that the value of m is: -5

Question 9.
5(3 – x) + 2(3 – x) = 14
Answer:
The value of x is: 1

Explanation:
The given equation is:
5 ( 3 – x ) + 2 ( 3 – x ) = 14
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive property,
5 ( 3 – x ) = 5 (3 ) -5 (x)
= 15 – 5x
2 ( 3 – x ) = 2 (3) – 2 ( x)
= 6 – 2x
So,
15 – 5x + 6 – 2x = 14
21 – 7x = 14
7x = 21 – 14
7x = 7
x = 7 ÷ 7
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Monitoring Progress

Question 10.
The formula d = \(\frac{1}{2}\)n + 26 relates the nozzle pressure n (in pounds per square inch) of a fire hose and the maximum horizontal distance the water reaches d (in feet). How much pressure is needed to reach a fire 50 feet away?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 28
Answer:
The pressure needed to reach a fire 50 feet away (n ) is: 48 pounds per square inch

Explanation:
It is given that the formula
d = \(\frac{1}{2}\)n + 26
relates the nozzle pressure n (in pounds per square inch) of a fire hose and the maximum horizontal distance the water reaches d (in feet).
So,
The given equation is:
d = \(\frac{1}{2}\)n + 26
Where,
d is the maximum horizontal distance
n is the pressure
It is also given that the maximum horizontal distance is: 50 feet
So,
50 = \(\frac{1}{2}\)n + 26
\(\frac{1}{2}\)n = 50 – 26
\(\frac{1}{2}\)n = 24
\(\frac{1}{2}\) × n = 24
n = 24 × 2
n = 48 pounds per square inch
Hence, from the above
We can conclude that the pressure needed to reach 50 feet away is: 48 pounds per square inch

Question 11.
Monitoring Progress
You have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for your dog to exercise, the pen should be three times as long as it is wide. Find the dimensions of the pen.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 29
Answer:
The dimensions of the pen are:
length of the pen: 12 feet
Width of the pen: 36 feet

Explanation;
It is given that you  have 96 feet of fencing to enclose a rectangular pen for your dog. To provide sufficient running space for your dog to exercise, the pen should be three times as long as it is wide.
So,
The perimeter of the rectangular pen is: 96 feet
We know that,
The perimeter of the rectangle = 2 (Length + Width )
It is also given that the pen is three times as long as it is wide
So,
Width = 3 × Length
So,
The perimeter of the rectangular pen =2 (  Length + ( 3 × Length ) )
96 = 2 ( 4 × Length )
4 × Length = 96 ÷ 2
4 × Length = 48
Length = 48 ÷ 4
Length = 12 feet
So,
Width = 3 × Length
= 3 × 12 = 36 feet
hence, from the above,
We can conclude that
The dimensions of the rectangular pen are:
Length of the pen is: 12 feet
Width of the pen is: 36 feet

Solving Multi-step Equations 1.2 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3−14, solve the equation. Check your solution.

Vocabulary and Core ConceptCheck

Question 1.
COMPLETE THE SENTENCE To solve the equation 2x + 3x = 20, first combine 2x and 3x because they are _________.
Answer:
The given equation is:
2x + 3x = 20
As 2x and 3x are combined by the symbol “+”, add 2x and 3x
So,
2x + 3x = 5x
So,
5x = 20
x = 20 ÷ 4
x = 5

Question 2.
WRITING Describe two ways to solve the equation 2(4x – 11) = 10.
Answer:
The given equation is:
2 (4x – 11) = 10
Way-1:
2 × (4x – 11) = 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4x – 11 = 10 ÷ 2
4x – 11 = 5
4x = 5 + 11
4x = 16
x = 16 ÷ 4
x = 4
Hence,
The value of x is: 4

Way-2:
By using the Distributive Property,
2 (4x – 11) = 2 (4x) – 2 (11)
= 8x – 22
So,
8x – 22 = 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8x = 10 + 22
8x = 32
x = 32 ÷ 8
x = 4
Hence,
The value of x is: 4

Question 3.
3w + 7 = 19

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q3

Question 4.
2g – 13 = 3
Answer:
The value of g is: 8

Explanation:
The given equation is:
2g – 13 = 3
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2g = 3 + 13
2g = 16
2 × g = 16
g = 16 ÷ 2
g = 8
Hence, from the above,
We can conclude that the value of g is: 8

Question 5.
11 = 12 – q

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q5

Question 6.
10 = 7 – m
Answer:
The value of m is: -3

Explanation:
The given equation is:
10 = 7 – m
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-m = 10 – 7
-m = 3
Multiply with “-” on both sides
– (-m ) = -3
m = -3
Hence, from the above,
We can conclude that the value of m is: -3

Question 7.
5 = \(\frac{z}{-4}\) – 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q7

Question 8.
\(\frac{a}{3}\) + 4 = 6
Answer:
The value of a is: 6

Explanation:
The given equation is:
\(\frac{a}{3}\) + 4 = 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{a}{3}\) = 6 – 4
\(\frac{a}{3}\) = 2
a = 2 × 3
a = 6
Hence, from the above,
We can conclude that the value of a is: 6

Question 9.
\(\frac{h + 6}{5}\) = 2

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q9

Question 10.
\(\frac{d – 8}{-2}\) = 12
Answer:
The value of d is: -16

Explanation:
The given equation is:
\(\frac{d – 8}{-2}\) = 12
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
d – 8 = 12 × (-2)
d – 8 = -24
d = -24 + 8
d = -16
Hence, from the above,
We can conclude that the value of d is: -16

Question 11.
8y + 3y = 44

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q11

Question 12.
36 = 13n – 4n
Answer:
The value of n is: 4

Explanation:
The given equation is:
36 = 13n – 4n
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
36 = 9n
9n = 36
n = 36 ÷ 9
n = 4
Hence, from the above,
We can conclude that the value of n is: 4

Question 13.
12v + 10v + 14 = 80

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q13

Question 14.
6c – 8 – 2c = -16
Answer:
The value of c is: -2

Explanation:
The given equation is:
6c – 8 – 2c = -16
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4c – 8 = -16
4c = -16 + 8
4 × c = -8
c = -8 ÷ 4
c = -2
Hence, from the above,
We can conclude that the value of c is: -2

Question 15.
MODELING WITH MATHEMATICS
The altitude a (in feet) of a plane in minutes after liftoff is given by a = 3400t + 600. How many minutes after liftoff is the plane at an altitude of 21,000 feet?
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 30

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q15

Question 16.
MODELING WITH MATHEMATICS
A repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour. Write and solve an equation to find the number of hours of labor spent repairing the car.
Answer:
The number of hours of labor spent repairing the car is: 6 hours

Explanation:
It is given that a repair bill for your car is $553. The parts cost $265. The labor cost is $48 per hour.
Let the number of hours of labor spent repairing the car be: x
So,
The total bill to repair your car = ( The labor cost per hour ) × ( The number of hours of labor spent repairing the car ) +  (The cost of the parts )
553 = 48x + 265
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
48x = 553 – 265
48x = 288
48 × x = 288
x = 288 ÷ 48
x = 6
Hence, from the above,
We can conclude that the number of hours of labor spent repairing the car is: 6 hours

In Exercises 17−24, solve the equation. Check your solution.

Question 17.
4(z + 5) = 32

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q17

Question 18.
-2(4g – 3) = 3018.
Answer:
The value of g is: 378

Explanation:
The given equation is:
-2 (4g – 3) = 3018
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2 × ( 4g – 3 ) = 3018
4g – 3 = 3018 ÷ 2
4g – 3 = 1,509
4g = 1,509 +3
4 × g = 1,512
g = 1,512 ÷ 4
g = 378
Hence, from the above,
We can conclude that the value of g is: 378

Question 19.
6 + 5(m + 1) = 26

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q19

Question 20.
5h+ 2(11 – h) = -5
Answer:
The value of h is: -9

Explanation:
The given equation is:
5h + 2 ( 11-h ) = -5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
By using the Distributive Property of Multiplication,
2 ( 11 – h ) = 2 (11 ) – 2 ( h )
= 22 – 2h
So,
5h + 22 – 2h = -5
3h + 22 = -5
3h = -5 – (+22)
3h = -5 -22
3h = -27
h = -27 ÷ 3
h = -9
Hence, from the above,
We can conclude that the value of h is: -9

Question 21.
27 = 3c – 3(6 – 2c)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q21

Question 22.
-3 = 12y – 5(2y – 7)
Answer:
The value of y is: -19

Explanation:
The given equation is:
-3 = 12y – 5 (2y – 7)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
12y – 5 (2y – 7) = -3
By using the Distributive Property of Multiplication,
5 ( 2y – 7 ) = 5 (2y ) – 5 (7 )
= 10y – 35
So,
12y – ( 10y – 35 ) = -3
12y – 10y + 35 = -3
2y + 35 = -3
2y = -3 – (+35 )
2y = -3 – 35
2y = -38
y = -38 ÷ 2
y = -19
Hence, from the above,
We can conclude that the value of y is: -19

Question 23.
-3(3 + x) + 4(x – 6) = -4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q23

Question 24.
5(r + 9) – 2(1 – r) = 1
Answer:
The value of r is: -6

Explanation:
The given equation is:
5 ( r + 9 ) – 2 ( 1 – r ) = 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
5 ( r + 9 ) = 5 ( r ) + 5 ( 9 )
= 5r + 45
2 ( 1 – r ) = 2 ( 1 ) – 2 ( r )
= 2 – 2r
So,
5r + 45 – ( 2 – 2r ) = 1
5r + 45 – 2 + 2r = 1
7r + 43 = 1
7r = 1 – 43
7r = -42
r = -42 ÷ 7
r = -6
Hence, from the above,
We can conclude that the value of r is: -6

USING TOOLS
In Exercises 25−28, find the value of the variable. Then find the angle measures of the polygon. Use a protractor to check the reasonableness of your answer.

Question 25.
Big Ideas Math Algebra 1 Answers Chapter 1 Solving Linear Equations 31

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q25

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 32
Answer:
The angle measures of the rhombus are:
60°, 60°, 120°, 120°

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 32
From the above figure,
The angle measures of the rhombus are: a°, 2a°, a°, 2a°
It is also given that the sum of all the angle measures is: 360°
So,
a° + 2a° + a° + 2a° = 360°
6a° = 360°
a = 360° ÷ 6
a = 60°
Hence, from the above,
We can conclude that the angle measures of the rhombus are:
a°, 2a°, a°, 2a° = 60°, 2 ( 60° ), 60°, 2 ( 60° )
= 60°, 60°, 120°, 120°

Question 27.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 33

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q27

Question 28.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 34
Answer:
The angle measures of the hexagon are:
120°, 120°, 100°, 120°, 250°, 260°

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 34
From the given figure,
The angle measures of the hexagon are:
120°, 120°, 100°, 120°, x°, (x + 10)°
It is also given that the sum of the angle measures of the hexagon is: 720°
So,
120° + 120° + 100° + 120° + x° + (x + 10)° = 720°
470° + x = 720°
x = 720° – 470°
x = 250°
Hence, from the above,
We can conclude that the angle measures of the hexagon are:
120°, 120°, 100°, 120°, x°, (x + 10)° = 120°, 120°, 100°, 120°, 250°, (250 + 10)°
= 120°, 120°, 100°, 120°, 250°, 260°

In Exercises 29−34, write and solve an equation to find the number.

Question 29.
The sum of twice a number and 13 is 75.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q29

Question 30.
The difference of three times a number and 4 is -19.
Answer:
The number is: -5

Explanation:
It is given that the difference of three times of a number and 4 is -19
Now,
Let the number be x
So,
The three times of a number = 3 (x) = 3x
So,
3x – 4 = -19
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3x = -19 + 4
3x = -15
x = -15 ÷ 3
x = -5
Hence, from the above,
We can conclude that the number is: -5

Question 31.
Eight plus the quotient of a number and 3 is -2.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q31

Question 32.
The sum of twice a number and half the number is 10.
Answer:
The number is: 4

Explanation:
It is given that the sum of twice of a number and half the number is 10.
Let the number be x.
So,
The twice of a number = 2 (x ) = 2x
Half of the number = x ÷ 2 = \(\frac{x}{2}\)
So,
2x + \(\frac{x}{2}\) = 10
2x can be rewritten as: \(\frac{4x}{2}\)
So,
\(\frac{4x}{2}\) + \(\frac{x}{2}\) = 10
\(\frac{4x + x}{2}\) = 10
\(\frac{5x}{2}\) = 10
5x = 10 × 2
5x = 20
x = 20 ÷ 5
x = 4
Hence, from the above,
We can conclude that the numebr is: 4

Question 33.
Six times the sum of a number and 15 is -42.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q33

Question 34.
Four times the difference of a number and 7 is 12.
Answer:
The number is: 4

Explanation:
It is given that the four times the difference of a number and 7 is 12
Let the number be x
So,
Four times of the number = 4 ( x ) = 4x
So,
4x – x = 12
3x = 12
x = 12 ÷ 3
x = 4
Hence, from the above,
We can conclude that the number is: 4

USING EQUATIONS
In Exercises 35−37, write and solve an equation to answer the question. Check that the units on each side of the equation balance.

Question 35.
During the summer, you work 30 hours per week at a gas station and earn $8.75 per hour. You also work as a landscaper for $11 per hour and can work as many hours as you want. You want to earn a total of $400 per week. How many hours must you work as a landscaper?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q35

Question 36.
The area of the surface of the swimming pool is 210 square feet. What is the length d of the deep end (in feet)?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 35

Answer:
The length d of the deep end is: 12 feet

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 35
It is given that the area of the surface of the swimming pool is 210 square feet
From the above figure,
We can observe that the shape of the swimming pool is a rectangle.
So,
Length of the swimming pool = 10 ft
Width of the swimming pool = d + 9 ft
So,
The area of the swimming pool = Length × Width
= 10 × ( d + 9 )
Now,
210 = 10 × ( d + 9 )
d + 9 = 210 ÷ 10
d + 9 = 21
d = 21 – 9
d = 12 feet
Hence, from the above,
We can conclude that the length d of the deep end is: 12 feet

Question 37.
You order two tacos and a salad. The salad costs $2.50. You pay 8% sales tax and leave a $3 tip. You pay a total of $13.80. How much does one taco cost?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q37

JUSTIFYING STEPS
In Exercises 38 and 39, justify each step of the solution.

Question 38.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 36

Answer:
–\(\frac{1}{2}\) ( 5x – 8 ) – 1 = 6                          Write the equation
–\(\frac{1}{2}\) ( 5x – 8 ) = 6 + 1                         Arrange the similar terms
–\(\frac{1}{2}\) ( 5x – 8 ) = 7                                  Simplify
– ( 5x – 8 ) = 7 × 2                                                              Divide by 2 on both sides
– ( 5x – 8 ) = 14                                                                    Simplify
5x – 8 = -14                                                                       Multiply with “-” on both sides
5x = -14 + 8                                                                         Arrange the similar terms
5x = -6                                                                               Divide by 6 on both sides
x = –\(\frac{6}{5}\)                                              The result
Hence,
The solution is: x = –\(\frac{6}{5}\)

Question 39.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 37

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q39

ERROR ANALYSIS
In Exercises 40 and 41, describe and correct the error in solving the equation.

Question 40.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 38

Answer:
The given equation is:
-2 ( 7 – y ) + 4 = -4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2 ( 7 – y ) = -4 –  (+4 )
-2 ( 7 – y ) = -4 – 4
-2 ( 7 – y ) = -8
Now,
By using the Distributive Property of Multiplication,
2 ( 7 – y ) = 2 ( 7 ) – 2 ( y )
= 14 – 2y
So,
– ( 14 – 2y ) = -8
2y – 14 = -8
2y = -8 + 14
2y = 6
y = 6 ÷ 2
y = 3
Hence,
The value of y is: 3

Question 41.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 39

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q41

MATHEMATICAL CONNECTIONS
In Exercises 42−44, write and solve an equation to answer the question.

Question 42.
The perimeter of the tennis court is 228 feet. What are the dimensions of the court?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 40

Answer:
The dimensions of the court are:
The Length of the court is: 36 feet
The width of the court is: 78 feet

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 40
It is given that the perimeter of the tennis court is: 228 feet
From the above figure,
We can observe that the shape of the tennis court is the rectangle
So,
The length of the tennis court = w feet
The width of the tennis court =  (2w + 6 ) feet
We know that,
The perimeter of the rectangle = 2 ( Length + Width )
So,
The perimeter of the tennis court = 2 ( Length + Width )
228 = 2 ( w + 2w + 6 )
By using the Distributive Property of Multiplication,
2 ( w + 2w + 6 ) = 2 ( 3w + 6 )
= 2 ( 3w ) + 2 ( 6 )
= 6w + 12
So,
228 = 6w + 12
6w = 228 – 12
6w = 216
w = 216 ÷ 6
w = 36
Hence, from the above,
We can conclude that
The length of the tennis court is: 36 feet
The width of the tennis court is: 2w + 6  = 2 ( 36 ) + 6 = 78 feet

Question 43.
The perimeter of the Norwegian flag is 190 inches. What are the dimensions of the flag?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 41

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q43

Question 44.
The perimeter of the school crossing sign is 102 inches. What is the length of each side?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 42

Answer:
The length of each side is: 15 inches

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 42
It is given that the perimeter of the crossing sign is 102 inches
We know that,
The perimeter of any polygon is the sum of all the sides of that polygon
So,
The perimeter of the crossing sign = s + ( s + 6 ) + ( s + 6 ) + s + 2s
102 = 6s + 12
102 – 12 = 6s
6s = 90
s = 90 ÷ 6
s = 15 inches
Hence, from the above,
We can conclude that the length of each side is: 15 inches

Question 45.
COMPARING METHODS
Solve the equation 2(4 – 8x) + 6 = -1 using (a) Method 1 from Example 3 and (b) Method 2 from Example 3. Which method do you prefer? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q45
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q45-i

Question 46.
PROBLEM – SOLVING
An online ticket agency charges the amounts shown for basketball tickets. The total cost for an order is $220.70. How many tickets are purchased?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 43

Answer:

Question 47.
MAKING AN ARGUMENT
You have quarters and dimes that total $2.80. Your friend says it is possible that the number of quarters is 8 more than the number of dimes. Is your friend correct? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q47

Question 48.
THOUGHT-PROVOKING
You teach a math class and assign a weight to each component of the class. You determine final grades by totaling the products of the weights and the component scores. Choose values for the remaining weights and find the necessary score on the final exam for a student to earn an A (90%) in the class, if possible. Explain your reasoning.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 44

Answer:
The completed table is:

From the above table,
The weights can be calculated by the difference between the total participation and the class participation and divide the total value by 100.
So,
The weight of homework = [ ( 100 – 95 ) ÷ 100]
= 5 ÷ 100
= 0.50
The weight of midterm exam = ( 100 – 88 ) ÷ 100
= 12 ÷ 100
= 0.12
So,
The necessary score of the final exam = ( 92 + 95 + 88 ) % ÷ 3
= 275 % ÷ 3
= 91.6 %

Question 49.
REASONING
An even integer can be represented by the expression 2n, where n is an integer. Find three consecutive even integers that have a sum of 54. Explain your reasoning.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q49

Question 50.
HOW DO YOU SEE IT?
The scatter plot shows the attendance for each meeting of a gaming club.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 45
a. The mean attendance for the first four meetings is 20. Is the number of students who attended the fourth meeting greater than or less than 20? Explain.
Answer:
The number of students who attended the fourth meeting is greater than 20

Explanation:
The given graph is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 45
We know that
The mean = ( Sum of the given numbers ) ÷ (The total number of the numbers)
From the above graph,
The attendance of the 1st meeting = 18
The attendance of the 2nd meeting = 21
The attendance of the 3rd meeting = 17
Let the attendance of the 4th meeting be: x
So,
The mean attendance  of the first four meetings = ( The attendance of the 4 meetings ) ÷  ( The total number of meetings )
= ( 18 + 21 + 17 + x ) ÷ 4
It is given that the mean attendance of the first four meetings is: 20
So,
20 =  ( 18 + 21 + 17 + x ) ÷ 4
( 56 + x ) ÷ 4 = 20
56 + x = 20 × 4
56 + x = 80
x = 80 – 56
x = 24
Hence, from the above,
We can conclude that the attendance of the 4th meeting is greater than 20

b. Estimate the number of students who attended the fourth meeting.
Answer:
The number of students who attended the fourth meeting is: 24

Explanation:
The mean attendance  of the first four meetings = ( The attendance of the 4 meetings ) ÷  ( The total number of meetings )
= ( 18 + 21 + 17 + x ) ÷ 4
It is given that the mean attendance of the first four meetings is: 20
So,
20 =  ( 18 + 21 + 17 + x ) ÷ 4
( 56 + x ) ÷ 4 = 20
56 + x = 20 × 4
56 + x = 80
x = 80 – 56
x = 24
Hence, from the above,
We can conclude that the number of students who attended the 4th meeting is: 24

c. Describe a way you can check your estimate in part (b).
Answer:
The estimate in part (b) can be checked by using the property of the mean
So,
The mean attendance of the four meetings = ( The attendance of the four meetings ) ÷ ( The total number of meetings )
= ( 18 + 21 + 17 + 24 ) ÷ 4
= 80 ÷ 4
= 20
Hence, from the above,
We can conclude that the mean attendance of the four meetings is the same as given above.

REASONING
In Exercises 51−56, the letters a, b, and c represent nonzero constants. Solve the equation for x.

Question 51.
bx = -7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q51

Question 52.
x + a = \(\frac{3}{4}\)
Answer:
The value of x is: \(\frac{3}{4}\) – a

Explanation:
The given equation is:
x + a = \(\frac{3}{4}\)
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x =  \(\frac{3}{4}\) – a
Hence, from the above,
We can conclude that the value of a is: \(\frac{3}{4}\) – a

Question 53.
ax – b = 12.5

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q53

Question 54.
ax + b = c
Answer:
The value of x is: \(\frac{c – b}{a}\)

Explanation:
The given equation is:
ax + b = c
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
ax = c – b
x = \(\frac{c – b}{a}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{c – b}{a}\)

Question 55.
2bx – bx = -8

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q55

Question 56.
cx – 4b = 5b
Answer
The value of x is: \(\frac{9b}{c}\)

Explanation:
The given equation is:
cx – 4b = 5b
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
cx = 5b + 4b
cx = 9b
x = \(\frac{9b}{c}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{9b}{c}\)

Maintaining Mathematical Proficiency

Simplify the expression.

Question 57.
4m + 5 – 3m

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q57

Question 58.
9 – 8b + 6b
Answer:
9 – 8b + 6b
= 9 – (8b – 6b )
= 9 – 2b

Question 59.
6t + 3(1 – 2t) – 5

Answer:

Determine whether (a) x = −1 or (b) x = 2 is a solution of the equation.

Question 60.
x – 8 = -9
Answer:
x = -1 is a solution to the given equation

Explanation:
The given equation is:
x – 8 = -9
a) Let x = -1
So,
-1 – 8 = -9
-9 = -9
As LHS is equal to RHS
x = -1 is a solution of the given equation
b) Let x = 2
So,
2 – 8 = -9
-6 = -9
As LHS is not equal to RHS,
x = 2 is not a solution of the given equation

Question 61.
x + 1.5 = 3.5

Answer:

Question 62.
2x – 1 = 3
Answer:
x = 2 is a solution to the given equation

Explanation:
The given equation is:
2x – 1 = 3
a) Let x = -1
So,
2 ( -1 ) – 1 = 3
-2 – 1 = 3
-3 = 3
As LHS is not equal to RHS
x = -1 is not a solution to the given equation
b) Let x = 2
So,
2 ( 2 ) -1 = 3
4 – 1 = 3
3 = 3
As LHS is equal to RHS,
x = 2  is a solution to the given equation

Question 63.
3x + 4 = 1

Answer:

Question 64.
x + 4 = 3x
Answer:
x = 2 is a solution to the given equation

Explanation:
The given equation is:
x + 4 = 3x
a) Let x = -1
So,
-1 + 4 = 3 ( -1 )
= 3 = -3
As LHS is not equal to RHS,
x = -1 is not a solution to the given equation
b) Let x = 2
So,
2 + 4 = 3 ( 2 )
6 = 6
As LHS is equal to RHS,
x = 2 is a solution to the given equation.

Question 65.
-2(x – 1) = 1 – 3x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.2-Q65

Lesson 1.3 Solving Equations with Variables on Both Sides

EXPLORATION 1
Perimeter

Work with a partner. The two polygons have the same perimeter. Use this information to write and solve an equation involving x. Explain the process you used to find the solution. Then find the perimeter of each polygon.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 46
Answer:
The perimeter of the hexagon is: 6
The perimeter of the square is: 6

Explanation:
The given figures are:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 46
Based on the number of sides (n) in the polygon,
We can say the name of that polygon
So,
In the first figure,
The number of sides is: 6
So,
It is Hexagon
In the second figure,
The number of sides is: 4
So,
It is square ( As all the angles are 90° )
It is given that the two polygons have the same perimeter
We know that,
“Perimeter” of a polygon is defined as the sum of all the sides in the polygon
So,
The sum of all sides in the hexagon = 5 + 2 + 5 + 2 + x + x
= 14 + 2x
The sum of all sides in the square = \(\frac{3x}{2}\) + 3 + 4 + 5
= \(\frac{3x}{2}\) + 12
It is given that the perimeter of both the polygons are equal
So,
14 + 2x = \(\frac{3x}{2}\) + 12
14 – 12 = \(\frac{3x}{2}\) – 2x
\(\frac{3x}{2}\) – 2x = 2
We can write 2x as \(\frac{4x}{2}\)
So,
\(\frac{3x}{2}\) – \(\frac{4x}{2}\) = 2
\(\frac{3x – 4x}{2}\) = 2
\(\frac{-x}{2}\) = 2
– \(\frac{x}{2}\) = 2
-x = 2 × 2
-x = 4
x = -4
Hence,
The perimeter of the Hexagon = 14 + 2x = 14 + 2 ( -4 )
= 14 – 8 = 6
The perimeter of the square = \(\frac{3x}{2}\) + 12
\(\frac{3 × -4}{2}\) + 12
= \(\frac{-12}{2}\) + 12
= \(\frac{-12}{2}\) + \(\frac{24}{2}\)
= \(\frac{24 – 12}{2}\)
= \(\frac{12}{2}\)
= 6
Hence, from the above,
We can conclude that
The perimeter of the Hexagon is: 6
The perimeter of the square is: 6

EXPLORATION 2
Perimeter and Area

Work with a partner.

  • Each figure has the unusual property that the value of its perimeter (in feet) is equal to the value of its area (in square feet). Use this information to write an equation for each figure.
  • Solve each equation for x. Explain the process you used to find the solution.
  • Find the perimeter and area of each figure.

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 47

Question 3.
How can you solve an equation that has variables on both sides?
Answer:
If the variable is the same on both sides in an equation, then rearrange the like terms
So,
Separate the variables and the numbers and simplify the variables and the numbers
In this way,
We can solve an equation with a single variable

Question 4.
Write three equations that have the variable x on both sides. The equations should be different from those you wrote in Explorations 1 and 2. Have your partner solve the equations.
Answer:
Let the three equations that have variable x on both sides and different from Explorations 1 and 2 are:
a) 6x + 2 = 5x-6
b) 16x = 18x – 2
c) 12x = 15x + 63
Now,
a)
The given equation is:
6x + 2 = 5x – 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6x – 5x = -6 – 2
x = -8
Hence,
The value of x is: -8
b) The given equation is:
9x = 18x – 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
18x – 16x = 2
2x = 2
x = 2 ÷ 2
x = 1
Hence,
The value of x is: 1
c) The given equation is:
12x = 15x + 63
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
12x -15x = 63
-3x = 63
x = 63 ÷ ( -3 )
x = -63 ÷ 3
x = -21
Hence,
The value of x is: -21

1.3 Lesson

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
-2x = 3x + 10
Answer:
The value of x is: -2

Explanation:
The given equation is:
-2x = 3x + 10
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-2x – 3x = 10
-5x = 10
x = 10 ÷ (-5)
x = -10 ÷ 5
x = -2
Hence, from the above,
We can conclude that the value of x is: -2

Question 2.
\(\frac{1}{2}\)(6h – 4) = -5h + 1
Answer:
The value of h is: \(\frac{3}{8}\)

Explanation:
The given equation is:
\(\frac{1}{2}\) ( 6h – 4 ) = -5h + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6h – 4 = 2 ( -5h + 1 )
6h – 4 = 2 ( -5h ) + 2 ( 1 ) [ By using the Distributive Property of Multiplication )
6h – 4 = -10h + 2
6h + 10h = 2 + 4
16h = 6
h = \(\frac{6}{16}\)
h= \(\frac{3}{8}\)
Hence, from the above,
We can conclude that the value of h is: \(\frac{3}{8}\)

Question 3.
–\(\frac{3}{4}\)(8n + 12) = 3(n – 3)
Answer:
The value of n is: 0

Explanation:
The given equation is:
–\(\frac{3}{4}\) ( 8n + 12 ) = 3 ( n – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8n + 12 = –\(\frac{4}{3}\) × 3 ( n – 3 )
8n + 12 = –\(\frac{4}{3}\) \(\frac{3}{1}\) ( n – 3 )
8n + 12 = –\(\frac{3 × 4}{3 × 1}\) ( n – 3 )
8n + 12 = -4 ( n – 3 )
8n + 12 = -4n – 4 ( -3 )
8n + 12 = -4n + 12
8n + 4n =12 – 12
12n = 0
n = 0
Hence, from the above,
We can conclude that the value of n is: 0

Monitoring Progress

Solve the equation.

Question 4.
4(1 – p) = 4p – 4
Answer:
The value of p is: 1

Explanation:
The given equation is:
4 ( 1 -p ) = 4p – 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4 ( 1 ) – 4 ( p ) = 4p – 4
4 – 4p = 4p – 4
4p + 4p = 4 + 4
8p = 8
p = 8 ÷ 8
p = 1
Hence, from the above,
We can conclude that the value of p is: 1

Question 5.
6m – m = –\(\frac{5}{6}\)(6m – 10)
Answer:
The value of m is: \(\frac{5}{6}\)

Explanation:
The given equation is:
6m – m = –\(\frac{5}{6}\) ( 6m – 10 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5m = –\(\frac{5}{6}\) ( 6m – 10 )
5m = –\(\frac{5}{6}\) ( 6m ) – ( –\(\frac{5}{6}\) ( 10 ) )
5m = –\(\frac{5}{6}\) × \(\frac{6m}{1}\) + \(\frac{5}{6}\) \(\frac{10}{1}\)
5m = –\(\frac{5 × 6m}{6 × 1}\) + \(\frac{5 × 10}{6 × 1}\)
5m = -5m + \(\frac{25}{3}\)
5m + 5m = \(\frac{25}{3}\)
10m = \(\frac{25}{3}\)
m = \(\frac{25}{3}\) ÷ \(\frac{10}{1}\)
m = \(\frac{25}{3}\) × \(\frac{1}{10}\)
m = \(\frac{25}{30}\)
m = \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the value of m is: \(\frac{5}{6}\)

Question 6.
10k + 7 = -3 – 10k
Answer:
The value of k is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
10k + 7 = -3 – 10k
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
10k + 10k = -3 – ( +7 )
20k = -3 –
20k = -10
k = -10 ÷ 20
k = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of k is: –\(\frac{1}{2}\)

Question 7.
3(2a – 2) = -2(3a – 3)
Answer:
The value of a is: 1

Explanation:
The given equation is:
3 ( 2a – 2 ) = -2 ( 3a – 3 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
3 ( 2a ) – 3 ( 2 ) = -2 ( 3a ) + 2 ( 3 )
6a – 6 = -6a + 6
6a + 6a = 6 + 6
12a = 12
a = 12 ÷ 12
a = 1
Hence, from the above,
We can conclude that the value of a is: 1

Concept Summary

Steps for Solving Linear Equations
Here are several steps you can use to solve a linear equation. Depending on the equation, you may not need to use some steps.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 48
Step 1
Use the Distributive Property to remove any grouping symbols.
Step 2
Simplify the expression on each side of the equation.
Step 3
Collect the variable terms on one side of the equation and the constant terms on the other side.
Step 4
Isolate the variable.
Step 5
Check your solution.

Monitoring Progress

Question 8.
A boat travels upstream on the Mississippi River for 3.5 hours. The return trip only takes 2.5 hours because the boat travels 2 miles per hour faster downstream due to the current. How far does the boat travel upstream? Answer:
The distance the boat travel upstream is: 17.5 miles

Explanation:
It is given that a boat travels upstream on the Mississippi River for 3.5 hours. The return trip only takes 2.5 hours because the boat travels 2 miles per hour faster downstream due to the current.
Now,
Let x be the speed of the boat traveled upstream
We know that,
Speed = Distance ÷ Time
Distance = Speed × Time
It is given that the time taken by the boat traveled upstream is: 3.5 hours
So,
Distance traveled upstream = 3.5 × x = 3.5x
Now,
It is also given that the speed of the boat is 2 miles per hour faster downstream
So,
Distance traveled downstream by boat = 2.5 ( x + 2 )
SO,
As both the distances are the same,
3.5x = 2.5 ( x + 2 )
By using the Distributive Property of Multiplication,
3.5x = 2.5 ( x) + 2.5 ( 2 )
3.5x = 2.5x + 5
3.5x – 2.5x = 5
x = 5
So,
The distance traveled upstream by boat = 3.5x = 3.5 ( 5 )
= 17.5 miles per hour
Hence, from the above,
We can conclude that the distance traveled upstream by boat is: 17.5 miles per hour

Solving Equations with Variables on Both Sides 1.3 Exercises

Monitoring Progress and Modeling with Mathematics

In Exercises 3–16, solve the equation. Check your solution.

Question 1.
VOCABULARY Is the equation -2(4 – x) = 2x + 8 an identity? Explain your reasoning.
Answer:
-2 ( 4 – x ) = 2x + 8 is not an identity

Explanation:
The given equation is:
-2 ( 4 – x ) = 2x + 8
By using the Distributive Property of Multiplication,
-2 ( 4 ) + 2 ( x ) = 2x + 8
-8 + 2x = 2x + 8
2x – 8 = 2x + 8
As
LHS ≠ RHS
-2 ( 4 – x ) = 2x + 8  is not an identity

Question 2.
WRITING Describe the steps in solving the linear equation 3(3x – 8) = 4x + 6
Answer:
The value of x is: 6

Explanation:
The given equation is:
3 ( 3x – 8 ) = 4x + 6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
Now,
By using the Distributive Property of Multiplication,
3 ( 3x ) – 3 ( 8 ) = 4x + 6
9x – 24 = 4x + 6
9x – 4x = 6 + 24
5x = 30
x = 30 ÷ 5
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 3.
15 – 2x = 3x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q3

Question 4.
26 – 4s = 9s
Answer:
The value of s is: 2

Explanation:
The given equation is:
26 – 4s = 9s
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
26 = 9s + 4s
13s = 26
s = 26 ÷ 13
s = 2
Hence, from the above,
We can conclude that the value of s is: 2

Question 5.
5p – 9 = 2p + 12

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q5

Question 6.
8g + 10 = 35 + 3g
Answer:
The value of g is: 5

Explanation:
The given equation is:
8g + 10 = 35 + 3g
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
8g – 3g = 35 – 10
5g = 25
g = 25 ÷ 5
g = 5
Hence, from the above,
We can conclude that the value of g is: 5

Question 7.
5t + 16 = 6 – 5t

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q7

Question 8.
-3r + 10 = 15r – 8
Answer:
The value of r is: 1

Explanation:
The given equation is:
-3r + 10 = 15r – 8
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-3r – 15r = -10 – 8
-18r = -18
r = -18 ÷ ( -18 )
r = 1 [ since – ÷ – = + ]
Hence, from the above,
We can conclude that the value of r is: 1

Question 9.
7 + 3x – 12x = 3x + 1

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q9

Question 10.
w – 2 + 2w = 6 + 5w
Answer:
The value of w is: -4

Explanation:
The given equation is:
w – 2 + 2w = 6 + 5w
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
w + 2w -5w =6 + 2
-2w = 8
w = -8 ÷ 2
w = -4
Hence, from the above,
We can conclude that the value of w is: -4

Question 11.
10(g + 5) = 2(g + 9)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q11

Question 12.
-9(t – 2) = 4(t – 15)
Answer:
The value of t is: 6

Explanation:
The given equation is:
-9 ( t – 2 ) = 4 ( t – 15 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
Now,
By using the Distributive Property of Multiplication,
-9 ( t ) +  9 ( 2 ) = 4 ( t ) – 4 ( 15 )
-9t + 18 = 4t – 60
-9t – 4t = -60 – 18
-13t = -78
t = -78 ÷ ( -13 )
t = 6 [ Since  -÷ – = + ]
Hence, from the above,
We can conclude that the value of t is: 6

Question 13.
\(\frac{2}{3}\)(3x + 9) = -2(2x + 6)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q13

Question 14.
2(2t + 4) = \(\frac{3}{4}\)(24 – 8t)
Answer:
The value of t is: 1

Explanation:
The given equation is:
2 ( 2t + 4 ) = \(\frac{3}{4}\) ( 24 – 8t )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
2 ( 2t ) + 2 ( 4 ) = \(\frac{3}{4}\) ( 24 ) – 8t  (\(\frac{3}{4}\) )
4t + 8 = \(\frac{3}{4}\) × \(\frac{24}{1}\) – \(\frac{8t}{1}\) × \(\frac{3}{4}\)
4t + 8 = \(\frac{3 × 24}{4 × 1}\) – \(\frac{3 × 8t}{4 × 1}\)
4t + 8 = \(\frac{18}{1}\) – \(\frac{6t}{1}\)
4t + 8 = 18 – 6t
4t + 6t = 18 – 8
10t = 10
t = 10 ÷ 10
t = 1
Hence, from the above,
We can conclude that the value of t is: 1

Question 15.
10(2y + 2) – y = 2(8y – 8)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q15

Question 16.
2(4x + 2) = 4x – 12(x – 1)
Answer:
The value of x is: \(\frac{1}{2}\)

Explanation:
The given equation is:
2 ( 4x + 2 ) = 4x – 12 ( x – 1 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
2 ( 4x ) + 2 ( 2 ) = 4x – 12 ( x ) + 12 ( 1 ) [ Since – × – = + ]
8x + 4 = 4x – 12x + 12
8x + 4 =12 – 8x
8x + 8x = 12 – 4
16x = 8
x = 8 ÷ 16
x = \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{1}{2}\)

Question 17.
MODELING WITH MATHEMATICS
You and your friend drive toward each other. The equation 50h = 190 – 45h represents the number h of hours until you and your friend meet. When will you meet?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q17

Question 18.
MODELING WITH MATHEMATICS
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store. How many movies must you rent to spend the same amount at each movie store?
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 49

Answer:
The number of movies you rent to spend the same amount at each movie store is: 20

Explanation:
It is given that
The equation 1.5r + 15 = 2.25r represents the number r of movies you must rent to spend the same amount at each movie store.
Now,
We have to find the value of r to find the number f movies you must rent to spend the same amount at each movie store
So,
The given equation is:
1.5r + 15 = 2.25r
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2.25r – 1.5r = 15
0.75r = 15
\(\frac{75}{100}\)r = 15
r = 15 × \(\frac{100}{75}\)
r = \(\frac{15}{1}\) × \(\frac{100}{75}\)
r = \(\frac{15 × 100}{1 × 75}\)
r = 20
Hence, from the above,
We can conclude that the number of movies you rent to spend the same amount at each movie store is: 20

In Exercises 19–24, solve the equation. Determine whether the equation has one solution, no solution, or infinitely many solutions.

Question 19.
3t + 4 = 12 + 3t

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q19

Question 20.
6d + 8 = 14 + 3d
Answer:
The value of d is: 2

Explanation:
The given equation is:
6d + 8 = 14 + 3d
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6d – 3d = 14 – 8
3d = 6
d = 6 ÷ 3
d = 2
Hence, from the above,
We can conclude that the value of d is: 2

Question 21.
2(h + 1) = 5h – 7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q21

Question 22.
12y + 6 = -6(2y + 1)
Answer:
The value of y is: –\(\frac{1}{2}\)

Explanation:
The given equation is:
12y + 6 = -6 ( 2y + 1 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
12y + 6 = -6 ( 2y ) – 6 ( 1 )
12y + 6 = -12y – 6
12y + 12y = -6 – ( +6 )
24y = -6 – 6
24y = -12
y = -12 ÷ 24
y = –\(\frac{1}{2}\)
Hence, from the above,
We can conclude that the value of y is: –\(\frac{1}{2}\)

Question 23.
3(4g + 6) = 2(6g + 9)

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q23

Question 24.
5(1 + 2m) = \(\frac{1}{2}\)(8 + 20m)
Answer:
m has indefinite solutions

Explanation:
The given equation is:
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
5 ( 1 ) + 5 ( 2m ) = \(\frac{1}{2}\) ( 8 ) + \(\frac{1}{2}\) ( 20m )
2 ( 5 + 10m ) = 8 + 20m
2 ( 5 ) + 2 ( 10m ) = 8 + 20m
10 + 20m = 8 + 20m
20m – 20m = 8 – 10
20m – 20m = -2
As  m has the same coefficients and have the opposite signs, m has indefinite solutions
Hence, from the above,
We can conclude that the equation has the indefinite solutions

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in solving the equation.

Question 25.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 50

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q25

Question 26.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 51

Answer:
The given equation is:
6 ( 2y + 6 ) = 4 ( 9 + 3y )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
Now,
By using the Distributive Property of Multiplication,
6 ( 2y ) + 6 ( 6 ) = 4 ( 9 ) + 4 ( 3y )
12y + 36 = 36 + 12y
12y – 12y = 36 – 36
0 = 0
As the coefficients of y are zero, the equation has no solution
Hence, from the above,
We can conclude that there is no error in the analysis of the equation.

Question 27.
MODELING WITH MATHEMATICS
Write and solve an equation to find the month when you would pay the same total amount for each Internet service.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 52

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q27

Question 28.
PROBLEM-SOLVING
One serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams of protein from other sources. How many grams of protein do you need daily?
Answer:
The number of grams of protein you need daily is: 50 grams

Explanation:
It is given that one serving of granola provides 4% of the protein you need daily. You must get the remaining 48 grams of protein from other sources.
So,
Let the number of grams of protein you need daily be: x
So,
The number of grams of protein you need daily = 4 % of x + 48
We know that,
100%  = 1
So,
4 % = 0.04
So,
The number of grams of protein you need daily = 0.04x + 48
x = 0.04x + 48
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x – 0.04x = 48
0.96x = 48
\(\frac{96}{100}\)x = 48
x = 48 × \(\frac{100}{96}\)
x = \(\frac{48}{1}\) × \(\frac{100}{96}\)
x = \(\frac{48 × 100}{1 × 96}\)
x = \(\frac{50}{1}\)
x = 50 grams
Hence, from the above,
We can conclude that the number of proteins you need daily is: 50 grams

USING STRUCTURE
In Exercises 29 and 30, find the value of r.

Question 29.
8(x + 6) – 10 + r = 3(x + 12) + 5x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q29

Question 30.
4(x – 3) – r + 2x = 5(3x – 7) – 9x
Answer:
The value of r is: 23

Explanation:
The given equation is:
4 ( x – 3 ) – r + 2x = 5 ( 3x – 7 ) – 9x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
By using the Distributive Property of Multiplication,
4x – 4 ( 3 ) – r + 2x = 5 ( 3x ) – 5 ( 7 ) – 9x
4x – 12 – r + 2x = 15x – 35 – 9x
6x – 12 – r = 6x – 35
r = 6x – 6x – 12 + 35
r = 23
Hence, from the above,
We can conclude that the value of r is: 23

MATHEMATICAL CONNECTIONS
In Exercises 31 and 32, the value of the surface area of the cylinder is equal to the value of the volume of the cylinder. Find the value of x. Then find the surface area and volume of the cylinder.

Question 31.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 53

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q31

Question 32.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 54
Answer:
The Surface Area of the cylinder is: 461.49 cm²
The volume of the cylinder is: 488.58 cm³

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 54
From the above figure,
The radius of the cylinder is: 7\(\frac{1}{5}\) feet
The height of the cylinder is: x feet
It is given that the Total Surface Area of the cylinder and the volume of the cylinder are equal.
We know that,
The Surface Area of the cylinder = 2πr² + 2πrh
The volume of the cylinder = πr²h
The value of π is: 3.1416
Now,
The representation of 7\(\frac{1}{5}\) in the improper fraction form is: \(\frac{36}{5}\)
So,
2πr² + 2πrh = πr²h
[ 2 × 3.1416 × \(\frac{36}{5}\) × \(\frac{36}{5}\) ] + [ 2 × 3.1416 × \(\frac{36}{5}\) × x ] = [3.1416 × latex]\frac{36}{5}[/latex] × \(\frac{36}{5}\) × x ]
325.72 + 45.23x = 162.86x
162.86x – 45.23x = 325.72
117.63x = 325.72
x = 2.76
x = 3
So,
The Surface Area of the cylinder = 2πr² + 2πrh
= [ 2 × 3.1416 × \(\frac{36}{5}\) × \(\frac{36}{5}\) ] + [ 2 × 3.1416 × \(\frac{36}{5}\) × x ]
= 325.72 + 45.23x
= 325.72 + 45.23 ( 3 )
= 461.49 cm²
The volume of the cylinder = πr²h
= [3.1416 × latex]\frac{36}{5}[/latex] × \(\frac{36}{5}\) × x ]
= 162.86x
= 162.86 ( 3 )
= 488.58 cm³
Hence, from the above,
We can conclude that
The Surface Area of the cylinder is: 461.49 cm²
The volume of the cylinder is: 488.58 cm³

Question 33.
MODELING WITH MATHEMATICS
A cheetah that is running 90 feet per second is 120 feet behind an antelope that is running 60 feet per second. How long will it take the cheetah to catch up to the antelope?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q33

Question 34.
MAKING AN ARGUMENT
A cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds, the antelope is probably safe. Your friend claims the antelope in Exercise 33 will not be safe if the cheetah starts running 650 feet behind it. Is your friend correct? Explain.
Answer:
Your friend is not correct

Explanation:
It is given that a cheetah can run at top speed for only about 20 seconds. If an antelope is too far away for a cheetah to catch it in 20 seconds, the antelope is probably safe. Your friend claims the antelope in Exercise 33 will not be safe if the cheetah starts running 650 feet behind it.
Let the distance of running antelope be x.
Let ‘t’ be the time taken
So,
The distance of running Antelope is:
x = 650 + 60t
The cheetah must arrive at the same position to catch the antelope
So,
x = 90t
Now,
90t = 650 + 60t
90t – 60t = 650
30t = 650
t = 650 ÷ 30
t = 21.7 seconds
But it is given that the cheetah has to reach the same position as the antelope in 20 seconds
But according to the calculation, it takes 21.7 seconds
So,
According to your friend, the antelope is not safe if the cheetah is running 650 meters behind it.
Hence, from the above,
We can conclude that your friend is not correct.

REASONING
In Exercises 35 and 36, for what value of a is the equation an identity? Explain your reasoning.

Question 35.
a(2x + 3) = 9x + 15 + x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q35

Question 36.
8x – 8 + 3ax = 5ax – 2a
Answer:
The given equation becomes an identity at a = 4

Explanation:
The given equation is:
8x – 8 + 3ax = 5ax – 2a
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5ax – 3ax = 8x – 8 – 2a
2ax = 8 ( x – 1 ) – 2a
2ax + 2a = 8x – 8
Equate the like coefficients of x and the like constants in both LHS and RHS
So,
2ax = 8x                                   2a = -8
a = 8x ÷ 2x                               a = -8 ÷ 2
a = 4                                         a = -4
Now,
At a = 4,
The equation becomes
8x – 8 + 3ax = 5ax – 2a
8x – 8 + 3x ( 4 ) = 5x ( 4 ) -2 ( 4 )
8x – 8 + 12x = 20x – 8
20x – 8 = 20x – 8
Hence,
At a =4,
The given equation is an Identity
At a = -4,
The equation becomes
8x – 8 + 3ax = 5ax – 2a
8x – 8 + 3x ( -4 ) = 5x ( -4 ) -2 ( -4 )
8x – 8 – 12x = -20x + 8
-4x – 8 = -20x + 8
Hence,
At a = -4, the given equation is not an Identity
Hence, from the above,
We can conclude that the given equation is an Identity at a = 4

Question 37.
REASONING
Two times the greater of two consecutive integers is 9 less than three times the lesser integer. What are the integers?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q37

Question 38.
HOW DO YOU SEE IT?
The table and the graph show information about students enrolled in Spanish and French classes at a high school.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 55
a. Use the graph to determine after how many years there will be equal enrollment in Spanish and French classes.
Answer:
The year where there will be equal enrollment in Spanish and French classes is: 6

Explanation:
The given table and the graph of the students for the Spanish and French classes are:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 55
So,
From the above table,
To find the year where there is equal enrollment in both Spanish and French, we have to find the point in the graph that the two points of Spanish and French intersect
So,
From the above graph,
The point that is intersecting is at 6th year
Hence, from the above,
We can conclude that there is an equal enrollment of students in the 6th year in both Spanish and French classes

b. How does the equation 355 – 9x = 229 + 12x relate to the table and the graph? How can you use this equation to determine whether your answer in part (a) is reasonable?
Answer:
The given equation
355 – 9x = 229 + 12x
represents the total number of students enrolled in the different years in both Spanish and French classes
Now,
In part (a),
We observed that there is an equal enrollment of the students at 6th year in both Spanish and French classes
So,
Here,
x is: The number of years
So,
In part (a),
x = 6
Now,
Substitute x = 6 in the given equation.
Now,
355 – 9x = 229 + 12x
355 – 9 ( 6 ) = 229 + 12 ( 6 )
355 – 54 = 229 + 72
301 = 301
As
LHS = RHS
We can say that the answer is reasonable in part (a)

Question 39.
WRITING EQUATIONS
Give an example of a linear equation that has (a) no solution and (b) infinitely many solutions. Justify your answers.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q39

Question 40.
THOUGHT-PROVOKING
Draw a different figure that has the same perimeter as the triangle shown. Explain why your figure has the same perimeter.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 56

Answer:

Maintaining Mathematical Proficiency

Order the values from least to greatest.

Question 41.
9, | -4|, -4, 5, | 2 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q41

Question 42.
| -32 |, 22, -16, -| 21 |, | -10 |
Answer:
We know that,
| -x | = x
| x | = x
So,
| -32 | = 32
| 21 | = 21
| -10 | = 10
Hence,
The order of the values from the least to the greatest is:
-21, -16, 10, 22, 32

Question 43.
-18, | -24 |, -19, | -18 |, | 22 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.3-Q43

Question 44.
-| – 3 |, | 0 |, -1, | 2 |, -2
Answer:
We know that,
| -x | = x
| x | = x
So,
| -3 | = 3
| 0 | = 0
| 2 | = 2
Hence,
The order of the numbers from the least to the greatest is:
-3, -2, -1, 0, 2

Solving Linear Equations Study Skills: Completing

1.1-1.3 What Did You Learn

Core Vocabulary

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 57

Core Concepts

Section 1.1
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 58

Section 1.2

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 59

Section 1.3

Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 60

Mathematical Practices

Question 1.
How did you make sense of the relationships between the quantities in Exercise 46 on page 9?
Answer:
In Exercise 46 on page 9,
There is a layout of the tatami mat which comprises the four identical rectangular mats and the one square mat.
it is also given that the area of the square mat is half of one of the rectangular mats
Now,
We know that,
The area of the square mat = Area² [ Since all the sides of the square are equal ]
The area of the rectangular mat = Length × Width
So,
According to the given condition,
The relation between the area of the square mat and one of the rectangular mat is:
Area of the square mat = \(\frac{1}{2}\) Area of one of the rectangular mat
Side² = \(\frac{1}{2}\) ( Length × Width )

Question 2.
What is the limitation of the tool you used in Exercises 25–28 on page 16?
Answer:
The limitations of the tool you used in Exercises 25 – 28 on page 16 are:
A) The calculated values and the values measured using the tool will be different
B) We won’t get the exact values of the angle measures using the tool

Question 3.
What definition did you use in your reasoning in Exercises 35 and 36 on page 24?
Answer:
The definition you used in your reasoning in Exercises 35 and 36 on page 24 is:
Make the like coefficients of the same variable in both LHS and RHS equal so that we get the value of the variable.

Study Skills

Completing Homework Efficiently

Before doing homework, review the Core Concepts and examples. Use the tutorials at BigIdeasMath.com for additional help.

Complete homework as though you are also preparing for a quiz. Memorize different types of problems, vocabulary, rules, and so on.
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 61

Solving Linear Equations 1.1-1.3 Quiz

Solve the equation. Justify each step. Check your solution. (Section 1.1)

Question 1.
x + 9 = 7
Answer:
The value of x is -2

Explanation:
The given equation is:
x + 9 = 7
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = 7 –  ( +9 )
x = 7 – 9
x = -2
Hence, from the above,
We can conclude that the value of x is: -2

Question 2.
8.6 = z – 3.8
Answer:
The value of z is: 12.4

Explanation:
The given equation is:
8.6 = z – 3.8
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
z = 8.6 + 3.8
z = 12.4
Hence, from the above,
We can conclude that the value of z is: 12.4

Question 3.
60 = -12r
Answer:
The value of r is:  -5

Explanation:
The given equation is:
60 = -12r
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
r = 60 ÷ ( -12 )
r = -60 ÷ 12
r = -5
Hence, from the above,
We can conclude that the value of r is: -5

Question 4.
\(\frac{3}{4}\)p = 18
Answer:
The value of p is: 24

Explanation:
The given equation is:
\(\frac{3}{4}\)p = 18
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
p = 18 × \(\frac{4}{3}\)
p = \(\frac{18}{1}\) × \(\frac{4}{3}\)
p = \(\frac{18 × 4}{1 × 3}\)
p = \(\frac{24}{1}\)
p = 24
Hence, from the above,
We can conclude that the value of p is: 24

Solve the equation. Check your solution. (Section 1.2)

Question 5.
2m – 3 = 13
Answer:
The value of m is: 8

Explanation:
The given equation is:
2m – 3 = 13
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2m = 13 + 3
2m = 16
m = 16 ÷ 2
m = 8
Hence, from the above,
We can conclude that the value of m is: 8

Question 6.
5 = 10 – v
Answer:
The value of v is: 5

Explanation:
The given equation is:
5 = 10 – v
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
v= 10 – 5
v = 5
Hence, from the above,
We can conclude that the value of v is: 5

Question 7.
5 = 7w + 8w + 2
Answer:
The value of w is: \(\frac{1}{5}\)

Explanation:
The given equation is:
5 = 7w + 8w + 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 – 2 = 7w + 8w
15w = 3
w = 3 ÷ 15
w = \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of w is: \(\frac{1}{5}\)

Question 8.
-21a + 28a – 6 = -10.2
Answer:
The value of a is: –\(\frac{3}{5}\)

Explanation:
The given equation is:
-21a + 28a  – 6 = -10.2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-21a + 28a = -10.2 + 6
7a = -4.2
a = -4.2 ÷ 7
a= –\(\frac{42}{10}\) ÷ 7
a = –\(\frac{42}{10}\) × \(\frac{1}{7}\)
a = –\(\frac{42 × 1}{10 × 7}\)
a = – \(\frac{6}{10}\)
a = –\(\frac{3}{5}\)
Hence, from the above,
We can conclude that the value of a is: –\(\frac{3}{5}\)

Question 9.
2k – 3(2k – 3) = 45
Answer:
The value of k is: -9

Explanation:
The given equation is:
2k – 3 ( 2k – 3 ) = 45
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
SO,
2k – 3 ( 2k ) + 3 ( 3 ) = 45
2k – 6k + 9 = 45
2k – 6k = 45 – 9
-4k = 36
k = 36 ÷ -4
k = -9
Hence, from the above,
We can conclude that the value of k is: -9

Question 10.
68 = \(\frac{1}{5}\)(20x + 50) + 2
Answer:
The value of x is: 14

Explanation:
The given equation is:
68 = \(\frac{1}{5}\) [ 20x + 50 ] + 2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
68 – 2 = \(\frac{1}{5}\) [ 20x + 50 ]
66 = \(\frac{1}{5}\) [ 20x + 50 ]
66 × 5 = 20x + 50
330 = 20x + 50
20x = 330 – 50
20x = 280
x = 280 ÷ 20
x = 14
Hence, from the above,
We can conclude that the value of x is: 14

Solve the equation. (Section 1.3)

Question 11.
3c + 1 = c + 1
Answer:
The value of c is: 0

Explanation:
The given equation is:
3c + 1 = c + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
3c – c = 1 – 1
2c = 0
c = 0
Hence, from the above,
We can conclude that the value of c is: 0

Question 12.
-8 – 5n = 64 + 3n
Answer:
The value of n is: -9

Explanation:
The given equation is:
-8 – 5n = 64 + 3n
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-8 – 64 = 3n + 5n
-72 = 8n
n = -72 ÷ 8
n = -9
Hence, from the above,
We can conclude that the value of n is: -9

Question 13.
2(8q – 5) = 4q
Answer:
The value of q is: \(\frac{5}{6}\)

Explanation:
Te given equation is:
2 ( 8q – 5 ) = 4q
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2 ( 8q ) – 2 ( 5 ) = 4q
16q – 10 = 4q
16q – 4q = 10
12q = 10
q = 10 ÷ 12
q = \(\frac{5}{6}\)
Hence, from the above,
We can conclude that the value of q is: \(\frac{5}{6}\)

Question 14.
9(y – 4) – 7y = 5(3y – 2)
Answer:
The value of y is: -2

Explanation:
The given equation is:
9 ( y – 4 ) – 7y = 5 ( 3y – 2 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
9 ( y ) – 9 ( 4 ) – 7y = 5 ( 3y ) – 5 ( 2 )
9y – 36 – 7y = 15y – 10
2y – 36 = 15y – 10
15y – 2y = 10 – 36
13y = -26
y = -26 ÷ 13
y = -2
Hence, from the above,
We can conclude that the value of y is: -2

Question 15.
4(g + 8) = 7 + 4g
Answer:
The given equation has no solution

Explanation:
The given equation is:
4 ( g + 8 ) = 7 + 4g
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,the given equation has no solution
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4 ( g ) + 4 ( 8 ) = 7 + 4g
4g + 32 = 7 + 4g
4g – 4g + 7 = 32
7 = 32
Hence, from the above,
We can conclude that the given equation has no solution.

Question 16.
-4(-5h – 4) = 2(10h + 8)
Answer:
The value of h is: 0

Explanation:
The given equation is:
-4 ( 5h – 4 ) = 2 ( 10h + 8 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-4 ( 5h ) + 4 ( 4 ) = 2 ( 10h ) + 2 ( 8 )
-20h + 16 = 20h + 16
-20h – 20h = 1 – 16
-40h =0
h = 0
Hence, from the above,
We can conclude that the value of h is: 0

Question 17.
To estimate how many miles you are from a thunderstorm, count the seconds between when you see lightning and when you hear thunder. Then divide by 5. Write and solve an equation to determine how many seconds you would count for a thunderstorm that is 2 miles away. (Section 1.1)
Answer:

Question 18.
You want to hang three equally-sized travel posters on a wall so that the posters on the ends are each 3 feet from the end of the wall. You want the spacing between posters to be equal. Write and solve an equation to determine how much space you should leave between the posters. (Section 1.2)
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 62

Answer:
The space you should leave between the posters is: \(\frac{3}{2}\) ft

Explanation:
The given figure is:
Big Ideas Math Answers Algebra 1 Chapter 1 Solving Linear Equations 62
It is given that you want to hang three equally-sized travel posters on a wall so that the posters on the ends are each 3 feet from the end of the wall. You want the spacing between posters to be equal.
So,
From the figure,
The total space = 15 ft
The total spacing covered at the ends = 3 + 3 = 6 ft
Let the space between the equally spaced posters be x
So,
The total spacing between the travel posters = 2x + 2x + 2x = 6x ft
SO,
The total space = ( The total spacing covered at the ends ) + ( The total spacing between the travel posters )
15 = 6 + 6x
6x = 15 – 6
6x = 9
x = 9 ÷ 6
x = \(\frac{3}{2}\) ft
Hence, from the above,
We can conclude that the spacing between the travel posters is: \(\frac{3}{2}\) ft

Question 19.
You want to paint a piece of pottery at an art studio. The total cost is the cost of the piece plus an hourly studio fee. There are two studios to choose from. (Section 1.3)
a. After how many hours of the painting are the total costs the same at both studios? Justify your answer.
b. Studio B increases the hourly studio fee by $2. How does this affect your answer in part (a)? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 63

Answer:
a)
It is given that,
The total cost = Cost of the vase + The hourly studio fee
Let the number of hours be x
So,
The total cost for studio A = 10 + 8x
The total cost of studio B = 16 + 6x
It is given that the total costs are the same
So,
10 + 8x = 16 + 6x
8x – 6x = 16 – 10
2x = 6
x = 6 ÷ 2
x = 3
Hence, from the above,
We can conclude that the total cost will be the same after 3 hours for both the studios

b)
It is given that the studio B increases the hourly studio fee by $2
So,
The total hourly studio fee for studio B = 6 + 2 = $8
So,
Now,
As in part (a), the same process will be repeated but in the studio B’s hourly fee of $6, we have to put $8
So,
10 + 8x = 16 + 8x
8x – 8x = 6 – 10
10 = 16
Hence, from the above,
We can conclude that the value of x has no solutions

Lesson 1.4 Solving Absolute Value Equations

Essential Question

How can you solve an absolute value equation?
EXPLORATION 1
Solving an Absolute Value Equation Algebraically
Work with a partner. Consider the absolute value equation
| x + 2 | = 3.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 64
a. Describe the values of x + 2 that make the equation true. Use your description to write two linear equations that represent the solutions of the absolute value equation.
Answer:
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
Now,
-| x + 2 | = 3
| x + 2  | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The values of x + 2 that make the equation true is: 3 and 3
The value of x is: 1 and 1

b. Use the linear equations you wrote in part (a) to find the solutions of the absolute value equation.
Answer:
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
Now,
-| x + 2 | = 3
-| x + 2  | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The solutions of | x + 2 | are: 1 and 1

c. How can you use linear equations to solve an absolute value equation?
Answer:
We use linear equations to solve an absolute value equation by using the following properties. They are:
A) | x | = x
B) -| x | = -x

EXPLORATION 2
Solving an Absolute Value Equation Graphically
Work with a partner.
Consider the absolute value equation
| x + 2 | = 3.
a. On a real number line, locate the point for which x + 2 = 0.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 64.1
Answer:
The given equation is:
x + 2 = 0
x = 0 – 2
x = -2
So,
On a real number line, we have to locate the point x = -2
Hence,
The point we have to locate on the real number line is:

b. Locate the points that are 3 units from the point you found in part (a). What do you notice about these points?
Answer:
From part (a).
We found that x = -2
Now,
To locate the points that are 3 units away or 3 units behind from the point you found in part (a), i.e., x = -2
We know that,
3 units away imply ” Add 3 ”
3 units behind imply ” Subtract 3 ”
Now,
We have to add 3 and subtract 3 to the point we obtained in part (a)
So,
When we add 3 to x = -2,
x = -2 + 3
x =1
When we subtract 3 from x = -2,
x = -2 – 3
x = -5
Hence,
The points we have to locate in the real number line are: 1 and -5
So,
The real number line with the located points is:

c. How can you use a number line to solve an absolute value equation?
Answer:
The given absolute value equation is:
| x + 2 | = 3
We know that,
| x | = x
-| x | = -x
So,
| x + 2 | = 3
x + 2 = 3
x = 3 – 2
x = 1
– | x + 2 | = -3
-x – 2 = -3
-x = -3 + 2
-x = -1
x = 1
So,
The values we have to locate in the number line is:

EXPLORATION 3
Solving an Absolute Value Equation Numerically

Work with a partner. Consider the absolute value equation
| x + 2 | = 3.
a. Use a spreadsheet, as shown, to solve the absolute value equation.
b. Compare the solutions you found using the spreadsheet with those you found in Explorations 1 and 2. What do you notice?
c. How can you use a spreadsheet to solve an absolute value equation?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 65

Communicate Your Answer

Question 4.
How can you solve an absolute value equation?
Answer:
We can solve the absolute equation by using the following properties. They are:
A) | x| = x
B) – | x | = -x

Question 5.
What do you like or dislike about the algebraic, graphical, and numerical methods for solving an absolute value equation? Give reasons for your answers.
Answer:
The algebraic, numerical, and graphical methods have their own advantages in their own perspective.
The algebraic methods used to solve the linear equations whereas the graphical method used to indicate the linear equations.
The numerical method is applicable for mathematical operations

1.4 Lesson

Monitoring Progress

Solve the equation. Graph the solutions, if possible.

Question 1.
| x | = 10
Answer:
The value of x is: 10

Explanation:
The given absolute value equation is:
| x | = 10
We know that,
| x | = x
So,
x = 10
Hence, from the above,
We can conclude that the value of x is: 10

Question 2.
| x – 1 | = 4
Answer:
The value of x is: 5

Explanation:
The given absolute value equation is:
| x – 1 | = 4
We know that,
| x | = x
So,
x – 1 = 4
x = 4 + 1
x = 5
Hence, from the above,
We can conclude that the value of x is: 5

Question 3.
| 3 + x | = -3
Answer:
The value of x is: -6

Explanation:
The given absolute value equation is:
| 3 + x | = -3
We know that,
| x | = x
So,
x + 3 = -3
x = -3 – 3
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Solve the equation. Check your solutions.

Question 4.
| x – 2 | + 5 = 9
Answer:
The value of x is: 6

Explanation:
The given absolute value equation is:
| x – 2 | + 5 = 9
| x – 2 | = 9 – 5
| x – 2 | = 4
We know that,
| x | = x
So,
x – 2 = 4
x = 4 + 2
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 5.
4 | 2x + 7 | = 16
Answer:
The value of x is: –\(\frac{3}{2}\)

Explanation:
The given absolute value equation is:
4 | 2x + 7 | = 16
| 2x + 7 | = 16 ÷ 4
| 2x + 7| = 4
We know that,
| x | = x
So,
2x + 7 = 4
2x = 4 – 7
2x = -3
x = –\(\frac{3}{2}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{3}{2}\)

Question 6.
-2 | 5x – 1 | – 3 = -11
Answer:
The value of x is: 1

Explanation:
The given absolute value equation is:
-2 | 5x – 1 | – 3 = -11
-2 | 5x – 1 | = -11 + 3
-2 | 5x – 1 | = -8
| 5x – 1 | = -8 ÷ ( -2 )
| 5x – 1 | = 4
We know that,
| x | = x
So,
5x – 1 = 4
5x = 4 + 1
5x = 5
x = 5 ÷ 5
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Question 7.
For a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines. Write an absolute value equation that represents the minimum and maximum lengths.
Answer:
The minimum value length is: 16
The maximum length is: 32

Explanation:
It is given that for a poetry contest, the minimum length of a poem is 16 lines. The maximum length is 32 lines.
So,
The absolute value equation that represents the minimum length of a poem = | The minimum length of a poem |
= | 16 |
= 16
The absolute value equation that represents the maximum length of a poem = | The maximum length of a poem |
= | 32 |
= 32
Hence, from the above,
We can conclude that
The minimum value length is: 16
The maximum length is: 32

Solve the equation. Check your solutions.

Question 8.
| x + 8 | = | 2x + 1 |
Answer:
The value of x is: 7

Explanation:
The given absolute value equation is:
| x + 8 | = | 2x + 1 |
We know that,
| x | = x
So,
x + 8 = 2x + 1
2x – x = 8 – 1
x = 7
Hence from the above,
We can conclude that the value of x is: 7

Question 9.
3 | x – 4 | = | 2x + 5 |
Answer:
The value of x is: 17

Explanation:
The given absolute equation is:
3 | x – 4 | = | 2x + 5 |
We know that,
| x |  = x
So,
3 ( x – 4 ) = 2x + 5
3 ( x ) – 3 ( 4 ) = 2x + 5
3x – 12 = 2x + 5
3x – 2x = 5 + 12
x = 17
Hence, from the above,
We can conclude that the value of x is: 17

Solve the equation. Check your solutions.

Question 10.
| x + 6 | = 2x
Answer:
The value of x is: 6

Explanation:
The absolute value equation is:
| x + 6 | = 2x
We know that,
| x | = x
So,
x + 6 = 2x
2x – x = 6
x = 6
Hence, from the above,
We can conclude that the value of x is: 6

Question 11.
| 3x – 2 | = x
Answer:
The value of x is: 1

Explanation:
The given absolute value equation is:
| 3x – 2 | = x
We know that,
| x | = x
So,
3x – 2 = x
Soo,
3x – x = 2
2x = 2
x = 2 ÷ 2
x = 1
Hence, from the above,
We can conclude that the value of x is: 1

Question 12.
| 2 + x | = | x – 8 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 2 + x | = | x – 8 |
We know that,
| x | = x
So,
2 + x = x – 8
2 = x – x – 8
2 = -8
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 13.
| 5x – 2 | = | 5x + 4 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 5x – 2 | = | 5x + 4 |
We know that,
| x | = x
So,
5x – 2 = 5x + 4
5x – 5x – 2 = 4
-2 = 4
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Solving Absolute Value Equations 1.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is an extraneous solution?
Answer:
Extraneous solutions are values that we get when solving equations that are not really solutions to the equation.
Example for extraneous solution:
| 5x – 2 | = | 5x + 4 |

Question 2.
WRITING
Without calculating, how do you know that the equation | 4x – 7 | = -1 has no solution?
Answer:
The given absolute value equation is:
| 4x – 7 | = -1
We know that,
An absolute value can never equal a negative number.
So,
By the above,
We can say that
| 4x – 7 | must not equal to a negative number.
Hence, from the above,
We can conclude that | 4x – 7 | = -1 has no solution without calculating its solution

Monitoring Progress and Modeling with Mathematics

In Exercises 3−10, simplify the expression.

Question 3.
| -9 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q3

Question 4.
– | 15 |
Answer:
The value of -| 15 | is: -15

Explanation:
The given absolute value is: -| 15 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
-| 15 | = -15
Hence, from the above,
We can conclude that the value of -| 15 | is: -15

Question 5.
| 14 | – | -14 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q5

Question 6.
| -3 | + | 3 |
Answer:

The value of | -3 | + | 3 | is: 6

Explanation:
The given absolute value expression is:
| -3 | + | 3 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
| -3 | + | 3 |
= 3 + 3
= 6
Hence, from the above,
We can conclude that the value of | -3 | + | 3 | is: 6

Question 7.
– | -5 • (-7) |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q7

Question 8.
| -0.8 • 10 |
Answer:
The value of | -0.8 ⋅ 10 | is: 8

Explanation:
The given absolute value expression is:
| -0.8 ⋅ 10 |
We know that,
| x | = x
| -x | = x
-| x | = -x
So,
| -0.8 ⋅ 10 |
= | – ( 8 ⁄ 10 ) ⋅ ( 10 ⁄ 1 ) |
= | – ( 8 × 10 ) ⁄ ( 10 × 1 ) |
= | -8 |
= 8
hence, from the above,
We can conclude that the value of | -0.8 ⋅ 10 | is: 8

Question 9.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 66

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q9

Question 10.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 67
Answer:
The value of | -12 ⁄ 4 | is: 3

Explanation:
The given absolute value expression is: | -12 ⁄ 4 |
We know that,
| x | = x
| -x | = x
– | x | = -x
So,
| -12 ⁄ 4 | = | -3 |
= 3
Hence, from the above,
We can conclude that the value of | -12 ⁄ 4 | is: 3

In Exercises 11−24, solve the equation. Graph the solution(s), if possible.

Question 11.
| w | = 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q11

Question 12.
| r | = -2
Answer:
The absolute value of a number must be greater than or equal to 0 and can not be equal to -2.
Hence,
The given absolute eqution has no solution

Question 13.
| y | = -18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q13

Question 14.
| x | = 13
Answer:
The value of x is: 13 or -13

Explanation:
The given absolute value equation is:
| x | = 13
We know that,
|x | = x
– | x | = -x
So,
| x | = 13 or – 13
Hence, from the above,
We can conclude that the value of x is: 13 or -13

Question 15.
| m + 3 | = 7

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q15

Question 16.
| q – 8 | = 14
Answer:
The value of q is: 22 or -6

Explanation:
The given absolute value equation is:
| q – 8 | = 14
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
q – 8 = 14                                                      q – 8 = -14
q = 14 + 8                                                     q = -14 + 8
q = 22                                                             q = -6
Hence, from the above,
We can conclude that the value of q is: 22 or -6

Question 17.
| -3d | = 15

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q17

Question 18.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 68
Answer:
The value of t is: 12 or -12

Explanation:
The given absolute value equation is:
| t / 2 | = 12
We know that,
| x | = x  for x > 0
| x | = -x for x < 0
So,
t / 2 = 6                                       t / 2 = -6
t = 6 × 2                                      t = 6 × -2
t = 12                                           t = -12
Hence, from the above,
We can conclude that the value of t is: 12 or -12

Question 19.
| 4b – 5 | = 19

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q19

Question 20.
| x – 1 | + 5 = 2
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| x – 1 | + 5 = 2
| x – 1 | = 2 – 5
| x – 1  | = -3
We know that,
The absolute value of an equation must be greater than or equal to zero
So,
| x – 1 | = -3 has no solution
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 21.
-4 | 8 – 5n | = 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q21

Question 22.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 69
Answer:
The value of y is: -3 or 6

Explanation:
The given absolute value equation is:
-3 | 1 – ( 2 / 3 ) y | = -9
| 1 – (2 / 3  ) y | = -9 ÷ ( -3 )
| 1 – ( 2 / 3 ) y | = 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
1 – ( 2 / 3 ) y = 3                                    1 – ( 2 /3 ) y = -3
2/3 y = 1 – 3                                           2/3 y = 1 + 3
2 / 3 y = -2                                              2 / 3 y = 4
2y = -2 × 3                                               2y = 4 × 3
2y = -6                                                      2y = 12
y = -6 ÷ 2                                                  y = 12 ÷ 2
y = -3                                                         y = 6
Hence, from the above,
We can conclude that the value of y is: -3 or 6

Question 23.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 70

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q23

Question 24.
9 | 4p + 2 | + 8 = 35
Answer:
The value of p is: 1 / 4 or -5 / 4

Explanation:
The given absolute value equation is:
9 | 4p + 2 | + 8 = 35
9 | 4p + 2 | = 35 – 8
9 | 4p + 2 | = 27
| 4p + 2 | = 27 ÷ 9
| 4p + 2 | = 3
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
4p + 2 = 3                    4p + 2 = -3
4p = 3 – 2                     4p = -3 – 2
4p = 1                           4p = -5
p = 1 / 4                         p = -5 / 4
Hence, from the above,
We can conclude that the value of p is: 1 / 4 or -5 / 4

Question 25.
WRITING EQUATIONS
The minimum distance from Earth to the Sun is 91.4 million miles. The maximum distance is 94.5 million miles.
a. Represent these two distances on a number line.
b. Write an absolute value equation that represents the minimum and maximum distances.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q25

Question 26.
WRITING EQUATIONS
The shoulder heights of the shortest and tallest miniature poodles are shown.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 71
a. Represent these two heights on a number line.
b. Write an absolute value equation that represents these heights.

Answer:
a)
The number line that represents the two heights on a number line is:

b)
The minimum shoulder height = ( 15 – 10 ) / 2
= 5 / 2
= 2.5 inches
The maximum shoulder height = 10 + 2.5
= 12.5 inches
Now,
Let the heights between poodles be x.
Hence,
The absolute value equation is:
| x – 12.5 | = 2.5

USING STRUCTURE In Exercises 27−30, match the absolute value equation with its graph without solving the equation.

Question 27.
| x + 2 | = 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q27

Question 28.
| x + 4 | = 2
Answer:
The given absolute value equation is:
| x + 4 | = 2
To find the halfway point, made the absolute value equation equal to 0.
So,
| x + 4  | = 0
So,
x = -4
From the given absolute value equation,
We can say that the distance from the halfway point to the minimum and maximum points is: 2

Question 29.
| x – 2 | = 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q29

Question 30.
| x + 4 | = 2
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 72

Answer:

In Exercises 31−34, write an absolute value equation that has the given solutions.

Question 31.
x = 8 and x = 18

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q31

Question 32.
x = -6 and x = 10
Answer:
The given absolute value equation is:
| x – 2 | = 8

Explanation:
The given values of x are:
x = -6 and x = 10
Now,
The halfway point between 10 and -6 = [ 10 – ( -6 ) ] / 2
= [ 10 + 6 ] / 2
= 16 / 2
= 8
The minimum distance from the halfway point = 8 – 6 = 2
Hence,
The absolute value equation is:
| x – 2 | = 5

Question 33.
x = 2 and x = 9

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q33

Question 34.
x = -10 and x = -5

Answer:
The given values of x are:
x = -10 and x = -5
Now,
The halfway point between -10 and -5 = [ 10 – ( 5 ) ] / 2
= [ 10 – 5 ] / 2
= 5 / 2
= 2.5
So,
The minimum value from the half-point = 2.5 + ( -10 )
= 2.5 – 10
= -7.5
Hence,
The absolute value equation is:
| x – ( -7.5 ) | = 2.5
| x + 7.5 | = 2.5

In Exercises 35−44, solve the equation. Check your solutions. 

Question 35.
| 4n – 15 | = | n |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q35

Question 36.
| 2c + 8 | = | 10c |
Answer:
The values of c are: 1 and 2 / 3

Explanation:
The given absolute value equation is:
| 2c + 8 | = | 10c |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
Now,
2c + 8 = 10c                                      2c + 8 = -10c
10c – 2c = 8                                       2c + 10c = 8
8c = 8                                                12c = 8
c = 8 / 8                                              c = 8 / 12
c = 1                                                    c = 2 /3
Hence, from the above,
We can conclude that the values of c are: 1 and 2 / 3

Question 37.
| 2b – 9 | = | b – 6 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q37

Question 38.
| 3k – 2 | = 2 | k + 2 |
Answer:
The values of k are: 6 and -2 / 5

Explanation:
The given absolute equation is:
| 3k – 2 | = 2 | k + 2 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 ( k + 2 ) = 3k – 2                                            2 ( k + 2 ) = – ( 3k – 2 )
2k + 4 = 3k – 2                                                 2k + 4 = -3k + 2
3k – 2k = 4 + 2                                                 2k + 3k = 2 – 4
k = 6                                                                 5k = -2
k = 6                                                                 k = -2 / 5
Hence, from the above,
We can conclude that the values of k are: 6 and -2 / 5

Question 39.
4 | p – 3 | = | 2p + 8 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q39

Question 40.
2 | 4w – 1 | = 3 | 4w + 2 |
Answer:
The value of w is: -2

Explanation:
The given absolute value equation is:
2 | 4w – 1 | = 3 | 4w+ 2 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
2 ( 4w – 1 ) = 3 ( 4w + 2 )                        -2 ( 4w – 1 ) = -3 ( 4w + 2 )
8w – 2 = 12w + 6                                      -8w + 2 = -12w -6
12w – 8w = -6 – 2                                      -12w + 8w = 6 + 2
4w = -8                                                       -4w = 8
w = -8 / 4                                                      w = 8 / -4
w = -2                                                            w = -2
Hence, from the above,
We can conclude that the value of w is: -2

Question 41.
| 3h + 1 | = 7h

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q41

Question 42.
| 6a – 5 | = 4a
Answer:
The value of a is: 5 / 2 and 1 / 2

Explanation:
The given absolute value equation is:
| 6a – 5 | = 4a
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
6a – 5 = 4a                                                6a – 5 = -4a
6a – 4a = 5                                                 6a + 4a = 5
2a = 5                                                         10a = 5
a = 5 / 2                                                       a = 5 / 10
a = 5 / 2                                                       a = 1 / 2
Hence, from the above,
We can conclude that the values of a are: 5 / 2 and 1 / 2

Question 43.
| f – 6 | = | f + 8 |

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q43

Question 44.
| 3x – 4 | = | 3x – 5 |
Answer:
The given absolute value equation has no solution

Explanation:
The given absolute value equation is:
| 3x – 4 | = | 3x – 5 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
3x – 4 = 3x – 5                                – ( 3x – 4 ) = – ( 3x – 5 )
4 = 5                                                 4 = 5
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 45.
MODELING WITH MATHEMATICS
Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance d (in feet) of the car from you after t seconds is given by the equation d = | 300 – 48t |. At what times is the car 60 feet from you?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q45

Question 46.
MAKING AN ARGUMENT
Your friend says that the absolute value equation | 3x + 8 | – 9 = -5 has no solution because the constant on the right side of the equation is negative. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
The given absolute value equation is:
| 3x + 8 | – 9 = -5
We know that,
The absolute value equation value must have greater than or equal to 0
But here
The value of the absolute value equation is less than 0
Hence,
The given absolute value equation has no solution.
Hence, from the above,
We can conclude that your friend is correct.

Question 47.
MODELING WITH MATHEMATICS
You randomly survey students about year-round school. The results are shown in the graph.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 73
The error given in the graph means that the actual percentage could be 5% more or 5% less than the percent reported by the survey.
a. Write and solve an absolute value equation to find the least and greatest percents of students who could be in favor of the year-round school.
b. A classmate claims that \(\frac{1}{3}\) of the student body is actually in favor of the year-round school. Does this conflict with the survey data? Explain.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q47

Question 48.
MODELING WITH MATHEMATICS
The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams.
a. Write and solve an absolute value equation to find the minimum and maximum acceptable soccer ball weights.
Answer:
It is given that the recommended weight of a soccer ball is 430 grams and the actual weight is allowed to vary up to 20 grams
Hence,
The absolute value equation that represents the minimum and maximum acceptable soccer ball weights is:
| x – 430 | = 20
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 430 = 20                               x – 430 = -20
x = 20 + 430                              x = -20 + 430
x = 460 grams                            x = 410 grams
Hence, from the above,
We can conclude that the maximum and minimum acceptable soccer weights respectively are: 460 grams and 410 grams

b. A soccer ball weighs 423 grams. Due to wear and tear, the weight of the ball decreases by 16 grams. Is the weight acceptable? Explain.
Answer:
The weight that caused due to wear and tear is not acceptable

Explanation:
From the above problem,
We get the maximum weight of the soccer ball to be 460 grams with 20 grams increase or decreased to the weight of the ball
Now,
It is given that the weight of the ball is decreased by 16 grams due to wear and tear
So,
The weight of the ball now = 460 – 16 = 444 grams
But it is given that the weight of the ball becomes 423 grams due to wear and tear.
Hence, from the above,
We can conclude that the weight is not acceptable

ERROR ANALYSIS
In Exercises 49 and 50, describe and correct the error in solving the equation.

Question 49.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 74

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q49

Question 50.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 75

Answer:
The values of x are: -2 and -4 / 3

Explanation:
The given absolute value equation is:
| 5x + 8 | = x
We know that,
| x | = x for x > 0
| x | = – x for x < 0
So,
5x + 8 = x                                                      5x + 8 = -x
5x – x = -8                                                      5x + x = -8
4x = -8                                                            6x = -8
x = -8 / 4                                                         x = -8 / 6
x = -2                                                               x = -4 / 3
Hence, from the above,
We can conclude that the values of x are: -2 and -4 / 3

Question 51.
ANALYZING EQUATIONS
Without solving completely, place each equation into one of the three categories.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 76

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q51

Question 52.
USING STRUCTURE
Fill in the equation Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 77 with a, b, c, or d so that the equation is graphed correctly.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 77.1

Answer:

ABSTRACT REASONING
In Exercises 53−56, complete the statement with always, sometimes, or never. Explain your reasoning.

Question 53.
If x2 = a2, then | x | is ________ equal to | a |.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q53

Question 54.
If a and b are real numbers, then | a – b | is _________ equal to | b – a |.
Answer:
If a and b are real numbers, then
| a – b | is equal to | b – a |

Explanation:
Let,
| a | = 5 and | b | = 9
We know that,
| x | =x for  x > 0
| x | = -x for x < 0
So,
| a – b | = | 5 – 9 |
= | -4 | = 4
| b – a | = | 9 – 5 |
=  | 4 |
= 4
Hence, from the above,
We can conclude that value of
| a – b | is equal to | b – a | if a and b are real numbers

Question 55.
For any real number p, the equation | x – 4 | = p will ________ have two solutions.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q55

Question 56.
For any real number p, the equation | x – p | = 4 will ________ have two solutions.
Answer:
For any real number,
| x – p | = 4 will have two solutions

Explanation:
The given absolute value equation is:
| x – p | = 4
Let the value of p be 1
We know that,
| x | = x for x > 0
| x | = – x for x < 0
So,
| x – 1 | = 4
| x – 1 | = 4                                        | x – 1 | = -4
x = 4 + 1                                           x = -4 + 1
x = 5                                                  x = -3
Hence, from the above,
We can conclude that
| x – p | = 4 will have two solutions for any real number p

Question 57.
WRITING
Explain why absolute value equations can have no solution, one solution, or two solutions. Give an example of each case.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q55

Question 58.
THOUGHT-PROVOKING
Describe a real-life situation that can be modeled by an absolute value equation with the solutions x = 62 and x = 72.
Answer:
Suppose in a school, an exam is conducted. In that examination, 67% of the students are passed. If the error of the pass percentage is 5 %, then what are the minimum and the maximum number of students passed in the examination?
Now,
The absolute value equation for the given real-life situation is:
| x – 67 | = 5
We know that,
| x | = x for x> 0
| x | =-x for x < 0
So,
x – 67 = 5                                         x – 67 = -5
x = 5 + 67                                        x = -5 + 67
x = 72                                               x = 62
Hence, from the above,
We can conclude that the minimum and maximum number of students passed in the examination  respectively are: 72 and 67

Question 59.
CRITICAL THINKING
Solve the equation shown. Explain how you found your solution(s).
8 | x + 2 | – 6 = 5 | x + 2 | + 3

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q59
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q59-i

Question 60.
HOW DO YOU SEE IT?
The circle graph shows the results of a survey of registered voters on the day of an election.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
The error given in the graph means that the actual percentage could be 2% more or 2% less than the percent reported by the survey.
a. What are the minimum and maximum percents of voters who could vote Republican?
Answer:
The minimum percentage of voters for Republicans is: 40%
The maximum percentage of voters for Republicans is: 44 %

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
From the graph,
The vote percentage for Republicans is: 42 %
The error percentage is: ±2%
So,
The absolute value equation for the maximum and the minimum number of voters is:
| x – 42 | = 2
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 42 = 2                                                                 x – 42 = -2
x = 2 + 42                                                                x = -2 + 42
x = 44                                                                       x = 40
Hence, from the above,
We can conclude that
The minimum percentage of voters for Republicans is: 40%
The maximum percentage of voters for Republicans is: 44 %

b. How can you use absolute value equations to represent your answers in part (a)?
Answer:
From the property of absolute values,
We know that,
| x | = x for x > 0
| x | = -x for x < 0
From the part ( a ),
The absolute value equation is:
| x – 42 | = 2
So,
x – 42 = 2                                                                 x – 42 = -2
x = 2 + 42                                                                x = -2 + 42
x = 44                                                                       x = 40
Hence, from the above,
We can conclude that we can use absolute values in the above way to represent the answers

c. One candidate receives 44% of the vote. Which party does the candidate belong to? Explain.
Answer:
The candidate of the Republican party receives 44 % of the vote.

Explanation:
The given graph is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 78
From the above graph,
We can say that,
The percentage of the vote received by the candidate of the Republican party = 42 %
The error percentage = ± 2 %
So,
Now,
The percentage of the vote received by the Republicans = 42 + 2  ( or ) 42 – 2
= 44 ( or ) 40
Hence, from the above,
We can conclude that the candidate of the Republican party received the 44 % of the vote

Question 61.
ABSTRACT REASONING
How many solutions does the equation a | x + b | + c = d have when a > 0 and c = d? when a < 0 and c > d? Explain your reasoning.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q61

Maintaining Mathematical Proficiency

Identify the property of equality that makes Equation 1 and Equation 2 equivalent. (Section 1.1)

Question 62.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 79

Answer:
The given equations are:
Equation 1: 3x + 8 = x – 1
Equation 2: 3x + 9= x
From Equation 1,
3x + 8 = x – 1
3x + 8 + 1 = x
3x + 9 = x
Hence, from the above,
We can conclude that we can get Equation2 by rearranging the Equation 1

Question 63.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 80

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q63

Use a geometric formula to solve the problem.

Question 64.
A square has an area of 81 square meters. Find the side length.
Answer:
The side length of the square is: 9 meters

Explanation:
The given area of the square is: 81 square meters
We know that,
Area of the square = Side × Side
81 = Side × Side
Side² = 81
Apply square root on both sides
√Side² = √81
Side = 9 meters
Hence, from the above,
We can conclude that the side of the square is: 9 meters

Question 65.
A circle has an area of 36π square inches. Find the radius.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q65

Question 66.
A triangle has a height of 8 feet and an area of 48 square feet. Find the base.
Answer:
The base of the triangle is: 12 feet

Explanation:
It is given that a triangle has a height of 8 feet and an area of 48 square feet
We know that,
The area of the triangle = ( 1 /  2 ) × Base × Height
48 = ( 1 / 2 ) × Base × 8
Base × 8 = 48 × 2
Base = ( 48 × 2 ) ÷ 8
Base = 96 ÷ 8
Base = 12 feet
Hence, from the above,
We can conclude that the base of the triangle is: 12 feet

Question 67.
A rectangle has a width of 4 centimeters and a perimeter of 26 centimeters. Find the length.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.4-Q67

Lesson 1.5 Rewriting Equations and Formulas

Essential Question
How can you use a formula for one measurement to write a formula for a different measurement?
Answer:
Write the formula for one measurement and then solve the formula for the different measurement you want to find and use this new formula to find that measurement
Hence, in the above way,
We can use a formula for one measurement to write a formula for a different measurement

EXPLORATION 1
Using an Area Formula
Work with a partner.

a. Write a formula for the area A of a parallelogram.
Answer:
We know that,
The area of the parallelogram ( A) = Base × Height

b. Substitute the given values into the formula. Then solve the equation for b. Justify each step.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 81
Answer:
The value of b is: 6 in

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 81
From the given figure,
Area ( A) = 30 in²
Height ( h ) = 5 in
Base = b
From part ( a),
Area of the parallelogram = Base × Height
30 = 5 × b
b = 30 ÷ 5
b = 6 in
Hence, from the above,
We can conclude that the value of b is: 6 in

c. Solve the formula in part (a) for b without first substituting values into the formula. Justify each step.
Answer:
From part ( a ),
Area of the parallelogram = Base × Height
Base = ( Area of the parallelogram ) ÷ Height of the parallelogram
From the given figure,
Base = b
So,
b = ( Area of the parallelogram ) ÷ Height of the parallelogram

d. Compare how you solved the equations in parts (b) and (c). How are the processes similar? How are they different?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 82
Answer:
We know that,
The area of the parallelogram = Base × Height
Using the above formula,
We solved parts (b ) and ( c )

EXPLORATION 2
Using Area, Circumference, and Volume Formulas
Work with a partner. Write the indicated formula for each figure. Then write a new formula by solving for the variable whose value is not given. Use the new formula to find the value of the variable.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 83
Answer:
The given geometrical figures are:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 83
Now,
a)
The given figure is:

We know that,
Area of the trapezoid = h ( a + b ) / 2
Where,
h is the height between the two sides
a + b is the sum of the parallel sides [ Here, b1 and b2 ]
From the given figure,
Area of the trapezoid = 63 cm²
1st parallel side ( b1 ) = 8 cm
2nd parallel side ( b2 ) = 10 cm
So,
Area of the trapezoid = h ( 8 + 10 ) / 2
63 = h ( 18 ) / 2
63 × 2 = h × 18
h = ( 63 × 2 ) ÷ ( 18 × 1 )
h = 7 cm
Hence, from the above,
We can conclude that the value of h is: 7 cm
b)
The given figure is:

We know that,
Circumference of a circle = 2πr
Where
r is the radius of the circle
From the above figure,
Circumference of the circle ( C ) = 24π ft
So,
24π = 2πr
r = ( 24π ) ÷ ( 2π )
r = 12 ft
Hence, from the above,
We can conclude that the value of r is: 12 ft
c)
The given figure is:

We know that,
The volume of the rectangular prism ( V ) = Length × Width × Height
The area of the rectangle ( B) = Length × Width
So,
The volume of the rectangular prism (V ) = B × Height
From the above figure,
The volume of the rectangular prism ( V ) = 75 yd³
The area of the rectangle ( B ) = 15 yd²
So,
75 = 15 × Height
Height = 75 ÷ 15
Height = 15 yd
Hence, from the above,
We can conclude that the value of h is: 15 yd
d)
The given figure is:

We know that,
The volume of cone ( V ) = πr²h / 3
The area of the circle ( B ) = πr²
Where,
r is the radius of the circle
h is the height of the cone
So,
The volume of the cone ( V ) = Bh / 3
From the above figure,
V = 24π m³
B = 12π m³
So,
24π = 12π × h / 3
h / 3 = 24π ÷ 12π
h / 3 = 2
h = 2 × 3
h = 6 m
Hence, from the above,
We can conclude that the height of the cone is: 6 m

Communicate Your Answer

Question 3.
How can you use a formula for one measurement to write a formula for a different measurement? Give an example that is different from those given in Explorations 1 and 2.
Answer:
Write the formula for one measurement and then solve the formula for the different measurement you want to find and use this new formula to find that measurement
Example:
The given figure is:

We know that,
The area of the rectangle ( A ) = Length × Width
From the above figure,
A = 20 cm²
L = 10 cm
Let,
W be the width of the rectangle
So,
20 = 10 × W
W = 20 ÷ 10
W = 2 cm
Hence, from the above,
We can conclude that the value of W is: 2 cm

1.5 Lesson

Monitoring Progress

Solve the literal equation for y.

Question 1.
3y – x = 9
Answer:
The value of y is: ( x + 9 ) / 3

Explanation:
The given equation is:
3y – x = 9
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3y = 9 + x
y = ( x + 9 ) / 3
Hence, from the above,
We can conclude that the value of y is: ( x + 9 ) / 3

Question 2.
2x – 2y = 5
Answer:
The value of y is: ( 2x – 5 ) / 2

Explanation:
The given equation is:
2x – 2y = 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
2y = 2x – 5
y = ( 2x – 5 ) / 2
Hence, from the above,
We can conclude that the value of y is: ( 2x – 5 ) / 2

Question 3.
20 = 8x + 4y
Answer:
The value of y is: 5 – 2x

Explanation:
The given equation is:
20 = 8x + 4y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 20 – 8x
y = ( 20 – 8x ) / 4
y = ( 20 ÷ 4 ) – ( 8x ÷ 4 )
y = 5 – 2x
Hence, from the above,
We can conclude that the value of y is: 5 – 2x

Solve the literal equation for x.

Question 4.
y = 5x – 4x
Answer:
The value of x is: y

Explanation:
The given equation is:
y = 5x – 4x
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x = y
Hence, from the above,
We can conclude that the value of x is: y

Question 5.
2x + kx = m
Answer:
The value of x is: m / ( k + 2  )

Explanation:
The given equation is:
2x + kx = m
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
x ( k + 2 ) = m
x = m / ( k + 2 )
Hence, from the above,
We can conclude that the value of x is: m / ( k + 2 )

Question 6.
3 + 5x – kx = y
Answer:
The value of x is: ( y – 3 ) / ( 5 – k )

Explanation:
The given equation is:
3 + 5x – kx = y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5x – kx = y – 3
x ( 5 – k ) = y – 3
x = ( y – 3 ) / ( 5 – k )
Hence, from the above,
We can conclude that the value of x is: ( y – 3 ) / ( 5 – k )

Solve the formula for the indicated variable. 

Question 7.
Area of a triangle: A = \(\frac{1}{2}\)bh; Solve for h.
Answer:
The value of h is: \(\frac{2A}{b}\)

Explanation:
The given area of a triangle is:
A = \(\frac{1}{2}\) bh
bh = 2A
h = \(\frac{2A}{b}\)
Hence, from the above,
We can onclude that the value of h is: \(\frac{2A}{b}\)

Question 8.
The surface area of a cone: S = πr2 + πrℓ; Solve for ℓ.
Answer:
The value of l is: \(\frac{S}{πr}\) – r

Explanation:
The given surface area of a cone is:
S = πr² + πrl
S= πr ( r + l )
r + l = \(\frac{S}{πr}\)
l = \(\frac{S}{πr}\) – r
Hence, from the above,
We can conclude that the value of l is: \(\frac{S}{πr}\) – r

Monitoring Progress

Question 9.
A fever is generally considered to be a body temperature greater than 100°F. Your friend has a temperature of 37°C. Does your friend have a fever?
Answer:
Your friend does not have a fever

Explanation:
It is given that a fever is generally considered to be a body temperature greater than 100°F.
We know that,
To convert Fahrenheit into Celsius,
°C = ( °F – 32 ) × \(\frac{5}{9}\)
°C = ( 100 – 32 ) × \(\frac{5}{9}\)
°C = 68 × \(\frac{5}{9}\)
°C = 37.7°
But it is given that your friend has a temperature of 37°C
So, for fever, the temperature has to be 37.7°C
Hence, from the above,
We can conclude that your friend does not have a fever

Question 10.
How much money must you deposit in a simple interest account to earn $500 in interest in 5 years at 4% annual interest?
Answer:
The money you deposit in simple interest is: $2,500

Explanation:
It is given that you earned $500 in a simple interest to earn in 5 years at 4% annual interest
Let,
The money you deposited be: $x
We know that,
Simple interest = ( Principle × Time × Rate ) / 100
The principle is the money you deposited
So,
500 = ( x × 5 × 4 ) / 100
( x × 5 × 4 ) = 500 × 100
x × 20 = 500 × 100
x = ( 500 × 100 ) ÷ 20
x = $2,500
Hence, from the above,
We can conclude that the money you deposited is: $2,500

Question 11.
A truck driver averages 60 miles per hour while delivering freight and 45 miles per hour on the return trip. The total driving time is 7 hours. How long does each trip take?
Answer:
The time taken for each trip is: 3 hours and 4 hours respectively

Explanation:
It is given that a truck driver averages 60 miles per hour while delivering freight and 45 miles per hour on the return trip. The total driving time is 7 hours.
We know that,
Speed = \(\frac{Distance}{Time}\)
Time = \(\frac{Distance}{Speed}\)
Let the distance be D
It is given that the total driving time is: 7 hours
So,
7 = \(\frac{D}{60}\) + \(\frac{D}{45}\)
7 / D = \(\frac{60 + 45}{60 × 45}\)
7 / D = \(\frac{105}{2,700}\)
D = 7 / \(\frac{105}{2,700}\)
D = 7 × \(\frac{2,700}{105}\)
D = \(\frac{7}{1}\) × \(\frac{2,700}{105}\)
D = \(\frac{7 × 2,700}{1 × 105}\)
D = 180 miles
So,
The time taken to deliver = \(\frac{180}{60}\) = 3 hours
The time taken to return = \(\frac{180}{45}\) = 4 hours
Hence, from the above,
We can conclude that the time taken for each trip is: 3 hours and 4 hours respectively

Rewriting Equations and Formulas 1.5 Exercices

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Is Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 84 a literal equation? Explain.
Answer:
The ” Literal equation” is an equation that contains only letters
Now,
The given equation is:
9r + 16 = π / 5
From the above definition,
We can say that the given equation is a ” Literal equation ”

Question 2.
DIFFERENT WORDS, SAME QUESTION?
Which is different? Find “both” answers.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 85
Answer:
The given problems are:
a) Solve 3x + 6y = 24 for x
b) Solve 24 – 3x = 6y for x
c) Solve 6y = 24 – 3x in terms of x
d) Solve 24 – 6y = 3x for x in terms of y
So,
From the above-given problems,
We can observe that d) is different as we have to find x in terms of y whereas in the remaining three problems, we have to find x

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the literal equation for y.

Question 3.
y – 3x = 13

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q3

Question 4.
2x + y = 7
Answer:
The value of y is: 7 – 2x

Explanation:
The given literal equation is:
2x + y = 7
Now,
y = 7 – 2x
Hence, from the above,
We can conclude that the value of y is: 7 – 2x

Question 5.
2y – 18x = -26

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q5

Question 6.
20x + 5y = 15
Answer:
The value of y is: 3 – 4x

Explanation:
The given literal equation is:
20x + 5y = 15
Now,
5y = 15 – 20x
y = ( 15 – 20x ) / 5
y = ( 15  / 5 ) – ( 20x / 5 )
y = 3 – 4x
Hence, from the above,
We can conclude that the value of y is: 3 – 4x

Question 7.
9x – y = 45

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q7

Question 8.
6 – 3y = -6
Answer:
The value of y is: 4

Explanation:
The given literal equation is:
6 – 3y = -6
-3y = -6 – ( +6 )
-3y = -6 -6
-3y = -12
y = -12 ÷ ( -3 )
y = 12 ÷ 3
y = 4
Hence, from the above,
We can conclude that the value of y is: 4

Question 9.
4x – 5 = 7 + 4y

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q9

Question 10.
16x + 9 = 9y – 2x
Answer:
The value of y is: 18x + 9

Explanation:
The given literal equation is:
16x + 9 = y – 2x
So,
16x + 2x + 9 = y
18x + 9 = y
y = 18x + 9
Hence, from the above,
We can conclude that the value of y is: 18x + 9

Question 11.
2 +\(\frac{1}{6}\)y = 3x + 4

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q11

Question 12.
11 – \(\frac{1}{2}\)y = 3 + 6x
Answer:
The value of y is: 16 – 12x

Explanation:
The given literal equation is:
11 – \(\frac{1}{2}\)y = 3 + 6x
So,
–\(\frac{1}{2}\)y = 3 + 6x – 11
-y = 2 ( 3 + 6x – 11 )
y = -2 ( 3 + 6x – 11 )
y = -2 ( 3 ) -2 ( 6x ) + 2 ( 11 )
y = -6 – 12x + 22
y = 16 – 12x
Hence, from the above,
We can conclude that the value of y is: 16 – 12x

In Exercises 13–22, solve the literal equation for x.

Question 13.
y = 4x + 8x

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q13

Question 14.
m = 10x – x
Answer:
The value of x is: m / 9

Explanation:
The given literal equation is:
m = 10x – x
m = 9x
x = m / 9
Hence, from the above,
We can conclude that the value of x is: m / 9

Question 15.
a = 2x + 6xz

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q15

Question 16.
y = 3bx – 7x
Answer:
The value of x is: y / ( 3b – 7 )

Explanation:
The given literal equation is:
y = 3bx – 7x
So,
y = x ( 3b – 7 )
x = y / ( 3b – 7 )
Hence, from the above,
We can conclude that the value of x is: y / ( 3b – 7 )

Question 17.
y = 4x + rx + 6

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q17

Question 18.
z = 8 + 6x – px
Answer:
The value of x is: ( z – 8 ) / ( 6 – p )

Explanation:
The given literal equation is:
z = 8 + 6x – px
So,
z – 8 = 6x – px
z – 8 = x ( 6 – p )
x = ( z – 8 ) / ( 6 – p )
Hence, from the above,
We can conclude that the value of x is: ( z – 8 ) / ( 6 – p )

Question 19.
sx + tx = r

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q19

Question 20.
a = bx + cx + d
Answer:
The value of x is: ( a – d ) / ( b + c )

Explanation:
The given literal equation is:
a = bx + cx + d
a – d = bx + cx
a – d = x ( b + c )
x = ( a – d ) / ( b + c )
Hence, from the above,
We can conclude that the value of x is: ( a – d ) / ( b + c )

Question 21.
12 – 5x – 4kx = y

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q21

Question 22.
x – 9 + 2wx = y
Answer:
The value of x is: ( y – 9 ) / ( 1 – 2w )

Explanation:
The given literal equation is:
x – 9 + 2wx = y
x – 2wx = y + 9
x ( 1 – 2w ) = y + 9
x = ( y – 9 ) / ( 1 – 2w )
Hence, from the above,
We can conclude that the value of x is: ( y – 9 ) / ( 1 – 2w )

Question 23.
MODELING WITH MATHEMATICS
The total cost C (in dollars) to participate in a ski club is given by the literal equation C = 85x + 60, where x is the number of ski trips you take.
a. Solve the equation for x.
b. How many ski trips do you take if you spend a total of $315? $485?
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 86

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q23

Question 24.
MODELING WITH MATHEMATICS
The penny size of a nail indicates the length of the nail. The penny size d is given by the literal equation d = 4n – 2, where n is the length (in inches) of the nail.
a. Solve the equation for n.
b. Use the equation from part (a) to find the lengths of nails with the following penny sizes: 3, 6, and 10.
Answer:
a)
The given literal equation is:
d = 4n – 2
Where,
n is the length ( in inches ) of the nail
So,
4n = d + 2
n = ( d + 2 ) / 4
b)
It is given that,
The penny sizes ( d ) are: 3, 6, and 10
From part ( a ),
The literal equation is:
n = ( d + 2  ) / 4
Put, d= 3, 6 and 10
So,
n = ( 3 + 2 ) /4 = 5 / 4 inches
n = ( 6 + 2 ) / 4 = 2 inches
n = ( 10 + 2 ) / 4 = 3 inches

ERROR ANALYSIS
In Exercises 25 and 26, describe and correct the error in solving the equation for x.

Question 25.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 87

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q25

Question 26.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 72.1

Answer:
The given literal equation is:
10 = ax – 3b
So,
ax = 10 + 3b
x = ( 10 + 3b ) / a

In Exercises 27–30, solve the formula for the indicated variable.

Question 27.
Profit: P = R – C; Solve for C.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q27

Question 28.
Surface area of a cylinder: S = 2πr2 + 2πrh; Solve for h.
Answer:
The given Surface area of a cylinder is:
S = 2πr² + 2πrh
So,
S = 2πr ( r + h )
S / 2πr = r + h
h = S / 2πr – r
Hence, from the above,
We can conclude that the value of h is: S / ( 2π

Question 29.
Area of a trapezoid: A = \(\frac{1}{2}\)h(b1 + b2); Solve for b2.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q29

Question 30.
The average acceleration of an object: Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 88; Solve for v1.
Answer:
The given average acceleration of an object is:
a = ( v1 – v0 ) / t
So,
at = v1 – v0
v1 = at + v0
Hence, from the above,
We can conclude that the value of v1 is: at + v0

Question 31.
REWRITING A FORMULA
A common statistic used in professional football is the quarterback rating. This rating is made up of four major factors. One factor is the completion rating given by the formula
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 88.1
where C is the number of completed passes and A is the number of attempted passes. Solve the formula for C.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q31

Question 32.
REWRITING A FORMULA
Newton’s law of gravitation is given by the formula
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 89
where F is the force between two objects of masses m1 and m2, G is the gravitational constant, and d is the distance between the two objects. Solve the formula for m1.
Answer:
The value of m1 is: Fd² / Gm2

Explanation:
The given Newton’s law of gravitation is given by:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 89
where
F is the force between two objects of masses m1 and m2
G is the gravitational constant
d is the distance between the two objects.
Now,
( m1m2 ) = Fd² / G
m1 = Fd² / Gm2
Hence, from the above,
We can conclude that the value of m1 is: Fd² / Gm2

Question 33.
MODELING WITH MATHEMATICS
The sale price S (in dollars) of an item is given by the formula S = L – rL, where L is the list price (in dollars) and r is the discount rate (in decimal form).
a. Solve the formula for r.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 90
b. The list price of the shirt is $30. What is the discount rate?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q33

Question 34.
MODELING WITH MATHEMATICS
The density d of a substance is given by the formula d = \(\frac{m}{V}\), where m is its mass and V is its volume.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91
a. Solve the formula for m?
Answer:
The given density d of a substance is given by:
d = \(\frac{m}{V}\)
So,
d × V = m
Hence, from the above,
We can conclude that the value of m is: d × V

b. Find the mass of the pyrite sample.
Answer:
The mass of the pyrite sample is: 6.012 gm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91
From the above figure,
The density of Pyrite = 5.01 g/cm³
The volume of Pyrite = 1.2 cm³
From part (a),
The mass of Pyrite = Density × Volume
So,
The mass of Pyrite ( m ) = 5.01 × 1.2
= 6.012 gm
Hence, from the above,
We can conclude that the mass of Pyrite is: 6.012 gm

Question 35.
PROBLEM-SOLVING
You deposit $2000 in an account that earns simple interest at an annual rate of 4%. How long must you leave the money in the account to earn $500 in interest?

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q35

Question 36.
PROBLEM-SOLVING
A flight averages 460 miles per hour. The return flight averages 500 miles per hour due to a tailwind. The total flying time is 4.8 hours. How long is each flight? Explain.
Big Ideas Math Algebra 1 Answer Key Chapter 1 Solving Linear Equations 91.1

Answer:
The time taken for flight is: 2.5 hours
The time taken for return is: 2.3 hours

Explanation:
It is given that a flight averages 460 miles per hour. The return flight averages 500 miles per hour due to a tailwind. The total flying time is 4.8 hours.
We know that,
Speed = Distance / Time
Time = Distance / Speed
It is also given that the total flying time is 4.8 hours
Let the distance be D
So,
\(\frac{D}{460}\) + \(\frac{D}{500}\) = 4.8
\(\frac{460 + 500}{230,000}\) = 4.8 / D
\(\frac{960}{230,000}\) = 4.8 / D
D = 4.8 × \(\frac{230,000}{960}\)
D = 1,150 miles
Hence,
The time taken for flight = 1,150 ÷ 460 = 2.5 hpurs
The time taken for return = 1,150 ÷ 500 = 2.3 hours

Question 37.
USING STRUCTURE
An athletic facility is building an indoor track. The track is composed of a rectangle and two semicircles, as shown.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 92
a. Write a formula for the perimeter of the indoor track.
b. Solve the formula for x.
c. The perimeter of the track is 660 feet, and r is 50 feet. Find x. Round your answer to the nearest foot.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q37

Question 38.
MODELING WITH MATHEMATICS
The distance d (in miles) you travel in a car is given by the two equations shown, where t is the time (in hours) and g is the number of gallons of gasoline the car uses.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 93
a. Write an equation that relates g and t.
Answer:
The given equations are:
d = 55t ——————— (1)
d = 20g ——————— (2)
By the law of Equality,
55t = 20g [ As the LHS for both the equations are equal, make the RHS equal ]
t / g = 20 / 55
t / g = 4 / 11

b. Solve the equation for g.
Answer:
From the given figure,
d = 20g
d = 55t
From part (a),
t / g = 4 / 11
11t = 4g
g = 11t / 4
Hence, from the above,
We can conclude that the value of g is: 11t / 4

c. You travel for 6 hours. How many gallons of gasoline does the car use? How far do you travel? Explain.
Answer:
From part (b),
g = 11t / 4
Where,
g is the number of gallons of gasoline
It is given that you travel for 6 hours
So,
t = 6 hours
Now,
g = ( 11 × 6 ) / 4
g = 66/4 gallons
Hence, from the above,
We can conclude that the number of gallons of gasoline is: 66 / 4 gallons

Question 39.
MODELING WITH MATHEMATICS
One type of stone formation found in Carlsbad Caverns in New Mexico is called a column. This cylindrical stone formation connects to the ceiling and the floor of a cave.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 94
a. Rewrite the formula for the circumference of a circle, so that you can easily calculate the radius of a column given its circumference.
b. What is the radius (to the nearest tenth of a foot) of a column that has a circumference of 7 feet? 8 feet? 9 feet?
c. Explain how you can find the area of a cross-section of a column when you know its circumference.

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q39

Question 40.
HOW DO YOU SEE IT?
The rectangular prism shown has bases’ with equal side lengths.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 95
a. Use the figure to write a formula for the surface area S of the rectangular prism.
Answer:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 95
From the above figure,
The surface area of the rectangular prism ( S) = 2 ( lb + bh + lh )
Where,
l is the length of the rectangular prism
b is the Width of the rectangular prism
h is the height of the rectangular prism

b. Your teacher asks you to rewrite the formula by solving for one of the side lengths, b or ℓ. Which side length would you choose? Explain your reasoning.
Answer:
From part (a),
The surface area of the rectangular prism ( S ) = 2 ( lb + bh + lh )
S / 2 = lb + bh + lh
S / 2 = b ( l + h ) + bh
S / 2 = b ( l + b + h )
b = S / 2 ( l + b + h )
Hence, from the above,
We can conclude that the value of b is: S / 2 ( l + b + h )

Question 41.
MAKING AN ARGUMENT
Your friend claims that Thermometer A displays a greater temperature than Thermometer B. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 96

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q41

Question 42.
THOUGHT-PROVOKING
Give a possible value for h. Justify your answer. Draw and label the figure using your chosen value of h.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 97

Answer:
The completed figure is:

The value of h is: 5 cm

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 97
From the given figure,
We can say that the geometrical figure is the parallelogram
We know that,
Area of the parallelogram = Base × Height
From the given figure,
Area of the parallelogram = 40 cm²
The base of the parallelogram = 8 cm
So,
40 = 8 × Height
Height = 40 ÷ 8
Height = 5 cm
Hence, from the above,
We can conclude that
The completed figure is:

The value of h is: 5 cm

MATHEMATICAL CONNECTIONS
In Exercises 43 and 44, write a formula for the area of the regular polygon. Solve the formula for the height h.

Question 43.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 98

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q43

Question 44.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 99

Answer:
The value of h is: A / 3b

Explanation:
The given figure is:
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 99
From the above figure,
We can say that the figure is Hexagon
Now,
The Hexagon with the six triangles is:

So,
From the figure,
There are 6 triangles
We know that,
The area of a triangle = \(\frac{1}{2}\) × Base × Height
So,
The area of the Hexagon = The area of the 6 triangles
= 6 ( \(\frac{1}{2}\) ) × Base × Height
Let,
The area of the Hexagon be A
The height of the hexagon be h
The Base of the hexagon be b
So,
A = 6 ( \(\frac{1}{2}\) ) × Base × Height
A = 3 × Base × Height
Base × Height = A / 3
Height = A / ( 3 × Base )
So,
h = A / 3b
Hence, from the above,
We can conclude that the value of h is: A / 3b

REASONING
In Exercises 45 and 46, solve the literal equation for a.

Question 45.
Big Ideas Math Algebra 1 Solutions Chapter 1 Solving Linear Equations 100

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q45

Question 46.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 101
Answer:
The value of a is: \(\frac{by}{y – bx}\)

Explanation:
The given literal equation is:
y = x [ \(\frac{ab}{a – b}\)
\(\frac{ab}{a – b}\) = y / x
x ( ab ) = y ( a – b )
abx = ay – by
by = ay – abx
by = a ( y – bx )
a = \(\frac{by}{y – bx}\)
Hence, from the above,
We can conclude that the value of a is: \(\frac{by}{y – bx}\)

Maintaining Mathematical Proficiency

Evaluate the expression.

Question 47.
15 – 5 + 52

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q47

Question 48.
18 • 2 – 42 ÷ 8
Answer:
The given expression is:
18 ⋅ 2 – 4² ÷ 8
We have to remember that,
When there is an expression to solve with multiple mathematical symbols, we have to follow the BODMAS rule
BODMAS indicates the hierarchy we have to follow when we will solve mathematical symbols
In BODMAS,
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
So,
18 ⋅ 2 – 4² ÷ 8 = 18 ⋅ 2 – ( 4 × 4 ) ÷ 8
= 18 ⋅ 2 – 2
= 36 – 2
= 34

Question 49.
33 + 12 ÷ 3 • 5

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q49

Question 50.
25(5 – 6) + 9 ÷ 3
Answer:
The given expression is:
25(5 – 6) + 9 ÷ 3
We have to remember that,
When there is an expression to solve with multiple mathematical symbols, we have to follow the BODMAS rule
BODMAS indicates the hierarchy we have to follow when we will solve mathematical symbols
In BODMAS,
B – Brackets
O – Of
D – Division
M – Multiplication
A – Addition
S – Subtraction
So,
25(5 – 6) + 9 ÷ 3 = ( 2 × 2 × 2 × 2 × 2 ) ( 5 – 6 ) + ( 9 ÷ 3 )
= ( 2 × 2 × 2 × 2 × 2 ) ( 5 – 6 ) + 3
= ( 2 × 2 × 2 × 2 × 2 ) ( -1 ) + 3
= -( 2 × 2 × 2 × 2 × 2 )  + 3
= -32 + 3
= -29

Solve the equation. Graph the solutions, if possible. (Section 1.4)

Question 51.
| x – 3 | + 4 = 9

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q51

Question 52.
| 3y – 12 | – 7 = 2
Answer:
The values of y are: 7 and 1

Explanation:
The given absolute value equation is:
| 3y – 12 | – 7 = 2
| 3y – 12 | = 2 + 7
| 3y – 12 | = 9
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
3y – 12 = 9                      3y – 12 = -9
3y = 9 + 12                     3y = -9 + 12
3y = 21                            3y = 3
y = 21 / 3                         y = 3 / 3
y = 7                                y = 1
Hence, from the above,
We can conclude that the values of y are: 7 and 1

Question 53.
2 | 2r + 4 | = -16

Answer:
Big-Ideas-Math-Algebra-1-Answers-Chapter-1-Solving-Linear-Equations-Lesson-1.5-Q53

Question 54.
-4 | s + 9 | = -24
Answer:
The value of s is: -3 and -15

Explanation:
The given absolute value equation is:
-4 | s + 9 | = -24
| s + 9 | = -24 ÷ ( -4 )
| s + 9 | = 6 [ Since – ÷ – = + ]
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
s + 9 = 6                                       s + 9 = -6
s = 6 – 9                                         s = -6 – 9
s = -3                                             s = -15
Hence, from the above,
We can conclude that the values of s are: -3 and -15

Solving Linear Equations Performance Task: Magic of Mathematics

1.4–1.5 What Did You Learn?

Core Vocabulary
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 102

Core Concepts
Section 1.4
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 103
Section 1.5
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 104

Mathematical Practices

Question 1.
How did you decide whether your friend’s argument in Exercise 46 on page 33 made sense?
Answer:
On page 33 in Exercise 46,
The given absolute equation is:
| 3x + 8 | – 9 = -5
| 3x + 8 | = -5 + 9
| 3x + 8 | = 4
So, from the absolute equation,
We can say that the given absolute value equation has a solution
But, according to your friend,
The argument is that the absolute value equation has no solution

Question 2.
How did you use the structure of the equation in Exercise 59 on page 34 to rewrite the equation?
Answer:
The given absolute value equation in Exercise 59 on page 34 is:
8 | x + 2 | – 6 = 5 | x + 2 | + 3
The above equation can be re-written as:
8 | x + 2  | – 5  | x + 2 | = 3 + 6
3 | x + 2  | = 9
Hence, from the above,
We can conclude that the re-written form of the given absolute value equation is:
3 | x + 2 | = 9

Question 3.
What entry points did you use to answer Exercises 43 and 44 on page 42?
Answer:
In Exercises 43 and 44 on page 42,
We used the triangles as an entry point
In Exercise 43,
The given figure is a pentagon
Using the above entry point,
We divided the pentagon into 5 triangles
In Exercise 44,
The given figure is a Hexagon
Using the above entry point,
We divided the hexagon into 6 triangles.

Performance Task

Magic of Mathematics

Have you ever watched a magician perform a number trick? You can use algebra to explain how these types of tricks work.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 105
To explore the answers to these questions and more, go to Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 106

Solving Linear Equations Chapter Review

1.1 Solving Simple Equations (pp. 3–10)

a. Solve x − 5 = −9. Justify each step.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 106.1

b. Solve 4x = 12. Justify each step.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 107

Solve the equation. Justify each step. Check your solution.

Question 1.
z + 3 = -6
Answer:
The value of z is: -9

Explanation:
The given equation is:
z + 3 = -6
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
z = -6 – ( +3 )
z = -6 – 3
z = -9
Hence, from the above,
We can conclude that the value of z is: -9

Question 2.
2.6 = -0.2t
Answer:
The value of t is: -13

Explanation:
The given equation is:
2.6 = -0.2t
\(\frac{26}{10}\) = –\(\frac{2}{10}\)t
t = \(\frac{26}{10}\) ÷ ( –\(\frac{2}{10}\) )
t = – \(\frac{26}{10}\) × \(\frac{10}{2}\)
t = -13
Hence, from the above,
We can conclude that the value of t is: -13

Question 3.
– \(\frac{n}{5}\) = -2
Answer:
The value of n is: 10

Explanation:
The given equation is:
–\(\frac{n}{5}\) = -2
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-n = -2 × 5
-n = -10
n = 10
Hence, from the above,
We can conclude that the value of n is: 10

1.2 Solving Multi-Step Equations (pp. 11–18)

Solve −6x + 23 + 2x = 15.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 108

Solve the equation. Check your Solution.

Question 4.
3y + 11 = -16
Answer:
The value of y is: -9

Explanation:
The given equation is:
3y + 11 = -16
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3y = -16 – 11
3y = -27
y = -27 ÷ 3
y = -9
Hence, from the above,
We can conclude that the value of y is: -9

Question 5.
6 = 1 – b
Answer:
The value of b is: -5

Explanation:
The given equation is:
6 = 1 – b
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
b = 1 – 6
b = -5
Hence, from the above,
We can conclude that the value of b is: -5

Question 6.
n + 5n + 7 = 43
Answer:
The value of n is: 6

Explanation:
The given equation is:
n + 5n + 7 = 43
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6n + 7 = 43
6n = 43 – 7
6n = 36
n = 36 ÷ 6
n = 6
Hence, from the above,
We can conclude that the value of n is: 6

Question 7.
-4(2z + 6) – 12 = 4
Answer:
The value of z is: -5

Explanation:
The given equation is:
-4 ( 2z + 6 ) – 12 = 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
-4 ( 2z + 6 ) = 4 + 12
-4 ( 2z + 6 ) = 16
-4 ( 2z ) – 4 ( 6 ) = 16
-8z – 24 = 16
-8z = 16 + 24
-8z = 40
z = 40 ÷ ( -8 )
z = -5
Hence, from the above,
We can conclude that the value of z is: -5

Question 8.
\(\frac{3}{2}\)(x – 2) – 5 = 19
Answer:
The value of x is: 18

Explanation:
The given equation is:
\(\frac{3}{2}\) ( x – 2 ) – 5 = 19
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
\(\frac{3}{2}\) ( x – 2 ) = 19 + 5
\(\frac{3}{2}\) ( x – 2 ) = 24
x – 2 = 24 × \(\frac{2}{3}\)
x – 2 = \(\frac{24}{1}\) × \(\frac{2}{3}\)
x – 2 = 16
x = 16 + 2
x = 18
Hence, from the above,
We can conclude that the value of x is: 18

Question 9.
6 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w – 4
Answer:
The value of w is: \(\frac{25}{4}\)

Explanation:
The given equation is:
6 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w – 4
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
6 + 4 = \(\frac{1}{5}\)w + \(\frac{7}{5}\)w
10 = w [ \(\frac{1 + 7}{5}\) ]
10 = \(\frac{8}{5}\)w
w = 10 × \(\frac{5}{8}\)
w = \(\frac{10}{1}\) × \(\frac{5}{8}\)
w = \(\frac{25}{4}\)
Hence, from the above,
We can conclude that the value of w is: \(\frac{25}{4}\)

Find the value of x. Then find the angle measures of the polygon.

Question 10.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 109

Answer:
The angle measures of the given polygon are: 110°, 50°, 20°

Explanation:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 109
From the above figure,
The angle measures are: 110°, 5x°, 2x°
It is also given that
The sum of the angle measures = 180°
So,
110 + 5x + 2x = 180°
7x = 180 – 110
7x = 70°
x = 70 / 7
x = 10°
Hence, from the above,
We can conclude that the angle measures of the given polygon are: 110°, 50°, 20°

Question 11.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 110

Answer:
The angle measures of the given polygon are: 126°, 126°, 96°, 96°, 96°

Explanation:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 110
From the above figure,
The angle measures of the given polygon are: ( x – 30 )°, x°, x°, ( x – 30 )°, ( x – 30 )°
It is also given that,
The sum of the angle measures of the given polygon = 540°
So,
( x – 30 )° + x° + x° + ( x – 30 )° + ( x – 30 )° = 540°
5x – 90° = 540°
5x = 540° + 90°
5x = 630°
x = 630 / 5
x = 126°
Hence,f rom the above,
We can conclude that the angle measures of the given polygon are: 126°, 126°, 96°, 96°, 96°

1.3 Solving Equations with Variables on Both Sides (pp. 19–24)

Solve 2( y − 4) = −4( y + 8).
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 111

Solve the equation.

Question 12.
3n – 3 = 4n + 1
Answer:
The value of n is: -4

Explanation:
The given equation is:
3n – 3 = 4n + 1
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4n – 3n = -1 – 3
n = -4
Hence, from the above,
We can conclude that the value of n is: -4

Question 13.
5(1 + x) = 5x + 5
Answer:
The given equation has no solution

Explanation:
The given equation is:
5 ( 1 + x ) = 5x + 5
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
5 ( 1 ) + 5 ( x ) = 5x + 5
5 + 5x = 5x + 5
5 = 5x – 5x + 5
5 = 5
Hence, from the above,
We can conclude that the given equation has no solution

Question 14.
3(n + 4) = \(\frac{1}{2}\)(6n + 4)
Answer:
The given equation has no solution

Explanation:
The given equation is:
3 ( n + 4 ) = \(\frac{1}{2}\) ( 6n + 4 )
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
3 ( n ) + 3 ( 4 ) = \(\frac{1}{2}\) ( 6n + 4 )
3n + 12 = \(\frac{1}{2}\) ( 6n + 4 )
2 ( 3n + 12 ) = 6n + 4
2 ( 3n ) + 2 ( 12 ) = 6n + 4
6n + 24 = 6n + 4
24 = 6n – 6n + 4
24 = 4
Hence, from the above,
We can conclude that the given equation has no solution

1.4 Solving Absolute Value Equations (pp. 27–34)

a. Solve | x − 5 | = 3.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 112

b. Solve | 2x + 6 | = 4x. Check your solutions.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 113

Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 114

Check the apparent solutions to see if either is extraneous.
The solution is x = 3. Reject x = -1 because it is extraneous.

Solve the equation. Check your solutions.

Question 15.
| y + 3 | = 17
Answer:
The value of y is: 14 or -20

Explaantion:
The given absolute value equation is:
| y + 3 | = 17
We know that,
| x | = x for x  0
| x | = -x for x < 0
So,
y + 3 = 17                               y + 3 = -17
y = 17 – 3                                y = -17 – 3
y = 14                                      y = -20
Hence, from the above,
We can conclude that the value of y is: 14 or -20

Question 16.
-2 | 5w – 7 | + 9 = -7
Answer:
The value of w is: 3 or –\(\frac{1}{5}\)

Explanation:
The given absolute value equation is:
-2 | 5w – 7 | + 9 = -7
-2 | 5w – 7 | = -7 – 9
-2 | 5w – 7 | = -16
| 5w – 7 | = -16 / ( -2 )
| 5w – 7 | = 8
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
5w – 7 = 8                                     5w – 7 = -8
5w = 8 + 7                                    5w = -8 + 7
5w = 15                                         5w = -1
w = 15 ÷ 5                                     w = –\(\frac{1}{5}\)
w = 3                                              w =-\(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of w is: 3 or –\(\frac{1}{5}\)

Question 17.
| x – 2 | = | 4 + x |
Answer:
The given absolute equation has no solution

Explanation:
The given absolute value equation is:
| x – 2 | = | 4 + x |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 2 = 4 + x                                                   – ( x – 2 ) = – ( 4 + x )
-2 = 4                                                                2 = -4
Hence, from the above,
We can conclude that the given absolute equation has no solution

Question 18.
The minimum sustained wind speed of a Category 1 hurricane is 74 miles per hour. The maximum sustained wind speed is 95 miles per hour. Write an absolute value equation that represents the minimum and maximum speeds.
Answer:
The absolute value equation that represents the minimum and maximum speeds is:
| x – 84.5 | = 9.5

Explanation:
It is given that the minimum sustained wind speed of a Category 1 hurricane is 74 miles per hour. The maximum sustained wind speed is 95 miles per hour.
So,
The average wind speed sustained = ( 74 + 95 ) /2
= 169 / 2
= 84.5 miles per hour
Now,
The minimum wind speed from the average speed point = 84.5 – 74
= 9.5 miles per hour
So,
The absolute value equation that represents the minimum and maximum wind speed is:
| x – 84.5 | = 9.5

1.5 Rewriting Equations and Formulas (pp. 35–42)
a. The slope-intercept form of a linear equation is y = mx + b. Solve the equation for m.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 115

b. The formula for the surface area S of a cylinder is S = 2πr2 + 2πrh. Solve the formula for the height h.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 116

Solve the literal equation for y.

Question 19.
2x – 4y = 20
Answer:
The value of y is: ( x / 2 ) – 5

Explanation:
The given literal equation is:
2x – 4y = 20
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 2x – 20
y = ( 2x – 20 ) / 4
y = ( 2x / 4 ) – ( 20 / 4 )
y = ( x / 2 ) – 5
Hence, from the above,
We can conclude that the value of y is: ( x / 2 ) – 5

Question 20.
8x – 3 = 5 + 4y
Answer:
The value of y is: 2x – 2

Explanation:
The given literal equation is:
8x – 3 = 5 + 4y
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
4y = 8x – 3 – 5
4y = 8x – 8
y = ( 8x – 8 ) / 4
y = ( 8x / 4 ) – ( 8 – 4 )
y = 2x – 2
Hence, from the above,
We can conclude that the value of y is: 2x – 2

Question 21.
a a = 9y + 3yx
Answer:
The value of y is: a² / ( 3x + 9 )

Explanation:
The given literal equation is:
a² = 9y + 3yx
When we convert any sign from LHS, then the sign will be converted into the opposite sign in RHS,
So,
+ in LHS is converted into – in RHS and vice-versa
× in LHS is converted into ÷ in RHS and vice-versa
So,
a² = y ( 3x + 9 )
y = a² / ( 3x + 9 )
Hence, from the above,
We can conclude that the value of y is: a² / ( 3x + 9 )

Question 22.
The volume V of a pyramid is given by the formula V = \(\frac{1}{3}\)Bh, where B is the area of the base and h is the height.
a. Solve the formula for h.
b. Find the height h of the pyramid.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117

Answer:
a)
The value of h is: \(\frac{3V}{B}\)

Explanation:
The given formula is:
V = \(\frac{1}{3}\)Bh
Where,
B is the area of the base
h is the height
Now,
3V = Bh
h = \(\frac{3V}{B}\)
Hence, from the above,
We can conclude that the value of h is: \(\frac{3V}{B}\)

b)
The value of h is: 18 cm

Explanation:
From the given figure,
Area of the base ( B ) = 36 cm²
Volume of the base ( V ) = 216 cm³
From part (a),
h = \(\frac{3V}{B}\)
h = \(\frac{3 × 216}{36}\)
h = \(\frac{3 × 216}{36 × 1}\)
h = 18 cm
Hence, from the above,
We can conclud ethat the value of h is: 18 cm

Question 23.
The formula F = \(\frac{9}{5}\)(K – 273.15) + 32 converts a temperature from kelvin K to degrees Fahrenheit F.
a. Solve the formula for K.
b. Convert 180°F to kelvin K. Round your answer to the nearest hundredth.
Answer:
a)
The formula for K is:
K = \(\frac{5}{9}\) ( F – 32 ) + 273.15

Explanation:
The given formula for F is:
F = \(\frac{9}{5}\) ( K – 273.15 ) + 32
Now,
F – 32 = \(\frac{9}{5}\) ( K – 273.15 )
\(\frac{5}{9}\) ( F – 32 ) = K – 273.15
K = \(\frac{5}{9}\) ( F – 32 ) + 273.15
Hence, from the above,
We can conclude that the value of K is: \(\frac{5}{9}\) ( F – 32 ) + 273.15

Solving Linear Equations Chapter Test

Solve the equation. Justify each step. Check your solution.

Question 1.
x – 7 = 15
Answer:
The value of x is: 22

Explanation:
The given equation is:
x – 7 = 15
Now,
x = 15 + 7
x = 22
Hence, from the above,
We can conclude that the value of x is: 22

Question 2.
\(\frac{2}{3}\)x = 5
Answer:
The value of x is: \(\frac{15}{2}\)

Explanation:
The given equation is:
\(\frac{2}{3}\) x = 5
Now,
x = 5 × \(\frac{3}{2}\)
x = \(\frac{5}{1}\) × \(\frac{3}{2}\)
x = \(\frac{15}{2}\)
Hence, from the above,
We can conclude that the value of x is: \(\frac{15}{2}\)

Question 3
11x + 1 = -1 + x
Answer:
The value of x is: –\(\frac{1}{5}\)

Explanation:
The given equation is:
11x + 1 = -1 + x
Now,
11x – x = -1 – 1
10x = -2
x = –\(\frac{2}{10}\)
x = –\(\frac{1}{5}\)
Hence, from the above,
We can conclude that the value of x is: –\(\frac{1}{2}\)

Solve the equation.

Question 4.
2 | x – 3 | – 5 = 7
Answer:
The value of x is: 9 or -3

Explanation:
The given absolute value equation is:
2 | x – 3 | – 5 = 7
2 | x – 3 | = 7 + 5
2 | x – 3 | = 12
| x – 3 | = \(\frac{12}{2}\)
| x – 3 | = 6
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
x – 3 = 6                             x – 3 = -6
x = 6 + 3                            x = -6 + 3
x = 9                                   x = -3
Hence, from the above,
We can conclude that the value of x is: 9 or -3

Question 5.
| 2x – 19 | = 4x + 1
Answer:
The value of x is: -10 or 3

Explanation:
The given absolute value equation is:
| 2x – 19 | = 4x + 1
We know that,
| x | = x for x > 0
| x | = -x for x < 0
4x + 1 = 2x – 19                         4x + 1 = – ( 2x – 19 )
4x – 2x = -19 – 1                         4x + 2x = 19 – 1
2x = -20                                      6x = 18
x = \(\frac{-20}{2}\)      x = \(\frac{18}{6}\)
x = -10                                         x = 3
Hence, from the above,
We can conclude that the value of x is: -10 or 3

Question 6.
-2 + 5x – 7 = 3x – 9 + 2x
Answer:
The given absolute equation has no solution

Explanation:
The given equation is:
-2 + 5x – 7 = 3x – 9 + 2x
5x – 9 = 5x – 9
Hence, from the above,
We can conclude that the given absolute value equation has no solution

Question 7.
3(x + 4) – 1 = -7
Answer:
The value of x is: -6

Explanation:
The given equation is:
3 ( x + 4 ) – 1 = -7
So,
3 ( x ) + 3 ( 4 ) = -7 + 1
3x + 12 = -6
3x = -6 – 12
3x = -18
x = –\(\frac{18}{3}\)
x = -6
Hence, from the above,
We can conclude that the value of x is: -6

Question 8.
| 20 + 2x | = | 4x + 4 |
Answer:
The value of x is: 8

Explanation:
The given absolute value equation is:
| 20 + 2x | = | 4x + 4 |
We know that,
| x | = x for x > 0
| x | = -x for x < 0
So,
20 + 2x = 4x + 4
4x – 2x = 20 – 4
2x = 16
x = \(\frac{16}{2}\)
x = 8
Hence, from the above,
We can conclude that the value of x is: 8

Question 9.
\(\frac{1}{3}\)(6x + 12) – 2(x – 7) = 19
Answer:
The given equation has no solution

Explanation:
The given equation is:
\(\frac{1}{3}\) ( 6x + 12 ) – 2 ( x – 7 ) = 19
Now,
\(\frac{1}{3}\) ( 6x – 12 ) = 19 + 2 ( x – 7 )
\(\frac{1}{3}\) ( 6x – 12 ) = 19 + 2x – 14
\(\frac{1}{3}\) ( 6x – 12 ) = 2x + 5
1 ( 6x – 12 ) = 3 ( 2x + 5 )
6x – 12 = 6x + 15
6x – 6x = 15 + 12
15 = -12
Hence, from the above,
We can conclude that the given equation has no solution

Describe the values of c for which the equation has no solution. Explain your reasoning.

Question 10.
3x – 5 = 3x – c
Answer:
The value of c is: 5

Explanation:
The given equation is:
3x – 5 = 3x – c
It is given that the equation has no solution
So,
3x – 3x – 5 =-c
-c = -5
c = 5
Hence, from the above,
We can conclude that the value of c is: 5

Question 11.
| x – 7 | = c
Answer:
The value of c is: -7

Explanation:
The given absolute value equation is:
| x – 7 | = c
It is given that the equation has no solution i.e., x = 0
So,
0 – 7 = c
c = -7
Hence, from the above,
We can conclude that the value of c is: -7

Question 12.
A safety regulation states that the minimum height of a handrail is 30 inches. The maximum height is 38 inches. Write an absolute value equation that represents the minimum and maximum heights.
Answer:
The absolute value expression that represents the minimum and maximum heights is:
| x – 64 | = 34

Explanation:
It is given that a safety regulation states that the minimum height of a handrail is 30 inches. The maximum height is 38 inches.
So,
The average height of a handrail = ( 30 + 38 ) / 2
= 68 / 2
= 34 inches
Now,
The minimum height from the average height of a handrail = 34 + 30
= 64 inches
Hence,
The absolute value equation that represents the minimum and maximum height of a handrail is:
| x – 64 | = 34

Question 13.
The perimeter P (in yards) of a soccer field is represented by the formula P = 2ℓ + 2w, where ℓ is the length (in yards) and w is the width (in yards).
a. Solve the formula for w.
Answer:
The given formula is:
P = 2l + 2w
Where,
P is perimeter ( in yards )
l is the length ( in yards )
w is the width ( in yards )
So,
2w = P – 2l
w = ( P – 2l ) / 2
Hence, from the above,
We can conclude that the formula for w is:
w = ( P – 2l ) / 2

b. Find the width of the field.
Answer:
The given figure is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117.1
From the above figure,
Perimeter ( P ) = 330 yd
Length ( l) = 100 yd
From part (a),
w = ( P – 2l ) /2
w = ( 330 – 100 ) / 2
w = 230 / 2
w = 115 yd
Hence, from the above,
We can conclude that the width of the field is: 115 yd

c. About what percent of the field is inside the circle?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 117.1
Answer:

Question 14.
Your car needs new brakes. You call a dealership and a local mechanic for prices.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 118
a. After how many hours are the total costs the same at both places? Justify your answer.
Answer:
From the given table,
Let the total labor hours be x
So,
The total cost at the Dealership = ( Cost of parts ) + ( Labor cost per hour ) × ( Total labor hours )
= 24 + 99x
The total cost at the local mechanic = ( Cost of parts ) + ( Labor cost per hour ) × ( Total labor hours )
= 45 + 89x
It is given that the total cost is the same in both places
So,
24 + 99x = 45 + 89x
99x – 89x = 45 – 24
10x = 21
x = 21 / 10
x = 2.1 hours
Hence, from the above,
We can conclude that after 2.1 hours, the total cost will be the same in both places

b. When do the repairs cost less at the dealership? at the local mechanic? Explain.
Answer:
The given table is:
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 118
From the above table,
Compare the labor cost per hour
By comparison,
We can say that the labor cost per hour is less at the local mechanic
Hence, from the above,
We can conclude that the repair cost less at the local mechanic

Question 15.
Consider the equation | 4x + 20 | = 6x. Without calculating, how do you know that x = -2 is an extraneous solution?
Answer:
We know that,
The absolute value equations only accept the values greater than or equal to 0
Hence,
For the given absolute value equation,
| 4x + 20 | = 6x
x = -2 is an extraneous solution

Question 16.
Your friend was solving the equation shown and was confused by the result “-8 = -8.” Explain what this result means.
4(y – 2) – 2y = 6y – 8 – 4y
4y – 8 – 2y = 6y – 8 – 4y
2y – 8 = 2y – 8
-8 = -8
Answer:
The result ” -8 = -8 ” means that the solved equation has no solution

Solving Linear Equations Cumulative Assessment

Question 1.
A mountain biking park has 48 trails, 37.5% of which are beginner trails. The rest are divided evenly between intermediate and expert trials. How many of each kind of trail are there?
A. 12 beginner, 18 intermediate, 18 expert
B. 18 beginner, 15 intermediate, 15 expert
C. 18 beginner, 12 intermediate, 18 expert
D. 30 beginner, 9 intermediate, 9 expert

Answer:
The correct option is: B
The number of beginner trials is: 18
The number of intermediate trials is: 15
The number of expert trials is: 15

Explanation:
It is given that a mountain biking park has 48 trails, 37.5% of which are beginner trails. The rest are divided evenly between intermediate and expert trials.
So,
The number of beginner trials is 3.5 % of the total number of trials
It is given that the total number of trials is: 48
We know that,
The value of 37.5 % is: \(\frac{3}{8}\) [ 37.5 % = 50 % – 12.5 % ]
So,
The number of beginner trials = \(\frac{3}{8}\) × 48
= \(\frac{3}{8}\) × \(\frac{48}{1}\)
= \(\frac{3 × 48}{8 × 1}\)
= 18
So,
The number  of intermediate and expert trials = ( The total number of trials ) – ( The number of beginner trials )
= 48 – 18
= 30 trials
It is also given that the intermediate trials and expert trails are divided evenly
So,
30 ÷ 2 = 15 trials each
Hence, from the above,
We can conclude that
The number of beginner trials is: 18
The number of intermediate trials is: 15
The number of expert trials is: 15

Question 2.
Which of the equations are equivalent to cx – a = b?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 119

Answer:
The given equations are:
a) cx – a + b = 2b
b) 0 = cx – a + b
c) 2cx – 2a = b / 2
d) x – a = b / 2
e) x = ( a + b ) / c
f) b + a = cx
Now,
We have to find the equations from above that is equivalent to the given equation cx – a = b
Now,
a)
The given equation is:
cx – a + b = 2b
So,
cx – a = b – b
cx – a = b
b)
The given equation is:
0 = cx – a + b
So,
cx – a = -b
c)
The given equation is:
2cx – 2a = b / 2
So,
cx – a = b / 4
d )
The given equation is:
cx – a = b / 2
So,
2 ( cx – a ) = b
e)
The given equation is:
x = ( a + b ) / c
So,
cx = a + b
cx – a = b
f)
The given equation is:
b + a = cx
So,
cx – a = b
Hence, from the above,
We can conclude that the equations that are equivalent to cx – a = b is: a, e, f

Question 3.
Let N represent the number of solutions of the equation 3(x – a) = 3x – 6. Complete each statement with the symbol <, >, or =.
a. When a = 3, N ____ 1.
b. When a = -3, N ____ 1.
c. When a = 2, N ____ 1.
d. When a = -2, N ____ 1.
e. When a = x, N ____ 1.
f. When a = -x, N ____ 1.

Answer:
The given equation is:
3 ( x – a ) = 3x – 6
So,
3x – 3a = 3x – 6
Now,
a) When a = 3,
3x – 3 ( 3 ) = 3x – 6
3x – 9 = 3x – 6
9 = 6
Hence,
When a = 3 there is no solution
Hence,
N < 1
b) When a = -3
3x + 3 ( 3 ) =3 x – 6
9 = -6
Hence,
When a = -3, there is no solution
Hence,
N < 1
c) When a = 2
3x – 3 ( 2 ) = 3x – 6
3x – 6 =3x –
6 = 6
Hence,
When a= 2, there is no solution
Hence,
N < 1
d) When a = -2
3x + 3 ( 2 ) = 3x – 6
3x + 6 = 3x – 6
6 = -6
Hence,
When a = -2, thereis no solution
Hence,
N < 1
e) When a = x
3x – 3 ( x ) = 3x – 6
3x = 6
x = 6 / 3
x = 2
Hence,
When a  x, theer is 1 solution
Hence,
N = 1
f) When a = -x
3x + 3 ( x ) = 3x – 6
6x – 3x = -6
3x = -6
x = -6 / 3
x = -2
Hence,
When a = -x, there is 1 solution
Hence,
N = 1

Question 4.
You are painting your dining room white and your living room blue. You spend $132 on 5 cans of paint. The white paint costs $24 per can, and the blue paint costs $28 per can.
a. Use the numbers and symbols to write an equation that represents how many cans of each color you bought.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 120
Answer:
The number of cans of white paint is: 2
The number of cans of blue paint is: 3

Explanation:
It is given that you spend $132 on 5 cans of paint
It is also given that the white paint costs $24 per can and the blue paint costs $28 per can
Now,
Let
The number of white cans is: x
The number of blue cans be: 5 – x
So,
The total cost of paint = ( The number of white cans ) × ( The cost of white paint per can ) + ( The number of blue cans ) × ( The cost of blue paint per can )
132 = 24x + 28 ( 5 – x )
24x + 28 ( 5 ) – 28x = 132
140 – 4x = 132
4x = 140 – 132
4x = 8
x = 8 ÷ 4
x = 2
Hence, from the above,
We can conclude that
The number of cans of white paint is: 2
The number of cans of blue paint is: 3

b. How much would you have saved by switching the colors of the dining room and living room? Explain.
Answer:
The money you have saved by switching the colors of the dining room and living room is: $0

Explanation:
From part (a),
The number of white cans is: 2
The number of blue cans is: 3
It is given that white color is used in the dining room and the blue color is used in the living room
So,
The cost of white paint after interchanging the color in the living room = 24 × 2
= $48
The cost of blue paint after interchanging the color in the dining room = 28 × 3
= $84
So,
The total cost of paint after interchanging the colors = 48 + 84
= $132
Hence,
The amount of money saved = ( The money you paid for the paint before interchanging ) – ( The money you paid for the paint after interchanging )
= 132 – 132
=$0

Question 5.
Which of the equations are equivalent?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 121

Answer:
The given equations are:
a ) 6x + 6 = -14
b ) 8x + 6 = -2x – 14
c ) 5x + 3 = -7
d ) 7x + 3 = 2x – 13
Now,
To find the equivalent equations, find the value of x
So,
a)
The given equation is:
6x + 6 = -14
6x = -14 – 6
6x = -20
x = -20 / 6
x = -10 / 3
b)
The given equation is:
8x + 6 = -2x – 14
8x + 2x = -14 – 6
10x = -20
x = -20 / 10
x = -2
c)
The given equation is:
5x + 3 = -7
5x = -7 -3
5x = -10
x= -10 / 5
x = -2
d)
The given equation is:
7x + 3 = 2x – 13
7x – 2x = -13 – 3
5x = -16
x = -16 / 5
Hence, from the above,
We can conclude that the equations c) and d) are equivalent

Question 6.
The perimeter of the triangle is 13 inches. What is the length of the shortest side?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 122
Answer:
The length of the shortest side is: 3 inches

Explanation:
We know that,
The perimeter is the sum of all the sides of the given figure
It is given that the perimeter of the triangle is: 13 inches
So,
The perimeter of the triangle = ( x – 5 ) + ( x / 2 ) + 6
13 = x + 1 + ( x / 2 )
13 = ( 2x / 2 ) + ( x / 2 ) + 1
3x / 2 = 13 – 1
3x / 2 = 12
3x = 12 × 2
3x = 24
x = 24 / 3
x = 8
So,
The lengths of all sides are: ( 8 – 5 ), 6, ( 8 / 2 ) = 3 inches, 6 inches, 4 inches
Hence, from the above,
We can conclude that the length of the shortest side is: 3 inches

Question 7.
You pay $45 per month for cable TV. Your friend buys a satellite TV receiver for $99 and pays $36 per month for satellite TV. Your friend claims that the expenses for a year of satellite TV are less than the expenses for a year of cable.
a. Write and solve an equation to determine when you and your friend will have paid the same amount for TV services.
Answer:
It is given that you pay $45 per month for cable TV. Your friend buys a satellite TV receiver for $99 and pays $36 per month for satellite TV
So,
Let the number of months be x
Now,
The time they paid the same amount for TV services is:
45x = 99 + 36x
45x – 36x = 99
9x= 99
x = 99 / 9
x = 11
Hence, from the above,
We can conclude that after 11 months, you and your friend will pay the same amount for TV services

b. Is your friend correct? Explain.
Answer:
Your friend is correct

Explanation:
We know that,
1 year = 12 months
So,
The expenses paid by you for TV services = 45x = 45 × 12 = $540
The expenses paid by your friend for TV services = 99 + 36x
= 99 + 36 ( 12 )
= 99 + 432
= $531
By comparing the expenses of you and your friend,
We can conclude that your friend is correct

Question 8.
Place each equation into one of the four categories.
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 123

Question 9.
A car travels 1000 feet in 12.5 seconds. Which of the expressions do not represent the average speed of the car?
Big Ideas Math Answer Key Algebra 1 Chapter 1 Solving Linear Equations 124

Answer:
We know that,
Average speed = ( Distance ) ÷ ( Time )
It is given that a car travels 1000 feet in 12.5 seconds
So,
Average speed = 1000 / 12.5
= 80\(\frac{feet}{second}\)
Now,
The given options are:
A) 80\(\frac{second}{feet}\) B) 80\(\frac{feet}{second}\) C) \(\frac{80 feet}{second}\)
D) \(\frac{second}{ 80 feet}\)
Hence, from the above,
We can conclude that option B) represents the average speed

Big Ideas Math Answers Grade 7 Chapter 7 Probability

In order to solve real-life mathematical problems, students must understand the concept behind the question. You have to know what is probability. This is the most interesting chapter among all the grade 7 chapters. Download Free Pdf of Big Ideas Math Answers Grade 7 Chapter 7 Probability to practice the exercise and homework problems.

Big Ideas Math Book 7th Grade Answer Key Chapter 7 Probability

Improve your math skills with the help of Big Ideas Math Answers Grade 7 Chapter 7 Probability. Unlimited practice with all the question and answers along with the practice test. It is mandatory to practice with the Big Ideas Math Answer Key for Grade 7 Chapter 7 Probability to secure maximum marks in the exams. Check out the links given the below sections before you start your preparation.

Performance Task

Lesson: 1 Probability

Lesson: 2 Experimental and Theoretical Probability

Lesson: 3 Compound Events

Lesson: 4 Simulations

Chapter: 7 – Probability 

Probability STEAM Video/Performance Task

STEAM Video

Massively Multiplayer Rock Paper Scissors.
You can use experimental probability to describe the percent of times that you win, lose, or tie in Rock Paper Scissors. Describe a real-life situation where it is helpful to describe the percent of times that a particular outcome occurs.

Watch the STEAM Video “Massively Multiplayer Rock Paper Scissors.”Then answer the following questions.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 1
1. The table shows the ways that you can win, lose, or tie in Rock Paper Scissors. You and your opponent throw the signs for rock, paper, or scissors at random. What percent of the time do you expect to win? lose? tie?
2. You play Rock Paper Scissors 15 times. About how many times do you expect to win? Explain your reasoning.

Performance Task

Fair and Unfair Carnival Games
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 2
You will be given information about a version of Rock Paper Scissors used at a carnival. Then you will be asked to design your own “unfair” carnival game using a spinner or a number cube, and test your game with a classmate.
In what ways can a game of chance be considered fair? unfair? Explain your reasoning.

Probability Getting Ready for Chapter 7

Chapter Exploration
Work with a partner.
1. Play Rock Paper Scissors 30 times. Tally your results in the table.
2. How many possible results are there?
3. Of the possible results, in how many ways can Player A win? In how many ways can Player B win? In how many ways can there be a tie?
4. Is one of the players more likely to win than the other player? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 3

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
probability
theoretical probability
simulation
relative frequency
sample space
experimental probability
compound event

Lesson 7.1 Probability

EXPLORATION 1

Determining Likelihood
Work with a partner. Use the spinners shown.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 1
a. For each spinner, determine which numbers you are more likely to spin and which numbers you are less likely to spin. Explain your reasoning.
b. Spin each spinner 20 times and record your results in two tables. Do the data support your answers in part(a)? Explain why or why not.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 2
c. How can you use percents to describe the likelihood of spinning each number? Explain.
Answer:

Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 3

Try It

Question 1.
You randomly choose one of the tiles shown from a hat.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 4
a. How many possible outcomes are there?

Answer:
There are 5 number of unique possible outcomes
They are { A,B,C,D,E}
Therefore, there are 5 possible outcomes for the event.

b. What are the favorable outcomes of choosing a vowel ?

Answer:
There are 2 favorable outcomes of choosing an vowel
They are {A,E}
Therefore, there are 2 favorable outcomes of choosing an vowel.

C. In how many ways can choosing a consonant occur?
Answer: 3 ways

Explanation:
There are 3 unique consonant tiles. They are {B,C,D}

Describe the likelihood of the event given its probability.
Question 2.
The probability that you land a jump on a snowboard is \(\frac{1}{10}\).
Answer:
Given, Probability of landing a jump on snowboard = 1/10 = 0.1
So, the likelihood of the event of jumping on to the snowboard after every jump is 1 out of 10 times.

Question 3.
There is a 100% chance that the temperature will be less than 120°F tomorrow.
Answer:
Certain event .
Given that , there is 100% chance that the temperature will be less than 120F tomorrow.
The probability for a certain event is P(E)= 1.
So, the probability that the temperature being 120F tomorrow is 1.

Question 4.
You attempt three-point shots on a basketball court and record the number of made and missed shots. Describe the likelihood of each event.̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 5
̇̇a. You make your next shot.
b. You miss your next shot.
Answer:
a.
Explanation:
Total number of shots = 15
Frequency of shots made = 9
The likelihood of making next shot is
Shots made/ Total = 9/15=0.6
So, the next shot is likely to occur

b.
Explanation:
Total number of shots = 15
Frequency of missed shots = 6
The likelihood of the next shot will miss is shots missed/ Total =6/15 0.4
So, the next shot is not likely to occur.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
IDENTIFYING OUTCOMES
You roll a number cube. What are the possible outcomes?
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 6

Answer: 6
A cube has 6 faces .
They are {1,2,3,4,5,6} .
There are 6 possible outcomes.
Therefore , there are 6 possible outcomes for a cube.

Question 6.
USING RELATIVE FREQUENCIES
A bag contains only red marbles and blue marbles. You randomly draw a marble from the bag and replace it. The table shows the results of repeating this experiment. Find the likelihood of each event.̇̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 7
a. The next marble you choose is red. ̇̇
b. The next marble you choose is neither red nor blue.
Answer:

a.
Explanation:
Total number of balls = 42
Number of red balls = 21
The likelihood of drawing the red marble in the next turn=
Number of red balls/ Total number of balls
= 21/42 =0.5
There is only 50% chance of chance of drawing a red ball in the next turn
So, the likelihood of the event is may or ,ay not occur evenly.

b.The likelihood of drawing neither red nor blue ball is 0
0%= Impossible
Because, there are only red and blue marbles
So, it is impossible to draw another ball other than red and blue.
Therefore, the likelihood of the event is cannot occur

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
The table shows the number of days you have a pop quiz and the number of days you do not have a pop quiz in three weeks of school. How many days can you expect to have a pop quiz during a 180-day school year? Explain.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 8
Answer:
Total number of days for pop quiz for 3 weeks = 2
Total frequency = 15
Now, the number of times to have a pop quiz in 180 days be x
Total number of days =180
2/15 = x/180
15x = 360
x =  360/ 15
x = 24
Therefore, we can expect 24 days to have a pop quiz during a 180-day school year.

Question 8.
In a football game, the teams pass the ball on 40% of the plays. Of the passes thrown, greater than 75% are completed. You watch the film of a randomly chosen play. Describe the likelihood that the play results in a complete pass. Explain your reasoning.
Answer:
Given ,
The team pass the ball on 40% of the plays
And also, 75% are completed
75% = 0.75
Therefore 75% = 3/4
Therefore , the likelihood is likely to happen of the plays results in a complete pass = 0.34

Probability Homework & Practice 7.1

Review & Refresh

An account earns simple interest. Find the interest earned.
Question 1
$700 at 3% for 4 years
Answer: $84
The formula for simple interest= Principal x rate x time
I = P x r x t
= $700 x0.03 x 4 =$84
Therefore , the interest earned =  $84

Question 2.
$650 at 2% for 6 years
Answer: $78
The formula for simple interest= Principal x rate x time
I = P x r x t
= $650 x0.02 x 6 =$78
Therefore , the interest earned =  $78

Question 3.
$480 at 1.5% for 5 years
Answer: $36
The formula for simple interest= Principal x rate x time
I = P x r x t
= $480 x0.015 x 5 =$36
Therefore , the interest earned =  $36

Question 4.
$1200 at 2.8% for 30 months
Answer: $84
The formula for simple interest= Principal x rate x time
I = P x r x t
= $1200 x0.028 x 2.5 =$84
Therefore , the interest earned =  $84

Write the indicated ratio. Then find and interpret the value of the ratio.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 9
Question 5.
rolled oats : chopped peanuts
Answer:
2 : 1/2
2/1 : 1/2
Multiplying with 2
2/1 x 2 : 1/2 x 2
4/2 : 2/4
2/1 : 1/2
2/1 x 2 : 1/2:2
4 : 1
Therefore, the value of ratio = 4: 1

Question 6.
sunflower seeds : pumpkin seeds
Answer:
1/3 :1/4
Multiplying with 12
1/3 x 12 : 1/4 x 12
12/36 : 12/ 48
1/3: 1 /4
1/3 x 12 : 1/4 x12
12/3 : 12/4
4 : 3
Therefore, the ratio = 4 : 3

Question 7.
pumpkin seeds : rolled oats
Answer:
1/4 :2
On multiplying with 4
1/4 x 4 : 2/1 x 4
4/16 : 8/4
1/4: 2
1/4 x 4 : 2 x 4
1 : 8
Therefore , ratio = 1:8

Solve the inequality. Graph the solution.
Question 8.
x + 5 < 9
Answer: x<4

Explanation:
x + 5 < 9
=x < 9-5
x<4
So, the inequality form = x<4
Graph:

Question 9.
b – 2 ≥ – 7
Answer: b≥-5

Explanation:
b – 2 ≥ – 7
= b ≥ -7+2
= b≥-5
Graph:

Question 10.
1 > – \(\frac{w}{3}\)
Answer:

Question 11.
6 ≤ – 2g
Answer: g≤ -3

Explanation:
= g≤ 6/-3
= g≤ -3
The inequality = g≤ -3
Graph:

Concepts, Skills, & Problem Solving

DETERMINING LIKELIHOOD Determine which numbers you are more likely to spin and which numbers you are less likely to spin. Explain your reasoning. (See Exploration 1, p. 283.)
Question 12.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 10

Answer:
The spinner is divided into 6 parts (unequal)
They are numbered as {1,2,3,4,5,6}
The numbers {2,4,5} are more likely to happen
Because, they occupy more space than remaining.
And , the numbers {1,3,6] are less likely to happen
Because , they occupy less space .

Question 13.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 11
Answer:
The spinner is divided into 4 equal parts
The 4 parts are numbered as {1,2,3,4}
So, each number have equal chance of spinning
Such events are also called as equally likely happen events.

IDENTIFYING OUTCOMES You spin the spinner shown.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 12
Question 14.
How many possible outcomes are there?
Answer: 8

Explanation:
The spinner is divided into 8 equal parts
The number of possible outcomes for the spinner are 8
{ 1,2,3,4,5,6,7,8}

Question 15.
What are the favorable outcomes of spinning a number not greater than 3?
Answer: 3

Explanation:
There are 3 numbers those are not greater than 3, they are  {1,2,3}.
There are 3 favorable outcomes.

Question 16.
In how many ways can spinning an even number occur?
Answer:  4

Explanation:
There are 4 even numbers , they are {2,4,6,8}
So, there are 4 ways of getting an even number.

Question 17.
In how many ways can spinning a prime number occur?

Answer:4

Explanation:
There are 4 prime numbers , they are {2,3,5,7}
So, there are 4 ways of getting a prime number.

IDENTIFYING OUTCOMES You randomly choose one marble from the bag.(a) Find the number of ways the event can occur. (b) Find the favorable outcomes of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 13

There are 9 colored balls in the bag in which ,
The number of red balls = 3 .
The number of blue balls = 2
The number of yellow balls = 1
The number of purple balls = 2.
The number of green balls =1
The number of favorable outcomes = 9.

Question 18.
Choosing blue
Answer: 2/9

Explanation :
The total number of out comes = 9.
The number of blue balls in the bag = 2
Consider B is the event of getting Blue ball
So, the probability of getting blue ball, P(B)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(B) = 2 / 9.
Therefore, the probability of getting blue ball is 2/9 = 0.22.

Question 19.
Choosing green
Answer: 1/9

Explanation :
The total number of out comes = 9.
The number of green balls in the bag = 1
Consider G is the event of getting Blue ball
So, the probability of getting green ball, P(G)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(G) = 2 / 9.
Therefore, the probability of getting green ball is 1/9 = 0.11.

Question 20.

Choosing purple

Answer: 2/9

Explanation :
The total number of out comes = 9.
The number of purple balls in the bag = 2
Consider P is the event of getting purple ball
So, the probability of getting purple ball, P(P)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(P) = 2 / 9.
Therefore, the probability of getting purple ball is 2/9 = 0.22.

Question 21.

Choosing yellow

Answer: 1/9

Explanation :
The total number of out comes = 9.
The number of yellow balls in the bag = 1
Consider Y is the event of getting yellow ball
So, the probability of getting yellow ball, P(Y)=  (Number of Favorable outcomes) / (Total number of outcomes)
P(Y) = 2 / 9.
Therefore, the probability of getting yellow ball is 1/9 = 0.11.

Question 22.

Choosing not red
Answer: 0.67
There are 3 red balls in the bag
The total number of colored balls in the bag are 9
The balls other than red balls are 9 – 3 = 6
Therefore, the probability of getting a ball other than red is
P(E) = Favorable outcomes / Total number of outcomes .
Where, E is the event of getting a ball other than red
P(E) = 6/9 = 2/3
Therefore , the probability of choosing not red ball is 2/3 = 0.67.

Question 23.
Choosing not blue

Answer: 7/9 = 0.78

Explanation:
There are 2 blue balls in the bag
The total number of colored balls in the bag are 9
The balls other than blue balls are 9 – 2 = 7
Therefore, the probability of getting a ball other than blue is
P(E) = Favorable outcomes / Total number of outcomes .
Where, E is the event of getting a ball other than blue
P(E) = 7/9 = 0.78
Therefore , the probability of choosing not blue ball is 7/9.

Question 24.
YOU BE THE TEACHER
Your friend finds the number of ways that choosing not purple can occur. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 14

Answer: Yes, he is correct
Given, the number of not purple colors=4
They are red, blue, green, yellow
Therefore the number of ways that choosing not purple can occur = 4 .

CRITICAL THINKING Tell whether the statement is true or false. If it is false, change the italicized word to make the statement true.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 15
Question 25.
Spinning blue and spinning green have the same number of favorable outcomes on Spinner A.
Answer: False
Spinning blue and spinning Red have the same number of favorable outcomes on spinner A.

Question 26.
There are three possible outcomes of spinning Spinner A.
Answer: True

Question 27.
Spinning red can occur in four ways on Spinner B.
Answer: True

Question 28.
Spinning not green can occur in three ways on Spinner B.
Answer: False

Spinning not green can occur in four ways on Spinner B.

DESCRIBING LIKELIHOOD Describe the likelihood of the event given its probability.
Question 29.
Your soccer team wins \(\frac{3}{4}\) of the time.
Answer:50
The likelihood of the event is may or may not happen .

Question 30.
There is a 0% chance that you will grow12 feet.
Answer:
The probability =0
Thus, the likelihood of the event is Impossible
Therefore , It is impossible to to grow 12 feet.

Question 31.
The probability that the sun rises tomorrow is 1.
Answer:
The probability 1=100%
The likelihood of the event is  certain
So, it is sure that the sun rises tomorrow

Question 32.
It rains on \(\frac{1}{5}\) of the days in June.
Answer:
The likelihood of the event is not likely to happen .

Question 33.
MODELING REAL LIFE
You have a 50% chance of being chosen to explain a math problem in front of the class. Describe the likelihood that you are chosen.
Answer:
Given,  there is 50% chance of being chosen to explain a math problem in front of the class
50% = 0.5
So, the likelihood of the event is may or may not occur .
there is 50% chance

Question 34.
MODELING REAL LIFE
You roll a number cube and record the number of times you roll an even number and the number of times you roll an odd number. Describe the likelihood of each event.̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 16
a. You roll an even number on your next roll.
b. You roll an odd number on your next roll.
Answer:

a.Explanation:
Total number times cube rolled = 50
The frequency for even number =26
Consider E be the event of rolling an even number
P(E)=(Number of times an event occurs)/ (Total number of trails)
P(E)= 26/50 = 0.72
Therefore , the chances of rolling an even number = 0.72

b.Explanation:
Total number times cube rolled = 50
The frequency for odd number =24
Consider O be the event of rolling a odd number
P(O)=(Number of times an event occurs)/ (Total number of trails)
P(O)= 24/50 = 0.48
Therefore , the chances of rolling a odd number = 0.48

Question 35.
REASONING
You want to determine whether a coin is fair. You flip the coin and record the number of times you flip heads and the number of times you flip tails.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 17
a Describe the likelihood that you flip heads on your next flip.
b. Describe the likelihood that you flip tails on your next flip.
c. Do you think the coin is a fair coin? Explain. ̇̇̇̇̇
Answer:

a.Total number of flips = 25
The total frequency for heads = 22
Now, the likelihood that you flip heads on your next flip
= Number of heads/ total
= 22 / 25 = 0.88
=88%
The likelihood of the event is more likely to occur

b.Total number of flips =25
The total frequency for tails = 3
Now, the likelihood that you flip tails on your next flip
= Number of tails/ Total
= 3/25 =0.12
Therefore, the likelihood of the event is not likely to occur

c.Every coin has both sides, { heads, tails}
But, the probability of tossing coin is not fair
They do not have 50/50 outcomes for both heads and tails .
So the coin is not fair in tossing.

Question 36.
LOGIC
At a carnival, each guest randomly chooses 1 of 50 rubber ducks and then replaces it. The table shows the numbers of each type of duck that have been drawn so far. Out of 150 draws, how many can you expect to not be a losing duck? Justify your answer.̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇̇
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 18

Answer:
Total number of ducks = 25
Total number of win ducks = 6
Now,
The probability to expect to not be a losing duck after 105 draws =
won ducks/ total =x/150
6/25 =x/150
25x= 900
x= 900/25
x= 36 =0.36
The chance to expect to not be a losing duck after 105 draws is ‘not likely to occur

Question 37.
CRITICAL THINKING
A dodecahedron has twelve sides numbered 1 through 12. Describe the likelihood that each event will occur when you roll the dodecahedron. Explain your reasoning.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 19
a. rolling a 1
b. rolling a multiple of 3
c. rolling a number greater than 6

a. rolling a 1

Answer: The probability of rolling a 1 is 1/12

Explanation :
There are 12 number of outcomes for a dodecahedron,
{1,2,3,4,5,6,7,8,9,10,11,12}
P{O}= favorable outcomes / Total number of outcomes.
Where O is the event of getting number 1
p{O}= 1 /12
Therefore , the probability of rolling 1 is 1/12.

b. Rolling a multiple of 3

Answer:  1/3

Explanation:
There are 4 multiples of 3 in a dodecahedron.
They are, {3,6,9,12}
There are 12 number of outcomes for a dodecahedron,
{1,2,3,4,5,6,7,8,9,10,11,12}
P{M}= Favorable outcomes/ Total number of outcomes
Where, M is the event of rolling a multiple of 3
P{M}= 4/12 = 1/3
Therefore, the probability of getting a multiple of 3 is 1/3

C. Rolling a number greater than 6.

Answer: 1/2

Explanation:
There are 6 numbers of which are greater than 6 . they are, {7,8,9,10,11,12}
There are 12 number of outcomes for a dodecahedron,
{1,2,3,4,5,6,7,8,9,10,11,12}
P{G}= Favorable outcome / Total number of outcomes
Where, G is the event of getting a number greater than 6
P{G} = 6/12=1/2
Therefore, the probability of getting a number greater than 6 is 1/2.

Question 38.
DIG DEEPER!
A bargain bin contains classical CDs and rock CDs. There are 60 CDs in the bin. Choosing a rock CD and not choosing a rock CD have the same number of favorable outcomes. How many rock CDs are in the bin?

Answer: 30 Rock CDs

Explanation:
Given that,
There are 60 CDs in the bin
So, There are 60 total number of outcomes
Also given that, choosing a rock CD and not choosing a rock CD have the same number of favorable outcomes
So, there is equal probability .
We can conclude that there are same number of classical CDs and rock CDs
Which means , 30 +30 = 60
Therefore, there are 30 rock CDs in the bin.

Question 39.
REASONING
You randomly choose one of the cards and set it aside. Then you randomly choose a second card. Describe how the number of possible outcomes changes after the first card is chosen.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 20
Answer:
Initially,
Total number cards =5
So, the number of possible outcome =5
If you choose one card and set it aside , the number of possible outcomes decreases by 1
Then when you randomly choose second card , there are only four cards left
Therefore, the number of possible outcomes decreases by 1 after the first card is chosen.

STRUCTURE A Punnett square is a grid used to show possible gene combinations for the offspring of two parents. In the Punnett square shown, a boy is represented by XY. A girl is represented by XX.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability 7.1 21
Question 40.
Complete the Punnett square. Explain why the events “having a boy” and “having a girl” are equally likely.
Answer:

There is a chance of 50% female and 50% male
The probability = 50%
Therefore , the likelihood of the event is equally likely.

Question 41.
Two parents each have the gene combination Cs. The gene C is for curly hair. The gene s is for straight hair. Any gene combination that includes a C results in curly hair. When all outcomes are equally likely, what is the probability of a child having curly hair?
Answer: 4
C is considered as X which results for curly hair
since, there are 4 outcomes includes X
Thus, the probability of child having curly hair = 4/4 =1
Therefore , the probability = 1
So, the likelihood or the chances of a child having curly hair is certain.

Lesson 7.2 Experimental and Theoretical Probability

EXPLORATION 1

Conducting Experiments
Work with a partner. Conduct the following experiments and find the relative frequencies.
Experiment 1
• Flip a quarter 25 times and record whether each flip lands heads up or tails up.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 1
Experiment 2
• Toss a thumbtack onto a table 25 times and record whether each toss lands point up or on its side.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 2
a. Combine your results with those of your classmates. Do the relative frequencies change? What do you notice?
b. Everyone in your school conducts each experiment and you combine the results. How do you expect the relative frequencies to change?
c. How many times in 1000 flips do you expect a quarter to land heads up? How many times in 1000 tosses do you expect a thumbtack to land point up? Explain your reasoning.
d. In a uniform probability model, each outcome is equally likely to occur. Can you use a uniform probability model to describe either experiment? Explain.
Answer:

Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 3

Try It

The table shows the results of rolling a number cube 50 times. Find the experimental probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 4
Question 1.
rolling a 3
Answer: 8/50 =  0.16

Explanation:
Total number of trails = 50
Frequency for  3 = 8
Consider T be the event of rolling a 3
Experimental probability = P(T)(Number of times an event occurs)/ (Total number of trails)
P(T)= 8/50= 0.16
Therefore , the probability of rolling a 3 is 0.16.

Question 2.
rolling an odd number.
Answer:
The probability of rolling an odd number = 0.58

Explanation:
Total number of odd numbers = 3
They are {1,3,5,}
The frequency for 1 = 10
The frequency for 3 =8
The frequency for 5= 11
Total frequency = 10 +8+11 = 29
Total number of trails = 50
Consider O be the event of rolling a Odd number
Experimental probability = P(O)=(Number of times an event occurs)/ (Total number of trails)
P(O)= 29/50= 0.58
Therefore , the probability of rolling an odd number = 0.58

Question 3.
What is the theoretical probability of randomly choosing an X?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 5
Answer: 1/7 =0.142

Explanation:
Number of favorable outcomes = 7
Consider X be the event of choosing an X
Theoretical probability P(X)=(Number of favorable outcomes)/(Total number of outcomes)
P(X)= 1/7 = 0.142
Therefore, the probability of randomly choosing X =0.142

Question 4.
How does the experimental probability of rolling a number greater than 1 compare with the theoretical probability?
Answer:
Theoretical probability  is what we expect to happen, where experimental probability is what actually happens when we try it out.
The experimental probability of an event cannot be greater than 1 since the number of trials in which the event can happen cannot be greater than the total number of trials.

Question 5.
An inspector randomly selects 200 pairs of jeans and finds 5 defective pairs. About how many pairs of jeans do you expect to be defective in a shipment of 5000?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 6
Answer: 125 defective jeans

Explanation:
Total number of pairs = 200
Number of defective pairs = 5
Let ,x be the number of defective pairs of jeans in a shipment of 5000
Now, the number of defective pair of jeans in a shipment of 5000 =
Defective / Total= x/5000
5/200 = X/5000
200 x x =  5000 x 5
200x = 25000
x = 25000/200
x = 125
Therefore there are 125 pairs of jeans are expected  to be defective in a shipment of 5000

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
Explain what it means for an event to have a theoretical probability of 0.25 and an experimental probability of 0.3.
Answer:
If you have a fair coin, and flip it twice,
The  theoretical probability of getting 2 heads = 0.25
There’s 4 outcomes: HH, TT, TH, and HT.
1 out of 4 is 0.25 chance of 2 heads.
But if you run the experiment and do actual flips, maybe 100 times,
The average times you get 2 heads might be 0.3 if you got 2 heads 30 times out of a 100.
It could be just random chance or maybe the coin is slightly heavier on the head side, causing slightly more heads than the theoretical probability.

Question 7.
DIFFERENT WORDS, SAME QUESTION
You flip a coin and record the results in the table. Which is different? Find “both” answers.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 7
Answer:

  1. Experimental probability of flipping heads = frequency of heads / Total number of trails = 32/60 =0.53

2.       The fraction of the flips you can expect a result of heads = 32/60 =0.53

3.        The percent of flips result in heads = 0.53 =53%

4.        The relative frequency of flipping heads =  32/60 = 0.53

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
Contestants randomly draw a ticket from a hat and replace it. The table shows the results after 40 draws. There are 7 winning tickets in the hat. Predict the total number of tickets in the hat. Explain.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 8
Answer:
Let, the total numbers of tickets in the hat = x
Given that , Number of winning tickets = 7
Number of winning won after 40 draws = 2
Total number of tickets =
2/40 = 7/x
2x = 280
x = 280/2
x= 140
Therefore, The total number of tickets in the hat = 140

Question 9.
DIG DEEPER!
You choose two different songs on a music play list at random. Out of 80 songs on the playlist, 36 are hip hop songs. The first song you choose is a hip hop song. What is the probability that the second song is also a hip hop song? Explain your reasoning.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 9
Answer:
Total number of songs = 80
Number of hip pop songs = 36
Given that the first song is hip pop
The probability of playing second song also hip pop =
Total number of hip pop songs / total number of songs
Consider H be the event of playing hip pop
P(H)= 36/ 80 =0.45
If there is no repeat :
Total number of hip pop songs after playing one song = 35
P(H)= 0.43
Therefore ,the likelihood of the event is unlikely to happen.

Experimental and Theoretical Probability Homework & Practice 7.2

Review & Refresh

Describe the likelihood of the event given its probability.
Question 1.
You randomly guess the correct answer of a multiple choice question \(\frac{1}{4}\) of the time.
Answer:
The likelihood of the event is not likely to happen .

Question 2.
There is a 95% chance that school will not be cancelled tomorrow.
Answer:
Given , 95% chance that school will be cancelled tomorrow
95%= 0.95
Probability = 0.95
So the likelihood of the event is more likely to happen

Find the annual interest rate.
Question 3.
I = $16, P = $200, t = 2 years
Answer: 4%
The formula for simple interest= Principal x rate x time
I = P x r x t
$16 = $200 x r x 2
$16= 400r
r = 16/400
r= 0.04
0.04 x 100= 4
Therefore, the annual interest rate = 4%

Question 4.
I = $26.25, P = $500, t = 18 months
Answer:3.5
The formula for simple interest= Principal x rate x time
I = P x r x t
$26.25 = $500 x r x 1.5
$26.25= 750r
r = 26.25/750
r= 0.035
0.035 x 100= 3.5
Therefore, the annual interest rate = 3.5%

Tell whether x and y are proportional.
Question 5.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 10
Answer:
The ratio of x and y for the given values is
1/8 = 0.12
3/24= 0.12
9/75 = 0.12
We get equal values for all the ratios
Therefore, the relationship given in the table is proportional.

Question 6.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 11
Answer:
The ratio of x and y for the given values is
0.75/0.3 =2.5
1.5/0.6 = 2.5
2.25/0.9 = 2.5
We get equal values for all the ratios
Therefore, the relationship given in the table is proportional.

Concepts, Skills, & Problem Solving

CONDUCTING AN EXPERIMENT Use the bar graph below to find the relative frequency of the event. (See Exploration 1, p. 291.)
Question 7.
spinning a 6
Answer: 0.14

Explanation:
Total number of spins =50
The number of times 6 spun =7
Consider X be the event of spinning a 6
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(X)= 7/50= 0.14
Therefore, the probability spinning a 6= 0.14

Question 8.
spinning an even number
Answer:0.48

Explanation:
Total number of spins =50
Numbers of even numbers = 3
They are {2,4,6}
Number of times 2 spun = 6
Number of times 4 spun = 11
Number of times 6 spun =7
Total : 24
Consider E be the event of spinning a number less than 3
Experimental probability  P(E)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 24/50=0.48
Therefore the probability of spinning a even number = 0.48

FINDING AN EXPERIMENTAL PROBABILITY Spinning a Spinner .Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 12
Question 9.
spinning a number less than 3
Answer:  14/50=0.28

Explanation:
Total number of spins =50
Numbers less than 3 are {1,2}
Number of times 1 spun =8
Number of times 2 spun =6
Total : 8 +6 = 14
Consider T be the event of spinning a number less than 3
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 14/50=0.28
Therefore, the probability spinning a number less than 3 =0.28

Question 10.
not spinning a 1
Answer: 42/50= 0.84

Explanation:
Total number of spins =50
The numbers other than 1 are{2,3,4,5,6}
The number of times 2 spun = 6
The number of times 3 spun= 9
The number of times 4 spun= 11
The number of times 5 spun= 9
The number of times 6 spun= 7
Total : 6+9+11+9+7=42
Consider O be the event of spinning a number less than 3
Experimental probability = P(O)=(Number of times an event occurs)/ (Total number of trails)
P(O)= 42/ 50= 0.84
Therefore, the probability of spinning a number other than 1 is 0.84.

Question 11.
spinning a 1 or a 3
Answer:
Total number of spins =50
Number of times 1 spun = 8
The number of times 3 spun= 9
Total : 8+9 = 17
Consider S be the event spinning a 1 or 3
Experimental probability = P(S)=(Number of times an event occurs)/ (Total number of trails)
P(S)= 17/ 50= 0.34
Therefore, the probability of spinning a  1 OR 3 is 0.34.

Question 12.
spinning a 7
Answer: 0

Explanation:
The bar graph consists 6 numbers
They are, {1,2,3,4,5,6}
So, it is impossible to spin 7
Therefore the probability of spinning is 0.

Question 13.
YOU BE THE TEACHER
Your friend uses the bar graph above to find the experimental probability of spinning. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 13
Answer:
No, he is wrong

Explanation:
The formula for finding experimental probability is
Experimental probability P(O)=(Number of times an event occurs)/ (Total number of trails)
Consider O be the event of spinning 4
P(O)=11/50
Therefore, the experimental probability of spinning a 4 is 0.22
So, he is wrong

Question 14.
MODELING REAL LIFE
You check 20 laser pointers at random. Three of the laser pointers are defective. What is the experimental probability that a laser pointer is defective?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 14
Answer: 0.3
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
Total number of lasers = 20
Number of defective lasers =3
P(T) = 3/20
= 0.3
Therefore, the experimental probability that a laser pointer is defective is 0.3

FINDING A THEORETICAL PROBABILITY Use the spinner to find the theoretical probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 15
Question 15.
spinning red
Answer: 0.33

Explanation:
The spinner is divided into 6 equal parts.
So, the total number of outcomes of a spinner = 6
In which 2 parts are red and remaining parts are blue, green , orange and purple
Consider R is the event of getting a red color.
So, the probability of spinning red color P(R) = ( number of favorable outcomes)/(Total number of outcomes)
P(R)= 2/6= 0.33

Question 16.
spinning a 1
Answer: 1/6 = 0.16

Explanation:
The spinner is divided into 6 equal parts.
So, the total number of outcomes of a spinner = 6
{1,2,3,4,5,6}
Consider ,O is the event of getting 1
So, the probability of getting 1 is P(O)=( number of favorable outcomes)/(Total number of outcomes)
P(O)= 1/6 = 0.16
Therefore the probability of spinning a 1= 0.16

Question 17.
spinning an odd number
Answer: 3/6= 0.5

Explanation:
The number of total outcomes of a spinner = 6
They are , {1,2,3,4,5,6}
There are 3 odd numbers, They are {1,3,5}
Consider N is the event of getting an odd number
So, the probability of spinning an odd number P(N) =( number of favorable outcomes)/(Total number of outcomes)
P(N)= 3/6= 0.5.
Therefore, the probability of spinning an odd number = 3/6 =0.5

Question 18.
spinning a multiple of 2
Answer: 3/6= 0.5

Explanation:
The total number of outcomes of a spinner = 6
They are , {1,2,3,4,5,6}
There are 3 multiples of 2 , They are {2,4,6}
Consider M is the event of getting an odd number
So, the probability of spinning a multiple of 2is  P(M) =( number of favorable outcomes)/(Total number of outcomes)
P(M)= 3/6= 0.5.
Therefore, the probability of spinning a multiple of 2 = 3/6 =0.5

Question 19.
spinning a number less than 7
Answer: 1

Explanation:
The spinner is divided into 6 equal parts.
So, the total number of outcomes of a spinner = 6
{1,2,3,4,5,6}
Consider, S is the event of spinning a number less than 7
So, the probability of spinning a number less than 7 P(S) =( number of favorable outcomes)/(Total number of outcomes)
P(S)= 6/6 =1
So, the probability of spinning a number less than 7 = 1

Question 20.
spinning a 9
Answer: 0
The spinner is divided into 6 equal parts numbering from 1 to 6
={1,2,3,4,5,6}
There is no 9 in the spinner
So, the probability of spinning 9 is 0
It is also called as impossible event .

Question 21.
REASONING
Each letter of the alphabet is printed on an index card. What is the theoretical probability of randomly choosing any letter except Z?
Answer: 25/26 =0.96

Explanation:
We know that, there are 26 alphabets.
Also, the number of  alphabets other than Z are: 25
Let us consider Z is the event of choosing any letter except Z .
P(Z)= (number of favorable outcomes) / (Total number of outcomes )
p(Z)= 25/ 26 =0.96
Therefore, the theoretical probability of choosing a letter except Z is 25/26 = 0.96

COMPARING PROBABILITIES The bar graph shows the results of spinning the spinner below 200 times. Compare the theoretical and experimental probabilities of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 16.
Question 22.
spinning a 4
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider F be the event of spinning 4
Theoretical probability P(F)= (Number of favorable outcome )/(Total number of outcomes)
P(F)= 1/5= 0.2
Experimental probability :
Total number of spins = 200
Number of times 4 spun =37
Consider F be the event of spinning 4
Experimental probability = P(F)=(Number of times an event occurs)/ (Total number of trails)
P(F)= 37/200= 0.185
Therefore, the theoretical probability = 0.2
The experimental probability = 0.18

Question 23.
spinning a 3
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider T be the event of spinning 3
Theoretical probability P(T)= (Number of favorable outcome )/(Total number of outcomes)
P(T)= 1/5= 0.2
Experimental probability :
Total number of spins = 200
Number of times 3 spun = 39
Consider T be the event of spinning 3
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 39/200= 0.195
Therefore, the theoretical probability = 0.16
The experimental probability = 0.195

Question 24.
spinning a number greater than 4
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider N be the event of spinning a number greater than 4
Numbers greater than 4 = {5}
Theoretical probability P(T)= (Number of favorable outcome )/(Total number of outcomes)
P(T)= 1/5= 0.2
Experimental probability :
Total number of spins = 200
Consider N be the event of spinning a number greater than 4
Number of times 5 spun =40
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 40/200= 0.2
Therefore, the theoretical probability = 0.2
The experimental probability = 0.2

Question 25.
spinning an odd number
Answer:
Theoretical probability:
Total number of outcomes = 5
Consider S be the event of spinning an odd number
Total odd numbers = 3
They are {1,3,5}
Theoretical probability P(T)= (Number of favorable outcome )/(Total number of outcomes)
P(T)= 3/5= 0.6
Experimental probability :
Total number of spins = 200
Consider S be the event of spinning an odd number
The number of time 1 spun=41
The number of time 3 spun=39
The number of time 5 spun=40
Total : 41+39+40= 120
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 120/200= 0.6
Therefore, the theoretical probability = 0.6
The experimental probability = 0.6

Question 26.
REASONING
Should you use theoretical or experimental probability to predict the number of times you will spin a 3 in 10,000 spins? Explain.
Answer: theoretical probability

Explanation:
Theoretical probability of event =
Theoretical probability = P(T)=(Number of favorable outcomes)/ (Total number of outcomes)
Now,
Consider T be the event of spinning 3
P(T)= 1/6.

Question 27.
MODELING REAL LIFE
A board game uses a bag of 105 lettered tiles. You randomly choose a tile and then return it to the bag. The table shows the number of vowels and the number of consonants after 50 draws. Predict the number of vowels in the bag.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 16
Answer:
let , X be the number of vowels in the bag.
Total number of tiles = 105
Number of vowels after 50 draws = 18
Now, the number of vowels in the bag =
X/150 = 18/50
50X=2700
X = 2700/50
X = 54
Therefore, 54  vowels are expected to be in the bag.

Question 28.
MODELING REAL LIFE
On a game show, a contestant randomly draws a chip from a bag and replaces it. Each chip says either win or lose. The theoretical probability of drawing a winning chip is \(\frac{3}{10}\). The bag 10contains 9 winning chips.
a. How many chips are in the bag?
b. Out of 20 contestants, how many do you expect to draw a winning chip?
Answer:
Given , the probability of drawing a winning chip = 3/10
Consider , x be the number of chips in the bag
The number of chips = 9
The ration of strikes in the bag = 9/x
3/10 = 9/x
3x = 90
x= 30
Therefore , there are 30 chips in the bag

Question 29.
PROBLEM SOLVING
There are 8 females and 10 males in a class.
a. What is the theoretical probability that a randomly chosen student is female?
b. One week later, there are 27 students in the class. The theoretical probability that a randomly chosen student is a female is the same as last week. How many males joined the class?
Answer:

a.Number of females=8
Total number of students = 8+10= 18
The theoretical probability that a randomly chosen student is female
Consider F be the event of chosen a female =
P(F)= Number of favorable outcomes / Total number of outcomes
P(F)= 8/18 =0.44

b.Initially,
Total number of students = 18
In which , number of females =8
Number of males =10
one week later , the total number of students = 27
Given, the theoretical probability for choosing girl is same
So, Number of males joined = 27 -18 = 9
Therefore, 9 males joined

Question 30.

NUMBER SENSE
The table at the right shows the results of flipping two coins 12 times each.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 17
a. What is the experimental probability of flipping two tails? Using this probability, how many times can you expect to flip two tails in 600 trials?
b. The table at the left shows the results of flipping the same two coins 100 times each. What is the experimental probability of flipping two tails? Using this probability, how many times can you expect to flip two tails in 600 trials?
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 18
c. Why is it important to use a large number of trials when using experimental probability to predict results?
Answer:

a.Explanation:
According to table,
Number of turns = 12
Frequency of flipping 2 tails = 1
The probability of flipping two tails in 600 trails =
1/12 = x/600
12x =600
x = 600/ 12
x = 50
In 600 trails , flipping 2 tails is expected to be 50 times.

b.Explanation:
According to table,
Number of turns = 100
Frequency of flipping 2 tails = 22
The probability of flipping two tails in 600 trails =
22/100= x/600
100x =13,200
x = 13200/100
x = 132
In 600 trails , flipping 2 tails is expected to be 132 times.

c.In experimental probability, as the number of trials increases, the experimental probability gets closer to the theoretical probability.
So, it is important to use a large number of trails.

Question 31.
COMPARING PROBABILITIES
The table shows the possible outcomes of rolling a pair of number cubes. You roll a pair of number cubes 60 times and record your results in the bar graph shown.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 19
a. Compare the theoretical and experimental probabilities of rolling each sum.
b. Which sum do you expect to be most likely after 500 trials? 1000 trials? Explain your reasoning.
c. Predict the experimental probability of rolling each sum after 10,000 trials. Explain your reasoning.
Answer:

a.

b. 6 sum is more likely to happen after after 500 trials, 1000 trials, and even after 10,000 trails because the the number of times 6 rolled is greater than other numbers sum.

Question 32.
PROJECT
When you toss a paper cup into the air, there are three ways for the cup to land: open-end up, open-end down, or on its side.
a. Toss a paper cup 100 times and record your results. Do the outcomes for tossing the cup appear to be equally likely? Explain.
b. Predict the number of times each outcome will occur in 1000 tosses. Explain your reasoning.
c. Suppose you tape a quarter to the bottom of the cup. Do you think the cup will be more likely less likely or to land open-end up? Justify your answer.
Big Ideas Math Answers 7th Grade Chapter 7 Probability 7.2 20
Answer:
The answer will vary

a. The results of tossing a cup 100 times are :
Open end up      = 18
Open end down = 31
On its side           = 51
The events of tossing the cup are not equally likely.

b.The probability of the events in 1000 tosses
open end up =18/ 100 x 1000 =180
Open end down = 31/100 x 1000 =310
On its side = 51 / 100 x1000 =510

c.If you tape quarter to the bottom, it will probably be more likely to land open end up since the weight of the bottom
will pull the bottom of the cup down and the top of the cap up.

Lesson 7.3 Compound Events

EXPLORATION 1

Comparing Combination Locks
Work with a partner. You are buying a combination lock. You have three choices.
a. One lock has 3 wheels. Each wheel is numbered from 0 to 9. How many possible outcomes are there for each wheel? How many possible combinations are there?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 1
b. How can you use the number of possible outcomes on each wheel to determine the number of possible combinations?
c. Another lock has one wheel numbered from 0 to 39. Each combination uses a sequence of three numbers.How many possible combinations are there?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 2
d. Another lock has 4 wheels as described. How many possible combinations are there?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 3
e. For which lock are you least likely to guess the combination? Why?
Answer:

The set of all possible outcomes of one or more events is called the sample space. You can use tables and tree diagrams to find the sample space of two or more events.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 4

Try It

Question 1.
WHAT IF?
The sandwich shop adds a multi-grain bread. Find the sample space. How many sandwiches are possible?
Answer: sample space = {Sandwich , multi grain bread }

Question 2.
Find the total number of possible outcomes of spinning the spinner and randomly choosing a number from 1 to 5.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 5
Answer:
Number of colors in the spinner(a) =4
They are {red, blue, green, yellow}
Total numbers (b)= 5
{1,2,3,4,5}
Total number of possible outcomes =
According to fundamental principle ,(a x b)=
4 x 5 = 20
Therefore , total number of outcomes = 20

Question 3.
How many different outfits can you make from 4 T-shirts, 5 pairs of jeans, and 5 pairs of shoes?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 6

Answer: 100

Explanation:
Number of T-shirts(a) = 4
Number of jeans pairs(b) = 5
Number of pair of shoes (C)= 5
According to fundamental principle,
Total number of outcomes = a x b x c
=4 x 5 x5
=100
Therefore , about 100 outfits can be made .

Question 4.
In Example 2, what is the probability of rolling at most 4 and flipping heads?
Answer:

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
FINDING THE SAMPLE SPACE
You randomly choose a flower and ornament for a display case. Find the sample space. How many different displays are possible?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 7

Answer:
Total number of flowers =3
They are {daffodil, hyacinth, tulip}
Total number of ornament =2
They are {figurine, trophy}
Total number of outcomes = 3 x 2 = 6
sample space =
{daffodil, figurine} {daffodil, trophy}
{hyacinth, figurine } {hyacinth, trophy }
{tulip, figurine}{tulip, trophy }.

Question 6.
FINDING THE TOTAL NUMBER OF POSSIBLE OUTCOMES
You randomly choose a number from 1 to 5 and a letter from A to D. Find the total number of possible outcomes.
Answer: 20
Total numbers (a) = 5
They are {1,2,3,4,5}
Total number of letters (b)=4
They are {A,B,C,D}
According to fundamental theory ,
Total number of possible outcomes = a x b
= 4 x 5
=20
Therefore , the total number of possible outcomes = 20

Question 7.
WHICH ONE DOESN’T BELONG?
You roll a number cube and flip a not coin. Which probability does belong with the other three? Explain your reasoning.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 8
Answer: P(greater than 2 and tails ) probability does not belong with the other three.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
A tour guide organizes vacation packages at a beach side town. There are 7 hotels, 5 cabins, 4 meal plans, 3 escape rooms, and 2 amusement parks. The tour guide chooses either a hotel or a cabin and then selects one of each of the remaining options. Find the total number of possible vacation packages.
Answer:
Given, the tour guide chooses either a hotel or a cabin and then selects one of each of the remaining options.
Event 1 :
If the tour guide choose hotel , the total number of possible vacation packages.
According to fundamental counting principle ,
7 hotels x 4 meal plans x 3 escape rooms x 2 amusement parks =168
Therefore, The total number of possible vacation packages = 168
Event 2 :
If the tour guide choose cabin , the total number of possible vacation packages.
According to fundamental counting principle ,
5 cabins x4 meal plans x 3 escape rooms x 2 amusement parks =120
Therefore, The total number of possible vacation packages = 120

Question 9.
DIG DEEPER!
A fitness club with 100 members offers one free training session per member in either running, swimming, or weightlifting. Thirty of the fitness center members sign up for the free session. The running and swimming sessions are each twice as popular as the weightlifting session. What is the probability that a randomly chosen fitness club member signs up for a free running session?
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 9
Answer:
Total number of fitness club = 100
The number of members signed for free up session = 30
Given , the running and swimming sessions are each twice as popular as the weightlifting session
let , x be the event of weight lifting and 2x be the event of running and swimming
2x + 2x + x = 30
5x = 30
x = 6
Now,
the probability that a randomly chosen fitness club member signs up for a free running session=
P(R) =2x /30
= 2×6/ 30
= 12/30
P(R) = 0.4
Therefore , the probability of that a randomly chosen fitness club member signs up for a free running session= 0.4

Compound Events Homework & Practice 7.3

Review & Refresh

Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 10
Question 1.
rolling a 5
Answer:

Explanation:
Total number of spins =100
Total number of times 5 rolled= 19
Consider T be the event of rolling a 5
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 19/100=0.19
Therefore, the probability of rolling 5 =19/100=0.19

Question 2.
rolling a 2 or 6
Answer:
Total number of spins =100
The number of times 2 rolled =16
The probability of rolling 2 = 16 /100 =0.16
The number of times 6 rolled = 20
The probability of rolling 6 = 20/100 =0.20

Question 3.
rolling at least a 3
Answer:

Explanation:
The number of spins = 100
The number of times 3 =15
Consider S be the event of spinning at least 3 =
P(S) = 15/100
P(S) = 0.15

Question 4.
rolling a number less than or equal to 4
Answer: 0.6

Explanation:
Total number of spins =100
Total number of times 1 rolled = 12
Total number of times 2 rolled= 16
Total number of times 3 rolled = 15
Total number of times 4 rolled= 17
Total : 12 + 16 + 15 +17 = 60
Consider L be the event of rolling a number less than or equal to 4
Experimental probability = P(T)=(Number of times an event occurs)/ (Total number of trails)
P(T)= 60 /100=0.6
Therefore, the probability of rolling a number less than or equal to 4 = 0.6

Find the product.
Question 5.
3 . 2
Answer: 6

Question 6.
5(- 3)
Answer:
= -12

Question 7.
– 6(- 2)
Answer:
= 12

Concepts, Skills, & Problem Solving

COMPARING PASSWORDS Determine which password is less likely to be guessed. (See Exploration 1, p. 299.)
Question 8.
a password with 3 numbers or a password with 3 capital letters
Answer: the likelihood of the event is certain to happen with the passwords

Question 9.
a password with 6 numbers or a password with 4 capital letters
Answer: The password with 4 capital letters is more likely to guess than the password with 6 numbers .

USING A TREE DIAGRAM Use a tree diagram to find the sample space and the total number of possible outcomes.
Question 10.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 11

Answer:

Total number of possible outcomes = 3x 2 =6

Question 11.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 12
Answer:

Question 12.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 13
Answer:

Question 13.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 14
Answer:

Question 14.
YOU BE THE TEACHER
Your friend finds the total number of ways that you can answer a quiz with five true-false questions. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 15
Answer: No,

Explanation ,
The number of outcomes for each question =
Possible answers for question number 1 = 2
Possible answers for question number 2 = 2
Possible answers for question number 3= 2
Possible answers for question number 4= 2
Possible answers for question number 5 = 2
According to fundamental counting principle ,
Total number of outcomes = 2 x 2 x 2 x 2 x 2 = 32
Therefore , You can answer the quiz in 32 different ways

USING THE FUNDAMENTAL COUNTING PRINCIPLE Use the Fundamental Counting Principle to find the total number of possible outcomes.
Question 15.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 16
Answer: 12

Explanation:
Total number of sizes (a) = 3
Number of flavors (b) = 4
By using fundamental counting principle ,(a x b)
The total number of possible outcomes = 3 x 4 = 12
Therefore, total number of outcomes = 12

Question 16.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 17
Answer:20

Explanation:
Number of batteries (a)= 4
Number of colors (b)= 5
By using fundamental counting principle ,( a x b)
The total number of possible outcomes = 4 x 5 = 20
Therefore, total number of outcomes = 20

Question 17.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 18
Answer:24

Explanation:
Number of suits (a)= 3
Number of wigs (b)= 2
Number of talents (c)= 4
By using fundamental counting principle ,( a x b x c)
The total number of possible outcomes = 3 x 2 x 4=24
Therefore, total number of outcomes = 24

Question 18.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 19
Answer:36

Explanation:
Number of appetizer (a)= 3
Number of Entre (b)= 4
Number of dessert (c)= 3
By using fundamental counting principle ,( a x b x c)
The total number of possible outcomes = 3 x 4 x 3 = 36
Therefore, total number of outcomes = 36

Question 19.
CHOOSE TOOLS
You randomly choose one of the marbles. Without replacing the first marble, you choose a second marble.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 20
a. Name two ways you can find the total number of possible outcomes.
b. Find the total number of possible outcomes.
Answer:

a. Explanation:
The tree diagram and fundamental counting principle

b.There are 12 number of possible outcomes
Total number of balls = 4
Therefore, there are 4 possible outcomes
without replacing the first ball, the number of possible outcomes =3
According to fundamental principle ,( a x b )
= 4 x 3 = 12
Therefore there are 12 number of outcomes
Tree diagram:

Question 20.
FINDING A PROBABILITY
You roll two number cubes. What is the probability of rolling double threes?
Answer:
Number of cubes =2
{1,2,3,4,5,6} , {1,2,3,4,5,6}
Total number of  possible outcomes = 6 x 6 = 36
The probability of rolling double threes =
P(T) = 2/36 = 0.05
Therefore, the probability of rolling double threes = 0.05

FINDING THE PROBABILITY OF A COMPOUND EVENT You spin the spinner and flip a coin. Find the probability of the compound event.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 21
Question 21.
spinning a 1 and flipping heads
Answer: 0.1

Explanation:
The spinner is divided into 5 equal parts
They are {12,3,4,5}
Total number of outcomes =5
Consider O be the event of spinning a 1
P(O)= (Number of favorable outcomes )/ (Total number of outcomes)
P(O)= 1/ 5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event =
1/5 x 1/2 = 1/10 =0.1
Therefore ,the probability of compound event = 1/10

Question 22.
spinning an even number and flipping heads
Answer: 0.2

Explanation:
Number of even numbers = 2
They are {2,4}
Total number of outcomes = 5
Consider E be the event of spinning an even number
P(E) = 2/5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event (a x b)=
2/5 x 1/2 = 2/10 =0.2
Therefore, the probability of compound event = 0.2

Question 23.
spinning a number less than 3 and flipping tails
Answer: 0.2

Explanation:
Total number of outcomes = 5
Number of numbers less than 3 =2
They are {1,2}
Consider, S be the event of spinning a number less than 3
P(S) = 2/5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event (a x b)=
2/5 x 1/2 =2/10 =0.2
Therefore the probability of compound event = 0.2

Question 24.
spinning a 6 and flipping tails
Answer: 0

Explanation:
The spinner is divided into 5 equal parts
They are {12,3,4,5}
Total number of outcomes =5
Consider O be the event of spinning a 6
P(O)= (Number of favorable outcomes )/ (Total number of outcomes)
P(O)= 0
Therefore it is impossible to spin 6
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a tails
P(H)= 1/2
The probability of compound event =
0 x 1/2 =0
Therefore ,the probability of compound event =0

Question 25.
not spinning a 5 and flipping heads
Answer:0.4

Explanation:
The spinner is divided into 5 equal parts
They are {12,3,4,5}
Total number of outcomes =5
Consider N be the event of  not spinning a 5
Favorable outcomes = 4 .They are {1,2,3,4}
P(N)= (Number of favorable outcomes )/ (Total number of outcomes)
P(N)= 4/ 5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of flipping a heads
P(H)= 1/2
The probability of compound event =
4/5 x 1/2 = 4/10 =0.4
Therefore ,the probability of compound event = 0.4

Question 26.
spinning a prime number and not flipping heads
Answer:0.3

Explanation:
Total number of outcomes = 5
Number of prime numbers = 3
They are {2,3,5}
Consider P be the event of spinning a prime number
P(P) = (Number of favorable outcomes )/ (Total number of outcomes)
P(P) = 3/5
Total number of outcomes for a coin =2
They are {Heads, Tails}
The probability of not flipping a heads
P(H)= 1/2
The probability of compound event =
3/5 x 1/2 = 3/10 =0.3
Therefore ,the probability of compound event = 0.3

FINDING THE PROBABILITY OF A COMPOUND EVENT You spin the spinner, flip a coin, and then spin the spinner again. Find the probability of the compound event.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 22
Question 27.
spinning blue, flipping heads, then spinning a 1
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider B be the event of spinning blue
P(B) = 1/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider H be the event of flipping heads
P(H) = 1/2
Consider O be the event of spinning 1
P(O)= 1/3
The compound probability of compound event =
= 1/3 x 1/2 x1/3
= 1/18
Therefore , the compound probability of the event is 1/18.

Question 28.
spinning an odd number, flipping heads, then spinning yellow
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider Y be the event of spinning yellow
P(Y) = 1/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider H be the event of flipping heads
P(H) = 1/2
Consider O be the event of spinning Odd  number
Number of odd numbers = 2
They are {1,3}
P(O)= 2/3
The compound probability of compound event =
= 1/3 x 1/2 x2/3
= 2/18 =0.11
Therefore , the compound probability of the event is 2/18.

Question 29.
spinning an even number, flipping tails, then spinning an odd number
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider E be the event of spinning an even number
Number of even numbers =1
They are {2}
P(B) = 1/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider T be the event of flipping Tails
P(T) = 1/2
Consider O be the event of spinning an odd number
Number of odd numbers =2
They are {1,3}
P(O)= 2/3
The compound probability of compound event =
= 1/3 x 1/2 x2/3
= 2/18 =0.11
Therefore , the compound probability of the event is 2/18.

Question 30.
not spinning red, flipping tails, then not spinning an even number
Answer:
The spinner is divided into 3 equal parts
Indicated numerally as {1,2,3} also colored in {red, blue , yellow}
Consider R be the event of  not spinning red
The colors other than red are {blue , yellow}
P(R) = 2/3
Number of outcomes for a coin =2
They are {Heads , Tails }
Consider T be the event of flipping tails
P(T) = 1/2
Consider E be the event of not spinning an even number
number of not even numbers =2
They are {1,3}
P(E)= 2/3
The compound probability of compound event =
= 2/3 x 1/2 x2/3
= 4/18 =0.22
Therefore , the compound probability of the event is 0.22

Question 31.
REASONING
You randomly guess the answers to two questions on a multiple-choice test. Each question has three choices: A,B, and C.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 23
a. What is the probability that you guess the correct answers to both questions?
b. Suppose you can eliminate one of the choices for each question. How does this change the probability that both of your guesses are correct?
Answer:
Probability of an event = Number of favorable outcomes / Total number of outcomes
There is one correct answer for each question.
So, the number of favorable outcomes = 2
Given that, there are 3 choices {a, b, c} for each question
So, total number of outcomes = 6
Consider, C be event that you guess the correct answers to both questions=
P(C)= 2/6 = 1/3
Therefore, the probability that you guess the correct answers to both questions =1/3

b.
Answer:
There is a correct answer for each question
So, the favorable outcomes =2
If you eliminate one of the choices for each question ,
Now, the choices for the questions are {a, b}
So, the total number of outcomes =4
Consider, S be the event of choosing correct answer
P(S) = 2/4
= 0.5
Therefore, the probability of choosing a correct answer is 0.5

Question 32.
REASONING
You forget the last two digits of your cell phone password.
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 24
a. What is the probability that you randomly choose the correct digits?
b. Suppose you remember that both digits are even. How does this change the probability that you choose the correct digits?
Answer:

a.Number of digits to be guessed = 2
There are 10 possible numbers from 0 to 9
By using fundamental counting principle,
Number of possible outcomes = 10 x 10 =100
Each digit has one correct answer
So, the number of favorable outcomes = 2
Probability = Number of favorable outcomes/ Total number of outcomes
Consider C be the event of choosing correct digits
P(C) = 2/100 =1/50
Therefore the probability of choosing correct digits is 1/50

b.Given that, the two digits to be guessed are even numbers
The number of even numbers from 0 to 9 = 5
They are {0,2,4,6,8}
According to fundamental counting principle ,
So, total number of possible outcomes = 5 x 5= 25
Each digit has only one correct answer,
So number of favorable outcomes = 2
Consider S be the event of choosing correct digits
P(S) = 2/25 =0.08
Therefore, the probability of choosing correct answer = 0.08

Question 33.
MODELING REAL LIFE
A combination lock has 3 wheels, each numbered from 0 to 9. You try to guess the combination by writing five different numbers from 0 to 999 on a piece of paper. Find the probability that the correct combination is written on the paper.
Answer:
The number of digits numbered from 0 to 9 = 10
You get to try 5 of the 1000 possible combinations
So, the probability of getting them all right is 5/1000 or 1/200
Therefore ,5 tries, with a 1/1000 chance each time = 5/1000 = 1/200

Question 34.

MODELING REAL LIFE
A train has one engine and six train cars. Find the total number of ways an engineer can arrange the train. (The engine must be first.)
Big Ideas Math Answers Grade 7 Chapter 7 Probability 7.3 25
Answer:
Total number of train cars = 6
The engine must be first So, the total outcomes for train cars
So , there are 6 possible places for the train cars
Number of possible outcomes for the first train car = 6
Number of possible outcomes for the second train car =5
Number of possible outcomes for the third train car = 4
Number of possible outcomes for the fourth train car = 3
Number of possible outcomes for the fifth train car = 2
Number of possible outcomes for the sixth train car = 1
According to fundamental principle ,
6 x 5 x 4 x 3 x 2 x 1 = 720
Therefore, the number of ways you can arrange the train = 720

Question 35.
REPEATED REASONING
You have been assigned a nine-digit identification number.
a. Should you use the Fundamental Counting Principle or a tree diagram to find the total number of possible identification numbers? Explain.
b. How many identification numbers are possible?
c. RESEARCH Use the Internet to find out why the possible number of Social Security numbers is not the same as your answer to part(b).
Answer:

a.It is difficult to use tree diagram to find the total number of possible identification numbers because, the total number of identification is very large . So, it is impossible to use tree diagram
Therefore , it is easy to easy to use the fundamental counting principle .

b.Total number of identification digits are from 0 to 9
They are {0,1,2,3,4,5,6,7,8,9}
The number of possible numbers from 0 to 9 for first digit = 10
The number of possible numbers from 0 to 9 for second digit = 10
The number of possible numbers from 0 to 9 for third  digit = 10
The number of possible numbers from 0 to 9 for fourth digit = 10
The number of possible numbers from 0 to 9 for fifth digit = 10
The number of possible numbers from 0 to 9 for sixth digit = 10
The number of possible numbers from 0 to 9 for seventh digit = 10
The number of possible numbers from 0 to 9 for eighth digit = 10
The number of possible numbers from 0 to 9 for ninth digit = 10
According to fundamental principle
10x10x10x10x10x10x10x10x10 = 1,000,000,000
Therefore there are 1,000,000,000 possible identification numbers

c.The possible number of Social Security numbers is not the same as your answer to part(b)
Because , some special numbers are never allocated.

Question 36
DIG DEEPER!
A social media account password includes a number from 0 to 9, an uppercase letter, a lowercase letter, and a special character, in that order.
a. There are 223,080 password combinations. How many special characters are there?
b. What is the probability of guessing the account password if you know the number and uppercase letter, but forget the rest?
Answer:

a.From 0 to 9, there are a total of 10 values.
For uppercase letters, there are a total of 26 values from A, B, C, D …Z
For lower case letters, there are also a total of 26 values from a, b, c, d …z.
So out of these three characters, we have a total of 10 × 26 × 26 = 6,760 different combinations.
If there are 223,080 password combinations
We need to divide this by 6,760 to calculate the possible values of the special characters.
6,760 × Number of possible special characters = 223,080,
Number of special characters = 223,080 / 6760= 33.
So there are 33 special characters.
If the number and uppercase values are known then the various lowercase letters and special characters are the unknown values.
The number of possible combinations = number of lowercase letters × number of special characters = 26 × 33 = 858.
So the probability of guessing the password is 1 out of 858 combinations.

Question 37.
PROBLEM SOLVING
From a group of 5 scientists, an environmental committee of 3 people is selected. How many different committees are possible?
Answer:
Let,
The 5 scientists be indicated as A,B,C,D,E
The environmental committee of three people =
ABC, ABD, ABE ,ACD,ACE , ADE
BCD,BCE, BDE,CDE .
Therefore, there are 10 different ways to form a committee.

Lesson 7.4 Simulations

EXPLORATION 1

Using a Simulation
Work with a partner. A basketball player makes 80% of her free throw attempts.
a. Is she likely to make at least two of her next three free throws? Explain your reasoning.
b. The table shows 30 randomly generated numbers from 0 to 999. Let each number represent three shots. How can you use the digits of these numbers to represent made shots and missed shots?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 1
c. Use the table to estimate the probability that of her next three free throws, she makes

  • exactly two free throws.
  • at most one free throw.
  • at least two free throws.
  • at least two free throws in a row.

d. The experiment used in parts (b) and (c) is called a simulation. Another player makes \(\frac{3}{5}\) of her free throws. Describe a simulation that can be 5used to estimate the probability that she makes three of her next four free throws.
Answer:
A simulation is an experiment that is designed to reproduce the conditions of a situation or process.Simulations allow you to study situations that are impractical to create in real life.

Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 2

Try It

Question 1.
You randomly guess the answers to four true-false questions.
a. Design a simulation that you can use to model the answers.
b. Use your simulation to find the experimental probability that you answer all four questions correctly.

Answer:a.  There are two answers in a true-false question.
There is an equal chance of being correct or incorrect.
Therefore, we can use a coin to simulate answers where heads is correct and tails is incorrect.
Flip 4 coins in each trial to simulate the four answers. Run the simulation 50 times.

b.The table below shows the  results of simulation 50 times
Out of 50 trails , there are only 4 times all the answers are correct
So, the probability that you answer all four questions correct =
Consider C be the event of answer all four questions correct
P(C) =4 /50 = 0.08

   

Question 2.
A baseball team wins 70% of the time. Design and use a simulation to estimate the probability that the team wins the next three games.
Answer:
Given that there is 70% chance of winning
let us use numbers to determine the win or lose of game
Assume 0-6 numbers to represent win and 7-9 numbers represent lose since, there is 70% chance of winning.
The table below shows the possible results after simulation 3 numbers 50 times on a random number on a graphing calculator.
Out of 50 trails , 19 of them had three wins
Therefore, the probability = 19/50 =0.38

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
SIMULATING OUTCOMES
Four multiple-choice questions on a quiz each have five answer choices. You randomly guess the answer to each question. Design and use a simulation to find the experimental probability that you answer all of the questions correctly.
Answer:
Given , the number of questions = 4
The number of answer choices for each question = 5
The experimental probability that you answer all of the questions correctly = 0.5%

Question 4.
SIMULATING OUTCOMES
You select a marble from a bag and a chip from a box. You have a 20% chance of choosing a green marble and a 90% chance of choosing a red chip. Estimate the probability that you choose a green marble and a red chip.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 3
Answer:
Given ,
The chance of  choosing green marble = 20%
The chance of choosing a red chip =90%
The simulation about the probability that you choose green marble and red marble
88  5  86  13  31  49  33  21  99  97
30  62  18  4  63  3  32  94  8  77
24  87  74  56  19  42  61  75  81
45  84  51  17  15  46  66  69  34  28
36  9  64  53  59  10  58  57  39  43  93
The digits from 1 through 2 in tens place indicate green marble
The digits from 1 through 9 in tens place indicates red chip
The experimental probability of choosing green marble and red marble
P(B) = 8/50 = 0.16

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
Each day there is a 50% chance that your tablet overheats. Estimate the probability that your tablet overheats on exactly 2 of the next 3 days.
Answer:
Given that ,
The percent of  tablet overheats each day = 50%
The digits from 1 through 5 in the numbers
460  677  330  467  654  942  875  217  734  692
646  421  222  804  399  380  103  977  991  996
367  337  760  495  280  359  726  349  605  649
573  836  413  196  475  470  315  911  122  970
369  666  908  431  396  230  885  863  333  762
The digits 1 through 5 at least two in a three digit number indicate that your tablet overheats on exactly 2 of the next 3 days.
The experimental probability that your tablet overheats on exactly 2 of the next 3 days
P(S) = 15/ 50 = 0.3

Question 6.
DIG DEEPER
The probability that a homeowner needs a plumber this year is 22%. The probability that the homeowner needs a septic tank specialist is 14%. Estimate the probability that the homeowner needs a plumber, but not a septic tank specialist.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 4
Answer:
Given that,
The probability home owner needs a plumber this year = 22%
The probability that home owner needs septic tank specialist = 14%
The probability that the home owner needs plumber but not septic tank specialist
80  52  96  94  77  49  74  16  79  68
45  98  34  47  48  57  12  62  21  71
60  82  78  11  64  28  43  22  75  19
76  65  17  81  87  14  38  27  6  42
37  40  51  15  2  13  92  90  73  7
The numbers 1 through 2 in tens place indicates owner needs plumber
The number 2 through 9 in once place  indicate owner does not needs tank specialist
P(S) = 11 / 50 = 0.22

Simulations Homework & Practice 7.4

Review & Refresh
You flip a coin and roll the 20-sided figure. Find the probability of the compound event.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 5
Question 1.
Flipping tails and rolling at least a 14
Answer:
The total numbers of outcomes for a coin = 2
They are {Tails, heads }
Consider T be the event of flipping tails
P(T) = 1/2
The number of outcomes for a 20 sided figure  = 20
The probability of rolling at least a 14
Consider F be the event of rolling at least a 14
P(F) = 1/ 20
Compound event ( a x b )
= 1/2 x 1/20 = 1/40

Question 2.
Flipping heads and rolling less than 3
Answer:
The total numbers of outcomes for a coin = 2
They are {Tails, heads }
Consider H be the event of flipping heads
P(H ) = 1/2
The number of outcomes for a 20 sided figure  = 20
Consider L be the event of rolling less than 3
Number of numbers less than 3 = 2
They are {1,2}
P(L) = 2/ 20 = 1/10
Compound event ( a x b )
= 1/2 x 1/10 = 1/20

Simplify the expression.
Question 3.
5(a – 2)
Answer:
5(a – 2) =0
5a -10 =0
5a =10
a=10/5
a =2

Question 4.
– 7(1 + 3x)
Answer:
– 7(1 + 3x) =0

-7 – 21x=0
-21x= 7
x= 7/-21
x = -0.33

Question 5.
– 1(3p – 8)
Answer:
– 1(3p – 8) =0
-3p +8 =0
p =-8/-3  =-2.6

Concepts, Skills, & Problem Solving

USING A SIMULATION A medicine is effective for 80% of patients. The table shows 30 randomly generated numbers from 0 to 999. Use the table to estimate the probability of the event. (See Exploration 1, p. 307.)
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 6
Question 6.
The medicine is effective on each of three patients.
Answer:
Let the, digits 1 through 8 in the tens place represent the medicine is effective
P(E) = 24/30

Question 7.
The medicine is effective on fewer than two of the next three patients.
Answer:
P(N) = 6/30 = 0.2

SIMULATING OUTCOMES Design and use a simulation to find the experimental probability.
Question 8.
In your indoor garden, 50% of seeds sprout. What is the experimental probability that at least one of your next three seeds sprouts?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 7
Answer:
Use the random number generator on a graphing calculator.
Randomly generate 50 numbers from 0 to 99. The table below shows the results.
Let the digits 1 through 5 in the tens place represent the seeds are sprout

P(rain both days) =6/50
The experimental probability is 6/50 = 0.12, or 12%.

Question 9.
An archer hits a target 50% of the time. What is the experimental probability that the archer hits the target exactly four of the next five times?
Answer:
Given , the archer hits the target = 50%
25  82  53  49  24  95  31  66  40  90
42  30  9  78  4  80  16  99  23  85  39
44  6  61  46  5  87  64  36  21  57  58
98  81  13  97  29  18  92  22  77  35
55  14  60  28  38  89  56  69
Because, the makes 50 % of shots 1 to 50 represents successful shots .
the experimental probability that the archer hits the target exactly four of the next five times
P(X) = 25/50 =0.5 = 50%.

Question 10.
A bank randomly selects one of four free gifts to send to each new customer. Gifts include a calculator, a key chain, a notepad, and a pen. What is the experimental probability that the next two new customers both receive calculators? that neither receives a calculator?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 8
Answer:
Given that ,
Gifts include a calculator, a key chain, a notepad, and a pen.
The experimental probability that the next two new customers both receive calculators=
P(C) =1/4 x1/4 = 1/16
The experimental probability that the next two new customers both neither receive calculators=
P(N) = 3/4 x 3/4 = 9/16

Question 11.
Employees spin a reward wheel. The wheel is equally likely to stop on each of six rewards labeled A–F. What is the experimental probability that fewer than two of the next three spins land on reward A?
Answer:
Given , The six rewards are labelled as {A,B,C,D,E,F}
What is the experimental probability that fewer than two of the next three spins land on reward A
92  72  41  33  83  4  60  32  6  81
1  12  61  57  93  27  46  29  42  47
21  79  23  45  16  63  26  87  14  68
75  38  94  24  20  86  82  70  8  97
39  59  19  64  55  25  77  9  88  37
the experimental probability that fewer than two of the next three spins land on reward A = 4/50

USING NUMBER CUBES Design and use a simulation with number cubes to estimate the probability.
Question 12.
Your lawnmower does not start on the first try \(\frac{1}{6}\) of the time. Estimate the probability that your lawnmower will not start on the first try exactly one of the next two times you move the lawn.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 9
Answer:
Given the probability of the event is 1/6 = 16
The simulation of random generate calculator about 50 numbers
The digits from 1 through 16 indicates the event
50  99  88  6  98  76  63  21  65  42
96  84  92  37  27  33  11  57  69  68
48  40  10  30  24  26  67  47  60  34
59  87  18  78  7  79  12  95  9  8  1
86  38  45  93  44  4  15  73  39
The experimental probability of the event is 10/50
P(S) = 0.2

Question 13.
An application on your phone correctly identifies four out of every six songs. Estimate the probability that at least three of the next four songs are correctly identified.
Answer:

SIMULATING OUTCOMES Design and use a simulation to find the experimental probability.
Question 14.
Two beakers are used in a lab test. What is the experimental probability that there are reactions in both beakers during the lab test?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 10
Answer:
The simulation of beaker 1 and beaker 2 have reactions
The digits 1 through 8 in tens place indicate probability of beaker 1
The digits 1 through 5 in once place indicate probability of beaker 2
52 66 73 68 75 28 35 47 48 2
16 68 49 3 77 35 92 78 6 6
58 18 89 39 24 80 32 41 77 21
32 40 96 59 86 1 12 0 94 73
40 71 28 61 1 24 37 25 3 25
The experimental probability that there are reactions in both beakers during the lab test
P(R) =32/50 = 0.64

Question 15.
You use a stain remover on two separate stains on a shirt. What is the experimental probability that the stain remover removes both the mud stain and the food stain?
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 11
Answer:
The simulation of random generator calculator is as follows:
The digits 1 through 9 in tens place indicate the event in mud
The digits 1 through 8 in once place indicate the event of stain removal in food
62  35  50  43  96  37  70  31  59  40
55  92  89  14  10  41  87  95  11  99
12  53  71  26  38  24  80  36  16  42
85  32  3  33  47  1  49  51  8  23  88
69  15  64  84  81  60  66  72  74
The experimental probability that the stain remover removes both the mud stain and the food stain
P(S) = 41/50 = 0.82

Question 16.
DIG DEEPER!
The probability that a computer crashes one or more times in a month is 10%. Estimate the probability that the computer crashes at least one or more times per month for two months in a row during the first half of the year.
Answer:

Question 17.
MODELING REAL LIFE
You visit an orchard. The probability that you randomly select a ripe apple is 92%. The probability that you randomly select a ripe cherry is 86%. Estimate the probability that you pick an apple that is ripe and a cherry that is not ripe.
Big Ideas Math Solutions Grade 7 Chapter 7 Probability 7.4 12
Answer:
Given , the probability that you randomly select a ripe apple = 92%
The probability  that you pick a ripe cherry = 86%
The digits from 1 through 8 in tens place indicates event of choosing a ripe apple
The digits 9  in once place indicates event of not choosing a ripe cherry
75  99  69  33  67  1  22  17  18  37
29  27  9  12  54  7  31  39  26  87
10  72  82  42  36  85  74  3  5  92
29  11  21  81  76  77  52  13  90
15  73  69  70  8  23  53  59  51  44  48
The probability that you pick an apple that is ripe and a cherry that is not ripe = 9/50

Question 18
CRITICAL THINKING
You use a simulation to find an experimental probability. How does the experimental probability compare to the theoretical probability as the number of trials increases?
Answer:
Experimental probability is the result of an experiment. Theoretical probability is what is expected to happen.
In experimental probability, as the number of trials increases, the experimental probability gets closer to the theoretical probability.

Question 19.
LOGIC
At a restaurant,30% of customers donate to charity in exchange for a coupon. Estimate the probability that it will take at least four customers to find one who donates.
Answer:

Probability Connecting Concepts

Using the Problem-Solving Plan
Question 1.
In an Internet contest, gift cards and bicycles are given as prizes in the ratio 9 : 1. Estimate the probability that at least two of three randomly selected winners receive bicycles.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 1
Understand the problem.
You know the ratio of gift cards to bicycles awarded in the contest. You want to find the probability that atleast two of three randomly selected winners receive bicycles.
Make a plan.
Use the ratio to find the theoretical probability that a randomly selected winner receives a bicycle. Then use a simulation involving 50 randomly generated three-digit numbers to estimate the probability that at least two of three randomly selected winners receive bicycles.
Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:

Question 2.
A board game uses the spinner shown.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 2
a. Use theoretical probability to predict the number of times you will spin a number greater than or equal to 8 in 30 spins.
b. You play the game and record the results of 30 spins. Find the percent error of your prediction in part(a).
Answer:
a. Theoretical probability = number of favorable/ outcomes by total number of outcomes
Total number of outcomes = 10
The probability of spinning a number greater than or equal to 8 P(S) = 3/8 = 0.375
b. The experimental probability =
P(S) = 12/30= 0.4

Question 3.
The tiles shown are placed in a bag. You randomly select one of the tiles, return it to the bag, and then randomly select another tile. What is the probability that the product of the numbers on the tiles selected is greater than zero? Justify your answer.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 3
Answer:
Total number of tiles = 5
= 5×5 = 25 total number of outcomes
The probability that the product of the numbers on the tiles selected is greater than zero
P(Z) = 2/ 25
They are {1 x 2= 2 } and {2x 1 = 2 }
Therefore the probability of the product of the numbers on the tiles selected is greater than zero = 2/25 = 0.08

Performance Task

Fair and Unfair Carnival Games
At the beginning of this chapter, you watched a STEAM Video called “Massively Multi player Rock Paper Scissors.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cc 4

Probability Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 1

Graphic Organizers

You can use a Four Square to organize information about a concept. Each of the four squares can be a category, such as definition, vocabulary, example, non-example, words, algebra, table, numbers, visual, graph,or equation. Here is an example of a Four Square for probability.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 2
Choose and complete a graphic organizer to help you study the concept.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 3
1. favorable outcomes
2. relative frequency
3. experimental probability
4. theoretical probability
5. Fundamental Counting Principle
6. compound event
7. simulation

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 4

7.1 Probability (pp. 283–290)
Learning Target: Understand how the probability of an event indicates its likelihood.

You randomly choose one toy race car.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 5
Question 1.
How many possible outcomes are there?
Answer:
There are 5 green cars , one blue car and one red car
So the possible outcomes are 7
They are {green , red, blue}

Question 2.
What are the favorable outcomes of choosing a car that is not green?
Answer: 2
Number of green cars = 5
Number of cars that is not green = 2
They are {blue , red}
Therefore the possible outcomes of choosing a car that is not green = 2

Question 3.
In how many ways can choosing a green car occur?
Answer: 5
Number of green cars = 5
So, choosing a green car can occur in 5 ways
You spin the spinner. (a) Find the number of ways the event can occur. (b) Find the favorable outcomes of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 6
Question 4.
spinning a 1
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a 1 = 2
Total number of outcomes = 8
Consider O be the event of spinning a 1
Probability P(O) = Number of favorable outcomes/ Total number of outcomes
P(O) = 2/8 =1/4
So, the probability of spinning a 1 is 0.25

Question 5.
spinning a 3
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a 3 = 3
Total number of outcomes = 8
Consider T be the event of spinning a 3
Probability P(T) = Number of favorable outcomes/ Total number of outcomes
P(T) = 3/8 = 0.37
Therefore, the probability of spinning a 3 = 0.37

Question 6.
spinning an odd number
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning an odd number  = 5
They are {1,1,3,3,3}
Total number of outcomes = 8
Consider O be the event of spinning an odd number
Probability P(O) = Number of favorable outcomes/ Total number of outcomes
P(O) = 5/8 =0.62
Therefore , the probability of spinning an odd number = 0.62

Question 7.
spinning an even number
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning an even number = 3
They are {2,2,2}
Total number of outcomes = 8
Consider E be the event of spinning an even number
Probability P(E) = Number of favorable outcomes/ Total number of outcomes
P(E) = 3/8 = 0.37

Question 8.
spinning a number greater than 0
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a number greater than 0 =8
They are {1,1,2,2,2,3,3,3}
Total number of outcomes = 8
Consider S be the event of spinning a 1
Probability P(S) = Number of favorable outcomes/ Total number of outcomes
P(S) = 8/8 =1

Question 9.
spinning a number less than 3
Answer:
The spinner is divided into 8 equal parts
numbered with two 1s , three 2s and three 3s
{1,1,2,2,2,3,3,3}
The number of favorable outcomes of spinning a number less than a 3 = 5
They are {1,1,2,2,2}
Total number of outcomes = 8
Consider T be the event of spinning a number less than a 3
Probability P(T) = Number of favorable outcomes/ Total number of outcomes
P(T) = 5/8 = 0.62

Describe the likelihood of the event given its probability.
Question 10.
There is a 0% chance of snow in July for Florida.
Answer:
Given , the probability = 0
So the likelihood of the event is impossible

Question 11.
The probability that you are called on to answer a question in class is \(\frac{1}{25}\).
Answer:

Question 12.
There is an 85% chance the bus is on time.
Answer:
Given, 85% chance = 0.85
So, the probability is more likely to happen

Question 13.
The probability of flipping heads on a coin is 0.5.
Answer:
Given 0.5 = 50%
So, the probability of flipping heads is may or may not occur

Question 14.
During a basketball game, you record the number of rebounds from missed shots for each team. (a)Describe the likelihood that your team rebounds the next missed shot. (b) How many rebounds should your team expect to have in 15 missed shots?
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 7
Answer:
a.Total number of shots = 10
Consider M be the event of rebounds the next missed shot
The probability  = Number of missed shots / total number of trails
P(M) = 7/10 = 0.7
Therefore , the likelihood that your team rebounds the next missed shot is likely to happen

b.Total number of shots = 15
Consider S be the event of rebounds the your team misses the shot
probability P(S) = Number of rebounds / total number of trails
P(S) = 7/15 = 0.46
Therefore, the likelihood that the rebounds are expected to happen in 15 shots is not likely to occur

7.2 Experimental and Theoretical Probability (pp. 291–298)
Learning Target :Develop probability models using experimental and theoretical probability.

The bar graph shows the results of spinning a spinner 100 times. Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 15
Question 15.
spinning a 2
Answer: 0.18

Explanation :
Total number of spins = 100
The number of times 2 spun = 18
Experimental probability = Number of times an event occur / Total number of spins
Consider T be the event of spinning a 2
P(T) = 18/100 =0.18
Therefore , experimental probability of spinning a 2 = 0.18

Question 16.
spinning an even number
Answer: 0.39

Explanation:
Total number of spins = 100
Number of even numbers = 2
They are {2,4}
The number of times 2 spun = 18
The number of times 4 spun = 21
Total : 18+21 = 39
Experimental probability = Number of times an event occur / Total number of spins
Consider T be the event of spinning an even number
P(T) = 38/100 =0.39
Therefore , experimental probability of spinning an even number = 0.39

Question 17.
not spinning a 5
Answer: 0.81
Total number of spins = 100
Number of numbers other than 5 = 4
They are {1,2,3,4}
The number of times 1 spun = 20
The number of times 2 spun = 18
The number of times 3 spun = 22
The number of times 4 spun = 21
Total : 81
Experimental probability = Number of times an event occur / Total number of spins
Consider F be the event of not spinning a 5
P(F) = 81/100 =0.81
Therefore , experimental probability of not spinning a 5 = 0.81

Question 18.
spinning a number less than 3
Answer:
Total number of spins = 100
The numbers less than 3 = 2
They are {1,2}
The number of times 1 spun = 20
The number of times 2 spun = 18
Total : 18 + 20 = 38
Experimental probability = Number of times an event occur / Total number of spins
Consider T be the event of spinning a number less than 3
P(T) = 38/100 =0.38
Therefore , experimental probability of spinning a number less than 3  = 0.38

Question 19.
In Exercise 16, how does the experimental probability of spinning an even number compare with the theoretical probability?
Answer:

Use the spinner to find the theoretical probability of the event.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 20
Question 20.
spinning blue
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
And colored in 2 red , 2 blue , 2 purple and 2 orange in color
Consider , B be the event of spinning a blue
P(B) = 2/8
P(B) = 0.25

Question 21.
spinning a 1
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
Consider , O be the event of spinning a 1
P(O) = 3/8
P(O) = 0.37

Question 22.
spinning an even number
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
Total number of even numbers = 5
They are {2,4,6,8,10}
Consider , E be the event of spinning an even number
P(E) = 5/8
P(E) = 0.62

Question 23.
spinning a 4
Answer:
The spinner is divided into 8 equal parts
They are numbered with three 1 s ,one 2 ,one 4 ,one 6 , one 8 , and one 10
Consider , F be the event of spinning a blue
P(F) = 1/8
P(F) = 0.125

Question 24.
The theoretical probability of choosing a red grape from a bag of grapes is \(\frac{2}{9}\). There are 8 red grapes in the bag. How many grapes are in the bag?
Answer:
Given theoretical probability = 2/9
Number of red grapes = 8
Total number of grapes = x
8/x = 2/9
72 = 2x
x = 72/2
x = 35

Question 25.
The theoretical probability of choosing Event A is \(\frac{2}{7}\). What is the theoretical probability of not choosing Event A? Explain your reasoning.
Answer:

7.3 Compound Events (pp. 299–306)
Learning Target: Find sample spaces and probabilities of compound events.

Question 26.
You have 6 bracelets and 15 necklaces. Find the number of ways you can wear one bracelet and one necklace.
Answer:
Number of bracelets = 6
Number of necklaces = 15
According to fundamental counting principle,
a x b = 6 x 15 = 90
Therefore, In 90 ways you can wear one bracelet and one necklace

Question 27.
Use a tree diagram to find how many different home theater systems you can make from 6 DVD players,8 TVs, and 3 brands of speakers.
Answer:
Number of DVDs = 6
Number of TVs = 8
Number of brands of speaker = 3
Total number of outcomes = 6 x 8 x 3 = 144

Question 28.
A red, green, and blue book are on a shelf. You randomly pick one of the books. Without replacing the first book, you choose another book. What is the probability that you picked the red and blue book?
Answer:
Number of books in shelf = 3

Question 29.
You flip two coins and roll a number cube. What is the probability of flipping two tails and rolling an even number?
Answer:
number of total outcomes for coin = 2
The probability of flipping tails P(T ) = 1/2
For two coins , probability of flipping tails = 1/2 x 2 = 1/4
The number of total outcomes for 6
even numbers = {2,4,6}
The probability for flipping an even number = 3/6 = 1/2
So, the probability of flipping two tails and rolling an even number
= 1/4 x 1/2 = 1/8

Question 30.
Describe a compound event that has a probability between 50% and 80%.
Answer:

Question 31.
Your science teacher sets up six flasks. Two of the flasks contain water and four of the flasks contain hydrogen peroxide. A reaction occurs when you add yeast to hydrogen peroxide. You add yeast to two of the flasks. What is the probability that at least one reaction will occur?
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 31
Answer:

7.4 Simulations (pp. 307–312)
Learning Target: Design and use simulations to find probabilities of compound events.

Question 32.
You select a marble from two different bags. You have a 30% chance of choosing a blue marble from the first bag and a 70% chance of choosing a blue marble from the second bag. Design and use a simulation to estimate the probability that you choose a blue marble from both bags.
Answer:
Given , the chance of choosing blue marble from first bag =30%
The chance of choosing blue marble from second bag = 70%
82  4  90  96  18  65  11  71  97  53
48  63 32  54  52  17  14  5  83  50
61  36  10  81 58  84  9  80  94  95
15  42  2  45  68  64  33 38  12  60
35  16  21  99  59  55  22  20  37  87
Let the digits 1 through 3 in the tens place represents choosing blue marble from first bag
the digits 1 and 7 in the ones place represent choosing blue marble from second bag
let P(S) be the event choosing both blue marbles from both bags = 18/50
Therefore, the experimental probability = 0.36 = 36%

Question 33.
A cereal company is including a prize in each box. There are 5 different prizes, all of which are equally likely.
a. Describe a simulation involving 50 trials that you can use to model the prizes in the next 3 boxes of cereal you buy.
b. Use your simulation to find the experimental probability that all three boxes contain a different prize.
Answer:
Given ,
There are 5 different prizes, all of which are equally likely.

Question 34.
In the past month, your cell phone has lost its entire charge on 40% of days. Design and use a simulation to estimate the experimental probability that your cell phone loses its entire charge on exactly 2 of the next 5 days.
Answer:
Given,
Cell phone has lost its entire charge on 40%
Let the numbers 1 through 40 represents possible
6  79  3  23  82
84  67  14  59 88
11  47  26 68  18
 69  91  51  56  42
 15  99  58  92  62
 13  41  55  17  35
 28  54  85  97  65
 98  57  71  73  38
 89  90  87  39  72
 19  8  21  45  31
There are 2 or more successes in trails {1,3,6,10 }
The experimental probability that your cell phone loses its entire charge on exactly 2 of the next 5 days = 4/50
P(S) = 0.08 = 8%

Question 35.
You and your friends form a team in gym class. You have an 80% chance of winning a game of basketball and a 10% chance of winning a game of soccer. Design and use a simulation involving 50 randomly generated numbers to estimate the probability of winning both games.
Big Ideas Math Answer Key Grade 7 Chapter 7 Probability cr 35
Answer:
Given , The chance of winning a game of basket ball = 80%
The probability of winning a soccer game = 10%
33  91  74  87  53  67  63  9  47  52  17  8
40  58  100  11  44  20  49  72  60  66  79  51
69  73  76  43  77  97  2  93  12  36  6  86  92 59
 84  29  15  85  88  96  32  38  64  71  98  34
Let the digits 1 through 8 in the tens place represents winning a basketball
Let digits 1 and 2 in the ones place represent winning a soccer game.
let P(S) be the event of winning both the games = 7/50
Therefore, the experimental probability = 0.14 = 14%

Probability Practice Test

You randomly choose one game piece. (a) Find the number of ways the event can occur. (b) Find the favorable outcomes of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 1
Question 1.
choosing green
Answer:
Number of game pieces = 10
In which , 3 are yellow , 2 blue , 2 red and 1 are green in color
Probability P(G)= Number of favorable outcomes / Total number of outcomes
Consider G be the event of choosing green
P(G) = 1 / 10= 0..1
Therefore, the probability of choosing green = 0.1

Question 2.
choosing not yellow
Answer:
Number of game pieces = 10
In which , 3 are yellow , 2 blue , 2 red and 1 are green in color
Number of not yellow colors = 5
They are { blue , green , red }
P(N) = Number of favorable outcomes / Total umber of outcomes
P(N) = 5/10 = 0.5
Therefore, the probability of not choosing yellow= 0.5

Find the sample space and the total number of possible outcomes.
Question 3.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 3
Answer:
Number of SPF = 5
Number of types = 3
The total number of outcomes = 5 x 3 =15
Therefore number of possible outcomes = 15
{10,lotion} {15,lotion} {30, lotion} {45, lotion} {50, lotion}
{10, spray } {15, spray} {30, spray } {45, spray} {50, spray}
{10,gel} {15, gel} {30 , gel} {45 ,gel} {50 , gel}

Question 4.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 4
Answer:
Number of types = 4
Number of colors = 3
Total number of possible outcomes = 4 x3 = 12
{Basic display, Black  } {Basic display, white } {Basic display , silver}
{Scientific, Black} {Scientific, white} {Scientific , silver}
{Graphing ,black } {Graphing , white} {Graphing , silver}

Use the bar graph to find the experimental probability of the event.
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 5
Question 5.
rolling a 1 or a 2
Answer:
Total Number of rolls = 90
The number of times 1 rolled = 12
The number of times 2 rolled = 18
Total = 30
The probability of rolling a 1 or a 2 =
P(R) = 30/90
P(R) = 0.33

Question 6.
rolling an odd number
Answer:
The number of odd numbers = 3
They are {1,3,5}
Number of times 1 rolled = 12
Number of times 3 rolled = 14
Number of times 5 rolled =  16
Total : 12 + 14 + 16 =42
Consider O be the event of spinning an odd number
P(O) = 42/90 = 0.46

Question 7.
not rolling a 5
Answer:
The total number of rolls = 90
The number of  numbers other than 5 are =5
They are 1,2,3,4,6
The number of times 1 rolled = 12
The number of times 2 rolled = 18
The number of times 3 rolled = 14
The number of times 4 rolled = 17
The number of times 6 rolled = 13
Total = 74
The probability of rolling not 5 = 74/ 90
P(G) = 0.82

Question 8.
rolling a number less than 7
Answer:
The total number of rolls = 90
The number of  numbers less than 7 are = 6
They are 1,2,3,4,5,6
The number of times 1 rolled = 12
The number of times 2 rolled = 18
The number of times 3 rolled = 14
The number of times 4 rolled = 17
The number of times 5 rolled = 16
The number of times 6 rolled = 13
Total = 90
The probability of rolling a number less than 7 = 90/ 90
P(L) =1

Use the spinner to find the theoretical probability 93of the event(s).
Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 9
Question 9.
spinning an even number
Answer:
The spinner is divided into 9 equal parts
Numbered as {1,2,3,4,5,6,7,8,9}
So, total number of outcomes = 9
Number of even numbers = 4
They are {2,4,6,8}
Theoretical probability = Number of favorable outcomes / Total number of outcomes
Consider , E be the event of choosing an even number
P(E) = 4/ 9 = 0.44
Therefore , the probability for spinning an even number =0.44

Question 10.
spinning a 1 and then a 2
Answer:
The spinner is divided into 9 equal parts
Numbered as {1,2,3,4,5,6,7,8,9}
So, total number of outcomes = 9
The probability of spinning 1 = 1/9
Then the probability of spinning 2 = 1/9
Total probability of the event spinning a 1 and then 2 = 1/9 x 1/9
Total P(B) = 1/ 81
Therefore , the probability of spinning a 1 and then 2 = 0.012

Big Ideas Math Answers 7th Grade Chapter 7 Probability pt 11
Question 11.
You randomly choose one of the pens shown. What is the theoretical probability of choosing a black pen?
Answer:
Theoretical probability = Number of favorable outcomes / Total number of outcomes
Total number of pens = 5
Number of black pens = 2
Consider B be the event of choosing a black pen
P(B) = 2/5 = 0.4
Therefore, the theoretical probability of choosing a black pen = 0.4

Question 12.
You randomly choose one of the pens shown. Your friend randomly chooses one of the remaining pens. What is the probability that you and your friend both choose a blue pen?
Answer:
Total number of pens = 5
Number of blue pens = 2
The probability that you and your friend both choose a blue pen
P(B) = 2/5
P(B) = 0.4

Question 13.
There is an 80% chance of a thunderstorm on Saturday. Describe the likelihood that there is not a thunderstorm on Saturday.
Answer:
Given probability 80% = 0.8
Not thunder storm probability = 20 % = 0.2
The likelihood of the event is not likely to happen

Probability Cumulative Practice

Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 1
Question 1.
A school athletic director asked each athletic team member to name his or her favorite professional sports team. The results are below:

  • D.C.United: 3
  • Florida Panthers: 8
  • Jacksonville Jaguars: 26
  • Jacksonville Sharks: 7
  • Miami Dolphins: 22
  • Miami Heat: 15
  • Miami Marlins: 20
  • Minnesota Lynx: 4
  • New YorkKnicks: 5
  • Orlando Magic: 18
  • Tampa Bay Buccaneers: 17
  • Tampa Bay Lightning: 12
  • Tampa Bay Rays: 28
  • Other: 6

One athletic team member is picked at random. What is the likelihood that this team member’s favorite professional sports team is not located in Florida?
A. certain
B. likely, but not certain
C. unlikely, but not impossible
D. impossible
Answer: B
13/14 = 0.92
The likelihood of the event is likely, but not certain

Question 2.
Each student in your class voted for his or her favorite day of the week. Their votes are shown in the circle graph:
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 2
A student from your class is picked at random. What is the probability that this student’s favorite day of the week is Sunday?
Answer:
Total number of students =30
Total number of Sunday = 6
The probability of that this student’s favorite day of the week is Sunday
P(S) = 6/30
= 0.2

Question 3.
What value makes the equation 11 – 3x = – 7 true?
F. – 6
G. \(-\frac{4}{3}\)
H. 6
I. 54
Answer:  F

Explanation:
11 – 3x = -7
Consider the value of x = -6
L.H.S  = 11-3X
11 – 3 (-6)
11 – 18
= -4
L.H.S = R.H.S
Therefore, the value of x =-6

Question 4.
Your friend solved the proportion in the box below.
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 4
What should your friend do to correct the error that he made?
A. Add 40 to 16 and 27 to p.
B. Subtract 16 from 40 and 27 from p.
C. Multiply 16 by 27 and p by 40.
D. Divide 16 by 27 and p by 40.
Answer: C
Multiply 16 by 27 and p by 40

Question 5.
Which value is a solution of the inequality?
3 – 2y < 7
F. – 6
G. – 3
H. – 2
I. – 1
Answer: H
3- 2y < 7 =0
3-2y >7
-2y > 7-3
-2y >4
y > -4/2
y > -2

Question 6.
A spinner is divided into eight equal sections, as shown. You spin the spinner twice. What is the probability that the arrow will stop in a yellow section both times?
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 6
Answer:
The spinner is divided 8 equal parts
In which , 3 are red , 2 are blue, 2 are yellow and 1 part is in pink color
so the probability of landing on yellow would be
P(Y) = 3/8 on first spun.
Therefore, for spinning the spinner twice then the probability of landing it on yellow would be
P(Y) = (3/8)/2
P(Y) =3/16

Question 7.
A pair of running shoes is on sale for 25% off the original price. Which price is closest to the sale price of the running shoes?
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 7
A. $93
B. $99
C. $124
D. $149
Answer: A

Explanation :
Original price = $122.76
Percent of discount = 25% = 0.25
The amount of discount :
Amount of discount = Discount x original price
= 0.25 x $125
= $30.39
Sale price :
Sale price = Original price – discount
= $122.76 – $30.39
= $92.07
It is very close to the sale price of the running shoes = $ 93

Question 8.
The value of a baseball card was $6 when it was sold. The value of this card is now $15. What is the percent increase in the value of the card?
F. 40%
G. 90%
H. 150%
I. 250%
Answer: H
Given,
Original price of the basket ball  = $6
New price = $15
The original price increases so, the percent of change is called percent of increase
Percent of increase = New price – original price/ Original price
On substituting ,
= 15 – 6/ 6
= 9 / 6
=3/2
On multiplying with 100
= 3/2 x 100
= 150%
Therefore , the percent increase in the value of the card = 150%

Question 9.
You roll a number cube twice. You want to roll two even numbers.
Big Ideas Math Answers 7th Grade Chapter 7 Probability cp 9
Part A Find the number of favorable outcomes and the number of possible outcomes of each roll.
Part B Find the probability of rolling two even numbers. Explain your reasoning.
Answer:
Part A :
The even numbers on a number cube are 2, 4, and 6
The favorable outcomes for 2 are =3
They are : 2 and 2 , 2 and 4 , 2 and 6
The favorable outcomes for 4 are =3
They are : 4 and 2 , 4 and 4 , 4 and 6
The favorable outcomes for 6 are =3
They are : 6 and 2 , 6 and 4 , 6 and 6
So,  total number favorable outcomes = 9
There are 6 outcomes for each cube so there are 6(6) = 36 total outcomes.

Part B :
There are 9 favorable outcomes of a total of 36 outcomes
So, the probability is 9/36 = 1/4

Question 10.
You put $600 into an account. The account earns 5% simple interest per year. What is the balance after 4 years?
A. $120
B. $720
C. $1800
D. $12,600
Answer: B
Simple interest = principle x annual rate of interest x Time
I = P x r x t
5% = 0.5
I = 600 x 0.05 x 4
I = $120
The interest earned = $120
The balance after 4 years = principle + Interest
= 600 + 120
=$720

Question 11.
You are comparing the prices of four boxes of cereal. Two of the boxes contain free extra cereal.

  • Box F costs $3.59 and contains 16 ounces.
  • Box G costs $3.79 and contains 16 ounces, plus an additional 10% for free.
  • Box H costs $4.00 and contains 500 grams.
  • Box  I costs $4.69 and contains 500 grams, plus an additional 20% for free.

Which box has the least unit cost?
F. Box F
G. Box G
H. Box H
I. Box I
Answer: Box G

Explanation :
Box F
The unit cost of box F = 3.59/16
The unit cost = $0.224 per ounces
Box G
Given 10% free
Which means, 10% = 16.01 = 0.6 ounces
The unit cost box G = $3.79 / 16 +0.6
=$0.215 ounces
Box H
Given, 500 grams
Which means , 500/28.35 = 17. 64 ounces

The unit cost for box H = $4.00/ 17.64
= 0.227 ounces
Box I
Given 500 grams and 20 % additional
Which means , 500/28.35 = 17. 64 ounces
20% = 17.64 .02 = 3.53 ounces
The unit cost = 4.69/ 17.64+3.54
= 0.222 ounces
Therefore, Box G has the least unit cost .
Final Words:
We wish the information prevailed here is helpful for all the students of grade 7. For any doubts post the comments in the below comment section. We will try to clarify your doubts as early as possible.

Big Ideas Math Answers Grade 2 Chapter 14 Money and Time

Students can browse Big Ideas Math Answers Grade 2 Chapter 14 Money and Time on CCSS Math Answers. Students can definitely score good marks in the exam with the help of the Big Ideas Math Book 2nd Grade Answer Key Chapter 14 Money and Time. The students can download and practice the questions from Big Ideas Math Answers Grade 2 Chapter 14 Money and Time for free of cost. Pupils can habituate practicing maths if they start their practice with Big Ideas Math Book Grade 2 Solution Key Chapter 14 Money and Time.

Big Ideas Math Book 2nd Grade Answer Key Chapter 14 Money and Time

So, make use of the Big Ideas Math Book 2nd Grade Chapter 14 Money and Time Solution Key links and go through the solutions. You can step by step explanation for all the questions in Big Ideas Math Grade 2 Answers Chapter 14 Money and Time. It is essential for all the students to learn the concepts of this chapter in-depth. Check out the list of the topics before you start your preparation from the below section.

Vocabulary

Lesson: 1 Find Total Values of Coins

Lesson: 2 Order to Find Total Values of Coins

Lesson: 3 Show Money Amounts in Different Ways

Lesson: 4 Make One Dollar

Lesson: 5 Make Change from One Dollar

Lesson: 6 Find Total Values of Bills

Lesson: 7 Problem Solving: Money

Lesson: 8 Tell Time to the Nearest Five Minutes

Lesson: 9 Tell Time Before and After the Hour

Lesson: 10 Relate A.M. and P.M.

Chapter: 10 – Money and Time

Money and Time Vocabulary

Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 1

Organize It
Use the review words to complete the graphic organizer.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 2

 

Answer:
1. Analog clock .
2. The long hand pointing minutes is the minutes hand.
3. The short hand pointing hours is the hour hand.

Define it
Use your vocabulary cards to complete the puzzle.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 3

Answer:

1. Penny
2.Dime
3. Quarter
4. Nickel

Chapter 14 Vocabulary Cards

Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 4
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 5
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 6
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 7
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 8
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 9

Lesson 14.1 Find Total Values of Coins

Explore and Grow

Sort your coins.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 10
Explain how you sorted.
__________________
__________________
__________________
__________________

 

Answer:

Show and Grow

Count on to find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 11

 

Answer:
36 cents

Explanation :
1 dime = 10 cents
That means, 3x 10 = 30 cents
1 nickel = 5 cents
That means,  1x 5= 5 cents
and 1 cent
So , 3 dime +1 nickel+1 cent = 30 cents+ 5 cents +1 cent = 36 cents.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 12

 

Answer:
42 cents

Explanation:
1 quarter = 25 cents
That means , 1 x25= 25 cents
1 nickel = 5 cents
That means , 3  x 5  = 15 cents & 2 cents additional
Total : 25 cents + 15 cents + 2 cents = 42 cents
Therefore , total value is 42 cents .

Apply and Grow: Practice

Count on to find the total value.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 13

 

Answer:
55 cents

Explanation :
1 quarter = 25 cents
Which means , 2 x 25 = 50
1 nickel = 5 cents
Total : 50 + 5 =55 cents
Total value of coins is 55 cents.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 14

 

Answer:
48 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents

Which means,  2 x 10 = 20 cents
and 3 cents
Total : 25 +20 +3 = 48 cents
Therefore, Total value is 48 cents.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 15

 

Answer:
66 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means , 3 x 10 = 30 cents
1 nickel = 5 cents
Which means , 2 x 5 = 10 cents
& 1 cent
Total : 25 +30+ 10 +1=66
Therefore , the total value of coins is 66 cents.

Question 6.
Reasoning
You have 27¢. Which groups of coins could you have?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 16

 

Answer:

This set of coins have total of 27 cents

Explanation:
1 dime = 10 cents
Which means, 2 x 10 = 20 cents
1nickel = 5 cents
and 2 cents
Total : 20 + 5 + 2 = 27 cents.

Think and Grow: Modeling Real Life

You have 2 quarters, 1 dime, 4 nickels, and 1 penny. How many cents do you have? Do you have enough money to buy the airplane?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 17
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 18

Answer:
I have 81 cents.
Yes, I have enough money to buy the airplane.

Explanation:
1 quarter = 25 cents
So 2 quarters = 25+25 =50 cents
1 dime =10 cents
1 nickel = 5 cents
So 4 nickels = 4 x 5 = 20 cents
1 penny =1 cent
Total amount is 50+10+20+1 =81 cents
The airplane costs 80 cents so, I have enough money to buy the plain.

Show and Grow

Question 7.
You have 5 dimes, 3 nickels, and 2 pennies. How many cents do you have? Do you have enough money to buy the coloring book?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 19
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 19.1

Answer:
I have 67 cents.
No, I do not have enough money to buy the coloring book.

Explanation:
1 dime =10 cents
so, 5 dimes = 5 x 10= 50 cents
1 nickel = 5 cents
So, 3 nickels = 3 x 5 = 15 cents
2 penny =2 cents
Total amount is 50+15+2 =67 cents
The coloring book costs 70 cents
So, I do not  have enough money to buy the book.

Question 8.
You have 4 dimes, 1 nickel, and 3 pennies. How many more cents do you need to buy the whistle? Draw and label the coins you need.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 20

Answer:
I have 48 cents.
No , I  do not have enough money to buy the whistle.

Explanation:
1 dime =10 cents
so, 4 dimes =4 x 10= 40 cents
1 nickel = 5 cents
3 pennies =3 cents
Total amount is 40+5+3 =48 cents
The whistle costs 59 cents so, I  do not have enough money to buy the whistle.

Question 9.
DIG DEEPER!
You have 3 quarters, 2 nickels, and 3 pennies. Your friend has 1 quarter and 5 dimes. Who has more money? How much more?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 21

Answer:
I have 13 cents money more than my friend .

Explanation:
My money :
1 quarter = 25 cents
So 3 quarters = 3 x 25 = 75 cents
1 nickel = 5 cents
So 2 nickel = 2 x 5 = 10 cents
3 pennies = 3 cents
Total amount: 75+10+3 = 88 cents
So, I have a total of 88 cents .
Friends money:
1 quarter= 25 cents
1 dime = 10 cents
So 5 dimes = 5 x 10 = 50 cents
Total amount = 25 +50= 75 cents
Friends money total is 75 cents
Now, the difference between my money and my friends money is
88 cents – 75 cents = 13 cents
That means, I have 13 cents more than my friend.

Find Total Values of Coins Homework & Practice 14.1

Count on to find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 22

Answer:
65 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means , 4 x 10 = 40 cents
Total : 25 + 40 = 65 cents
Therefore, the total value of coins is 65 cents.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 23

Answer:
38 cents

Explanation:
1 dime = 10 cents
Which means , 3 dime = 3 x 10 = 30 cents
1 nickel = 5 cents
and 3 cents
Total : 30+ 5 + 3 = 38 cents

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24

Answer:
72 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
Which means, 4 x 10 = 40 cents
1 nickel = 5 cents
and 2 cents
Total : 25 + 40+ 5 + 2 =72 cents

Question 4.
DIG DEEPER!
You had 52¢. You lost a coin. Now you have the 5 coins shown. What coin did you lose?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24.1

 

Answer:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 2 cents
Total :  25+10+5+2 = 42 cents
Total amount i have is 52 cents
Now , 52 – 42 = 10
1 dime = 10 cents
So , I lost 1 dime.

Question 5.
Precision
Circle coins to show 80¢.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 24.2

 

Answer:
1 dime = 10 cents
80 cents = 8 dimes

Question 6.
Modeling Real Life
You have 3 quarters, 1 nickel, and 4 pennies. How many cents do you have? Do you have enough money to buy the boomerang?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 25

 

Answer:
No, I do not have enough money to buy the boomerang.

Explanation:
The cost of boomerang is 94 cents
1 quarter = 25 cents
3 x 25 =75 cents
1 nickel =5 cents
4 pennies = 4 cents
Total amount : 75+5+4 =84
I have 84 cents .
No, I do not have enough money to buy the boomerang

Question 7.
Modeling Real Life
You have 1 quarter, 3 dimes, and 1 nickel. How many more cents do you need to buy the toy bird? Draw the coins you need.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 26

Answer:
The cost of bird is 75 cents.

Explanation:
1 quarter = 25 cents.
1 dime = 10 cents,
3 dimes = 3 x 10 = 30 cents
1 nickel = 5 cents
Total amount : 25 +30+ 5 =60 cents
Now,
75- 60 = 15cents
I need 15 more cents to buy the toy bird .

Review & Refresh

Compare.

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 27

Answer:
324 > 317
327 is greater 317

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 28

Answer:
426 > 206
426 is greater than 206

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 29

Answer:
546 < 564
564 is greater than 546

Question 11.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 30

Answer:
931 > 842
931 is greater than 846

Lesson 14.2 Order to Find Total Values of Coins

Explore and Grow

Order your coins from the greatest value to the least value. Draw and label each coin with its value. What is the total value of all the coins?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 31
Explain how ordering the coins helped you find the total.
_____________________________________
_____________________________________
_____________________________________
_____________________________________

Answer:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value of coins :
25+10+5+1=41 cents

>>>

Show and Grow

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 32

 

Answer:
45 cents

Explanation:
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 2 x 5 = 10 cents
Total value :
25 +10 + 10 = 45 cents

>

Quarter       >         Dime     >   Nickel

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 33

 

Answer:
1 quarter = 25 cents
which means, 2quarters = 2 x 25= 50 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cents
Total : 50 +10+5+1 = 66 cents
Therefore total value of coins is 66 cents .

>>>

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 34

Answer: 31 cents

Explanation:
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
which means , 2 nickel = 2 x5 = 10 cents
and 1 penny = 1 cent
Total : 20 +10 +1 = 31
Therefore, total value of coins is 31 cents

>>

Apply and Grow: Practice

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 35

 

Answer:
41 cents
Explanation :
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value = 25 +10 +5 +1 =41
Therefore total value of coins is 41 cents

>>>

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 36

 

Answer:
27 cents

Explanation :
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
and 2 pennies = 2 cents
total : 20 +5 +2= 27
Therefore total value of coins is 27 cents .

>>

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 37

Answer:
85 cents

Explanation :
1 quarter = 25 cents
Which means, 2 quarter = 2 x25 = 50 cents
1 dime = 10 cents
Which means, 3 dimes = 3 x 10 = 30 cents
And 1 nickel = 5 cents
Total :
50 +30 +5 = 85 cents
Therefore total value of coins is 85 cents.

>>

Question 7.
Reasoning
You have a dime, a nickel, and one other coin. The total value is 40¢. What is your third coin?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 38

Answer:
The total value of  3 coins is 40 cents
1 dime = 10 cents
1 nickel = 5 cents
10 cents + 5 cents = 15 cents
Now , 40 cents – 15 cents =  25 cents
So the other coin is quarter
1 quarter = 25 cents
Therefore, 25 +10+ 5 = 40 cents.

Think and Grow: Modeling Real Life

You have 3 nickels and 4 pennies in one pocket. You have 2 dimes and 2 quarters in your other pocket. How much money do you have in all? Do you have enough money to buy the car?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 39
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 40

Answer:
The cost of car is 9 cents .
The amount i have in one pocket is 3 nickels and 4 pennies
1 nickel = 5 cents
which means , 3 nickels = 3 x 5 = 15 cents
4 pennies = 4 cents
Total = 15 cents + 4 cents = 19 cents
The amount i have in another pocket is 2 dimes , 2 quarters
1 dime = 10 cents
in which , 2 dimes = 2 x 10 = 20 cents
1 quarter = 25 cents
2 quarters = 2 x 25 = 50 cents
Total : 20 cents + 50 cents = 70 cents
Total amount in both the pockets is 70 cents + 19 cents =89 cents
No, I do not have enough money to  buy the car.

Show and Grow

Question 8.
You have 30¢. You find 2 nickels, 1 dime, and 3 pennies in your room. How much money do you have now? Do you have enough money to buy the yo-yo?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 41
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 42

Answer:
I have 30 cents
Amount I found in my room =
1 nickel = 5 cents ,
2 nickel x 5 = 10 cents.
1 dime =10 cents.
3 pennies = 3 cents.
Total  : 10 +10 + 3
=23 cents
Now, total amount i have is
30 cents +23 cents = 53 cents
The cost of yo-yo is 50 cent
So , now I have enough money to buy the yo- yo

Question 9.
You have 1 nickel, 1 quarter, and 4 dimes. How many more cents do you need to buy the stuffed animal? Draw and label the coin you need.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 43

Answer:
1 nickel = 5 cents
1 quarter= 25 cents
1 dime =10 cents
Which means 4 dimes = 4 x 10=40 cents
Total amount = 5+25+40 =70 cents
But the cost of stuffed animal is 80 cents
So , I do not have enough cents to buy the stuffed animal.
I need 10 more cents to buy the stuffed animal.

Question 10.
DIG DEEPER!
You have 65¢. You give your friend a dime. You have 3 coins left. Draw and label the coins you have left.

Answer:
Given that,
I have 65 cents
The amount I gave to my friend is 1 dime
1 dime = 10 cents
Now,
65 cents – 10 cents = 55 cents
55 cents can be shown in 3 coins as
2 quarters and 1 nickel

Order to Find Total Values of Coins Homework & Practice 14.2

Draw and label the coins from the greatest value to the least value. Then find the total value.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 44

 

Answer: 35 cents

Explanation :
1 dime = 10 cents
which means, 2 dimes = 2 x 10 = 20 cents
and 1 nickel = 5 cents
which means, 3 nickels = 3 x 5 = 15 cents
Total:  20 +15 = 35 cents
Therefore total value of coins is 35 cents.

>

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 45

Answer: 63 cents

Explanation:
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
and 3 pennies = 3 cents
Total : 50 +10+3 = 63 cents

>>

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 46

Answer: 46 cents

Explanation :
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 2 nickel = 2 x 5 = 10 cents
and 1 penny = 1 cents
Total :
25 +10 +5 +5+1 = 46 cents

>>>

Question 4.
Open-Ended
Draw and label four coins that have a total value of 40¢

Answer: 40 cents can be written as 4 dimes
1 dime = 10 cents
Which means , 4 x 10 = 40 cents.

Question 5.
Modeling Real Life
You have 46¢. You find 4 pennies and 1 nickel in your room. How much money do you have now? Do you have enough money to buy the app?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 47
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 48

Answer:
I have 46 Cents

Explanation:
The amount I found is
4 pennies =  4 cents
1 nickel = 5 cents
Total  : 5 + 4 = 9 cents
Now,
Total amount I have is 46 cents +9 cents = 55 cents
The cost of app is 50 cents
yes , I have enough money to buy the app.

Question 7.
DIG DEEPER!
You have some nickels and dimes. You have 1 more nickel than dimes. The total value of your coins is 50¢. How many nickels and dimes do you have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 49

Answer:
I have  4 nickels and  3 dimes.

Explanation:
1 nickel =5 cents
That means, 4 nickels= 4 x 5 = 20 cents
1 dime = 10 cents
That means , 3 dimes = 3 x 10 = 30
Total: 20 + 30 = 50 cents
So, I have  4 nickels and  3 dimes.

Review & Refresh

Question 8.
Which fruit is the least favorite?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 50

Answer:
Cherry is the least favorite.

Explanation :
In the given table ,
Total lines for orange fruit are 5
Total lines for cherry fruit are 2
Total lines for apple fruit are  4
On comparing them , 2 is smaller than 4 and 5
so, Cherry is the least favorite.

Lesson 14.3 Show Money Amounts in Different Ways

Explore and Grow

Use your coins to show 25 cents in two different ways. Draw and label the coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 51

Answer :
25 cents can be written as
1 quarter
1 quarter = 25 cents

Another way :
25 cents can be written as
2 dime , 1 nickel
1 dime = 10 cents
Which means , 2 dime = 2 x 10= 20
1 nickel = 5 cents.

Did everyone in your class use the same coins?
_____________________________________
_____________________________________
_____________________________________
_____________________________________

Answer:
No, everyone used different coins to make a total value of 25 cents
For example, my friend used 1 dime, 3 nickels
1 dime = 10 cents
1 nickel = 5 cents
Which means , 3 x 5 = 15 cents
Total : 10 + 15 = 25 cents

Show and Grow

Show the amount in two different ways.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 52

 

Answer:
45 cents can be written as
4 dime + 1 nickel
Which means  one dime = 10 cents
4 dime = 4 x 10 = 40
1 nickel = 5 cents
Total: 40+5 = 45
Another way: 1 quarter +2 dime
Which means , 1 quarter =25 cents
1 dime= 10 cents
2 dime = 2 x 10 = 20
Total=: 25+20= 45
Therefore, 45 can be written as  4 dime 1 nickel and 1 quarter 2 dime

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 53

Answer:
27 cents can be written as
2 dime +1 nickel+2 penny
Which means , 1 dime = 10 cents
2 dime = 2 x 10= 20 cents
1 nickel = 5 cents2 penny = 2 cents
Total: 20+5+ 2= 27 cents
Another way:
1 quarter+2 pennies
Which means , 1 quarter = 25 cents
2 penny = 2 cents
Total= 25 + 2 = 27 cents
Therefore 27 cents can be written as 2 dime +1 nickel+2 penny and 1 quarter +2 pennies

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 54

Answer:
80 cents can be written as
8 dimes
Which means 1 dime = 10 cents
So 8 dimes = 80 cents
Another way:
3 quarter + 1 nickel
Which means , 1 quarter = 25 cents
3 quarters = 3 x 25 = 75 cents
1 nickel = 5 cents
Total : 75 + 5 = 80 cents
Therefore, 80 cents can e written as 8 dimes and 3quarter and 1 nickel .

Apply and Grow: Practice

Show the amount in two different ways.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 55

Answer:
23 cents can be written as
2 dime  3 pennies
Which means , 1 dime = 10 cents
2 dimes =2 x 10 = 20 cents
3 pennies = 3 cents
Total : 20 + 3 = 23 cents
Another way:
4 nickel ,3 pennies
Which means, 1 nickel = 5 cents
4 nickel = 4 x 5 =20 cents
3 pennies = 3 cents
Total= 20 + 3 =23 cents
Therefore 23 cents can be written as 2 dimes + 3 pennies and 4 nickel + 3 pennies

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 56

Answer:
49 cents can be written as
1 quarter ,2 dime ,4 pennies
Which means , 1 quarter = 25 cents
1 dime = 10 cents
So,2 dime = 2 x 10 = 20 cents
4 pennies = 4 cents
Total : 25 +20+4 =49 cents
Another way :
49 Pennies
1 penny = 1 cent
49 pennies = 49 cents
Therefore, 49cents can be written as 1 Quarter 2 dime 4 pennies and 49 pennies.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 57

 

Answer:
75 cents can be written as
3 quarters
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75
Another way:
1 quarter, 5 dimes
1 quarter = 25 cents
1 dime = 10 cents
Which means , 5 dimes = 5 x 10 = 50 cents
Total :  25 + 50 = 75 cents
Therefore , 75 cents can be written as 3 quarters and 1 quarter , 5 dimes

Question 7.
Structure
You have 55¢. You have no quarters. Draw to show what coins you might have.

Answer:
55 cents be written as 5 dimes , 1 nickel .
1 dime = 10 cents
Which means , 5 dimes = 5 x 10 = 50 cents
1 nickel = 5 cents
Total : 50 cents + 5 cents = 55 cents.

Question 8.
YOU BE THE TEACHER
Newton says he drew the fewest number of coins to show 66¢. Is he correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 58

Answer :
No, he is wrong
Newton used 6 coins to show 66 cents .
66 cents can also be shown as
2 quarters , 1 dime , 1 nickel and 1 cents
Total number of coins is 5
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
1 nickel = 5 cents
and cent
Total value of coins is 50+ 10 + 5 +1 = 66
Therefore ,66 cents can be shown in 5 coins .
so , Newton is wrong.

Think and Grow: Modeling Real Life

Newton has 2 dimes, 1 nickel, and 1 penny. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 59

 

Answer:

Total number of coins newton have is 4
2 dimes , 1 nickel and 1 penny
1 dime = 10 cents
Which means, 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
and 1 penny = 1 cent

Total value of coins is
20 +5+1 = 26 cents.
Given that ,
Descartes uses fewest number of coins to make same amount
26 can also be shown with 2 coins
1 quarter = 25 cents
1 penny = 1 cents
Total value is
25 + 1 = 26 cents

So, 26 cents can be shown with 2 coins
2 is less than 6 so , Descartes uses few coins than Newton to show the same amount.

Show and Grow

Question 9.
Newton has 3 dimes and 2 pennies. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 60

Answer:
Total number of coins newton have is 5
3 dimes , 2 pennies
1 dime = 10 cents
Which means , 3 dimes = 3 x 10 = 30 cents
2 pennies = 2 cents.
Total value of coins is
30 + 2 = 32 cents

Given that,
Descartes used the fewest number of coins to make the same amount
32 cents can be shown in 4 coins
1 quarter, 1 dime and 2 pennies
1 quarter = 25 cents
1 dime = 10 cents
and 2 pennies = 2 cents
Total : 25 + 10 + 2 = 37 cents

 

So, 37 cents can be shown in 4 coins
4 is less than 5.
Descartes used few coins than newton to show the same amount

Question 10.
You use fewer than 5 coins to buy the pack of gum. Draw and label coins to show how you pay.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 60.1

Answer:
The cost of gum is 70 cents
70 cents can be shown in 4 coins as
2 quarters , 2 dime
1 quarter = 25 cents
Which means , 2 x 25 = 50 cents
1 dime = 10 cents
Which means , 2 x 10 = 20 cents
Total amount = 50 + 20 = 70

Question 11.
DIG DEEPER!
You have 2 quarters. Newton and Descartes each have 5 coins and the same amount of money as you. Their coins are different. Draw and label their coins.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 61

Answer:
1 quarter = 25 cents
Which means, 2 quarters= 2 x 25 = 50 cents
50 cents can be shown in 5 coins as 5 dime
1 dime = 10 cents
Which means 50 cents = 5 dimes

Show Money Amounts in Different Ways Homework & Practice 14.3

Show money amounts in different ways.

Question 1.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 62

Answer:
15 cents can be written as:
1 dime +1 Nickel
which means 1 dime = 1o cents
1 nickel =5 cents
10+5 = 15 cents
Another way:
15 cents can also be written as:
1 dime +5 pennies
1 dime =10 cents
5 pennies = 5 cents
Total: 10 +5 = 15 cents
So, 15 cents can be written as 1 dime 1 nickel and 1 dime 5 pennies

Question 2.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 63

 

Answer:
58 cents can be written as
2 quarter 8 pennies
1 quarter = 25 cents
Which means 2 quarters = 2 x 25 = 50 cents
8 pennies = 8 cents
Total : 50 + 8 = 58 cents
Another way :
5 dime 1 nickel 3 pennies
Which means,
1 dime = 10 cents
=5 dime = 5 x 10 = 50cents
1 nickel = 5 cents
1 pennies = 3 cents
Total : 50+ 5+3= 58
So, 58 cents can be written as 2 quarter 8 pennies and 5 dime 1 nickel 3 pennies.

Question 3.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 64

Answer:
90 cents can be written as
3 quarters 1 nickel 1 dime
Which means,
1quarter =25 cents
3 quarters = 3 x 25 = 75 cents
1 nickel =5 cents
1 dime = 10 cents
Total : 75+5+10=  90 cents
Another way :
9 dimes
1dime = 10 cents
9 dimes = 9 x 10 = 90 cents .
So , 90 cents can be written as 3 quarters1 dime 1 nickel and 9 dimes.

Question 4.
Reasoning
Draw to show 60¢ with only 3 coins.

Answer:
60  cents can be drawn in 3 coins as
2 quarters and 1 dime
1 quarter = 25 cents
Which means , 2 x 25 = 50 cents
1 dime = 10 cents
Total : 50 + 10 = 60.

Question 5.
Structure
Draw to show 42¢ without using dimes.

Answer:
42 cents can be drawn as
1 quarter , 3 nickels and 2 pennies
1 quarter = 25 cents
1 nickel = 5 cents
Which means 3 x 5 = 15 cents
2 pennies = 2 cents
Total : 25 +15 + 2 = 42 cents

Question 6.
Modeling Real Life
Newton has 6 dimes and 1 nickel. Descartes uses the fewest number of coins to make the same amount. Draw and label their coins.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 65

Answer :
6 dimes and 1 nickel
Total number of coins is 7
1 dime = 10 cents
Which means, 6 dimes = 6 x 10 = 60 cents
1 nickel = 5 cents
Total : 60 +5 = 65 cents
65 cents can be shown in fewest number of coins as
2 quarters , 1 dime and 1 nickel
Total number of coins is 4

Question 7.
Modeling Real Life
You use fewer than 5 coins to buy the pen. Draw and label coins to show how you pay.
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 66

Answer:
The cost of pen is 52 cents
52 cents be shown in fewer than 5 coins as
2 quarter and 2 pennies
1 quarter = 25 cents
Which means, 2 quarters = 2 x 25 = 50 cents
and 2 pennies = 2 cents
Total 50 +2 = 52 cents.

Review & Refresh

Question 8.
A green scarf is 50 inches long. An orange scarf is 40 inches long. A red scarf is 38 inches long. How much longer is the green scarf than the red scarf?
Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 67

Answer:
22 inches

Explanation:
The length of green scarf is 50 inches
The length of red scarf is 38 inches
Now ,
50 inches – 38 inches = 22 inches
Therefore , Green scarf is 22 inches longer than red scarf.

Lesson 14.4 Make One Dollar

Explore and Grow

Newton has 4 coins. The total value is 100¢. Draw and label his coins

Answer:
Given ,
The total value of money is 100 cents
100 cents can be shown in 4 coins as 4 quarters
1 quarter = 25 cents
Which means 4 quarters = 4 x 25 = 100 cents
So, Newton has 4 quarters

Big Ideas Math Answers 2nd Grade Chapter 14 Money and Time 68
Descartes has 10 coins. The total value is 100¢. Draw and label his coins.

Answer:
Given ,
Total value of money is 100 cents
100 cents be shown in 10 coins as 10 dimes
1 dime = 10 cents
Which means , 10 dimes = 10 x 10 = 100 cents
So, Descartes have 10 dimes .

Show and Grow

Draw coins to make $1. How many cents do you need?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 69

Answer:
$1 = 100 cents
So, 100 cents – 25 cents = 75 cents
I need 75 cents to make $1.
75 Cents can e shown as 3 quarters
1 quarter = 25 cents
3 quarters = 3 x 25 = 7 cents

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70

Answer:
$1 = 100 cents
So, 100 cents – 80 cents = 20 cents
I need 20 cents to make $1.
20 cents can be shown as 2 dimes
1 dime = 10 cents
2 dimes = 2 x 10 = 20 cents

Apply and Grow: Practice

Draw coins to make $1. How many cents do you need?

Question 3.
35¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70.1

Answer:  $1 = 100 cents
So, 100 cents – 35 cents = 65 cents
I need 65 cents to make $1.
65 cents can be shown as
2 quarters , 1 dime and 1 nickel
1 quarter = 25 cents
Which means , 2 quarters =  2 x 25 = 50 cents
1 dime = 10 cents
and 1 nickel = 5 cents
Total : 50 +10 +5 = 65 cents

Question 4.
72¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 70.2

Answer:
$1 = 100 cents
So, 100 cents – 72 cents = 28 cents
I need 28 cents to make $1.
28 cents can be shown as
1 quarter , 3 pennies
1 quarter = 25 cents and
3 pennies = 3 cents
Total : 25 +3 = 28 cents

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 71

Answer:
$1 = 100 cents
1 dime = 10 cents
so 2 dimes = 2 x 10 = 20 cents
so, 100 cents – 20 cents = 80 cents
I need 20 cents to make $1.
20 cents can be shown as 2 dimes
1 dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 72

Answer:  $1 = 100 cents
1 quarter = 25 cents
Which means , 3 x 25 = 75 cents
1 dime = 10 cents
Total : 75 + 10 = 85 cents
Now, 100 – 85 = 15 cents
So, I need 15 cents to make $1.
15 cents can be shown as
1 dime and 1 nickel
1 dime = 10 cents
1 nickel = 5 cents
Total : 10 +5 = 15 cents

Question 7.
Number Sense
Circle coins to make $1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 73

 

Answer:

Explanation:
$1 is equals to 100  cents
1 quarter = 25 cents
Which means
2 quarters = 2x 25 = 50 cents
1 dime = 10 cents
5 dimes = 5 x 10 = 50 cents
Total : 50 +50 = 100 cents
So, 2 quarters and 5 dimes make $1.

Think and Grow: Modeling Real Life

You have 1 quarter, 3 pennies, and 1 dime in one pocket. You have 2 pennies, 2 nickels, and 4 dimes in your other pocket. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 74

Answer:
The amount i have in one pocket is
1 quarter =25 cents
3 pennies = 3 cents
and 1 dime = 10 cents
Total : 25 +10 +3 = 38 cents
The amount I have in another pocket :
2 pennies = 2 cents
1 nickel = 5 cents
which means, 2 nickels = 2 x5 = 10 cents
1 dime = 10 cents
Which means, 4 dimes = 4 x 10 = 40 cents .
Total : 2 + 10 +40 =52 cents
Now , the total amount i have in both pockets is
38+52 = 90 cents
We know that $1 is equal to 100 cents
100 – 90 = 10
10 cents can be shown as 1 dime

.

Show and Grow

Question 8.
You have 2 dimes and 1 nickel in your desk. You have a quarter and 10 pennies in your backpack. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 74.1

Answer:
Total amount I have in my desk is
2 dimes and 1 nickel
1 dime = 10 cents
Which means, 2 dime =2 x 10 = 20 cents
1 nickel = 5 cents
Total = 20 +5 = 25 cents
Total amount I have in my pocket is
a quarter and 10 pennies
1 quarter = 25 cents
and 10 pennies = 10 cents
Total : 25 +10 = 35 cents
Total amount I have is
The amount in my desk + the amount in my pocket
Which means, 25 cents +35 cents
=60 cents
We know that ,
$1 is equal to 100 cents
100 cents – 60 cents
= 40 cents.
So, I need 40 cents more to make $1.

Question 9.
A notebook costs $1. You have 5 dimes and 4 pennies. How much more money do you need to buy the notebook?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 75

Answer:
1 dime = 10 cents
Which means 5 dimes = 5 x 10 = 50 cents
and 4 pennies = 4 cents
Total : 50 cents +4 cents = 54 cents
we know that,
$1 is equal to 100 cents
Now , 100 – 54 = 46 cents
So ,I need 46 cents more to buy the book.

Question 10.
DIG DEEPER!
You have a $1 bill. You have 33 more cents than your friend. How much money does your friend have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 76

Answer:
I have $1 bill
Given that I have 33 cents more than my friend
We know that ,$1 is equal to 100 cents
Now,
100 cents – 33 cents = 67 cents
So , my friend have 67 cents.

Make One Dollar Homework & Practice 14.4

Draw coins to make $1. How many cents do you need?

Question 1.
54¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 77

Answer:  $1 = 100 cents
so, 100 cents – 54 cents = 46 cents
I need 46 cents to make $1.
46 cents can be shown as
1 quarter , 2 dime and 1 penny
1 quarter = 25 cents
1 dime = 10 cents
Which means, 2 dimes = 2 x 10 = 20 cents
and 1 penny = 1 cent

Question 2.
38¢
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 78

Answer:
$1 = 100 cents
So, 100 cents – 38 cents = 62 cents
I need 62 cents to make $1.
62 cents can be shown as
2 quarters, 1 dime and 2 pennies.
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50
1 dime= 10 cents
2 pennies = 2 cents

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 79

Answer:
$1 =100 cents
Given coins are 1 quarter and 2 pennies
1 quarter = 25 cents
and 2 pennies = 2cents
Total value : 25 +2 = 27
So, 100 – 27 =73
Therefore , I need 73 cents to make $1
73 cents can be shown as
2 quarters , 2 dime and 3 pennies
1 quarter = 25 cents
Which means 2 quarters = 2 x 25 = 50 cents
1 dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents
and 3 pennies = 3 cents
Total : 50 +20 +3 = 73.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 80

Answer:
$1 = 100 cents
Given coins are 1 dime and 1 nickel
1 dime = 10 cents
1 nickel = 5 cents
Total : 10 +5 = 15 cents
Now , 100 cents – 15 cents =85 cents
So , I need 85 cents to make $1.
85 cents can be written as
3 quarters , 1 dime
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75 cents
1 dime = 10 cents
Total : 75 +10 = 85 cents

Question 5.
Structure
Show $1 using only nickels and dimes.

Answer:
1 dime = 10 cents  and
1 nickel = 5 cents
$1 is equals to 100 cents
100 cents can be shown as 8 dimes 4 nickels
8dimes = 8 x 10 = 80
4 nickels = 4 x 5 = 20
Total value = 100 cents

 

Question 6.
Structure
How many nickels make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 81

Answer :
20 nickels
1 nickel = 5 cents
Which means ,20 nickels =
20 x 5 = 100
100 cents =$1
So 20 nickels make $1.

Question 7.
Modeling Real Life
You have 1 dime and 4 nickels in a jar. You have 1 quarter and 3 pennies in your pocket. How many more cents do you need to make $1?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 81.1

Answer:
The amount I have in jar is
1 dime = 10 cents
and 1 nickel = 5 cents
Which means , 4 nickels = 4 x 5 = 20 cents
Total : 10 + 20 = 30 cents
The amount i found in my pocket is
1 quarter = 25 cents
and 3 pennies = 3 cents
Total : 25+ 3 = 28 cents
Now , the total amount i have is
Amount in jar +amount in my pocket
= 30 +28 = 58cents
We know that, $1 is equal to 100 cents
100 – 58 = 42 cents
So, I need 42 cents more to make $1.

Question 8.
Modeling Real Life
A snack costs $1. You have 2 quarters and 2 dimes. How much more money do you need to buy the snack?

Answer:
The cost of snack is $1.
1 quarter = 25 cents
2 quarters = 2x 25= 50 cents
1 dimes = 10 cents
2 dimes= 2 x 10 = 20 cents
Total : 50 +20= 70 cents
$1 = 100 cents
100 – 70 = 30 cents
So , I need 30 cents more to buy the snack.

Review & Refresh

Question 9.
100 – 54 = ____

Answer : 46

By regrouping, 0 in units place becomes 10 and the 0 in tens place becomes 9 after the second regrouping.

Question 10.
200 – 134 = ____

Answer: 64

By regrouping, 0 in units place becomes 10 and the 0 in tens place becomes 9 after the second regrouping.

Lesson 14.5 Make Change from One Dollar

Explore and Grow

Model the story.
Newton buys a bag of fish crackers for 45¢. He pays with a $1 bill. What is his change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 82
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 83

Explain how you solved.
_______________________________
_______________________________
_______________________________
_______________________________

 

Answer: 55 cents

Explanation :
$1 is equals to 100 cents
The cost of fire crackers is 45 cents
Now his change is
$1 bill – cost of fire crackers
100 cents -45 cents
=55 cents
So, Newtons change is 55 cents.

Show and Grow

You buy the item shown. You pay with a $1 bill. What is your change?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 84

 

Answer:
65 Cents

Explanation:
$1 is equal to 100 cents
So, 100 cents – 35cents  = 65
Therefore, 65 cents is the change.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 85

Answer: 51 Cents

Explanation:
$1 is equals to 100 cents
So, 100 cents – 49 cents = 51 cents
Therefore: 51 cents is the change.

Apply and Grow: Practice

You buy the item shown. You pay with a $1 bill. What is your change?

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 86

Answer: 85 Cents

Explanation:
$1 = 100 cents
So, 100 cents -15 cents =85 cents
Therefore 85 cents is the change.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 87

Answer: 39 Cents

Explanation:
$1 = 100 cents
So, 100 cents -61 cents =39 cents
Therefore 39 cents is the change.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 88

 

Answer: 13 Cents

Explanation:
$1 = 100 cents
So, 100 cents -87 cents =13 cents
Therefore ,13 cents is the change.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 89

 

Answer: 41 Cents

Explanation:
$1 = 100 cents
So, 100 cents -59 cents =41 cents
Therefore ,41 cents is the change.

Question 7.
Reasoning
Newton buys a notebook for 34¢. Descartes buys one for 52¢. You buy one for 48¢. You each pay with $1. Who gets back the most amount of money? How do you know?
__________________
__________________
__________________

Answer:
Newton gets back the most amount of money.

Explanation:
$1 is equals to 100 cents
Given that ,
The cost of newtons book is 34 cents
So, 100 cents – 32 cents = 68 cents .
The cost of Descartes book is 52 cents.
So , 100 cents – 52 cents = 48 cents.
The cost of my book is 48cents .
So, 100 cents – 48 cents = 52 cents
Therefore, newton gets back the most amount of money.

Think and Grow: Modeling Real Life

You pay for some school supplies with $1. Your change is 17¢. How much money did you spend?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 90
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 91

Answer: 83 Cents

Explanation:
$1 is equals to 100 cents
My change is 17 cents
Now , 100 cents – 17 cents =83 cents
So, the cost of school supplies is 83 cents.

Show and Grow

Question 8.
You pay for some erasers with $1. Your change is 38¢. How much money did you spend?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 92

 

Answer: 62 Cents

Explanation:
$1 is equals to 100 cents
My change is 38 cents
Now , 100 cents – 38 cents =62 cents
So the cost of erasers is 62 cents.

Question 9.
You buy a toy ring. You pay with $1. You get back 1 quarter, 2 dimes, 1 nickel, and 1 penny. How much does the toy ring cost?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 93

Answer: 59 Cents.

Explanation:
$1 is equals to 100 cents
My change is 1 quarter = 25 cents
1dime = 10 cents
Which means , 2 dimes = 2 x 10 = 20 cents
1 nickel = 5 cents
1 penny = 1 cent
Total = 25 +10 + 5 +1 =41 cents
100 cents – 41 cents = 59 cents
Therefore , the cost of toy ring is 59 cents.

Question 10.
You buy a banana for 25¢ and an orange for 45¢. You pay with $1. What is your change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 94

Answer:  Change is 30 Cents.

Explanation:
The cost of banana is 25 cents and the cost of orange is 45 cents
Total cost : 25 cents + 45 cents = 70 cents
$1 is equals to 100 cents
Now, 100 – 70 = 30 cents
Therefore ,30 cents is the change.

Make Change from One Dollar Homework & Practice 14.5

You buy the item shown. You pay with a $1 bill. What is your change?

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 95

Answer: 55 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -45 cents =55 cents
Therefore 55 cents is the change.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 96

Answer: 24 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -76 cents =24 cents
Therefore, 24 cents is the change.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 97

Answer: 87 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -13 cents =87 cents
Therefore, 87 cents is the change.

Question 4.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 98

Answer: 63 Cents.

Explanation:
$1 = 100 cents
So, 100 cents -37 cents = 63 cents
Therefore, 63 cents is the change.

Question 5.
Reasoning
A puzzle costs 68¢. Newton pays for it with a $1 bill. Draw to show his change in two ways.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 99

Answer: Change is 32 Cents.

Explanation:
$1 is equals to 100 cents
The cost of puzzle is 68 cents
Now, 100 – 68 = 32 cents
Therefore, 32 is the change.
32 cents can be written as 3 dime, 2 pennies

1 dime = 10 cents
3 dimes = 3 x 10 = 30 cents
2 pennies = 2 cents
Total : 30 + 2 = 32 cents
Another way ,
1 quarter 7 pennies

Which means, 1 quarter = 25 cents
7 pennies = 7 cents
Total: 25 + 7 = 32 cents
Therefore 32 cents can be written as 3 dime 2 pennies and 1 quarter 7 pennies.

Question 6.
Modeling Real Life
You buy a pencil sharpener. You give the cashier $1. You get back 2 quarters, 1 nickel, and 3 pennies. How much does the pencil sharpener cost?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 100

Answer: 42 Cents.

Explanation:
$1 is equals to 100 cents
My change is 2 quarters 1 nickel 3 pennies
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 =50 cents
1 nickel = 5 cents
3 pennies = 3 pennies
Total: 50 +5+3=58 cents
Now,
100 – 58 = 42 cents
Therefore, the cost of the pencil sharpener is 42 cents .

Question 7.
Modeling Real Life
You buy an onion for 51¢ and a pepper for 22¢. You pay with a $1 bill. What is your change?

Answer: 27 Cents.

Explanation:
The cost of an onion is 51 cents and the cost of 22 cents
Total cost : 51 + 22 =73 cents
$1 is equals to 100 cents
100 cents – 73cents = 27 cents
Therefore, 27 cents is the change.

Review & Refresh

Question 8.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 101

Answer: There are three straight sides and three vertices.

Question 9.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 102

Answer: There are six straight sides and six vertices.

Lesson 14.6 Find Total Values of Bills

Explore and Grow

Model the story.
Descartes has three $5 bills and three $1 bills. How much money does he have in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 103

Answer: $18.

Explanation:
Given that ,
Descartes have Three $5 bills and three $1 bill
Three $5 bills = $5+$5+$5 = $15.
and three $1 bills = $3
Total amount =$15+$3= $18.

Show and Grow

Count on to find the total value.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 104

Answer: $45.

Explanation:
Three $5 bills =$5+$5+$5 = $15
$10 and $20
Total value :
$15 +$10 +$20= $45
Therefore, total value = $45.

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 105

Answer: $32.

Explanation:
Two $1 bills = $2
Two $5 bills = $5+$5= $10
and two $10 = $10 +$10= $20
Total value = $2+$10+$20=$32
Therefore ,the total value is $32.

Apply and Grow: Practice

Count on to find the total value.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 106

Answer: $16.

Explanation:
$10 ,$5 and $1
Total value :
$10 +$5 +$1 =$16

Question 4.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 107

Answer: $36.

Explanation:
Total value of $20 , $10 ,$5 and $1
=$20 +$10 +$5 +$1 =$36 .

Question 5.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 108

Answer: $75.

Explanation:
Total value of a $5 bill, $10 bill and three$20 bills is
$5+$10+$20+$20+$20= $75
Therefore the total value is $75.

Question 6.
Which One Doesn’t Belong?
Which group of bills does not belong with the other two?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 109

Answer:

Explanation:
The total value of dollar bills is $61
But the remaining dollar sets total value is $65
So ,this set is different from the other two sets.

Think and Grow: Modeling Real Life

You buy some T-shirts for $39. Draw and label bills to show two different ways to pay for the T-shirts. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 110

 

Answer:
$39 can be shown as
A $20 bill ,a $10 bill ,a $5 bill and four $1 bills

Another way:
$39 can be shown as
a $20 bill , three $5 bills and four $1 bills.

Show and Grow

Question 7.
You buy a pair of sneakers for $24. Draw and label bills to show two different ways to pay for the sneakers. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 111

Answer:
$24 can be shown as

a $20 bill and four $1 bill

    

$24 can also be shown as
A$20 bill and two $2 bills

Question 8.
Newton has three $20 bills, one $10 bill, one $5 bill, and three $1 bills. Does he have enough money to buy a new dog house that costs $80? Explain.

Answer: No , he do not have enough money to buy a dog house

Explanation:
Three $20 bills = $20+$20+$20= $60
$10 , $5 and three $1
Total value :
$60 +$10 +$5+$1 = $76
But, the cost of dog house is $80
So, he do not have enough money to buy a dog house.

Question 9.
DIG DEEPER!
Explain why you would order a group of bills from the greatest value to the least value to find the total value.
_____________________________________
_____________________________________

Find Total Values of Bills Homework & Practice 14.6

Count on to find the total value.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 112

Answer:
$10 +$ 1= $11
Therefore the total value is $11

Question 2.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 113

Answer: $50

Explanation:
Two $10 bills = $10 +$10 = $20
Two $5 bills = $5+$5 = $10
Total value = $20+$20+$10 = $50

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 114

Answer: $91

Explanation:
Four $20 bills = $20 +$20 +$20 +$20 = $80
$10 and $ 1
Total value = $80 +$10 +$1 = $91.

Question 4.
YOU BE THE TEACHER
Newton says he drew the fewest number of bills to show $35. Is he correct? Explain.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 115

Answer:
No , he is wrong.

Explanation:
$35 can be shown in few bills as
a $20 bill, a $10  and a $5 bill

Question 5.
Modeling Real Life
A pair of headphones costs $88. Draw and label bills to show two different ways to pay for the headphones. One way should use the fewest number of bills.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 116

Answer:
The cost of head phones is $88 .
$ 88 can be shown as
four $20 bills, a $5 bill ,a $2 bill and a$1 bill

   

Question 6.
Modeling Real Life
Descartes has two $20 bills, three $10 bills, and four $1 bills. Does he have enough money to buy a scratching post that costs $62? Explain.
__________________________

__________________________

Answer:
Yes ,he have enough money to buy a scratching post

Explanation:
Descartes have
Two $20 bills
Which means , 2 x 20 = 40
Three $10 bills
Which means, 3 x 10 = 30
Four $ 1bills
Total : $40 +$30 +$4 =
$74
The cost of scratching post is $62
So ,Descartes have enough money to but the post.

Review & Refresh

Question 7.
A photo album has 3 rows of photos. There are 4 photos in each row. How many photos are there in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 117

Answer: 12 Photos.

Explanation:
Given that,
A photo album has 3 rows of photos
and there are 4 photos in each row
Now, the total number of phone are
4 +4+4 = 12
So , there are 12 photos in all.

Lesson 14.7 Problem Solving: Money

Explore and Grow

Model the story.
You buy a book for 60¢. Your friend buys a book for 33¢. How much do you and your friend spend in all?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 118
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 119
Explain how you solved.
_______________________________
_______________________________
_______________________________

_______________________________

Answer: 93 Cents.

Explanation:
The cost of my book is 60 cents and
The cost of my friends book is 33 cents
Total value of our money is
60 cents + 33 cents
= 93 cents.
Therefore, me and my friend spent 93 cents in all.

Show and Grow

Question 1.
Descartes has two $10 bills and two $5 bills. He has $21 more than Newton. How much money does Newton have?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 120

 

Answer: $9

Explanation:
Total amount of Descartes is
Two $10 bills
Which means, 2 x $10 bills = $20
Two $5 bills
Which means , 2 x $5 = $10
Total value = $20 +$10 = $30
Descartes have $30
Given that, Descartes have $21 more than newton
now , Newtons money = $30 – $21 = $9
Therefore ,Newton have $9

Question 2.
Descartes has some coins in a jar. He puts in 4 dimes, 1 nickel, and 1 penny. Now he has $1. How many cents were in the jar to start?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 121

Answer: 54 Cents.

Explanation:
1 dime = 10 cents
Which means , 4 dimes = 4 x 10 = 40
1 nickel = 5 cents
and 1 penny = 1 cents
total value :
40 +5+1= 46 cents
We know that,
$1 is equals to 100 cents
now , 100 cents -46 cents = 54 cents
Therefore, there are 54 cents in the start.

Apply and Grow: Practice

Question 3.
Newton has some money. He loses a $10 bill and three $1 bills. Now he has $19. How much money did he have to start?

Answer: $32.

Explanation:
The total money newton lost is
$10 bill
and three $ 1 bill which means, $3
Total : $10 +$3 = $13
Given that ,
He have $19 after losing some money
Now, the total money newton have on the beginning is
lost money +leftover money
$13 +$19 = $32
Therefore , Newton have $32 in the starting.

Question 4.
Descartes has one $20 bill, three $10 bills, and three $5 bills. He spends $50. How much money does he have left?

Answer: $15

Explanation:
Total money Descartes have is
$20 bill
Three $10 bills = 3 x 10 = 30  and
Three $5 bills = 3 x 5 = 15
Total : $20 +$30 +$15= $65
Given that,
He spent $50
Therefore , Leftover money =
$65 – $50 = $15
So, Descartes is left with $15 .

Question 5.
A joke book costs $1. You have 2 quarters and 1nickel. How much more money do you need to buy the joke book?

Answer: 45 Cents.

Explanation:
$1 is equal to 100 cents
1 quarter = 25 cents
Which means, 2 quarters = 2 x 25 = 50 cents
1 nickel = 5 cents = 5 cents
Total : 50 cents +5 cents = 55 cents
now ,
100 cents – 55 cents = 45 cents
So , I need 45 cents more to buy the book.

Question 6.
YOU BE THE TEACHER
Your friend says that 3 dimes and 2 nickels is 50¢. Is your friend correct? Explain.
_______________________________
_______________________________

Answer: No, He is wrong.

Explanation:
1 dime = 10 cents
Which means , 3 dimes = 3 x 10 = 30 cents
1 nickel = 5 cents
Which means, 2 nickels = 2 x 5 = 10
Total value = 30 + 10 = 40
So, he is wrong

Think and Grow: Modeling Real Life

You have a $20 bill and a $5 bill. Your friend has $10 less than you. Do you and your friend have enough money to buy a $38 skateboard? Explain.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 122

Answer: No.

Explanation :
The  amount I have is
$20 and $ 5
Total : $20+$5 = $25
Given that , my friend have $10 less than me
Which means,
$25 – $10= $15
But the cost of skateboard is $38
So , me and my friend do not have enough value to buy the skateboard.

Show and Grow

Question 7.
You have 1 quarter, 2 dimes, and 3 pennies. Your friend has 4 nickels and 2 pennies. Do you and your friend have enough money to buy a 75¢ bottle of orange juice? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 123
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 124

Answer:
My money:
1 quarter = 25 cents
1 dimes = 10 cents
Which means , 2 x 10 = 20 cents
3 pennies = 3 cents
Total : 25 + 20 + 3 = 48 cents
I have 48 cents
My friends money:
1 nickel = 5 cents
Which means , 4 nickels = 4 x 5 = 20 cents
2 pennies = 2 cents
Total : 20 + 2 = 22 cents
My friend have 22 cents
The cost of orange juice is 75 cents.
Therefore, me and my friend do not have enough money to buy the orange juice.

Question 8.
Descartes buys a board game for $19. He has three $5 bills and two $1 bills left over. How much money did he have before he bought the game?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 125

Answer: $26.

Explanation:
The cost of board game = $19
Total value of leftover money is
Three $5 bills = 3 x 5 = $15
Two $1 bills = 2 x 1 = $2
Total : $15 +$2
= $17
The total money Descartes have in the beginning is
leftover money + cost of board game
=$17 +$19
= $26
Therefore , Descartes have $26 before he bought the game.

Question 9.
DIG DEEPER!
You have 25¢ in your desk, 18¢ in your backpack, and 50¢ in your pocket. You spend 43¢ and lose a quarter. How much money do you have left?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 126

Answer: 25 cents.

Explanation:
The amount I have in my desk is 25 cents
The amount have in my backpack is 18 cents
The amount I have in my pocket is 50 cents
Now,  The total amount I have with me is:
25 cents +18 cents + 50 cents = 93 cents.
1 quarter = 25 cents
The amount I spend = 43 cents
total : 25 + 43 = 68 cents
Now, The money I have left is
93 cents – 68 cents = 25 cents
Therefore , I am left with 25 cents  .

Problem Solving: Money Homework & Practice 14.7

Question 1.
Newton has $30. Descartes has a $20 bill, a $10 bill, and two $1 bills. How much more money does Descartes have?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 127

Answer:
The money newton has is $30 .
Descartes have $20 bill ,$10 bill and two $1 bill
Total money : 20 +10+1+1 =32
So Descartes have $32
Now, $32 – $30 = $2
Therefore, Descartes have $2 more than Newton.

Question 2.
You have some money. You spend 2 quarters and 3dimes at the cafeteria. Now you have 20¢. How much money did you have to start?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 128

Answer: $1.

Explanation:
1 quarter = 25 cents
Which means , 2 quarters = 2 x 25 = 50 cents
1 dime =10 cents
Which means, 3 dimes = 3 x 10 = 30 cents
Total amount : 50 +30 = 80
After spending 80 cents at cafeteria , I have 20 cents
Now , the money i have at start is
80 cents + 20 = 100 cents
100 cents =  $1
Therefore , I have $ 1 in the start.

Question 3.
YOU BE THE TEACHER
You buy a sandwich for 75¢. You pay with $1. Your friend says your change will be 1 quarter. Is your friend correct? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 129

Answer:
Yes, he is correct .
The cost of sandwich is 75 cents
$1 is equals to 100 cents
100 cents – 75 cents = 25 cents
As we know that ,
1 quarter = 25 cents
so , he is correct.

Question 4.
Modeling Real Life
You have 12 pennies, 2 dimes, and 1 nickel. Your friend has 20¢ more than you. Do you and your friend have enough money to buy a toy car that costs $1? Explain.
_______________________________
_______________________________

Answer: No , me and my friend do not have enough money to buy a toy car

Explanation :
The amount I have is
12 pennies, 2 dimes and 1 nickel
12 pennies = 12 cents
1 dime = 10 cents
Which means 2 dime = 2 x 10 = 20 cents
and 1 nickel = 5 cents
Total amount =
12 +20 +5 = 37 cents
Therefore I have 37 cents
Friends money :
He has 20 cents more than me
Which means, 37 +20 = 57 cents
But , we know that $1 = 100 cents
So we do not have enough money to buy a toy car

Question 5.
Modeling Real Life
Descartes has $45. He spends a $20 bill and a $1 bill. He earns two $5 bills and a $10 bill. How much money does he have now?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 130

Answer:
Given that , Descartes has $45
He spends $ 20 and $1 bill
total : 20 + 1 = 21
now, $45 -$21= $24
Then,
He earns 2 $5 bills and a $10 bill
Total : $24 +$5 +$5+$10= $44
Therefore ,Descartes have $44 .

Review & Refresh

Question 6.
Which time does not belong with the other three?
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 131

Answer:

Explanation :
In the above clock the time is 4 hours 30 minutes.
But the time in remaining three is 3 hours 3 minutes.
So, it does not belong with the other three.

Lesson 14.8 Tell Time to the Nearest Five Minutes

Explore and Grow

Label the missing minutes around the clock. Then tell the time.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 132

Answer:

  1. 15
  2. 30
  3. 55.
  4. The time is 3 hours 30 minutes

Show and Grow

Write the time.

Question 1.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 133

Answer: 1 hours 25 minutes.

Explanation:
As the hours hand is in between 1 and 2 and the minutes hand on 5. So, the time is on 25 minutes.
That means, the time is 1 hours and 25 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 134

Answer:8 hours 10 minutes.

Explanation:
As the hours hand is on 8 and the minutes hand on 10. So, the time is on 10 minutes.
That means, the time is 8 hours and 10 minutes.

Question 3.
Big Ideas Math Answers Grade 2 Chapter 14 Money and Time 135

Answer: 6 hours 45 minutes.

Explanation:
As the hours hand is in between 6 and 7 and the minutes hand on 9. So, the time is on 45 minutes.
That means, the time is 6 hours and 45 minutes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 136

Answer: 5 hours 55 minutes.

Explanation:
As the hours hand is in between 5 and 6 and the minutes hand on 11. So, the time is on 55 minutes.
That means, the time is 5 hours and 55 minutes.

Apply and Grow: Practice

Write the time.

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 137

Answer: 3 hours 50 minutes.

Explanation:
As the hours hand is in between 3 and 4 and the minutes hand on 10. So, the time is on 50 minutes.
That means, the time is 3 hours and 50 minutes.

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 138

Answer: 11 hours 20 minutes.

Explanation:
As the hours hand is in between 11 and 12 and the minutes hand on 4. So, the time is on 20 minutes.
That means, the time is 11 hours and 20 minutes.

Question 7.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 139

 

Answer: 5 hours 30 minutes.

Explanation:
As the hours hand is in between 5 and 6 and the minutes hand on 6. So, the time is on 30 minutes.
That means, the time is 5 hours and 30 minutes.

Draw to show the time.

Question 8.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 140

Answer:

Question 9.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 141

Answer:

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 142

 

Answer:

Question 11.
Patterns
Write the next time in the pattern.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 143

Answer:
8:30
The sequence is differ  by 5 minutes

Question 12.
Precision
The hour hand points between the 4 and the 5. The minute hand points to the 4. What time is it?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 144

Answer:
As the hour hand points between 4 and 5 .the minute hand points to the 4
The time is 4 hours 20 minutes

Think and Grow: Modeling Real Life

Baseball practice lasts 40 minutes. Show and write the time practice ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 145

Answer: 3 hours 55 minutes

Show and Grow

Question 13.
Recess lasts 25 minutes. Show and write the time recess ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 146

Answer: 12 hours 35 minutes

Question 14.
DIG DEEPER!
A train ride starts at 6:40. The ride lasts 45 minutes. What time does the ride end?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 147
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 148

Answer:  7 hours 25 minutes

Tell Time to the Nearest Five Minutes Homework & Practice 14.8

Write the time.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 149

Answer:
10 hours 35 minutes.
As the hours hand is in between 10 and 11 and the minutes hand is near to 7. So, the time is near to 35 minutes.
That means, the time is 10 hours and 35 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 150

Answer:
6 hours 10 minutes.
As the hours hand is on 6, the minutes hand is on 2. So, the time is near to 10 minutes.
That means, the time is 6 hours and 10 minutes.

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 151

Answer:
7 hours 25 minutes.
As the hours hand is in between 7 and 8 and the minutes hand is near to 5. So, the time is near to 25 minutes.
That means, the time is 7 hours and 25 minutes.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 152

 

Answer: 

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 153

Answer:

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 154

Answer:

Question 7.
Reasoning
The minute hand points to the 7. What number will it point to in 10 minutes?

Answer :  9
Each number the minutes hand points, indicates 5 minutes.
So,10 minutes indicates 2 more numbers, the minutes hand moves on to.
That means, 7 + 2 = 9.
So, after 10 minutes, the minutes hand points on to 9.

Question 8.
Precision
The hour hand points between the 11 and the 12. In 25 minutes it will be the next hour. What time is it now?

Answer: 11 hours 35 minutes.

Explanation:
Each number the minutes hand point, indicates 5 minutes.
So, 25  minutes indicates 5  more numbers, the minutes hand moves on to.
that means ,5 x 5 = 25
So, after 25 minutes, the time is 12: 00
now, the time is 11 hours 35 minutes .

Question 9.
Modeling Real Life
Your walk to school lasts 15 minutes. Show and write the time your walk ends.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 157

Answer: 8 hours 25 minutes

Question 10.
DIG DEEPER!
Your swimming lesson starts at 5:30. It lasts 35 minutes. What time does the lesson end?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 158

Answer:
6 hours 5 minutes.

Explanation:
Each number the minutes hand points, indicates 5 minutes.
So,35 minutes indicates 7 more numbers, the minutes hand moves on to.
That means, 7 x 5= 35
So, after 35 minutes, the lesson ends on 6 hours 5 minutes.

Review & Refresh

Question 11.
The crayon is about 7 centimeters long. What is the best estimate of the length of the toothpick?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 159

Answer: 4 Centimeters.

Explanation:
Given,
The length of crayon is 7 meters long.
4 centimeters is the best estimate of length of the toothpick .

Lesson 14.9 Tell Time Before and After the Hour

Explore and Grow

Write each time on the digital clocks. How much time has passed?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 160

Answer:
1) the time is  11hours 45 minutes
2)  the time is 12: 00
on comparing both the clocks 15 minutes time has been passed.

Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 161

Answer:
1) In the first clock the time is 1:00
2) In the second clock the time is 1:15 on comparing both the clocks 15 minutes has been passed

Show and Grow

Write the time. Circle another way to say the time.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 162

Answer:
4 hours 15 minutes.
As the hours hand is in between 4 and 5 and the minutes hand on 3 . So, the time is near to 15 minutes.
That means, the time is 4 hours and 15 minutes.
Also known as quarter past 4.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 163

Answer:
11 hours 30 minutes.
As the hours hand is in between 11 and 12 and the minutes hand is on 6. So, the time is near to 30 minutes.
That means, the time is 11 hours and 30 minutes.
Also known as half past 11.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 164

 

Answer:
2 hours 45 minutes.
As the hours hand is in between 2 and 3 and the minutes hand is near to 9. So, the time is near to 45 minutes.
That means, the time is 2 hours and 45 minutes.
Also known as quarter to 3.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 165

Answer:
9 hours 20 minutes.
As the hours hand is in between 9 and 10 and the minutes hand is near to 4. So, the time is near to 20 minutes.
That means, the time is 9 hours and 20 minutes.
also known as 20 minutes after 9.

Apply and Grow: Practice

Write the time. Circle another way to say the time.

Question 5.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 166

Answer: 1 hours 30 minutes.
As the hours hand is in between 1 and 2 and the minutes hand is near to 6. So, the time is near to 30 minutes.
That means, the time is 1 hours and 30 minutes.
Also known as half past 1.

Question 6.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 167

Answer:
7 hours 10 minutes.
As the hours hand is in between 7 and 8 and the minutes hand is near to 2. So, the time is near to 10 minutes.
That means, the time is 7 hours and 10 minutes.
also known as ,10 minutes after 7

Show and write the time.

Question 7.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 168

Answer: 10 : 45

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 169

Answer: 5 hours 15 minutes

Question 9.
Which One Doesn’t Belong? Which time does not belong with the other three?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 170

Answer: Quarter past 7

Explanation:
6 : 45
45 minutes after 6 means 6 :45
Quarter to 7 means 6 : 45
These three are same .
But , quarter past 7 means 7 : 30
So, quarter past 7 is different.

Question 10.
Precision
Is it time for homework or dinner?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 171

Answer: Dinner
Quarter after 6 is also known as 6 hours 15 minutes.
6: 15 p.m.- dinner

Think and Grow: Modeling Real Life

School starts at quarter past 8. Are you early or late to school? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 172
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 173

Answer:
Early to school
now, the time is 7 hours 55 minutes
the school starts at 8 hours 15 minutes ;it means quarter past 8 .
so, I am early to school

Show and Grow

Question 11.
A movie starts at quarter to 6. Are you early or late to the movie? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 174
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 175

Answer:
I am late to the movie .
Now, the time is 6 hours 5 minutes .
but, the movie starts at quarter to 6
which means , 5 hours 45 minutes
so I am late to the movie .

Question 12.
DIG DEEPER!
You arrive at the bus station 20 minutes before 12. Which is the first bus you can board? How many minutes are there until it leaves?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 176Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 177

Answer:
The time when i arrived bus station is 11 hours 40 minutes
So, I can board blue bus.
Blue bus leaves at 11 hours 55 minutes .
So , there are 15 more minutes until the blue bus leaves the bus station.

Tell Time Before and After the Hour Homework & Practice 14.9

Write the time. Circle another way to say the time.

Question 1.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 178

Answer: quarter to 12 means 11 hours 45 minutes

 

As the hours hand is between 11 and 12, the minutes hand is on 9. So, the time is near to  45 minutes.
That means, the time is 11 hours and 45 minutes.

Question 2.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 179

Answer: 40 minutes after 5 which means 5 hours 40 minutes

As the hours hand is between 5 and 6, the minutes hand is on 8. So, the time is near to  40 minutes.
That means, the time is 5 hours and 40 minutes.
Show and write the time.

Question 3.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 180

Answer: 3 hours 30 minutes

As the hours hand is between 3 and 4, the minutes hand is on 6. So, the time is near to 30 minutes.
That means, the time is 3 hours and 30 minutes.

Question 4.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 181

Answer: 12 hours 15 minutes

As the hours hand is between 12 and 1, the minutes hand is on 3. So, the time is near to  15 minutes.
That means, the time is 12 hours and 15 minutes.

Question 5.
YOU BE THE TEACHER
Newton says it is 2:45, or quarter to 3. Is he correct? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 182

Answer:
Yes, he is correct.

Explanation:
quarter means 15 minutes .
Each number the minutes hand points, indicates 5 minutes.
That means, 3 x 5= 15
so, 2:45 and quarter to 3 are both same
So, newton is correct

Question 6.
Modeling Real Life
A show starts at quarter to 7. Are you early or late to the show? Explain.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 183

Answer:
I am  early to the show
Now ,the time is 6 hours 30 minutes
but the show starts at quarter to 7 which means 6 hours 45 minutes
So ,I am 15 minutes early to the show

Question 7.
DIG DEEPER!
You arrive at the metro station 10minutes after 2. Which is the first train you can board? How many minutes are there until it leaves?
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 184

Answer:
The time 10 minutes after 2 is 2 hours 10 minutes
So ,I can board the yellow train which will leaves the railway station at 2: 30
There are 20 minutes more until the yellow train leaves the railway station.

Review & Refresh

Question 8.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 185

Answers: 162.

 

Question 9.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 186

Answer: 177

Explanation:

Question 10.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 187

Answer: 446.

Explanation:

Lesson 14.10 Relate A.M. and P.M.

Explore and Grow

Describe what you do in the morning. Show and write the time. Describe what you do in the evening. Show and write the time.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 188

Show and Grow

Write the time. Circle a.m. or p.m.

Question 1.
Eat breakfast
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 189

Answer: 7 hours 30 minutes a.m.
As the hours hand is on 7, the minutes hand is on 6. So, the time is near to 30  minutes.
That means, the time is 7 hours and 30 minutes.

Question 2.
Eat dinner
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 190

Answer:
6 hours 15 minutes p.m.
As the hours hand is between 6 and 7, the minutes hand is on 3. So, the time is near to  15 minutes.
That means, the time is 6 hours and 15 minutes.

Question 3.
Go to art class
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 191

Answer :
10 hours 25 minutes a.m.
As the hours hand is between 10 and11, the minutes hand is on 5. So, the time is near to 25  minutes.
That means, the time is 10 hours and 25 minutes.

Question 4.
Do homework
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 192

Answer:
4 hours 40 minutes p.m.
As the hours hand is between 4 and 5, the minutes hand is on 8. So, the time is near to  40  minutes.
That means, the time is 4 hours and 40 minutes.

Apply and Grow: Practice

Write the time. Circle a.m. or p.m.

Question 5.
Ride the bus to school
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 193

Answer:
8:00 a.m.
As the hours hand is on 8, the minutes hand is on 12.
That means, the time is 8 hours

Question 6.
Go to a party
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 194

Answer:
6hours 30 minutes p.m.
As the hours hand is between 6 and 7, the minutes hand is on 6. So, the time is 30  minutes.
That means, the time is 6 hours and 30 minutes.

Draw to show the time. Circle a.m. or p.m.

Question 7.
Read before bed
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 195

Answer:
7 hours and 50 minutes p.m.

Question 8.
Sunrise
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 196

Answer:
5 hours 20 minutes a.m.

Question 9.
Reasoning
Use the times in the list to complete the story.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 197

Answer:
You arrive at school at 8: 30 a.m. Your class goes to music at 10:15 a.m. After school,
you read a book at 5:20 p.m.

Think and Grow: Modeling Real Life

Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 198

Answer:
1) 8: 45 a.m. –  School time
2)2:50 p.m. –  Music class time
3) 4:10 p.m. –  Home work

Show and Grow

Question 10.
Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Solutions Grade 2 Chapter 14 Money and Time 199
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 200

Answer:
1) 7: 30 a.m. – Yoga class
2) 9:55 a.m.-  Art class
3) 7:35 p.m. – Dinner

Question 11.
DIG DEEPER!
Use the times to complete the timeline. Then rewrite each time digitally below, including a.m. or p.m
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 201
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 202

Answer:
1) 20 minutes before 7 –      6:40 a.m
2) 10 minutes after 11   –     11:10 a.m
3) noon                         –      12:00 p.m
4) quarter past 3           –        3:15 p.m
5)  half past 6                 –        6:30 p.m
6)2  0 minutes after 8    –        8:20 p.m

Relate A.M. and P.M. Homework & Practice 14.10

Write the time. Circle a.m. or p.m.

Question 1.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 203

Answer:
9 hours 25 minutes p.m
As the hours hand is between 9 and 10, the minutes hand is on 5. So, the time is near to 25 minutes.
That means, the time is 9 hours and 25 minutes.

Question 2.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 204

 

Answer:
7hours 15 minutes A.m.
As the hours hand is between 7 and 8, the minutes hand is on 3. So, the time is 15  minutes.
That means, the time is 7 hours and 15 minutes.

Draw to show the time. Circle a.m. or p.m.

Question 3.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 205

Answer: 12:00 P.m.

Question 4.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 206

Answer: 4: 40 P.m.

Question 5.
Reasoning
Right now, it is p.m. In 10 minutes it will be a.m. What time is it now? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 207

 

Answer:
Time is 11 hours 50 minutes.
Each minute hand point indicates 5 minutes
That means 2x 5= 10 minutes
a.m. starts imediatly after 12:00 pm
So, now the time is 11: 50 p.m.

Question 6.
Modeling Real Life
Use the times to complete the timeline. Write something you might do at those times.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 208

Answer:
1 ) 8:15 a.m. – break fast
2) 11: 55 a.m. – art class
3)  8: 15 p.m.- drawing time

Question 7.
DIG DEEPER!
Use the times to complete the timeline. Then rewrite each time digitally below, including a.m. or p.m.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 209

Answer:
1 ) Quarter to 8 –            7:45 a.m.
2) 20 minutes after 8 –  8: 20 a.m.
3) Quarter past 11 –      11: 15 a.m.
4) Noon –                       12:00 p.m.
5) 10 minutes after 2 –   2 : 10 p.m.
6) Half past 9-                 9:30 p.m.

Review & Refresh

Question 8.
65 + 36 = ____

Answer:
65 +36 = 101

Question 9.
56 + 18 = ____
Answer: 74

Money and Time Performance Task 14

Question 1.
a. You have two $1 bills, 1 quarter, 5 dimes, 3 nickels, and 2 pennies. How much more money do you need to buy a subway pass?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 210

Answer:
The cost of  subway pass is $3
$3 is equals to 300 cents
$1 is equals to 100 cents
Two $1 = 100 + 100 = 200cents
1 quarter = 25 cents
1 dime = 10 cents
Which means , 5 x 10 = 50
1 nickel = 5 cents
Which means, 3 x 5 = 15
and 2 pennies = 2  cents
Total :  200 +25+50+15+2 = 292
300 – 292 = 8 cents
I need 8 more cents to buy a pass.
So, I do not have enough money to buy a subway pass.

b. You find a dime. Do you have enough money to buy the pass now?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 211

Answer: Yes

Explanation:
1 dime = 10 cents
292 +10 = 302 cents
The cost of subway pass is $3 which means 300 cents
If I find a dime, I have enough money to buy a pass

Question 2.
A weekly subway pass is $32. A customer pays with a $50 bill. Use tally marks to show three different ways that the customer can receive change. What is the total change?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 212Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 212.2

Answer:
The cost of weekly subway pass is $32 .
Given that ,
A customer paid a $50 bill
Total change :
$50 -$32 = $18
$18 can be shown as
1) 1 $10 bill , 1 $5 bill and 3 $1 bill
2) 1 $10 bill, 8 $1 bills
3) 3 $ 5 bills and 3 $1 bill .

Question 3.
You arrive at the subway station at quarter to 3. What times will the subways arrive?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 213

Answer: car A arrives subway

Money and Time Activity

To Play: Place the Flip and Find Cards face down in the boxes. Take turns flipping 2 cards. If your cards show the same time or value, keep the cards. If your cards show different times or values, flip the cards back over. Play until all matches are made.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 214

Money and Time Chapter Practice

4.1 Find Total Values of Coins

Question 1.
Count on to find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 215

Answer: 46 cents

Explanation
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
Which means , 5 x 2 =10 cents
1 penny = 1 cent
Total : 25 +10+5+5+1 = 46 cents
Therefore the total value of coins 46 cents.

Question 2.
Modeling Real Life
You have 2 quarters, 2 dimes, and 1 penny. How many cents do you have? Do you have enough money to buy the frozen fruit bar?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 216
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 217

Answer:
The cost of frozen fruit bar is 75 cents
1 quarter = 25 cents
which means , 2 quarter = 2 x 25 = 50 cents
1 dime = 10 cents
which means , 2 dimes = 2 x 10 = 20 dimes
1 penny = 1 cent
total amount I have is 50+20+1 = 71 cents
therefore I have 71 cents
So ,I do not have enough money to buy the frozen fruit bar.

14.2 Order to Find Total Values of Coins

Question 3.
Draw and label the coins from the greatest value to the least value. Then find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 218

Answer:
43 cents

Explanation
1 quarter = 25 cents
1 dime = 10 cents
1 nickel = 5 cents
3 pennies = 3cents
Total value = 25 +10 +5 + 3 = 43 cents

>>>

14.3 Show Money Amounts in Different Ways

Question 4.
Draw and label coins to show the amount in two different ways.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 219

Answer:
56 cents can be shown as
2 quarter, 1 nickel and 1 penny
1 quarter = 25 cents
Which means, 2 x 25 = 50 cents
1 nickel = 5 cents
and 1 penny = 1 cents
Total : 50+5+1 = 56 cents

Another way :
56  cents can be shown as 5 dime, 1 nickel and 1 penny
1 dime = 10 cents
5 dimes = 5 x 10 = 50 cents
1 nickel = 5 cents
and 1 penny = 1 cent
Total value :
50 + 5+1 = 56 cents

14.4 Make One Dollar

Draw coins to make $1. How many cents do you need?

Question 5.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 220

Answer: $ 1 is equals to 100 cents
100 cents – 79 cents = 21 cents.
So , I need 21 cents to make $ 1
21 cents can be shown in coins as 2 dime , 1 penny
1 dime = 10 cents
Which means , 2 dime = 2 x 10 = 20 cents
1 penny = 1 cent
Total : 20 + 1 = 21 cents

Question 6.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 221

Answer:
$1 is equals to 100 cents
1 dime = 10 cents
Which means , 2 cents= 2 x 10 = 20 cents
3 pennies = 3 cents
Total value = 20+3 = 23 cents
Now, 100 cents- 23 cents = 77 cents
77 cents can be shown as 3 quarters , 2 pennies
1 quarter = 25 cents
Which means , 3 quarters = 3 x 25 = 75 cents
2 pennies = 2 cents
Total value : 75 + 2 = 77 cents

14.5 Make Change from One Dollar

Question 7.
Reasoning
Newton buys a toy for 21¢. Descartes buys one for 94¢. You buy one for 57¢. You each pay with $1. Who gets back the least amount of money? How do you know?
_____________________________________
_____________________________________

Answer:
$1 is equals to 100 cents
The cost of newtons toy is 21 cents
100 – 21 =79 cents
So, 79 cents is change newtons gets back.
Now, the cost of descartes toy is 94 cents
100 – 97 = 3 cents
So Descartes change is 3 cents
On comparing both newtons and Descartes money
79 cents is greater than 3cents;
79 > 3
Therefore, Descartes gets back the least amount of money.

14.6 Find Total Values of Bills

Question 8.
Count on to find the total value.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 222

Answer: $ 38

Explanation :
We have , Three $ 1 bills
which means , $1 +$1 $1 = $3
One $ 5 bill
One $ 10 bill
One $ 20 bill
Total value :
$ 3+ $5 +$10 + $20 =$38
Therefore the total value of the dollar bills is $ 38.

Question 9.
Newton has five $10 bills. He has $32 more than Descartes. How much money does Descartes have?
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 223

Answer:  $18

Explanation :
five $ 10 bills is equals to
5 x 10 = $ 50
Given that , Newton has $32 more than Descartes
We know that, Newton have $50
Now, $50 – $32 = $18
Therefore , Descartes have $18

14.8 Tell Time to the Nearest Five Minutes

Write the time.

Question 10.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 224

Answer: 7 hours 20 minutes

Explanation :
As the hours hand is between 7 and 8, the minutes hand is on 4. So, the time is 20 minutes.
That means, the time is 7 hours and 20 minutes.

Question 11.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 225

Answer: 3 hours 45 minutes

Explanation :
As the hours hand is between 3 and 4, the minutes hand is on 9. So, the time is 45 minutes.
That means, the time is 3 hours and 45 minutes.

Question 12.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 226

Answer: 10 hours 15 minutes

Explanation :
As the hours hand is between 10 and 11, the minutes hand is on 3. So, the time is 15 minutes.
That means, the time is 10 hours and 15 minutes.

14.9 Tell Time Before and After the Hour

Question 13.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 227

Answer: quarter past 5
which means 5 hours 15 minutes
quarter to 5 means 4 hours 45 minutes

Question 14.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 228

Answer: quarter to 10
which means ,9 hours 45 minutes
half past 9 means 9 hours 30 minutes

Question 15.
Modeling Real Life
Soccer practice starts at half past 1. Are you early or late to soccer practice? Explain.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 229

Answer:  I am late to soccer practice.

Explanation:
now the time is 1 hour 45 minutes .
The class starts at half past one
Which means ,1 hour 30 minutes
So ,I am late by 15 minutes to the class.

14.10 Relate A.M. and P.M.

Draw to show the time. Circle a.m. or p.m.

Question 16.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 230

Answer : 9hours 30 minutes p.m

Question 17.
Big Ideas Math Answer Key Grade 2 Chapter 14 Money and Time 231

 

Answer:7 hours 25 minutes a.m.

Conclusion:

We wish the information provided in the Big Ideas Math Answers Grade 2 Chapter 14 Money and Time for all the students. Go through the solved examples to have a complete grip on the subject and also on the way of solving each problem. Big Ideas Math Grade 2 Chapter 14 Money and Time Solution Key will help the students to score the highest marks in the exam.

Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions

Trying to Score Better Scores in Grade 5 Maths? Utilize Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions and make the most out of them. Begin your practice right away using the Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions and understand the concepts. Attain more knowledge using the Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions and cross-check the Solutions from it after your practice sessions.

Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions

Apply Maths in your Real-Time and get the Tips & Tricks to Solve Various Problems using Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions. Try to utilize the resource available Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions and make your preparation way more effective. You just need to tap on the quick links available in order to access them and learn all the topics.

Lesson: 1 Multiply Whole Numbers by Fractions

Lesson: 2 Use Models to Multiply Fractions by Whole Numbers

Lesson: 3 Multiply Fractions and Whole Numbers

Lesson: 4 Use Models to Multiply Fractions

Lesson: 5 Multiply Fractions

Lesson: 6 Find Areas of Rectangles

Lesson: 7 Multiply Mixed Numbers

Lesson: 8 Compare Factors and Products

Chapter: 9 – Multiply Fractions

Lesson 9.1 Multiply Whole Numbers by Fractions

Explore and Grow

Write any proper fraction that is not a unit fraction. Draw a model to represent your fraction. Draw a model to find a multiple of your fraction?
Answer:
The proper fraction that is not a unit fraction is:  \(\frac{5}{8}\)
Let the proper fraction be multiplied by 5
So,
We have to find the value of 5 × \(\frac{5}{8}\)
Now,
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{5}{8}\)
= \(\frac{5}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 5}{1 × 8}\)
= \(\frac{25}{8}\)
Hence, from the above,
We can conclude that the multiple of your proper fraction is: \(\frac{25}{8}\)

Reasoning
How can you use a model to multiply a whole number by a fraction? Explain.
Answer:
We can multiply a whole number by a fraction using the properties of multiplication. They are:
A) a = \(\frac{a}{1}\)
B) a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
Hence, by using the above properties, we can multiply the whole number by a fraction.

Think and Grow: Multiply Whole Numbers by Fractions

Example
Find 3 × \(\frac{2}{5}\)

Show and Grow

Multiply.
Question 1.
2 × \(\frac{3}{4}\) = ______
Answer:
2 × \(\frac{3}{4}\) = \(\frac{6}{4}\)

Explanation:
The given numbers are: 2 and \(\frac{3}{4}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
2 × \(\frac{3}{4}\)
= \(\frac{2}{1}\) × \(\frac{3}{4}\)
= \(\frac{2 × 3}{1 × 4}\)
= \(\frac{6}{4}\)
Hence,
2 × \(\frac{3}{4}\) = \(\frac{6}{4}\)

Question 2.
4 × \(\frac{5}{8}\) = ____
Answer:
4 × \(\frac{5}{8}\) = \(\frac{20}{8}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{5}{8}\)
= \(\frac{4}{1}\) × \(\frac{5}{8}\)
= \(\frac{4 × 5}{1 × 8}\)
= \(\frac{20}{8}\)
Hence,
4 × \(\frac{5}{8}\) = \(\frac{20}{8}\)

Apply and Grow: Practice

Multiply.
Question 3.
5 × \(\frac{7}{10}\) = ______
Answer:
5 × \(\frac{7}{10}\) = \(\frac{35}{10}\)

Explanation:
The given numbers are: 5 and \(\frac{7}{10}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{7}{10}\)
= \(\frac{5}{1}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{1 × 10}\)
= \(\frac{35}{10}\)
Hence,
5 × \(\frac{7}{10}\) = \(\frac{35}{10}\)

Question 4.
8 × \(\frac{2}{3}\) = ______
Answer:
8 × \(\frac{2}{3}\) = \(\frac{16}{3}\)

Explanation:
The given numbers are: 8 and \(\frac{2}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
8 × \(\frac{2}{3}\)
= \(\frac{8}{1}\) × \(\frac{2}{3}\)
= \(\frac{8 × 2}{1 × 3}\)
= \(\frac{16}{3}\)
Hence,
8 × \(\frac{2}{3}\) = \(\frac{16}{3}\)

Question 5.
7 × \(\frac{5}{6}\) = ______
Answer:
7 × \(\frac{5}{6}\) = \(\frac{35}{6}\)

Explanation:
The given numbers are: 7 and \(\frac{5}{6}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
7 × \(\frac{5}{6}\)
= \(\frac{7}{1}\) × \(\frac{5}{6}\)
= \(\frac{7 × 5}{1 × 6}\)
= \(\frac{35}{6}\)
Hence,
7 × \(\frac{5}{6}\) = \(\frac{35}{6}\)

Question 6.
9 × \(\frac{1}{2}\) = ______
Answer:
9 × \(\frac{1}{2}\) = \(\frac{9}{2}\)

Explanation:
The given numbers are: 9 and \(\frac{1}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
9 × \(\frac{1}{2}\)
= \(\frac{9}{1}\) × \(\frac{1}{2}\)
= \(\frac{9 × 1}{1 × 2}\)
= \(\frac{9}{2}\)
Hence,
9 × \(\frac{1}{2}\) = \(\frac{9}{2}\)

Question 7.
6 × \(\frac{3}{100}\) = ______
Answer:
6 × \(\frac{3}{100}\) = \(\frac{18}{100}\)

Explanation:
The given numbers are: 6 and \(\frac{3}{100}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
6 × \(\frac{3}{100}\)
= \(\frac{6}{1}\) × \(\frac{3}{100}\)
= \(\frac{6 × 3}{1 × 100}\)
= \(\frac{18}{100}\)
Hence,
6 × \(\frac{3}{100}\) = \(\frac{18}{100}\)

Question 8.
15 × \(\frac{4}{7}\) = ______
Answer:
15 × \(\frac{4}{7}\) = \(\frac{60}{7}\)

Explanation:
The given numbers are: 15 and \(\frac{4}{7}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
15 × \(\frac{4}{7}\)
= \(\frac{15}{1}\) × \(\frac{4}{7}\)
= \(\frac{15 × 4}{1 × 7}\)
= \(\frac{60}{7}\)
Hence,
15 × \(\frac{4}{7}\) = \(\frac{60}{7}\)

Question 9.
10 × \(\frac{5}{3}\) = ______
Answer:
10 × \(\frac{5}{3}\) = \(\frac{50}{3}\)

Explanation:
The given numbers are: 10 and \(\frac{5}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
10 × \(\frac{5}{3}\)
= \(\frac{10}{1}\) × \(\frac{5}{3}\)
= \(\frac{10 × 5}{1 × 3}\)
= \(\frac{50}{3}\)
Hence,
10 × \(\frac{5}{3}\) = \(\frac{50}{3}\)

Question 10.
4 × \(\frac{5}{2}\) = ______
Answer:
4 × \(\frac{5}{2}\) = 10

Explanation:
The given numbers are: 4 and \(\frac{5}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{5}{2}\)
= \(\frac{4}{1}\) × \(\frac{5}{2}\)
= \(\frac{4 × 5}{1 × 2}\)
= \(\frac{20}{2}\)
= 10
Hence,
4 × \(\frac{5}{2}\) = 10

Question 11.
3 × \(\frac{11}{8}\) = ______
Answer:
3 × \(\frac{11}{8}\) = \(\frac{33}{8}\)

Explanation:
The given numbers are: 3 and \(\frac{11}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
3 × \(\frac{11}{8}\)
= \(\frac{3}{1}\) × \(\frac{11}{8}\)
= \(\frac{11 × 3}{1 × 8}\)
= \(\frac{33}{8}\)
Hence,
3 × \(\frac{11}{8}\) = \(\frac{33}{8}\)

Find the unknown number.
Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 2
Answer:
The missing number is: 3

Explanation:
The given fractions are: \(\frac{3}{20}\) and \(\frac{9}{20}\)
Let the missing number be X
So,
X × \(\frac{3}{20}\) = \(\frac{9}{20}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{9}{20}\) ÷ \(\frac{3}{20}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{9}{20}\) ÷ \(\frac{3}{20}\)
= \(\frac{9}{20}\) × \(\frac{20}{3}\)
= \(\frac{9 × 20}{20 × 3}\)
= 3
Hence,
The missing number is: 3

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 3
Answer:
The missing number is: 6

Explanation:
The given fractions are: \(\frac{4}{9}\) and \(\frac{24}{9}\)
Let the missing number be X
So,
X × \(\frac{4}{9}\) = \(\frac{24}{9}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{24}{9}\) ÷ \(\frac{4}{9}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{24}{9}\) ÷ \(\frac{4}{9}\)
= \(\frac{24}{9}\) × \(\frac{9}{4}\)
= \(\frac{9 × 24}{9 × 4}\)
= 6
Hence,
The missing number is: 6

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 4
Answer:
The missing number is: 10

Explanation:
The given fractions are: \(\frac{5}{12}\) and \(\frac{50}{12}\)
Let the missing number be X
So,
X × \(\frac{5}{12}\) = \(\frac{50}{12}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{50}{12}\) ÷ \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{50}{12}\) ÷ \(\frac{5}{12}\)
= \(\frac{50}{12}\) × \(\frac{12}{5}\)
= \(\frac{12 × 50}{12 × 5}\)
= 10
Hence,
The missing number is: 10

Question 15.
A recipe calls for \(\frac{3}{4}\) cup of dried rice noodles. You make 44 batches of the recipe. How many cups of dried rice noodles do you use?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 5
Answer:
The number of cups of dried rice noodles you used is: 33

Explanation:
It is given that a recipe calls for \(\frac{3}{4}\) cup of dried rice noodles.
It is also given that you make 44 batches of the recipe
So,
The number of cups of dried rice noodles you used = ( The number of batches of the recipe ) × ( The number of dried rice noodles for each batch )
= 44 × \(\frac{3}{4}\)
Now,
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
44 × \(\frac{3}{4}\)
= \(\frac{44}{1}\) × \(\frac{3}{4}\)
= \(\frac{44 × 3}{1 × 4}\)
= \(\frac{33}{1}\)
= 33
Hence, from the above,
We can conclude that the number of cups of dried noodles you used is: 33 cups

Question 16.
YOU BE THE TEACHER
Your 5 × \(\frac{3}{5}\) friend says that 5 is equal to \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\). Is your friend correct? Explain.
Answer:
Yes, your friend is correct.

Explanation:
It is given that your 5 × \(\frac{3}{5}\) friend says that 5 is equal to \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\).
Now,
We know that,
n × a = a + a + a + a + …………….. + n times
Where,
‘n’ is the multiple of a
Here,
‘a’ will be the whole number or the fraction
So,
According to the above property,
5 × \(\frac{3}{5}\) = \(\frac{3}{5}\) + \(\frac{3}{5}\) + \(\frac{3}{5}\) +\(\frac{3}{5}\) + \(\frac{3}{5}\)
Hence, from the above,
We can conclude that your friend is correct.

Question 17.
Patterns
Describe and complete the pattern.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 6
Answer:
The completed pattern is:

From the above pattern,
We can observe that
The whole number is constant i.e., 9
The fraction part is changing i.e., the numerator part is increasing by 1 till 4 and the denominator is constant
So,
Now,
9 × \(\frac{1}{4}\)
= \(\frac{9}{1}\) × \(\frac{1}{4}\)
= \(\frac{9 × 1}{1 × 4}\)
= \(\frac{9}{4}\)
9 × \(\frac{2}{4}\)
= \(\frac{9}{1}\) × \(\frac{2}{4}\)
= \(\frac{9 × 2}{1 × 4}\)
= \(\frac{18}{4}\)
So,
The remaining two multiplication equations will also be solved in the same way.

Think and Grow: Modeling Real Life

Example
Your goal is to make a waterslide that is at least 10 meters long. You make the waterslide using 10 plastic mats that are each \(\frac{3}{2}\) meters long. Do you reach your goal?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 7
Find the length of the waterslide by multiplying the number of mats by the length of each mat.

So,
You reached your goal.

Show and Grow

Question 18.
An excavator is moving 4 piles of dirt that are the same size. Each pile requires \(\frac{3}{4}\) hour to move. Can the excavator move all of the piles in 2 hours?
Answer:
No, the excavator can’t move all of the piles in 2 hours

Explanation:
It is given that an excavator is moving 4 piles of dirt that are of the same size and each pile requires \(\frac{3}{4}\) hour to move.
So,
We know that,
1 hour = 60 minutes
So,
\(\frac{3}{4}\) of 1 hour = \(\frac{3}{4}\) × 60
= \(\frac{3}{4}\) × \(\frac{60}{1}\)
= \(\frac{3 × 60}{4 × 1}\)
= 45 minutes
So,
The time is taken to move 4 piles of dirt = ( The time is taken to move each pile )  × ( The total number of piles )
= 45 × 4
=180 minutes
It is given that you have to move the piles of dirt in 2 hours but the time taken is 4 hours.
Hence, from the above,
We can conclude that the excavator can’t move all the piles in 2 hours.

Question 19.
You walk dogs \(\frac{5}{4}\) miles two times each day. How far do you walk the dogs in 1 week?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 9
Answer:
The number of miles you walk the dogs in 1 week is: \(\frac{35}{2}\)

Explanation:
It is given that you walk dogs \(\frac{5}{4}\) miles two times each day
So,
The total number of miles you walk dogs in 1 day = 2 × \(\frac{5}{4}\)
= \(\frac{5}{4}\) × \(\frac{2}{1}\)
= \(\frac{5 × 2}{4 × 1}\)
= \(\frac{5}{2}\) miles
We know that,
1 week = 7 days
So,
The total number of miles you walk dogs in 1 week = ( The number of miles you walk dogs in 1 day ) × 7
= \(\frac{5}{2}\) × 7
= \(\frac{5}{2}\) × \(\frac{7}{1}\)
= \(\frac{5 × 7}{2 × 1}\)
= \(\frac{35}{2}\)
Hence, from the above,
We can conclude that the numebr of miles you walk dogs in 1 week is: \(\frac{35}{2}\) miles

Question 20.
DIG DEEPER
You have10 feet of string. You need \(\frac{5}{3}\) feet of string to make 1 necklace. You make 5 necklaces. Do you have enough string to make another necklace? Explain.
Answer:
Yes, we have enough string to make another necklace

Explanation:
It is given that you have 10 feet of string and you need \(\frac{5}{3}\) feet of string to make 1 necklace. and you make 5 necklaces.
So,
The length of the string required for 5 necklaces = ( The length of string required for 1 necklace ) × ( The total number of necklaces )
= \(\frac{5}{3}\) × 5
= \(\frac{5}{3}\) × \(\frac{5}{1}\)
= \(\frac{5 × 5}{3 × 1}\)
= \(\frac{25}{3}\) feet
We can write 10 feet as \(\frac{30}{3}\) feet
So,
When we compare the total length of the string and the length of the string to make 5 necklaces,
We can observe that the length of the string to make 5 necklaces is less than the total length of the string.
Hence, from the above,
We can conclude that we have enough string to make another necklace.

Multiply Whole Numbers by Fractions Homework & Practice 9.1

Multiply
Question 1.
5 × \(\frac{2}{3}\) = ______
Answer:
5 × \(\frac{2}{3}\) = \(\frac{10}{3}\)

Explanation:
The given numbers are: 5 and \(\frac{2}{3}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
5 × \(\frac{2}{3}\)
= \(\frac{5}{1}\) × \(\frac{2}{3}\)
= \(\frac{2 × 5}{1 × 3}\)
= \(\frac{10}{3}\)
Hence,
5 × \(\frac{2}{3}\) = \(\frac{10}{3}\)

Question 2.
9 × \(\frac{7}{8}\) = ______
Answer:
9 × \(\frac{7}{8}\) = \(\frac{63}{8}\)

Explanation:
The given numbers are: 9 and \(\frac{7}{8}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
9 × \(\frac{7}{8}\)
= \(\frac{9}{1}\) × \(\frac{7}{8}\)
= \(\frac{9 × 7}{1 × 8}\)
= \(\frac{63}{8}\)
Hence,
9 × \(\frac{7}{8}\) = \(\frac{63}{8}\)

Question 3.
4 × \(\frac{11}{12}\) = ______
Answer:
4 × \(\frac{11}{12}\) = \(\frac{44}{12}\)

Explanation:
The given numbers are: 4 and \(\frac{11}{12}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
4 × \(\frac{11}{12}\)
= \(\frac{4}{1}\) × \(\frac{11}{12}\)
= \(\frac{4 × 11}{1 × 12}\)
= \(\frac{44}{12}\)
Hence,
4 × \(\frac{11}{12}\) = \(\frac{44}{12}\)

Question 4.
8 × \(\frac{35}{100}\) = ______
Answer:
8 × \(\frac{35}{100}\) = \(\frac{280}{100}\)

Explanation:
The given numbers are: 8 and \(\frac{35}{100}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
8 × \(\frac{35}{100}\)
= \(\frac{8}{1}\) × \(\frac{35}{100}\)
= \(\frac{8 × 35}{1 × 100}\)
= \(\frac{280}{100}\)
Hence,
8 × \(\frac{35}{100}\) = \(\frac{280}{100}\)

Question 5.
3 × \(\frac{1}{2}\) = ______
Answer:
3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Explanation:
The given numbers are: 3 and \(\frac{1}{2}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
3 × \(\frac{1}{2}\)
= \(\frac{3}{1}\) × \(\frac{1}{2}\)
= \(\frac{1 × 3}{1 × 2}\)
= \(\frac{3}{2}\)
Hence,
3 × \(\frac{1}{2}\) = \(\frac{3}{2}\)

Question 6.
7 × \(\frac{2}{5}\) = ______
Answer:
7 × \(\frac{2}{5}\) = \(\frac{14}{5}\)

Explanation:
The given numbers are: 7 and \(\frac{2}{5}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
7 × \(\frac{2}{5}\)
= \(\frac{7}{1}\) × \(\frac{2}{5}\)
= \(\frac{2 × 7}{1 × 5}\)
= \(\frac{14}{5}\)
Hence,
7 × \(\frac{2}{5}\) = \(\frac{14}{5}\)

Question 7.
6 × \(\frac{7}{4}\) = ______
Answer:
6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)

Explanation:
The given numbers are: 6 and \(\frac{7}{4}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
6 × \(\frac{7}{4}\)
= \(\frac{6}{1}\) × \(\frac{7}{4}\)
= \(\frac{6 × 7}{1 × 4}\)
= \(\frac{42}{4}\)
Hence,
6 × \(\frac{7}{4}\) = \(\frac{42}{4}\)

Question 8.
12 × \(\frac{8}{7}\) = ______
Answer:
12 × \(\frac{8}{7}\) = \(\frac{96}{7}\)

Explanation:
The given numbers are: 12 and \(\frac{8}{7}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
12 × \(\frac{8}{7}\)
= \(\frac{12}{1}\) × \(\frac{8}{7}\)
= \(\frac{12 × 8}{1 × 7}\)
= \(\frac{96}{7}\)
Hence,
12 × \(\frac{8}{7}\) = \(\frac{96}{7}\)

Question 9.
25 × \(\frac{10}{9}\) = ______
Answer:
25 × \(\frac{10}{9}\) = \(\frac{250}{9}\)

Explanation:
The given numbers are: 25 and \(\frac{10}{9}\)
We know that,
a = \(\frac{a}{1}\)
a × \(\frac{a}{b}\) = \(\frac{a × a}{b × 1}\)
So,
25 × \(\frac{10}{9}\)
= \(\frac{25}{1}\) × \(\frac{10}{9}\)
= \(\frac{25 × 10}{1 × 9}\)
= \(\frac{250}{9}\)
Hence,
25 × \(\frac{10}{9}\) = \(\frac{250}{9}\)

Find the unknown number.
Question 10.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 10
Answer:
The missing number is: 5

Explanation:
The given fractions are: \(\frac{5}{7}\) and \(\frac{25}{7}\)
Let the missing number be X
So,
X × \(\frac{5}{7}\) = \(\frac{25}{7}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{25}{7}\) ÷ \(\frac{5}{7}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{25}{7}\) ÷ \(\frac{5}{7}\)
= \(\frac{25}{7}\) × \(\frac{7}{5}\)
= \(\frac{7 × 25}{7 × 5}\)
= 5
Hence,
The missing number is: 5

Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 11
Answer:
The missing number is: 7

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{63}{10}\)
Let the missing number be X
So,
X × \(\frac{9}{10}\) = \(\frac{63}{10}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{63}{10}\) ÷ \(\frac{9}{10}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{63}{10}\) ÷ \(\frac{9}{10}\)
= \(\frac{63}{10}\) × \(\frac{10}{9}\)
= \(\frac{63 × 10}{10 × 9}\)
= 7
Hence,
The missing number is: 7

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 12
Answer:
The missing number is: 9

Explanation:
The given fractions are: \(\frac{3}{5}\) and \(\frac{27}{5}\)
Let the missing number be X
So,
X × \(\frac{3}{5}\) = \(\frac{27}{5}\)
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = \(\frac{27}{5}\) ÷ \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{27}{5}\) ÷ \(\frac{3}{5}\)
= \(\frac{27}{5}\) × \(\frac{5}{3}\)
= \(\frac{5 × 27}{5 × 3}\)
= 9
Hence,
The missing number is: 9

Question 13.
You make 5 servings of pancakes. You top each serving with \(\frac{1}{4}\) cup of strawberries. How many cups of strawberries do you use?
Answer:
The number of cups of strawberries you used is: 20 cups

Explanation:
It is given that you make 5 servings of pancakes and you top each serving with \(\frac{1}{4}\) cup of strawberries.
So,
The number of cups of strawberries = \(\frac{The number of servings of pancakes}{Each serving in pancake}\)
= 5 ÷ \(\frac{1}{4}\)
= \(\frac{5}{1}\) ÷ \(\frac{1}{4}\)
= \(\frac{5}{1}\) × \(\frac{4}{1}\)
= \(\frac{5 × 4}{1 × 1}\)
= 20 cups
hence, from the above,
We can conclude that the number of cups of strawberries is: 20 cups

Question 14.
Which One Doesn’tBelong?
Which one does not belong with the other three?
4 × \(\frac{3}{8}\)
3 × \(\frac{1}{8}\)
\(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\)
1 + \(\frac{1}{2}\)
Answer:
Let the expresseons be named A., B., C., D.
So,
A) 4 × \(\frac{3}{8}\)
B) 3 × \(\frac{1}{8}\)
C) \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\) + \(\frac{3}{8}\)
D) 1 + \(\frac{1}{2}\)
So, from the above four expressions,
Expressions A), B), C) are in the normal form and C) is the expanded form of expression A)
Hence, from the above,
We can conclude that expression D) does not belong to the other three.

Question 15.
Modeling Real Life
You complete \(\frac{5}{2}\) inches of weaving each day for 5 days. The weaving needs to be at least 11 inches long. Is your weaving complete?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 13
Answer:
Yes, your weaving is complete

Explanation:
It is given that you complete \(\frac{5}{2}\) inches of weaving each day for 5 days.
So,
The total number of inches you completed in 5 days = ( The number of inches you weaved each day ) × 5
= \(\frac{5}{2}\) × 5
= \(\frac{5}{2}\) × \(\frac{5}{1}\)
= \(\frac{5 × 5}{2 × 1}\)
= \(\frac{25}{2}\)
It is also given that the weaving needs to be atleast 11 inches long.
So,
11 inches can also be written as \(\frac{22}{2}\)
So,
When we compare the total number of inches needed for weaving and the total number of inches you weaved in 5 days, we can observe that
You weaved more than the amount of the weaving.
Hence, from the above,
We can conclude that your weaving is completed.

Question 16.
DIG DEEPER!
You spend \(\frac{7}{2}\) hours playing drums each day for 2 days. Your friend spends \(\frac{5}{4}\) hours playing drums each day for 6 days. Who spends more time playing drums? How much more?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 14
Answer:
Your friend spend more time playing drums than you
The amount you played drums more than your friend is: \(\frac{1}{2}\) hours

Explanation:
It is given that you spend \(\frac{7}{2}\) hours each day playing drums for 2 days
So,
The time you played drums for 2 days = The time you played drums for each day × 2
= \(\frac{7}{2}\) × 2
= \(\frac{7}{2}\) × \(\frac{2}{1}\)
= \(\frac{7 × 2}{2 × 1}\)
= \(\frac{14}{2}\)
= \(\frac{28}{4}\)
= 7 hours
It is also given that Your friend spends \(\frac{5}{4}\) hours playing drums each day for 6 days.
So,
The time your friend played drums for 6 days = The time your friend played drums for each day × 6
= \(\frac{5}{4}\) × 6
= \(\frac{5}{4}\) × \(\frac{6}{1}\)
= \(\frac{5 × 6}{4 × 1}\)
= \(\frac{30}{4}\) hours
Now,
When we compare the time played drums by you and your friends, your friend played more time than you
So,
The amount of time more your friend played than you = \(\frac{30}{4}\) – \(\frac{28}{4}\)
= \(\frac{30 – 28}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\) hour
Hence, from the above,
We can conclude that
Your friend spend more time playing drums than you
The amount you played drums more than your friend is: \(\frac{1}{2}\) hours

Review & Refresh

Find the product.
Question 17.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 15
Answer:
0.6 × 0.4 = 0.24

Explanation:
The given decimal numbers are: 0.6 and 0.4
the representation of the decimal numbers in the fraction form is: \(\frac{6}{10}\) and \(\frac{4}{10}\)
Now,
\(\frac{6}{10}\) × \(\frac{4}{10}\)
= \(\frac{6 × 4}{10 × 10}\)
= \(\frac{24}{100}\)
So,
The representation of \(\frac{24}{100}\) in the decimal form is: 0.24
Hence, 0.6 × 0.4 = 0.24

Question 18.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 16
Answer:
2.37 × 1.9 = 4.503

Explanation:
The given decimal numbers are: 2.37 and 1.9
the representation of the decimal numbers in the fraction form is: \(\frac{237}{100}\) and \(\frac{19}{10}\)
Now,
\(\frac{237}{100}\) × \(\frac{19}{10}\)
= \(\frac{237 × 19}{100 × 10}\)
= \(\frac{4,503}{1000}\)
So,
The representation of \(\frac{4503}{1000}\) in the decimal form is: 4.503
Hence, 2.37 × 1.9 = 4.503

Question 19.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.1 17
Answer:
52.8 × 0.75 = 39.6

Explanation:
The given decimal numbers are: 52.8 and 0.75
the representation of the decimal numbers in the fraction form is: \(\frac{6}{10}\) and \(\frac{4}{10}\)
Now,
\(\frac{528}{10}\) × \(\frac{75}{100}\)
= \(\frac{528 × 75}{10 × 100}\)
= \(\frac{3960}{1000}\)
So,
The representation of \(\frac{3960}{1000}\) in the decimal form is: 39.6
Hence, 52.8 × 0.75 = 39.6

Lesson 9.2 Use Models to Multiply Fractions by Whole Numbers

You need to give water to \(\frac{2}{3}\) of the dogs at a shelter. There are 12 dogs at the shelter. How many dogs need water? Draw a model to support your answer.
Answer:
The number of dogs that need water is: 8 dogs

Explanation:
It is given that you need to give water to \(\frac{2}{3}\) of the dogs at a shelter and there are 12 dogs at the shelter
So,
The number of dogs that need water = ( The fraction of dogs that need water ) × ( The total number of dogs )
= \(\frac{2}{3}\) × 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 12
= \(\frac{2}{3}\) × \(\frac{12}{1}\)
= \(\frac{2 × 12}{3 × 1}\)
= \(\frac{8}{1}\)
= 8
Hence, from the above,
We can conclude that there are 8 dogs that need water

Reasoning
How can you use a model to multiply a fraction by a whole number? Explain.
Answer:
We can multiply a fraction by a whole number by using the following multiplication properties. They are:
A) \(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
B) x = \(\frac{x}{1}\)

Think and Grow: Multiply Fractions by Whole Numbers

You can use models to multiply a fraction by a whole number.
Example
Find \(\frac{3}{4}\) of 8.

Example
Find \(\frac{5}{8}\) × 4.

Show and Grow

Question 1.
Find \(\frac{2}{5}\) of 10.
Answer:
\(\frac{2}{5}\) × 10 = 4

Explanation:
The given numbers are: \(\frac{2}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 10 = \(\frac{2}{5}\) × \(\frac{10}{1}\)
= \(\frac{2 × 10}{5 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence,
\(\frac{2}{5}\) × 10 = 4

Question 2.
Find \(\frac{7}{12}\) × 6.
Answer:
\(\frac{7}{12}\) × 6 = \(\frac{7}{2}\)

Explanation:
The given numbers are: \(\frac{7}{12}\) and 6
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{7}{12}\) × 6 = \(\frac{7}{12}\) × \(\frac{6}{1}\)
= \(\frac{7 × 6}{12 × 1}\)
= \(\frac{7}{2}\)
Hence,
\(\frac{7}{12}\) × 6 = \(\frac{7}{2}\)

Apply and Grow: Practice

Multiply. Use a model to help.
Question 3.
\(\frac{5}{6}\) of 12
Answer:
\(\frac{5}{6}\) × 12 = 10

Explanation:
The given numbers are: \(\frac{5}{6}\) and 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{5}{6}\) × 12 = \(\frac{5}{6}\) × \(\frac{12}{1}\)
= \(\frac{5 × 12}{6 × 1}\)
= \(\frac{10}{1}\)
= 10
Hence,
\(\frac{5}{6}\) × 12 = 10

Question 4.
\(\frac{2}{3}\) × 9
Answer:
\(\frac{2}{3}\) × 9 = 6

Explanation:
The given numbers are: \(\frac{2}{3}\) and 9
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 9 = \(\frac{2}{3}\) × \(\frac{9}{1}\)
= \(\frac{2 × 9}{3 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{2}{3}\) × 9 = 6

Question 5.
\(\frac{1}{5}\) × 10
Answer:
\(\frac{1}{5}\) × 10 = 2

Explanation:
The given numbers are: \(\frac{1}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{5}\) × 10 = \(\frac{1}{5}\) × \(\frac{10}{1}\)
= \(\frac{1 × 10}{5 × 1}\)
= \(\frac{2}{1}\)
= 2
Hence,
\(\frac{1}{5}\) × 10 = 2

Question 6.
\(\frac{3}{5}\) of 5
Answer:
\(\frac{3}{5}\) × 5 = 3

Explanation:
The given numbers are: \(\frac{3}{5}\) and 5
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{5}\) × 5 = \(\frac{3}{5}\) × \(\frac{5}{1}\)
= \(\frac{3 × 5}{5 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence,
\(\frac{3}{5}\) × 5 = 3

Question 7.
\(\frac{1}{6}\) of 3
Answer:
\(\frac{1}{6}\) × 3 = \(\frac{1}{2}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 3
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{6}\) × 3 = \(\frac{1}{6}\) × \(\frac{3}{1}\)
= \(\frac{1 × 3}{6 × 1}\)
= \(\frac{1}{2}\)
Hence,
\(\frac{1}{6}\) × 3 = \(\frac{1}{2}\)

Question 8.
\(\frac{3}{8}\) × 4
Answer:
\(\frac{3}{8}\) × 4 = \(\frac{3}{2}\)

Explanation:
The given numbers are: \(\frac{3}{8}\) and 4
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{8}\) × 4 = \(\frac{3}{8}\) × \(\frac{4}{1}\)
= \(\frac{3 × 4}{8 × 1}\)
= \(\frac{3}{2}\)
Hence,
\(\frac{3}{8}\) × 4 = \(\frac{3}{2}\)

Question 9.
You have 25 beads. You use \(\frac{2}{5}\) of the beads to make a bracelet. How many beads do you use?
Answer:
The number of beads you used is: 10 beads

Explanation:
It is given that you have 25 beads and you use \(\frac{2}{5}\) of the beads to make a bracelet.
So,
The number of beads you used = ( The fraction of beads used to make 1 bracelet ) × ( The total number of beads )
= \(\frac{2}{5}\) × 25
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 25 = \(\frac{2}{5}\) × \(\frac{25}{1}\)
= \(\frac{2 × 25}{5 × 1}\)
= \(\frac{10}{1}\)
= 10 beads
Hence, from the above,
We can conclude that the number of beads you used to make bracelets are: 10 beads

Question 10.
Writing
Write and solve a real-life problem for the expression.
\(\frac{3}{4}\) × 20
Answer:
Suppose there are 20 passengers in a bus and they occupied the place \(\frac{3}{4}\) of the bus.
Hence,
The total number of seats in the bus = ( The fraction of the place occupied by the passengers ) × ( The total number of passengers )
= \(\frac{3}{4}\) × 20
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{4}\) × 20 = \(\frac{3}{4}\) × \(\frac{20}{1}\)
= \(\frac{3 × 20}{4 × 1}\)
= \(\frac{15}{1}\)
= 15 seats
Hence, from the above,
We can conclude that there are 15 seats in the bus.

Question 11.
YOU BE THE TEACHER
Descartes finds \(\frac{2}{3}\) × 6. Is he correct? Explain.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 3
Answer:
No, Descartes is not correct

Explanation:
According to Descartes,
The given numbers are: \(\frac{2}{3}\) and 6
So,
According to Descartes, 6 is divided into 3 equal parts
So,
6 ÷ 3 = 2 equal parts
So,
According to Descartes, these 2 equal parts also divided into 2 parts.
So,
The total number of parts = 2 equal parts + 2 equal parts = 4 parts
So,
The product of \(\frac{2}{3}\) and 6 is 4
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 6 = \(\frac{2}{3}\) × \(\frac{6}{1}\)
= \(\frac{6 × 2}{3 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence, from the above,
We can conclude that Descartes is not correct.

Think and Grow: Modeling Real Life

Example
A recipe calls for 2 cups of rice. You only have \(\frac{3}{4}\) of that amount. How much more rice do you need?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 4
Find the number of cups of rice that you have by finding \(\frac{3}{4}\) of 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 5
So,
You have \(\frac{3}{2}\) cups of rice.
Subtract the amount of rice you have from the amount of rice you need.
2 – \(\frac{3}{2}\) = \(\frac{2}{1}\) – \(\frac{3}{2}\)
= \(\frac{(2 × 2) – 3}{2}\)
= \(\frac{1}{2}\)
So,
you need \(\frac{1}{2}\) cup more of rice.

Show and Grow

Question 12.
You have 12 tokens. You use \(\frac{3}{4}\) of them to play a pinball game. How many tokens do you have left?
Answer:
The number of tokens you have left is: 3 tokens

Explanation:
It is given that you have 12 tokens and you use \(\frac{3}{4}\) of them to play a pinball game.
So,
The number of tokens used to play a pinball game = ( The fraction of tokens used to play a pinball game ) × ( The total number of tokens )
= \(\frac{3}{4}\) × 12
= \(\frac{3}{4}\) × \(\frac{12}{1}\)
= \(\frac{3 × 12}{4 × 1}\)
= \(\frac{9}{1}\)
= 9 tokens
So,
The number of tokens left = ( The total number of tokens ) – ( The number of tokens used to play the pinball )
= 12 – 9
= 3 tokens
Hence, from the above,
We can conclude that the number of tokens left are: 3 tokens

Question 13.
A male lion sleeps \(\frac{5}{6}\) × 6 hours of each day. How many hours does the lion sleep in 1 week?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 6
Answer:
The number of hours the lion sleeps in 1 week is: 35 hours

Explanation:
It is given that a male lion sleeps \(\frac{5}{6}\) × 6 hours of each day
So,
The number of hours the lion sleeps in 1 day = \(\frac{5}{6}\) × 6 hours
= \(\frac{5}{6}\) × \(\frac{6}{1}\)
= \(\frac{5 × 6}{6 × 1}\)
= \(\frac{5}{1}\)
= 5 hours
We know that,
1 week = 7 days
So,
The total number of hours the lion sleeps in 1 week = ( The number of hours the lion sleeps in 1 day ) × ( The number od fays in 1 week )
= 5 × 7
= 35 hours
Hence, from the above,
We can conclude that the lion sleeps for 35 hours in 1 week

Question 14.
DIG DEEPER!
In a class of 20 students, \(\frac{1}{10}\) of the students are 10 years old, \(\frac{4}{5}\)of the students are 11 years old, and the rest are 12 years old. How many more 11-year-olds than 12-year-olds are in the class?
Answer:
The number of 11-year-olds more than 12-year-olds is: 14

Explanation:
It is given that in a class of 20 students, \(\frac{1}{10}\) of the students are 10 years old, \(\frac{4}{5}\)of the students are 11 years old, and the rest are 12 years old.
So,
The number of 10-year-olds = ( The fraction of 10-year-olds ) × ( The total number of students )
= \(\frac{1}{10}\) × 20
= \(\frac{1 × 20}{10 × 1}\)
= 2 10-year-olds
The number of 11-year-olds = ( The fraction of 11-year-olds ) × ( The total number of students )
= \(\frac{4}{5}\) × 20
= \(\frac{4}{5}\) × \(\frac{20}{1}\)
= \(\frac{4 × 20}{5 × 1}\)
= \(\frac{16}{1}\)
= 16 11-year-olds
So,
The number of 12-year-olds = ( The total number of students ) – ( The number of 10-year-olds + The number of 11-year-olds )
= 20 – ( 16 + 2 )
= 2 12-year-olds
So,
The number of 11-year-olds more than 12-year-olds = ( The number of 11-year-olds ) – ( The numebr of 12-year-olds )
= 16 – 2
= 14 students
Hence, from the above,
We can conclude that there are 14 students who are 11-years-old more than 12-years-old.

Use Models to Multiply Fractions by Whole Numbers Homework & Practice 9.2

Multiply. Use a model to help.
Question 1.
\(\frac{2}{3}\) × 6
Answer:
\(\frac{2}{3}\) × 6 = 4

Explanation:
The given numbers are: \(\frac{2}{3}\) and 6
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{3}\) × 6 = \(\frac{2}{3}\) × \(\frac{6}{1}\)
= \(\frac{2 × 6}{3 × 1}\)
= \(\frac{4}{1}\)
= 4
Hence,
\(\frac{2}{3}\) × 6 = 4

Question 2.
\(\frac{3}{5}\) of 10
Answer:
\(\frac{3}{5}\) × 10 = 6

Explanation:
The given numbers are: \(\frac{3}{5}\) and 10
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{5}\) × 10 = \(\frac{3}{5}\) × \(\frac{10}{1}\)
= \(\frac{3 × 10}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{3}{5}\) × 10 = 6

Question 3.
\(\frac{1}{2}\) of 4
Answer:
\(\frac{1}{2}\) × 4 = 2

Explanation:
The given numbers are: \(\frac{1}{2}\) and 4
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{2}\) × 4 = \(\frac{1}{2}\) × \(\frac{4}{1}\)
= \(\frac{1 × 4}{2 × 1}\)
= \(\frac{2}{1}\)
= 2
Hence,
\(\frac{1}{2}\) × 4 = 2

Question 4.
\(\frac{1}{4}\) × 12
Answer:
\(\frac{1}{4}\) × 12 = 3

Explanation:
The given numbers are: \(\frac{1}{4}\) and 12
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{4}\) × 12 = \(\frac{1}{4}\) × \(\frac{12}{1}\)
= \(\frac{1 × 12}{4 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence,
\(\frac{1}{12}\) × 4 = 3

Question 5.
\(\frac{5}{6}\) × 3
Answer:
\(\frac{5}{6}\) × 3 = \(\frac{5}{2}\)

Explanation:
The given numbers are: \(\frac{5}{6}\) and 3
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{5}{6}\) × 3 = \(\frac{5}{6}\) × \(\frac{3}{1}\)
= \(\frac{5 × 3}{6 × 1}\)
= \(\frac{5}{2}\)
Hence,
\(\frac{5}{6}\) × 3 = \(\frac{5}{2}\)

Question 6.
\(\frac{3}{4}\) of 2
Answer:
\(\frac{3}{4}\) × 2 = \(\frac{3}{2}\)

Explanation:
The given numbers are: \(\frac{3}{4}\) and 2
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{3}{4}\) × 2 = \(\frac{3}{4}\) × \(\frac{2}{1}\)
= \(\frac{3 × 2}{4 × 1}\)
= \(\frac{3}{2}\)
Hence,
\(\frac{3}{4}\) × 2 = \(\frac{3}{2}\)

Question 7.
You have 27 foam balls. You use \(\frac{1}{3}\) of the balls for a model. How many balls do you use?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 7
Answer:
The number of balls you used is: 9 balls

Explanation:
It is given that there are 27 foam balls and you use \(\frac{1}{3}\) of the balls for a model.
So,
The number of balls you used = ( The fraction of balls you used ) × ( The total number of balls )
= \(\frac{1}{3}\) × 27
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{3}\) × 27 = \(\frac{1}{3}\) × \(\frac{27}{1}\)
= \(\frac{1 × 27}{3 × 1}\)
= \(\frac{9}{1}\)
= 9 balls
Hence, from the above,
We can conclude that the number of foam balls you used are: 27

Question 8.
An object that weighs 1 pound on Earth weighs about \(\frac{1}{15}\) pound on Pluto. A man weighs 240 pounds on Earth. How many pounds does he weigh on Pluto?
Answer:
The number of pounds the man weigh in Pluto is: 16 pounds

Explanation:
It is given that an object that weighs 1 pound on Earth weighs about \(\frac{1}{15}\) pound on Pluto.
It is also given that a man weighs 240 pounds on Earth.
So,
The weight of a man on pluto = ( The fraction of weight of a man on pluto when compared to earth ) × ( The weight of the man on earth )
= \(\frac{1}{15}\) × 240
Now,
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{1}{15}\) × 240 = \(\frac{1}{15}\) × \(\frac{240}{1}\)
= \(\frac{1 × 240}{15 × 1}\)
= \(\frac{16}{1}\)
= 16 pounds
Hence, from the above,
We can conclude that the weight of the man on pluto is: 16 pounds

Question 9.
Structure
Write a multiplication equation represented by the model.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 8
Answer:

Question 10.
DIG DEEPER!
Find each missing number.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 9
Answer:
Let the expression be named A), B), C), and D)
So,
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5

Explanation:
Let the expressions be named as A), B), C), and D)
So,
A) The given fractions are: \(\frac{1}{3}\) and 4
Let the missing number be X
So,
X × \(\frac{1}{3}\) = 4
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = 4 ÷ \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) ÷ \(\frac{x}{y}\) = \(\frac{a}{b}\) × \(\frac{y}{x}\)
So,
\(\frac{4}{1}\) ÷ \(\frac{1}{3}\)
= \(\frac{4}{1}\) × \(\frac{3}{1}\)
= \(\frac{4 × 3}{1 × 1}\)
= 12
Hence, the missing number is 12
B) The given numbers are: \(\frac{2}{5}\) and 15
We know that,
\(\frac{a}{b}\) × x = \(\frac{a}{b}\) × \(\frac{x}{1}\)
= \(\frac{a × x}{b × 1}\)
So,
\(\frac{2}{5}\) × 15 = \(\frac{2}{5}\) × \(\frac{15}{1}\)
= \(\frac{2 × 15}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
\(\frac{2}{5}\) × 15 = 6
Like the above two expressions, the remaining two expressions can also be solved.
hence, from the above,
We can conclude that
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5

Question 11.
Modeling Real Life
You have 28 craft sticks. You use \(\frac{4}{7}\) of them for a project. How many craft sticks do you have left?
Answer:
The number of craft sticks you have left is: 12 craft sticks

Explanation:
It is given that you have 28 craft sticks and you used \(\frac{4}{7}\) of them for a project.
So,
The number of craft sticks you used = ( The fraction of craft sticks you used ) × ( The total number of craft sticks )
= \(\frac{4}{7}\) × 28
= \(\frac{4}{7}\) × \(\frac{28}{1}\)
= \(\frac{4 × 28}{7 × 1}\)
= \(\frac{16}{1}\)
= 16
So,
The number of craft sticks you left = ( The total number of craft sticks ) – ( The number of craft sticks you used )
= 28 – 16
= 12 craft sticks
Hence, from the above,
We can conclude that the number of craft sticks you left is: 12 craft sticks

Question 12.
Modeling Real Life
A mother otter spends \(\frac{1}{3}\) of each day feeding her baby. How many hours does the mother otter spend feeding her baby in 1 week?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.2 10
Answer:
The number of hours the mother otter spend feeding her baby in 1 week is: 56 hours

Explanation:
It is given that a mother otter spends \(\frac{1}{3}\) of each day feeding her baby.
We know that,
1 day = 24 hours
So,
The number of hours the mother otter spends feeding her baby in 1 day = ( The fraction of time the mother otter spends in feeding milk ) × 24
= \(\frac{1}{3}\) × 24
= \(\frac{1}{3}\) × \(\frac{24}{1}\)
= \(\frac{1 × 24}{3 × 1}\)
= \(\frac{8}{1}\)
= 8 hours
Now,
We know that,
1 week = 7 days
So,
The number of hours the mother otter spends feeding her milk in 1 week = ( The number of hours the mother otter feeding her milk in 1 day ) × 7
= 8 × 7
= 56 hours
Hence, from the above,
We can conclude that the mother otter spends feeding her milk for 56 hours in 1 week

Review & Refresh

Estimate the quotient.
Question 13.
5,692 ÷ 5
Answer:
5,692 ÷ 5 =1,138 R 2

Explanation:
By using the partial quotients method,
5,692 ÷ 5 = ( 5,000 + 600 + 90 ) ÷ 5
= ( 5,000 ÷ 5 ) + ( 600 ÷ 5 ) + ( 90 ÷ 5 )
= 1,000 + 120 + 18
= 1,138 R 2
Hence, 5,692 ÷ 5 = 1,138 R 2

Question 14.
309 ÷ 12
Answer:
309 ÷ 12 = 25 R 9

Explanation:
By using the partial quotients method,
309 ÷ 12 = ( 240 + 60 ) ÷ 12
= ( 240 ÷ 12 ) + ( 60 ÷ 12 )
= 20 + 5
= 25 R 9
Hence, 309 ÷ 12 = 25 R 2

Question 15.
2,987 ÷ 53
Answer:
2,987 ÷ 53 = 2,987 ÷ ( 50 + 3 )
= ( 2,985 ÷ 50 ) + ( 2,985 ÷ 3 )
= 995 R 2 + 59.7
= 1,054.7 R 2

Lesson 9.3 Multiply Fractions and Whole Numbers

Explore and Grow

Use models to help you complete the table. What do you notice about each expression and its product?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 1
Answer:
The completed table is:

From the completed table,
We can observe the multiplication of the whole number and the fraction.
The product of a whole number and a fraction may be a whole number or a fraction.

Construct Arguments
Explain how to multiply fractions and whole numbers without using models.
Answer:
We can multiply the whole numbers and fractions by using the properties of the multiplication. They are:
A) \(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
B) p = \(\frac{p}{1}\)

Think and Grow: Multiply Fractions and Whole Numbers

Key Idea
You can find the product of a fraction and a whole number by multiplying the numerator and the whole number. Then write the result over the denominator.
Example
Find 2 × \(\frac{5}{6}\).
Multiply the numerator and the whole number.

Example
Find \(\frac{5}{6}\) × 2
Multiply the numerator and the whole number.

Show and Grow

Multiply.
Question 1.
3 × \(\frac{5}{8}\) = _______
Answer:
3 × \(\frac{5}{8}\) = \(\frac{15}{8}\)

Explanation:
The given numbers are: 3 and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
3 × \(\frac{5}{8}\) = \(\frac{3}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 3}{8 × 1}\)
= \(\frac{15}{8}\)
Hence,
3 × \(\frac{5}{8}\) = \(\frac{15}{8}\)

Question 2.
6 × \(\frac{4}{9}\) = _______
Answer:
6 × \(\frac{4}{9}\) = \(\frac{8}{3}\)

Explanation:
The given numbers are: 6 and \(\frac{4}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
6 × \(\frac{4}{9}\) = \(\frac{6}{1}\) × \(\frac{4}{9}\)
= \(\frac{6 × 4}{9 × 1}\)
= \(\frac{24}{9}\)
= \(\frac{8}{3}\)
Hence,
6 × \(\frac{4}{9}\) = \(\frac{8}{3}\)

Question 3.
\(\frac{2}{5}\) × 15 = _______
Answer:
15 × \(\frac{2}{5}\) = 6

Explanation:
The given numbers are: 15 and \(\frac{2}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
15× \(\frac{2}{5}\) = \(\frac{15}{1}\) × \(\frac{2}{5}\)
= \(\frac{15 × 2}{5 × 1}\)
= \(\frac{6}{1}\)
= 6
Hence,
15 × \(\frac{2}{5}\) = 6

Apply and Grow: Practice

Multiply.
Question 4.
\(\frac{3}{5}\) × 2 = _______
Answer:
2 × \(\frac{3}{5}\) = \(\frac{6}{5}\)

Explanation:
The given numbers are: 2 and \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
2 × \(\frac{3}{5}\) = \(\frac{2}{1}\) × \(\frac{3}{5}\)
= \(\frac{2 × 3}{5 × 1}\)
= \(\frac{6}{5}\)
Hence,
2 × \(\frac{3}{5}\) = \(\frac{6}{5}\)

Question 5.
5 × \(\frac{2}{9}\) = _______
Answer:
5 × \(\frac{2}{9}\) = \(\frac{10}{9}\)

Explanation:
The given numbers are: 5 and \(\frac{2}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
5 × \(\frac{2}{9}\) = \(\frac{5}{1}\) × \(\frac{2}{9}\)
= \(\frac{5 × 2}{9 × 1}\)
= \(\frac{10}{9}\)
Hence,
5 × \(\frac{2}{9}\) = \(\frac{10}{9}\)

Question 6.
\(\frac{5}{6}\) × 4 = _______
Answer:
4 × \(\frac{5}{6}\) = \(\frac{10}{3}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
4 × \(\frac{5}{6}\) = \(\frac{4}{1}\) × \(\frac{5}{6}\)
= \(\frac{5 × 4}{6 × 1}\)
= \(\frac{20}{6}\)
= \(\frac{10}{3}\)
Hence,
4 × \(\frac{5}{6}\) = \(\frac{10}{3}\)

Question 7.
8 × \(\frac{3}{10}\) = _______
Answer:
8 × \(\frac{3}{10}\) = \(\frac{12}{5}\)

Explanation:
The given numbers are: 8 and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
8 × \(\frac{3}{10}\) = \(\frac{8}{1}\) × \(\frac{3}{10}\)
= \(\frac{8 × 3}{10 × 1}\)
= \(\frac{24}{10}\)
= \(\frac{12}{5}\)
Hence,
8 × \(\frac{3}{10}\) = \(\frac{12}{5}\)

Question 8.
\(\frac{1}{5}\) × 7 = _______
Answer:
7 × \(\frac{1}{5}\) = \(\frac{7}{5}\)

Explanation:
The given numbers are: 7 and \(\frac{1}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
7 × \(\frac{1}{5}\) = \(\frac{7}{1}\) × \(\frac{1}{5}\)
= \(\frac{7 × 1}{5 × 1}\)
= \(\frac{7}{5}\)
Hence,
7 × \(\frac{1}{5}\) = \(\frac{7}{5}\)

Question 9.
9 × \(\frac{5}{12}\) = _______
Answer:
9 × \(\frac{5}{12}\) = \(\frac{15}{4}\)

Explanation:
The given numbers are: 9 and \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
9 × \(\frac{5}{12}\) = \(\frac{9}{1}\) × \(\frac{5}{12}\)
= \(\frac{9 × 5}{12 × 1}\)
= \(\frac{45}{12}\)
= \(\frac{15}{4}\)
Hence,
9 × \(\frac{5}{12}\) = \(\frac{15}{4}\)

Question 10.
15 × \(\frac{5}{8}\) = _______
Answer:
15 × \(\frac{5}{8}\) = \(\frac{75}{8}\)

Explanation:
The given numbers are: 15 and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
15 × \(\frac{5}{8}\) = \(\frac{15}{1}\) × \(\frac{5}{8}\)
= \(\frac{15 × 5}{8 × 1}\)
= \(\frac{75}{8}\)
Hence,
15 × \(\frac{5}{8}\) = \(\frac{75}{8}\)

Question 11.
\(\frac{3}{4}\) × 20 = _______
Answer:
20 × \(\frac{3}{4}\) = 15

Explanation:
The given numbers are: 20 and \(\frac{3}{4}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
20 × \(\frac{3}{4}\) = \(\frac{20}{1}\) × \(\frac{3}{4}\)
= \(\frac{20 × 3}{4 × 1}\)
= \(\frac{60}{4}\)
= \(\frac{15}{1}\)
= 15
Hence,
20 × \(\frac{3}{4}\) = 15

Question 12.
\(\frac{7}{9}\) × 5 = _______
Answer:
5 × \(\frac{7}{9}\) = \(\frac{35}{9}\)

Explanation:
The given numbers are: 5 and \(\frac{7}{9}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
5 × \(\frac{7}{9}\) = \(\frac{5}{1}\) × \(\frac{7}{9}\)
= \(\frac{5 × 7}{9 × 1}\)
= \(\frac{35}{9}\)
Hence,
5 × \(\frac{7}{9}\) = \(\frac{35}{9}\)

Question 13.
One-tenth of the 50 states in the United States of America have a mockingbird as their state bird. How many states have a mockingbird as their state bird?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 4
Answer:
The number of states that have mockingbird as their state bird is: 5 states

Explanation:
It is given that one-tenth of the 50 states in the United States of America have a mockingbird as their state bird
So,
The number of states that have a mockingbird as their state bird = ( The fraction of the states that have a mockingbird as their state bird ) × ( The total number of states in the United States of America )
= \(\frac{1}{10}\) × 50
Now,
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
50 × \(\frac{1}{10}\) = \(\frac{50}{1}\) × \(\frac{1}{10}\)
= \(\frac{50 × 1}{10 × 1}\)
= \(\frac{50}{10}\)
= \(\frac{5}{1}\)
= 5 states
Hence, from the above,
We can conclude that there are 5 states that have mockingbird as their state bird

Question 14.
Writing
Explain why 9 × \(\frac{2}{3}\) is equivalent to \(\frac{2}{3}\) × 9.
Answer:
By using the Commutative property of multiplication,
a × b = b × a
So,
By using the above property,
In 9 × \(\frac{2}{3}\),
‘a’ is: 9
‘b’ is: \(\frac{2}{3}\)
Hence, from the above,
We can conclude that  9 × “\(\frac{2}{3}\)” is equivalent to “\(\frac{2}{3}\) × 9″ by using the Commutative property of multiplication.

Question 15.
Reasoning
Without calculating, determine which product is greater. Explain.
\(\frac{1}{8}\) × 24
\(\frac{7}{8}\) × 24
Answer:
The product of “\(\frac{7}{8}\) × 24″ is greater than “\(\frac{1}{8}\) × 24″

Explanation:
The given products are: \(\frac{1}{8}\) × 24 and \(\frac{7}{8}\) × 24
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So, from the above fractions,
We can observe that except numerator, the denominator and the whole number is all the same.
So,
To find which product is greater, we just have to compare the numerators of the two fractions without calculating the value of the product.
In the comparison of the numerators,
We can observe that,
1 < 7
So,
First fraction numerator < Second fraction numerator
Hence, from the above,
We can conclude that the product of “\(\frac{7}{8}\) × 24″ is greater than “\(\frac{1}{8}\) × 24″

Think and Grow: Modeling Real Life

Example
Newton buys 27 songs. Two-thirds of them are classical songs. Descartes buys 16 songs. Seven-eighths of them are classical songs. Who buys more classical songs? How many more?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 5
Multiply \(\frac{2}{3}\) by 27 to find the number of classical songs Newton buys. Multiply \(\frac{7}{8}\) by 16 to find the number Descartes buys.

So, Newton buys more classical songs.
Subtract the products to find how many more.
18 – 14 = 4
Hence,
Newton buys 4 more classical songs than Descartes.

Show and Grow

Question 16.
You take 48 pictures on a walking tour. Five-twelfths of them is of buildings. Your friend takes 45 pictures. Six-fifteenths of them are of buildings. Who takes more pictures of buildings? How many more?
Answer:
You take more pictures
You take 2 pictures more than your friend

Explanation:
It is given that you take 48 pictures on a walking tour and five-twelfths if them is of buildings.
So,
The number of buildings taken by you = ( The total number of pictures taken by you ) × ( The fraction of pictures that is of buildings )
= 48 × \(\frac{5}{12}\)
= \(\frac{48}{1}\) × \(\frac{5}{12}\)
= \(\frac{48 × 5}{1 × 12}\)
= 20 pictures
It is also given that your friend taken 45 pictures and six-fifteenth of them is buildings
So,
The number of buildings taken by your friend = ( The total number of pictures taken by your friend ) × ( The fraction of pictures that is of buildings )
= 45 × \(\frac{6}{15}\)
= \(\frac{45}{1}\) × \(\frac{6}{15}\)
= \(\frac{45 × 6}{1 × 15}\)
= 18 pictures
So,
In the comparison of the pictures of buildings,
You take more pictures than your friend.
Now,
The number of pictures more taken by you than your friend = ( The number of pictures taken by you ) – ( The number of pictures taken by your friend )
= 20 – 18
= 2 pictures
Hence, from the above,
We can conclude that
You take more pictures
You take 2 pictures more than your friend

Question 17.
You have 72 rocks in your rock collection. Five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic. How many of your rocks are metamorphic?
Answer:
The number of rocks that are metamorphic is: 15 metamorphous rocks

Explanation:
It is given that you have 72 rocks in your rock collection and five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic.
So,
The number of rocks that are sedimentary = ( The total number of rocks ) × ( The fraction of rocks that are sedimentary )
= 72 × \(\frac{5}{8}\)
= \(\frac{72}{1}\) × \(\frac{5}{8}\)
= \(\frac{72 × 5}{1 × 8}\)
= 45 sedimentary rocks
The number of rocks that are igneous = ( The total number of rocks ) × ( The fraction of rocks that are igneous )
= 72 × \(\frac{1}{6}\)
= \(\frac{72}{1}\) × \(\frac{1}{6}\)
= \(\frac{72 × 1}{1 × 6}\)
= 12 igneous rocks
So,
The number of metamorphous rocks = ( The total number of rocks ) – ( The number of sedimentary rocks + The number of igneous rocks)
= 72 – ( 45 + 12 )
= 15 metamorphous rocks
Hence, from the above,
We can conclude that there are 15 metamorphous rocks

Question 18.
DIG DEEPER!
Each day, you spend \(\frac{3}{4}\) hour reading and \(\frac{1}{2}\) hour writing in a journal. How many total hours do you spend reading and writing in 1 week? Describe two ways to solve the problem.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 7
Answer:
The number of hours you spend on reading and writing in 1 week is:

Explanation:
It is given that, each day you spent \(\frac{3}{4}\) hour reading and \(\frac{1}{2}\) hour writing in a journal.
So,
The number of hours you spent on reading and writing a journal in 1 day = ( The number of hours spent on reading ) + ( The number of hours spent on writing )
= \(\frac{3}{4}\) + \(\frac{1}{2}\)
Multiply \(\frac{1}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{3}{4}\) + \(\frac{2}{4}\)
= \(\frac{3 + 2}{4}\)
= \(\frac{5}{4}\) hours
We know that,
1 day = 24 hours
1 week = 7 days
So,
1 week = 7 × 24 hours
So,
The number of hours you spent on reading and writing in 1 week = ( The number of hours you spent on reading and writing in 1 day ) × ( The number of hours in 1 week )
= \(\frac{5}{4}\) × 7 × 24
= \(\frac{5}{4}\) × \(\frac{7}{1}\) × \(\frac{24}{1}\)
= \(\frac{5 × 7 × 24}{4 × 1}\)
= 5 × 7 × 6
= 210 hours
Hence, from the above,
We can conclude that the number of hours you spent on reading and writing a journal in 1 week is: 210 hours

Multiply Fractions and Whole Numbers Homework & Practice 9.3

Multiply.
Question 1.
\(\frac{5}{6}\) × 3 = _______
Answer:
3 × \(\frac{5}{6}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 3 and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
3 × \(\frac{5}{6}\) = \(\frac{3}{1}\) × \(\frac{5}{6}\)
= \(\frac{3 × 5}{6 × 1}\)
= \(\frac{5}{6}\)
= \(\frac{5}{2}\)
Hence,
3 × \(\frac{5}{6}\) = \(\frac{5}{2}\)

Question 2.
\(\frac{2}{3}\) × 6 = _______
Answer:
6 × \(\frac{2}{3}\) = 4

Explanation:
The given numbers are: 6 and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
6 × \(\frac{2}{3}\) = \(\frac{6}{1}\) × \(\frac{2}{3}\)
= \(\frac{6 × 2}{3 × 1}\)
= \(\frac{12}{3}\)
= \(\frac{4}{1}\)
= 4
Hence,
6 × \(\frac{2}{3}\) = 4

Question 3.
7 × \(\frac{1}{8}\) = _______
Answer:
7 × \(\frac{1}{8}\) = \(\frac{7}{8}\)

Explanation:
The given numbers are: 7 and \(\frac{1}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
7 × \(\frac{1}{8}\) = \(\frac{7}{1}\) × \(\frac{1}{8}\)
= \(\frac{7 × 1}{8 × 1}\)
= \(\frac{7}{8}\)
Hence,
7 × \(\frac{1}{8}\) = \(\frac{7}{8}\)

Question 4.
2 × \(\frac{1}{2}\) = _______
Answer:
2 × \(\frac{1}{2}\) = 1

Explanation:
The given numbers are: 2 and \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
2 × \(\frac{1}{2}\) = \(\frac{2}{1}\) × \(\frac{1}{2}\)
= \(\frac{2 × 1}{2 × 1}\)
= \(\frac{2}{2}\)
= 1
Hence,
2 × \(\frac{1}{2}\) = 1

Question 5.
\(\frac{4}{5}\) × 9 = _______
Answer:
9 × \(\frac{4}{5}\) = \(\frac{36}{5}\)

Explanation:
The given numbers are: 9 and \(\frac{4}{5}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
9 × \(\frac{4}{5}\) = \(\frac{9}{1}\) × \(\frac{4}{5}\)
= \(\frac{9 × 4}{5 × 1}\)
= \(\frac{36}{5}\)
Hence,
9 × \(\frac{4}{5}\) = \(\frac{36}{5}\)

Question 6.
4 × \(\frac{5}{12}\) = _______
Answer:
4 × \(\frac{5}{12}\) = \(\frac{5}{3}\)

Explanation:
The given numbers are: 4 and \(\frac{5}{12}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
4 × \(\frac{5}{12}\) = \(\frac{4}{1}\) × \(\frac{5}{12}\)
= \(\frac{4 × 5}{12 × 1}\)
= \(\frac{20}{12}\)
= \(\frac{5}{3}\)
Hence,
4 × \(\frac{5}{12}\) = \(\frac{5}{3}\)

Question 7.
\(\frac{1}{4}\) × 24 = _______
Answer:
24 × \(\frac{1}{24}\) = 6

Explanation:
The given numbers are: 24 and \(\frac{1}{4}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
24 × \(\frac{1}{4}\) = \(\frac{24}{1}\) × \(\frac{1}{4}\)
= \(\frac{24 × 1}{4 × 1}\)
= \(\frac{24}{4}\)
= 6
Hence,
24 × \(\frac{1}{4}\) = 6

Question 8.
16 × \(\frac{3}{8}\) = _______
Answer:
16 × \(\frac{3}{8}\) = 6

Explanation:
The given numbers are: 16 and \(\frac{3}{8}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
16 × \(\frac{3}{8}\) = \(\frac{16}{1}\) × \(\frac{3}{8}\)
= \(\frac{16 × 3}{8 × 1}\)
= \(\frac{48}{8}\)
= 6
Hence,
16 × \(\frac{3}{8}\) = 6

Question 9.
\(\frac{7}{10}\) × 25 = _______
Answer:
25 × \(\frac{7}{10}\) = \(\frac{35}{2}\)

Explanation:
The given numbers are: 25 and \(\frac{7}{10}\)
We know that,
\(\frac{a}{b}\) × p = \(\frac{a}{b}\) × \(\frac{p}{1}\)
= \(\frac{a × p}{ b × 1}\)
p = \(\frac{p}{1}\)
So,
25 × \(\frac{7}{10}\) = \(\frac{25}{1}\) × \(\frac{7}{10}\)
= \(\frac{7 × 25}{10 × 1}\)
= \(\frac{175}{10}\)
= \(\frac{35}{2}\)
Hence,
25 × \(\frac{7}{10}\) = \(\frac{35}{2}\)

Question 10.
You spend \(\frac{3}{4}\) hour jumping rope every week for 8 weeks. How many hours do you jump rope altogether?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 8
Answer:
The number of hours you jump rope altogether in 8 weeks is: 6 hours

Explanation:
It is given that you spent \(\frac{3}{4}\) hour jumping rope every week for 8 weeks.
So,
The number of hours you spent jumping rope in 8 weeks = ( The number of hours you jump rope in 1 week ) × ( The total number of weeks )
= \(\frac{3}{4}\) × 8
= \(\frac{3}{4}\) × \(\frac{8}{1}\)
= \(\frac{3 × 8}{4 × 1}\)
= 6 hours
hence, from the above,
We can conclude that you spent 6 hours on jumping rope in 8 weeks.

Question 11.
Logic
Your friend finds 25 items that are either insects or flowers. She says that \(\frac{1}{6}\) of the items are insects. Can this be true? Explain.
Answer:
No, this is not true

Explanation:
It is given that your friend finds 25 items that are either insects or flowers and she says that \(\frac{1}{6}\) of the items are insects.
But,
The number of either insects or flowers will always be a whole number.
But, from the given information,
25 will not be divided by \(\frac{1}{6}\)
So,
The statement “\(\frac{1}{6}\) of the items are insects” is false.

Question 12.
Open-Ended
Write two different pairs of fractions that could represent the insects and flowers your friend finds in Exercise 11.
Answer:
The different pairs of fractions that could represent the insects and flowers are:
\(\frac{1}{5}\) and \(\frac{1}{25}\)

Explanation:
These different pairs of fractions have to divide 25
So,
We have to take the fractions the multiples of 5 i.e., 5 and 25
Hence, from the above,
We can conclude that
The different pairs of fractions that could represent the insects and flowers are:
\(\frac{1}{5}\) and \(\frac{1}{25}\)

Question 13.
Modeling Real Life
Newton bakes 56 treats. Five-eighths of them contains peanut butter. Descartes bakes 120 treats. Five-sixths of them contain peanut butter. Who bakes more peanut butter treats? How many more?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.3 9
Answer:
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65

Explanation:
It is given that Newton bakes 56 treats and five-eights of them contains peanut butter
So,
The number of peanut butter treats made by Newton = \(\frac{5}{8}\) × 56
= \(\frac{5}{8}\) × \(\frac{56}{1}\)
= \(\frac{5 × 56}{8 × 1}\)
= 35 peanut butter treats
It is also given that Descartes bakes 120 treats and five-sixths of them contain peanut butter
So,
the number of peanut butter treats made by Descartes = \(\frac{5}{6}\) × 120
= \(\frac{5}{6}\) × \(\frac{120}{1}\)
= \(\frac{5 × 120}{6 × 1}\)
= 100 peanut butter treats
So,
In the comparison of peanut butter treats,
Descartes bakes more
Now,
The number of peanut butter treats baked by Descartes more than Newton = 100 – 35
= 65 peanut butter treats
Hence, from the above,
We can conclude that
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65

Question 14.
Modeling Real Life
Your class conducts an egg-dropping experiment with 60 eggs. Three-fifths of the eggs break open, one-sixth of the eggs crack, and the rest do not break at all. How many of the eggs do not crack or break open?
Answer:
The number of eggs that do not break is:

Explanation:
It is given that your class conducts an egg-dropping experiment with 60 eggs and three-fifths of the eggs break open, one-sixth of the eggs crack and the rest do not break.
So,
The number of eggs that break open = 60 × \(\frac{3}{5}\)
= \(\frac{60}{1}\) × \(\frac{3}{5}\)
= 36
The number of eggs that cracked = 60 × \(\frac{1}{6}\)
= \(\frac{60}{1}\) × \(\frac{1}{6}\)
= 10
So,
The number of eggs that do not breaked = ( The total number of eggs ) – ( The number of eggs that break open + The number of eggs that cracked )
= 60 – ( 36 + 10 )
= 16
Hence, from the above,
We can conclude that the number of eggs that do not break is: 16 eggs

Review & Refresh

Add.
Question 15.
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = _______
Answer:
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = \(\frac{99}{8}\)

Explanation:
The give mixed fractions are: 6\(\frac{3}{4}\) and 5\(\frac{5}{8}\)
The representation of mixed fractions in the improper form is: \(\frac{45}{8}\) and \(\frac{27}{4}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{27}{4}\) with \(\frac{2}{2}\)
So,
\(\frac{45}{8}\) + \(\frac{54}{8}\) = \(\frac{45 + 54}{8}\)
= \(\frac{99}{8}\)
Hence,
5\(\frac{5}{8}\) + 6\(\frac{3}{4}\) = \(\frac{99}{8}\)

Question 16.
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = ________
Answer:
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = \(\frac{119}{12}\)

Explanation:
The give mixed fractions are: 1\(\frac{5}{6}\) and 8\(\frac{1}{12}\)
The representation of mixed fractions in the improper form is: \(\frac{11}{6}\) and \(\frac{97}{12}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{11}{6}\) with \(\frac{2}{2}\)
So,
\(\frac{97}{12}\) + \(\frac{22}{12}\) = \(\frac{97 + 22}{12}\)
= \(\frac{119}{12}\)
Hence,
1\(\frac{5}{6}\) + 8\(\frac{1}{12}\) = \(\frac{119}{12}\)

Question 17.
3\(\frac{1}{2}\) + \(\frac{3}{5}\) + 2\(\frac{7}{10}\) = _______
Answer:
3\(\frac{1}{2}\) + 2\(\frac{7}{10}\) + \(\frac{3}{5}\)  = \(\frac{68}{10}\)

Explanation:
The give mixed fractions are: 3\(\frac{1}{2}\), \(\frac{3}{5}\) and 2\(\frac{7}{10}\)
The representation of mixed fractions in the improper form is: \(\frac{7}{2}\) and \(\frac{27}{10}\)
In addition,
We have to make the denominators equal.
So,
Multiply \(\frac{7}{2}\) with \(\frac{5}{5}\)
Multiply \(\frac{3}{5}\) with \(\frac{2}{2}\)
So,
\(\frac{27}{10}\) + \(\frac{35}{10}\) + \(\frac{6}{10}\)  = \(\frac{27 + 35 + 6}{10}\)
= \(\frac{68}{10}\)
Hence,
3\(\frac{1}{2}\) + 2\(\frac{7}{10}\) + \(\frac{3}{5}\)  = \(\frac{68}{10}\)

Lesson 9.4 Use Models to Multiply Fractions

Explore and Grow

Fold a sheet of paper in half. Shade \(\frac{1}{4}\) of either half. What fraction of the entire sheet of paper did you shade? Draw a model to support your answer.
Answer:
The fraction of the entire sheet of paper you shaded is: \(\frac{1}{8}\)

Explanation:
Take a full sheet of paper and fold in half
So,
The number of parts of full sheet of paper = \(\frac{1}{2}\)
Now,
Shade \(\frac{1}{4}\) of the half of the paper
So,,
The fraction of the paper you shaded = \(\frac{1}{4}\) × \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{4 × 2}\)
= \(\frac{1}{8}\)
Hence, from the above,
We can conclude that the \(\frac{1}{8}\) part of the full sheet is shaded

Reasoning
What multiplication expression does your model represent? Explain your reasoning.
Answer:
The multiplication expression your model represents is: \(\frac{1}{8}\)
We can obtain the multiplication expression by using the following multiplication properties. They are:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
B) a = \(\frac{a}{1}\)

Think and Grow: Use Models to Multiply Fractions

You can use models to multiply a fraction by a fraction.
Example
Find \(\frac{1}{2}\) × \(\frac{1}{3}\).

Show and Grow

Multiply. Use a model to help.
Question 1.
\(\frac{1}{3}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{4}\) = \(\frac{1}{12}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{4 × 3}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{3}\) = \(\frac{1}{12}\)

Question 2.
\(\frac{2}{3}\) × \(\frac{1}{2}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{1}{2}\) = \(\frac{1}{3}\)

Explanation:
The given fractions are: \(\frac{2}{3}\) and \(\frac{1}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{2}{3}\)
= \(\frac{1 × 2}{2 × 3}\)
= \(\frac{1}{3}\)
Hence,
\(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\)

Apply and Grow: Practice

Multiply. Use a model to help.
Question 3.
\(\frac{1}{2}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{6}\) = \(\frac{1}{12}\)

Question 4.
\(\frac{1}{5}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{40}\)

Explanation:
The given fractions are: \(\frac{1}{5}\) and \(\frac{1}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{5 × 8}\)
= \(\frac{1}{40}\)
Hence,
\(\frac{1}{5}\) × \(\frac{1}{8}\) = \(\frac{1}{40}\)

Question 5.
\(\frac{1}{4}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{24}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{4 × 6}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{6}\) = \(\frac{1}{24}\)

Question 6.
\(\frac{2}{3}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{2}{9}\)

Explanation:
The given fractions are: \(\frac{2}{3}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{3}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{3 × 3}\)
= \(\frac{2}{9}\)
Hence,
\(\frac{2}{3}\) × \(\frac{1}{3}\) = \(\frac{2}{9}\)

Write a multiplication equation represented by the model.
Question 7.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 2
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{2}\) × \(\frac{1}{8}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 2
From the above model,
The number of rows is: 2
The number of columns is: 8
So,
The value of 1 row = \(\frac{1}{2}\)
The value of 1 column = \(\frac{1}{8}\)
So,
Rows × Columns = \(\frac{1}{8}\) × \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{8}\) × \(\frac{1}{2}\)

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 3
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{4}\) × \(\frac{1}{5}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 3
From the above model,
The number of rows is: 4
The number of columns is: 5
So,
The value of 1 row = \(\frac{1}{4}\)
The value of 1 column = \(\frac{1}{5}\)
So,
Rows × Columns = \(\frac{1}{4}\) × \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{4}\) × \(\frac{1}{5}\)

Question 9.
One-fifth of the students in your school have tried skating. Of those students, \(\frac{1}{7}\) have tried ice skating. What fraction of students in your school have tried ice skating?
Answer:
The fraction of students that have tried ice skating is: \(\frac{1}{35}\)

Explanation:
It is given that one-fifth of the students in your school have tried skating and of these students, \(\frac{1}{7}\) have tried ice skating
So,
The fraction of students that have tried ice skating = ( The fraction of students that have tried skating ) × ( The fraction of students that tried ice skating out of the total students )
= \(\frac{1}{5}\) × \(\frac{1}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{7}\)
= \(\frac{1 × 1}{5 × 7}\)
= \(\frac{1}{35}\)
Hence, from the above,
We can conclude that \(\frac{1}{35}\) of the total students tried ice skating.

Question 10.
DIG DEEPER!
Are both Newton and Descartes correct? Explain.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 4
Answer: Yes, both Newton and Descartes are correct.

Explanation:
The given models of Newton and Descartes are:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 4
From the model of Newton,
The number of rows is: 3
The number of columns is: 5
From the model of Descartes,
The number of rows is: 5
The number of columns is: 3
From the above models,
We can observe that the Descartes model is obtained by reversing the rows and columns of Newton i.e., the rows of Newton’s model become the columns of Descartes’s model and the columns of Newton’s model becomes the rows of Descartes’ model.
So,
The multiplication equation represented by Newton is:
\(\frac{1}{3}\) × \(\frac{1}{5}\)
The multiplication equation represented by Descartes is:
\(\frac{1}{5}\) × \(\frac{1}{3}\)
Hence, from the above,
We can conclude that Newton and Descartes are correct

Think and Grow: Modeling Real Life

Example
A recipe calls for \(\frac{3}{4}\) teaspoon of cinnamon. You4want to halve the recipe. What fraction of a teaspoon of cinnamon do you need?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 5
Because you want to halve the recipe, multiply \(\frac{1}{2}\) by \(\frac{3}{4}\) to find how many teaspoons of cinnamon you need.

So,
You need \(\frac{3}{8}\) teaspoon of cinnamon.

Show and Grow

Question 11.
The mass of mango is \(\frac{2}{5}\) kilogram. The mass of guava is \(\frac{1}{4}\) as much as the mango. What is the mass of the guava?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 7
Answer:
The mass of guava is: \(\frac{1}{10}\) kilograms

Explanation:
It is given that the mass of mango is \(\frac{2}{5}\) kilogram and the mass of guava is \(\frac{1}{4}\) as much as the mango.
So,
The mass of guava = ( The mass of mango ) × ( The fraction of mass of guava in the mass of mango )
= \(\frac{2}{5}\) × \(\frac{1}{4}\)
= \(\frac{1 × 2}{4 × 5}\)
= \(\frac{2}{20}\)
= \(\frac{1}{10}\) kilograms
Hnce, from the above,
We can conclude that the mass of guava is: \(\frac{1}{10}\) kilogram

Question 12.
A giant panda spends \(\frac{2}{3}\) of 1-day eating and foraging. It spends \(\frac{3}{4}\) of that time for eating bamboo. What fraction of 1 day does the panda spend eating bamboo?
Answer:
The fraction of 1 day the panda spent on eating bamboo is: \(\frac{1}{2}\)

Explanation:
It is given that a giant panda spends \(\frac{2}{3}\) of 1-day eating and foraging and it spends \(\frac{3}{4}\) of that time for eating bamboo.
So,
The fraction of 1 day the panda spent on eating bamboo = ( The fraction of time the panda spent on eating ) × ( The fraction of time the panda spent time on eating bamboo out of the total time )
= \(\frac{2}{3}\) × \(\frac{3}{4}\)
= \(\frac{3 × 2}{4 × 3}\)
= \(\frac{6}{12}\)
= \(\frac{1}{2}\) hour
Hence, from the above,
We can conclude that the time spent by the panda on eating bamboo is: \(\frac{1}{2}\) hour

Question 13.
DIG DEEPER!
You have a half-gallon carton of milk that you use only for cereal. You use the same amount each day for 5 days. There is \(\frac{3}{8}\) of the carton left. How many cups of milk do you use each day? Explain.
Answer:

Use Models to Multiply Fractions Homework & Practice 9.4

Multiply. Use a model to help.
Question 1.
\(\frac{1}{3}\) × \(\frac{1}{7}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{7}\) = \(\frac{1}{21}\)

Explanation:
The given fractions are: \(\frac{1}{7}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{7}\) × \(\frac{1}{3}\) = \(\frac{1}{21}\)

Question 2.
\(\frac{1}{2}\) × \(\frac{1}{9}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{9}\) = \(\frac{1}{18}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{1}{9}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{2 × 9}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{9}\) = \(\frac{1}{18}\)

Question 3.
\(\frac{2}{5}\) × \(\frac{1}{6}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{1}{6}\) = \(\frac{1}{15}\)

Explanation:
The given fractions are: \(\frac{2}{5}\) and \(\frac{1}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{5}\) × \(\frac{1}{6}\)
= \(\frac{2 × 1}{5 × 6}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{1}{6}\) = \(\frac{1}{15}\)

Question 4.
\(\frac{3}{4}\) × \(\frac{2}{7}\) = _______
Answer:
\(\frac{3}{4}\) × \(\frac{2}{7}\) = \(\frac{3}{14}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{2}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\) = \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) × \(\frac{2}{7}\)
= \(\frac{3 × 2}{4 × 7}\)
= \(\frac{3}{14}\)
Hence,
\(\frac{3}{4}\) × \(\frac{2}{7}\) = \(\frac{3}{14}\)

Write a multiplication equation represented by the model.
Question 5.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 8
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{2}\) × \(\frac{1}{5}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 8
From the above model,
The number of rows is: 2
The number of columns is: 5
So,
The value of 1 row = \(\frac{1}{2}\)
The value of 1 column = \(\frac{1}{5}\)
So,
Rows × Columns = \(\frac{1}{2}\) × \(\frac{1}{5}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{2}\) × \(\frac{1}{5}\)

Question 6.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 9
Answer:
The multiplication equation represented by the model is:
\(\frac{1}{3}\) × \(\frac{1}{6}\)

Explanation:
The given model is:
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 9
From the above model,
The number of rows is: 3
The number of columns is: 6
So,
The value of 1 row = \(\frac{1}{3}\)
The value of 1 column = \(\frac{1}{6}\)
So,
Rows × Columns = \(\frac{1}{3}\) × \(\frac{1}{6}\)
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
\(\frac{1}{3}\) × \(\frac{1}{6}\)

Question 7.
One-sixth of the animals at a zoo are birds. Of the birds, \(\frac{1}{3}\) are female. What fraction of the animals at the zoo are female birds?
Answer:
The fraction of the animals at the zoo that is female birds is: \(\frac{1}{18}\)

Explanation:
It is given that one-sixth of the animals at the zoo are birds and of the birds, \(\frac{1}{3}\) is female.
So,
The fraction of the animals at the zoo are female birds = ( The fraction of animals in the zoo that is birds ) × ( The fraction of the animals in the zoo that is female birds )
= \(\frac{1}{3}\) × \(\frac{1}{6}\)
\(\frac{1 × 1}{3 × 6}\)
= \(\frac{1}{18}\)
Hence, from the above,
We can conclude that the number of animals that are female birds at the zoo is: \(\frac{1}{18}\)

Question 8.
Writing
Write and solve a real-life problem for the expression.
\(\frac{2}{3}\) × \(\frac{3}{7}\)
Answer:
Suppose in a school, there are boys ad girls.
In school, \(\frac{2}{3}\) of the boys and \(\frac{3}{7}\) of the girls passed the PET test
So,
The fraction of the students that passed the PET test out of the total number of students = ( The fraction of the boys that passed the PET test ) × ( The fraction of the girls that passed the PET test )
= \(\frac{2}{3}\) × \(\frac{3}{7}\)
= \(\frac{3 × 2}{7 × 3}\)
= \(\frac{6}{21}\)
= \(\frac{2}{7}\)
Hence, from the above,
We can conclude that the number of students who passed the PET test is: \(\frac{2}{7}\)

Question 9.
Modeling Real Life
A Gouldian finch is \(\frac{1}{2}\) the length of the sun conure. How long is the Gouldian finch?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.4 10
Answer:
The length of Gouldian finch is: \(\frac{11}{24}\) feet

Explanation:
It is given that a Gouldian finch is \(\frac{1}{2}\) the length of the sun conure.
From the given figure,
The length of the sun conure = \(\frac{11}{12}\) feet
So,
The length of Gouldian finch = ( The length of Sun Conure ) ÷ 2
= \(\frac{11}{12}\) ÷ 2
= \(\frac{11}{12}\) ÷ \(\frac{2}{1}\)
= \(\frac{11}{12}\) × \(\frac{1}{2}\)
= \(\frac{11 × 1}{12 × 2}\)
= \(\frac{11}{24}\) feet
Hence, from the above,
We can conclude that the length of the Gouldian finch is: \(\frac{11}{24}\) feet

Question 10.
DIG DEEPER!
A recipe calls for \(\frac{2}{3}\) cup of chopped walnuts. You chop 4 walnuts and get \(\frac{1}{4}\) of the amount you need. How much more of a cup of chopped walnuts do you need? All of your walnuts are the same size. How many more walnuts should you chop? Explain.
Answer:
The cup of chopped walnuts more you need is: \(\frac{1}{3}\)
The number of walnuts you should chop is: 6

Explanation:
It is given that a recipe calls for \(\frac{2}{3}\) cup of chopped walnuts.
It is also given that you chop 4 walnuts and get \(\frac{1}{4}\) of the amount you need.
So,
The cup more of chopped walnuts you need = 1- ( The cup of chopped walnuts you need for recipe )
= 1 – \(\frac{2}{3}\)
= \(\frac{3}{3}\) – \(\frac{2}{3}\)
= \(\frac{3 – 2}{3}\)
= \(\frac{1}{3}\) more cup
Now,
The number of walnuts you need more = ( The number of walnuts you needed to get \(\frac{1}{4}\) of cup of chopped walnuts ) ÷ ( The amount of chopped walnuts you need for recipe )
= 4 ÷ \(\frac{2}{3}\)
= \(\frac{4}{1}\) ÷ \(\frac{2}{3}\)
= \(\frac{4}{1}\) × \(\frac{3}{2}\)
= \(\frac{4 × 3}{1 × 2}\)
= \(\frac{12}{2}\)
= 6
Hence, from the above,
We can conclude that
The cup of chopped walnuts more you need is: \(\frac{1}{3}\)
The number of walnuts you should chop is: 6

Review & Refresh

Subtract.
Question 11.
5 – 1\(\frac{3}{4}\) = ______
Answer:
5 – 1\(\frac{3}{4}\) = \(\frac{13}{4}\)

Explanation:
The given numbers are: 5 and 1\(\frac{3}{4}\)
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
So,
5 – \(\frac{7}{4}\) = \(\frac{20}{4}\) – \(\frac{7}{4}\)
= \(\frac{20 – 7}{4}\)
= \(\frac{13}{4}\)
Hence,
5 – 1\(\frac{3}{4}\) = \(\frac{13}{4}\)

Question 12.
13\(\frac{1}{4}\) – 7\(\frac{5}{8}\) = ________
Answer:
13\(\frac{1}{4}\) – 7\(\frac{5}{8}\) = \(\frac{45}{8}\)

Explanation:
The given mixed fractions are: 13\(\frac{1}{4}\) and 7\(\frac{5}{8}\)
The representation of 13\(\frac{1}{4}\) in the improper fraction form is: \(\frac{53}{4}\)
The representation of 13\(\frac{1}{4}\) in the improper fraction form is: \(\frac{61}{8}\)
To subtract the fractions, we have to make the denominators equal.
So,
Multiply \(\frac{53}{4}\) by \(\frac{2}{2}\)
So,
\(\frac{53}{4}\) = \(\frac{106}{8}\)
So,
\(\frac{106}{8}\) – \(\frac{61}{8}\)
= \(\frac{106 – 61}{8}\)
= \(\frac{45}{8}\)
Hence,
13\(\frac{1}{4}\) – 13\(\frac{1}{4}\) = \(\frac{45}{8}\)

Question 13.
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = ________
Answer:
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = \(\frac{62}{10}\)

Explanation:
The given mixed fractions are: 12\(\frac{7}{10}\), 5\(\frac{3}{10}\) and 1\(\frac{1}{5}\)
The representation of 12\(\frac{7}{10}\) in the improper fraction form is: \(\frac{127}{10}\)
The representation of 5\(\frac{3}{10}\) in the improper fraction form is: \(\frac{53}{10}\)
The representation of 1\(\frac{1}{5}\) in the improper fraction form is: \(\frac{6}{5}\)
To subtract the fractions, we have to make the denominators equal.
So,
Multiply \(\frac{6}{5}\) by \(\frac{2}{2}\)
So,
\(\frac{6}{5}\) = \(\frac{12}{10}\)
So,
\(\frac{127}{10}\) – \(\frac{53}{10}\) – \(\frac{12}{10}\)
= \(\frac{127 – 53 – 12}{8}\)
= \(\frac{62}{10}\)
Hence,
12\(\frac{7}{10}\) – 5\(\frac{3}{10}\) – 1\(\frac{1}{5}\) = \(\frac{62}{10}\)

Lesson 9.5 Multiply Fractions

Explore and Grow

Use models to help you complete the table. What do you notice about each expression and its product?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 1
Answer:
The completed table is:

From the given table,
We can observe the product of fractions is also a product.

Construct Arguments
Explain how to multiply two fractions without using a model.
Answer:
We can multiply the fractions by using the properties of multiplication. They are:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
B) a = \(\frac{a}{1}\)

Think and Grow: Multiply Fractions

Key Idea
You can find the product of a fraction and a fraction by multiplying the numerators and multiplying the denominators.
Example
Find \(\frac{1}{2}\) × \(\frac{3}{2}\).
Multiply the numerators and multiply the denominators.

Example
Find \(\frac{5}{6}\) × \(\frac{3}{5}\).
Multiply the numerators and multiply the denominators.

Show and Grow

Multiply.
Question 1.
\(\frac{1}{2}\) × \(\frac{4}{3}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{4}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{4}{3}\)
= \(\frac{1 × 4}{2 × 3}\)
= \(\frac{4}{6}\)
= \(\frac{2}{3}\)
Hence,
\(\frac{1}{2}\) × \(\frac{4}{3}\) = \(\frac{2}{3}\)

Question 2.
\(\frac{2}{5}\) × \(\frac{2}{3}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

Explanation:
The given fractions are: \(\frac{2}{5}\) and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{2}{5}\) × \(\frac{2}{3}\)
= \(\frac{2 × 2}{5 × 3}\)
= \(\frac{4}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{2}{3}\) = \(\frac{4}{15}\)

Question 3.
\(\frac{3}{4}\) × \(\frac{5}{8}\) = _______
Answer:
\(\frac{3}{4}\) × \(\frac{5}{8}\) = \(\frac{15}{32}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{5}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) × \(\frac{5}{8}\)
= \(\frac{3 × 5}{4 × 8}\)
= \(\frac{15}{32}\)
Hence,
\(\frac{3}{4}\) × \(\frac{5}{8}\) = \(\frac{15}{32}\)

Apply and Grow: Practice

Multiply.
Question 4.
\(\frac{1}{4}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{4}\)
= \(\frac{1 × 1}{4 × 4}\)
= \(\frac{1}{16}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)

Question 5.
\(\frac{5}{6}\) × \(\frac{7}{10}\) = _______
Answer:
\(\frac{5}{6}\) × \(\frac{7}{10}\) = \(\frac{7}{12}\)

Explanation:
The given fractions are: \(\frac{5}{6}\) and \(\frac{7}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{6}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{6 × 10}\)
= \(\frac{35}{60}\)
= \(\frac{7}{12}\)
Hence,
\(\frac{5}{6}\) × \(\frac{7}{10}\) = \(\frac{7}{12}\)

Question 6.
\(\frac{6}{9}\) × \(\frac{8}{2}\) = _______
Answer:
\(\frac{6}{9}\) × \(\frac{8}{2}\) = \(\frac{8}{3}\)

Explanation:
The given fractions are: \(\frac{6}{9}\) and \(\frac{8}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{6}{9}\) × \(\frac{8}{2}\)
= \(\frac{6 × 8}{2 × 9}\)
= \(\frac{48}{18}\)
= \(\frac{8}{3}\)
Hence,
\(\frac{6}{9}\) × \(\frac{8}{2}\) = \(\frac{8}{3}\)

Question 7.
\(\frac{21}{100}\) × \(\frac{3}{5}\) = _______
Answer:
\(\frac{21}{100}\) × \(\frac{3}{5}\) = \(\frac{63}{500}\)

Explanation:
The given fractions are: \(\frac{21}{100}\) and \(\frac{3}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{21}{100}\) × \(\frac{3}{5}\)
= \(\frac{21 × 3}{100 × 5}\)
= \(\frac{63}{500}\)
Hence,
\(\frac{21}{100}\) × \(\frac{3}{5}\) = \(\frac{63}{500}\)

Question 8.
\(\frac{1}{12}\) × \(\frac{9}{4}\) = _______
Answer:
\(\frac{1}{12}\) × \(\frac{9}{4}\) = \(\frac{3}{16}\)

Explanation:
The given fractions are: \(\frac{1}{12}\) and \(\frac{9}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{12}\) × \(\frac{9}{4}\)
= \(\frac{1 × 9}{12 × 4}\)
= \(\frac{9}{48}\)
= \(\frac{3}{16}\)
Hence,
\(\frac{1}{12}\) × \(\frac{9}{4}\) = \(\frac{3}{16}\)

Question 9.
\(\frac{4}{7}\) × \(\frac{8}{8}\) = _______
Answer:
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Explanation:
The given fractions are: \(\frac{4}{7}\) and \(\frac{8}{8}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{4}{7}\) × \(\frac{8}{8}\)
= \(\frac{8 × 4}{7 × 8}\)
= \(\frac{4}{7}\)
Hence,
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Evaluate
Question 10.
\(\left(\frac{1}{2} \times \frac{7}{8}\right)\) × 2 = _______
Answer:
\(\left(\frac{1}{2} \times \frac{7}{8}\right)\) × 2 = \(\frac{7}{8}\)

Explanation:
The given fractions are: \(\frac{7}{8}\) and \(\frac{1}{2}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{1}{2}\) × \(\frac{7}{8}\)  ) × 2
= ( \(\frac{1 × 7}{2 × 8}\) ) × \(\frac{2}{1}\)
= \(\frac{7}{16}\) × \(\frac{2}{1}\)
= \(\frac{7 × 2}{16 × 1}\)
= \(\frac{7}{8}\)
Hence,
\(\frac{4}{7}\) × \(\frac{8}{8}\) = \(\frac{4}{7}\)

Question 11.
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = _______
Answer:
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = \(\frac{2}{9}\)

Explanation:
The given fractions are: \(\frac{7}{6}\) , \(\frac{5}{6}\), and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{7}{6}\) – \(\frac{5}{6}\)  ) × \(\frac{2}{3}\)
= ( \(\frac{7 – 5}{6}\) ) × \(\frac{2}{3}\)
= \(\frac{2}{6}\) × \(\frac{2}{3}\)
= \(\frac{2 × 2}{6 × 3}\)
= \(\frac{4}{18}\)
= \(\frac{2}{9}\)
Hence,
\(\left(\frac{7}{6}-\frac{5}{6}\right)\) × \(\frac{2}{3}\) = \(\frac{2}{9}\)

Question 12.
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = _________
Answer:
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = \(\frac{7}{10}\)

Explanation:
The given fractions are: \(\frac{9}{10}\), \(\frac{4}{9}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) × ( \(\frac{4}{9}\) + \(\frac{1}{3}\)  )
= ( \(\frac{4 + 3}{9}\) ) × \(\frac{9}{10}\)
= \(\frac{7}{9}\) × \(\frac{9}{10}\)
= \(\frac{7 × 9}{9 × 10}\)
= \(\frac{7}{10}\)
Hence,
\(\frac{9}{10}\) × \(\left(\frac{4}{9}+\frac{1}{3}\right)\) = \(\frac{7}{10}\)

Question 13.
At a school, \(\frac{3}{4}\) of the students play a sport. Of the students that play a sport, \(\frac{1}{5}\) play baseball. What fraction of the students at the school play baseball?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 4
Answer:
The fraction of the students at the school that play baseball is: \(\frac{3}{20}\)

Explanation:
It is given that at a school, \(\frac{3}{4}\) of the students play a sport and of the students that play a sport, \(\frac{1}{5}\) play baseball.
So,
The fraction of the students at the school that play baseall = ( The fraction of the students that play a sport ) × ( The fraction of the students that play baseball out of thetotal number of students )
= \(\frac{3}{4}\) × \(\frac{1}{5}\)
= \(\frac{3 × 1}{4 × 5}\)
= \(\frac{3}{20}\)
Hence, from the above,
We can conclude that the fraction of the students that play baseball is: \(\frac{3}{20}\)

Question 14.
Reasoning
Descartes says he can find the product of a whole number and a fraction the same way he finds the product of two fractions. Explain why his reasoning makes sense.
Answer:
The product of a whole number and a fraction follows the same procedure as that of the product of the two fractions because of the following properties of multiplication:
A) \(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
B) a= \(\frac{a}{1}\)
Hence, from the above two properties of multiplication,
We can conclude that the reasoning of Descartes makes sense

Question 15.
Writing
Explain how multiplying fractions is different than adding and subtracting fractions.
Answer:
In the multiplication of the fractions, we multiply numerators and denominators.
Example:
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
In addition of the fractions, we add only numerators making the denominators equal
Example:
\(\frac{a}{b}\) + \(\frac{p}{b}\)
= \(\frac{a + p}{b}\)
In subtraction of the fractions, we subtract only numerators making the denominators equal
Example:
\(\frac{a}{b}\) – \(\frac{p}{b}\)
= \(\frac{a – p}{b}\)

Think and Grow: Modeling Real Life

Example
A tourist is walking from the Empire State Building to Times Square. She is \(\frac{2}{3}\) of the way there. What fraction of a mile does she have left to walk?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 5
Find the distance she has walked. Because she has walked \(\frac{2}{3}\) of \(\frac{3}{4}\) mile, multiply \(\frac{2}{3}\) by \(\frac{3}{4}\).

Show and Grow

Question 16.
At a zoo, \(\frac{3}{5}\) of the animals are mammals. Of the mammals, \(\frac{5}{12}\) are primates.What fraction of the animals at the zoo are not primates?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 7
Answer:
The fractions of the animals at the zoo that are not primates is: \(\frac{3}{4}\)

Explanation:
It is given that at the zoo, \(\frac{3}{5}\) of the animals are mammals and of the animals, \(\frac{5}{12}\) are primates
So,
The fraction of the animals that are primates =( The fraction of the animals that are primates ) × ( The fraction of the animals that are primates )
= \(\frac{3}{5}\) × \(\frac{5}{12}\)
= \(\frac{3 × 5}{5 × 12}\)
= \(\frac{15}{60}\)
= \(\frac{1}{4}\)
Now,
Let the total number of animals that are mammals =1
So,
The fraction of the animals that are not primates = ( The total number of mammals ) – ( The fraction of the animals that are primates )
= 1 – \(\frac{1}{4}\)
= \(\frac{4}{4}\) – \(\frac{1}{4}\)
= \(\frac{4 – 1}{4}\)
= \(\frac{3}{4}\)
Hence, from the above,
We can conclude that the fraction of the animals that are not primates is: \(\frac{3}{4}\)

Question 17.
You have an album of 216 trading cards. One page contains \(\frac{1}{24}\) of the cards. On that page, \(\frac{2}{3}\) of the cards are epic. You only have one page with any epic cards. How many epic cards do you have?
Answer:
The number of epic cards is: 6 epic cards

Explanation:
It is given that you have an album of 216 trading cards
It is also given that one page contains \(\frac{1}{24}\) of the cards. On that page, \(\frac{2}{3}\) of the cards are epic.
So,
The fraction of the epic cards = ( The total number of cards on that page ) × ( The fraction of the epic cards that is on that page )
= \(\frac{1}{24}\) × \(\frac{2}{3}\)
= \(\frac{2 × 1}{3 × 24}\)
= \(\frac{1}{36}\)
So,
The number of epic cards = ( The total number of cards ) × ( The fraction of the epic cards )
= 216 × \(\frac{1}{36}\)
= \(\frac{216}{1}\) × \(\frac{}{36}\)
= \(\frac{216}{36}\)
= 6 epic cards
Hence, from the above,
We can conclude that there are 6 epic cards

Question 18.
DIG DEEPER!
In a class, \(\frac{2}{5}\) of the students play basketball and \(\frac{7}{10}\) play soccer.Of the students who play basketball, \(\frac{2}{3}\) also play soccer. There are 30 students in the class. How many students play soccer but do not play basketball?
Answer:
The number of students that play soccer but do not play basketball is: 4 students

Explanation:
It is given that in a class, \(\frac{2}{5}\) of the students play basketball and \(\frac{7}{10}\) play soccer.Of the students who play basketball, \(\frac{2}{3}\) also play soccer.
So,
The number of students who play basketball and soccer = ( The number of students who play basketball ) × ( The number of students playing soccer who are also playing basketball )
= \(\frac{2}{5}\) × \(\frac{2}{3}\)
= \(\frac{4}{15}\)
Now,
The number of students who only play soccer but not basketball = ( The number of students who play soccer only ) – ( The number of students who play both basketball and soccer )
= \(\frac{2}{5}\) – \(\frac{4}{15}\)
In subtraction, the denominators must be equal.
So,
Multiply \(\frac{2}{5}\) by \(\frac{3}{3}\)
So,
\(\frac{6}{15}\) – \(\frac{4}{15}\)
= \(\frac{4}{15}\)
It is also given that the total number of students are: 30
So,
The number of students who play soccer only = ( The fraction of the students who play soccer only ) × ( The total number of students )
= \(\frac{2}{15}\) × 30
= \(\frac{2}{15}\) × \(\frac{30}{1}\)
= 4 students
Hence, from the above,
We can conclude that there are 4 students who play soccer only.

Multiply Fractions Homework & Practice 9.5

Multiply.
\(\frac{1}{4}\) × \(\frac{1}{5}\) = _______
Answer:
\(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{1}{20}\)

Explanation:
The given fractions are: \(\frac{1}{4}\) and \(\frac{1}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) × \(\frac{1}{5}\)
= \(\frac{1 × 1}{4 × 5}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{4}\) × \(\frac{1}{5}\) = \(\frac{1}{20}\)

Question 2.
\(\frac{2}{7}\) × \(\frac{1}{2}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{2}{7}\) = \(\frac{1}{7}\)

Explanation:
The given fractions are: \(\frac{1}{2}\) and \(\frac{2}{7}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1 × 2}{2 × 7}\)
= \(\frac{2}{14}\)
= \(\frac{1}{7}\)
Hence,
\(\frac{1}{2}\) × \(\frac{2}{7}\) = \(\frac{1}{7}\)

Question 3.
\(\frac{9}{10}\) × \(\frac{2}{3}\) = _______
Answer:
\(\frac{9}{10}\) × \(\frac{2}{3}\) = \(\frac{3}{5}\)

Explanation:
The given fractions are: \(\frac{9}{10}\) and \(\frac{2}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) × \(\frac{2}{3}\)
= \(\frac{9 × 2}{10 × 3}\)
= \(\frac{18}{30}\)
= \(\frac{3}{5}\)
Hence,
\(\frac{9}{10}\) × \(\frac{2}{3}\) = \(\frac{3}{5}\)

Question 4.
\(\frac{5}{8}\) × \(\frac{5}{6}\) = _______
Answer:
\(\frac{5}{8}\) × \(\frac{5}{6}\) = \(\frac{25}{48}\)

Explanation:
The given fractions are: \(\frac{5}{8}\) and \(\frac{5}{6}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{8}\) × \(\frac{5}{6}\)
= \(\frac{5 × 5}{8 × 6}\)
= \(\frac{25}{48}\)
Hence,
\(\frac{5}{8}\) × \(\frac{5}{6}\) = \(\frac{25}{48}\)

Question 5.
\(\frac{9}{7}\) × \(\frac{3}{4}\) = _______
Answer:
\(\frac{9}{7}\) × \(\frac{3}{4}\) = \(\frac{27}{28}\)

Explanation:
The given fractions are: \(\frac{9}{7}\) and \(\frac{3}{4}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{9}{7}\) × \(\frac{3}{4}\)
= \(\frac{9 × 3}{7 × 4}\)
= \(\frac{27}{28}\)
Hence,
\(\frac{9}{7}\) × \(\frac{3}{4}\) = \(\frac{27}{28}\)

Question 6.
\(\frac{11}{100}\) × \(\frac{2}{5}\) = _______
Answer:
\(\frac{11}{100}\) × \(\frac{2}{5}\) = \(\frac{11}{250}\)

Explanation:
The given fractions are: \(\frac{11}{100}\) and \(\frac{2}{5}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{11}{100}\) × \(\frac{2}{5}\)
= \(\frac{11 × 2}{100 × 5}\)
= \(\frac{22}{500}\)
= \(\frac{11}{250}\)
Hence,
\(\frac{11}{100}\) × \(\frac{2}{5}\) = \(\frac{11}{250}\)

Question 7.
\(\frac{7}{20}\) × \(\frac{6}{2}\) = _______
Answer:
\(\frac{7}{20}\) × \(\frac{6}{2}\) = \(\frac{21}{20}\)

Explanation:
The given fractions are: \(\frac{7}{20}\) and \(\frac{6}{2}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{7}{20}\) × \(\frac{6}{2}\)
= \(\frac{7 × 6}{20 × 2}\)
= \(\frac{42}{40}\)
= \(\frac{21}{20}\)
Hence,
\(\frac{7}{20}\) × \(\frac{6}{2}\) = \(\frac{21}{20}\)

Question 8.
\(\frac{15}{16}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{15}{16}\) × \(\frac{1}{3}\) = \(\frac{5}{16}\)

Explanation:
The given fractions are: \(\frac{15}{16}\) and \(\frac{1}{3}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{15}{16}\) × \(\frac{1}{3}\)
= \(\frac{15 × 1}{16 × 3}\)
= \(\frac{15}{48}\)
= \(\frac{5}{16}\)
Hence,
\(\frac{15}{16}\) × \(\frac{1}{3}\) = \(\frac{5}{16}\)

Question 9.
\(\frac{5}{12}\) × \(\frac{3}{10}\) = _______
Answer:
\(\frac{5}{12}\) × \(\frac{3}{10}\) = \(\frac{1}{8}\)

Explanation:
The given fractions are: \(\frac{5}{12}\) and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
\(\frac{5}{12}\) × \(\frac{3}{10}\)
= \(\frac{5 × 3}{12 × 10}\)
= \(\frac{15}{120}\)
= \(\frac{1}{8}\)
Hence,
\(\frac{5}{12}\) × \(\frac{3}{10}\) = \(\frac{1}{8}\)

Evaluate.
Question 10.
3 × \(\frac{3}{10}\) = _______
Answer:
3 ×  \(\frac{3}{10}\) = \(\frac{9}{10}\)

Explanation:
The given numbers are: 3 and \(\frac{3}{10}\)
We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
3 × \(\frac{3}{10}\)
= \(\frac{3}{1}\) × \(\frac{3}{10}\)
= \(\frac{3 × 3}{10 × 1}\)
= \(\frac{9}{10}\)
Hence,
3 × \(\frac{3}{10}\) = \(\frac{9}{10}\)

Question 11.
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = _______
Answer:
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

Explanation:
The given fractions are: \(\frac{1}{3}\), \(\frac{1}{3}\) and \(\frac{4}{5}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{1}{3}\) + \(\frac{1}{3}\)  ) × \(\frac{4}{5}\)
= ( \(\frac{1 + 1}{3}\) ) × \(\frac{4}{5}\)
= \(\frac{2}{3}\) × \(\frac{4}{5}\)
= \(\frac{4 × 2}{3 × 5}\)
= \(\frac{8}{15}\)
Hence,
\(\left(\frac{1}{3}+\frac{1}{3}\right)\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

Question 12.
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = _______
Answer:
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = \(\frac{2}{7}\)

Explanation:
The given fractions are: \(\frac{6}{7}\), \(\frac{3}{4}\) and \(\frac{5}{12}\)

We know that,
\(\frac{a}{b}\) × \(\frac{p}{q}\)
= \(\frac{a × p}{b × q}\)
a = \(\frac{a}{1}\)
So,
( \(\frac{3}{4}\) – \(\frac{5}{12}\)  ) × \(\frac{6}{7}\)
= ( \(\frac{9}{12}\) – \(\frac{5}{12}\)  ) × \(\frac{6}{7}\)
= \(\frac{4}{12}\) × \(\frac{6}{7}\)
= \(\frac{4 × 6}{12 × 7}\)
= \(\frac{2}{7}\)
Hence,
\(\frac{6}{7}\) × \(\left(\frac{3}{4}-\frac{5}{12}\right)\) = \(\frac{2}{7}\)

Question 13.
A pancake recipe calls for \(\frac{1}{3}\) cup of water. You want to halve the recipe. What fraction of a cup of water do you need?
Answer:
The fraction of a cup of water you need is: \(\frac{1}{6}\)

Explanation:
It is given that a pancake recipe calls for \(\frac{1}{3}\) cup of water and you want to halve the recipe.
So,
The fraction of a cup of water you need for half of the recipe = ( The cup of water you need for a recipe ) × \(\frac{1}{2}\)
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{3 × 2}\)
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that the fraction of water you need for half of the recipe is: \(\frac{1}{6}\)

Question 14.
Number Sense
Which is greater, \(\frac{3}{4}\) × \(\frac{1}{5}\) or \(\frac{3}{4}\) × \(\frac{1}{8}\)? Explain.
Answer:
\(\frac{3}{4}\) × \(\frac{1}{5}\)” is greater than “\(\frac{3}{4}\) × \(\frac{1}{8}\)

Explanation:
The given fractions are: \(\frac{3}{4}\), \(\frac{1}{5}\) and \(\frac{1}{8}\)
Now,
\(\frac{3}{4}\) × \(\frac{1}{5}\)
= \(\frac{3 × 1}{4 × 5}\)
= \(\frac{3}{20}\)
Now,
\(\frac{3}{4}\) × \(\frac{1}{8}\)
= \(\frac{3 × 1}{4 × 8}\)
= \(\frac{3}{32}\)
So,
For the comparison of the products, equate the numerators.
Multiply \(\frac{3}{20}\) by \(\frac{32}{32}\)
Multiply \(\frac{3}{32}\) by \(\frac{20}{20}\)
So,
\(\frac{3}{20}\) = \(\frac{96}{640}\)
\(\frac{3}{32}\) = \(\frac{60}{640}\)
So,
By comparison of the products,
We can observe that
96 > 60
Hence, from the above,
We can conclude that “\(\frac{3}{4}\) × \(\frac{1}{5}\)” is greater than “\(\frac{3}{4}\) × \(\frac{1}{8}\)”

Question 15.
Reasoning
Is \(\frac{17}{24}\) × \(\frac{7}{8}\) equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)? Explain.
Answer:
\(\frac{17}{24}\) × \(\frac{7}{8}\) is equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)

Explanation:
By the commutative property of multiplication,
a × b = b × a
So,
By the above property of multiplication,
We can interchange the numbers
Hence, from the above,
We can conclude that \(\frac{17}{24}\) × \(\frac{7}{8}\) is equal to \(\frac{17}{8}\) × \(\frac{7}{24}\)

Question 16.
Number Sense
In which equations does k = \(\frac{5}{6}\)?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 8
Answer:
Let the multiplication equations be represented A), B), C) and D)
The given expressions are:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 9.5 8
So,
The product of the given expressions are:
A) \(\frac{1}{2}\) × \(\frac{5}{3}\) = \(\frac{5}{6}\)
B) \(\frac{1}{10}\) × k = \(\frac{1}{12}\)
C) \(\frac{1}{3}\) × k = \(\frac{2}{9}\)
D) \(\frac{5}{4}\) × \(\frac{2}{3}\) = k
Hence, from the above expressions,
We can conclude that equation A) fits the value of k perfectly.

Question 17.
Modeling Real Life
At a town hall meeting, \(\frac{37}{50}\) of the members are present. Of those who are present, \(\frac{1}{2}\) vote in favor of a new park. What fraction of the members do not vote in favor of the new park?
Answer:
The fraction of the members that do not vote in favor of the new park is: \(\frac{13}{50}\)

Explanation:
It is given that at a town hall meeting, \(\frac{37}{50}\) of the members are present. Of those who are present, \(\frac{1}{2}\) vote in favor of a new park.
So,
The fraction of the members that do not vote in favor of the new park = ( The total number of members at the town hall meeting ) – ( The fraction of the members that vote in the favor of a new park )
= 1 – \(\frac{37}{50}\)
= \(\frac{50}{50}\) – \(\frac{37}{50}\)
= \(\frac{50 – 37}{50}\)
= \(\frac{13}{50}\)
Hence, from the above,
We can conclude that the number of members that do not vote in the favor of a new park is: \(\frac{13}{50}\)

Question 18.
Modeling Real Life
There are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico.Of those states, \(\frac{2}{7}\) share a border with Mexico. How many states share a border with Canada?
Answer:
The number of states that share a border with Canada is: 4 states

Explanation:
It is given that there are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico and of those states, \(\frac{2}{7}\) share a border with Mexico.
So,
The number of states that share a border with Canada or Mexico = \(\frac{7}{25}\) × 50
= \(\frac{7}{25}\) × \(\frac{50}{1}\)
= 14 states
So,
The number of states that share a border with Canada = \(\frac{2}{7}\) × 14
= \(\frac{2}{7}\) × \(\frac{14}{1}\)
= 4 states
Hence, from the above,
We can conclude that there are 4 states that share a border with Canada

Review & Refresh

Evaluate. Check whether your answer is reasonable.
Question 19.
15.67 + 4 + 6.5 = _____
Answer:
15.67 + 4 + 6.5 = 26.17

Explanation:
The given numbers are: 15.67, 4, and 6.5
The representation of the numbers in the fraction form is: \(\frac{1567}{100}\), \(\frac{65}{10}\) and 4
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 4 with 100 in the denominator is: \(\frac{400}{100}\)
The representation of 6.5 with 100 in the denominator is: \(\frac{650}{100}\)
So,
\(\frac{1567}{100}\) + \(\frac{400}{100}\) + \(\frac{650}{100}\)
= \(\frac{  1567 + 400 + 650 }{100}\)
= \(\frac{2,617}{100}\)
= 26.17
Hence, 15.67 + 4 + 6.5 = 26.17

Question 20.
20.7 – 9.54 + 25.81 = _______
Answer:
20.7 – 9.54 + 25.81 = 36.97

Explanation:
The given numbers are: 20.7, 9.54, and 25.81
The representation of the numbers in the fraction form is: \(\frac{2581}{100}\), \(\frac{207}{10}\) and \(\frac{954}{100}\)
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 20.7 with 100 in the denominator is: \(\frac{2070}{100}\)
So,
\(\frac{2581}{100}\) + \(\frac{2070}{100}\) – \(\frac{954}{100}\)
= \(\frac{  2581 + 2070 – 954 }{100}\)
= \(\frac{3,697}{100}\)
= 36.97
Hence, 20.7 – 9.54 + 25.81 = 36.97

Lesson 9.6 Find Areas of Rectangles

Explore and Grow

Draw and cut out a rectangle that has any two of the side lengths below.
\(\frac{1}{2}\) ft
\(\frac{1}{3}\) ft
\(\frac{1}{4}\) ft
Use several copies of your rectangle to create a unit square. What is the area (in square feet) of each small rectangle? Explain your reasoning.
Answer:
Let there are two small rectangles with the following lengths:
\(\frac{1}{2}\) ft and \(\frac{1}{3}\) ft
\(\frac{1}{2}\) ft and \(\frac{1}{4}\) ft
So,
The area of the unit square with lengths \(\frac{1}{2}\) ft and \(\frac{1}{3}\) ft is:
\(\frac{1}{2}\) ft × \(\frac{1}{3}\) ft
= \(\frac{1 × 1}{2 × 3}\) ft
= \(\frac{1}{6}\) ft
The area of the unit square with lengths \(\frac{1}{2}\) ft and \(\frac{1}{4}\) ft is:
\(\frac{1}{2}\) ft × \(\frac{1}{4}\) ft
= \(\frac{1 × 1}{2 × 4}\) ft
= \(\frac{1}{8}\) ft
Hence, from the above,
We can conclude that the areas ofthe smaller rectngles with unit lengths is: \(\frac{1}{6}\) ft and \(\frac{1}{8}\) ft

Reasoning
How can you use a rectangle with unit fraction side lengths to find the area of the rectangle below? Explain your reasoning.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 1
Answer:
The given rectangle is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 1
To find the area of the rectangle by the unit fractions, multiply the values of the length and the breadth
So,
\(\frac{3}{4}\) × \(\frac{3}{2}\)
= 3  (\(\frac{1}{4}\) ) × 3 ( \(\frac{1}{2}\) )
= 9 ( \(\frac{1}{4}\) × \(\frac{1}{2}\) )
= 9 × \(\frac{1 × 1}{4 × 2}\)
= \(\frac{9}{1}\) × \(\frac{1}{8}\)
= \(\frac{9}{8}\)
The unit fractions are the fractions that contain the value 1 in the numerator.
Here,
\(\frac{1}{4}\) and \(\frac{1}{2}\) are the unit fractions

Think and Grow: Find Areas of Rectangles

One way to find the area of a rectangle with fractional side lengths is to fill it with smaller rectangles.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 2
Example
Find the area of the rectangle.

Show and Grow

Question 1.
Find the area of the shaded region.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 4
Answer:
The area of the shaded region is: \(\frac{3}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 4
From the given figure,
The area of the shaded region = \(\frac{3}{4}\) × \(\frac{1}{3}\)
= \(\frac{3 × 1}{4 × 3}\)
= \(\frac{3}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{3}{12}\)

Apply and Grow: Practice

Find the area of the shaded region.
Question 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 5
Answer:
The area of the shaded region is: \(\frac{2}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 5
From the given figure,
The area of the shaded region = \(\frac{2}{4}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{4 × 3}\)
= \(\frac{2}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{2}{12}\)

Question 3.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 6
Answer:
The area of the shaded region is: \(\frac{3}{16}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 6
From the given figure,
The area of the shaded region = \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{3 × 1}{4 × 4}\)
= \(\frac{3}{16}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{3}{16}\)

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 4.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 7
Answer:
The area of the rectangle is: \(\frac{6}{10}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 7
From the given figure,
The area of the rectangle = \(\frac{3}{2}\) × \(\frac{2}{5}\)
= \(\frac{3 × 2}{2 × 5}\)
= \(\frac{6}{10}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{6}{10}\)

Question 5.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 8
Answer:
The area of the rectangle is: \(\frac{10}{18}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 8
From the given figure,
The area of the rectangle = \(\frac{5}{6}\) × \(\frac{2}{3}\)
= \(\frac{5 × 2}{6 × 3}\)
= \(\frac{10}{18}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{10}{18}\)

Question 6.
Find the area of a rectangle with side lengths of \(\frac{5}{8}\) and \(\frac{4}{3}\)?
Answer:
The area of a rectangle is: \(\frac{20}{24}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{5}{8}\) and \(\frac{4}{3}\)
So,
The area of the rectangle = \(\frac{5}{8}\) × \(\frac{4}{3}\)
= \(\frac{5 × 4}{8 × 3}\)
= \(\frac{20}{24}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{20}{24}\)

Question 7.
Find the area of a rectangle with side lengths of \(\frac{7}{9}\) and \(\frac{1}{2}\)?
Answer:
The area of a rectangle is: \(\frac{7}{18}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{7}{9}\) and \(\frac{1}{2}\)
So,
The area of the rectangle = \(\frac{7}{9}\) × \(\frac{1}{2}\)
= \(\frac{7 × 1}{9 × 2}\)
= \(\frac{7}{18}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{7}{18}\)

Question 8.
Reasoning
Can you find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths? Explain.
Answer:
Yes, we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.
Example:
Let the fractional side lengths be: \(\frac{1}{5}\) and \(\frac{1}{9}\)
Let the whole number side lengths be: 3 and 4
So,
The area of the rectangle with the fractional side lengths = \(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{9 × 5}\)
= \(\frac{1}{45}\)
The area of the rectangle with the whole number side lengths = 3 × 4
= \(\frac{3}{1}\) × \(\frac{4}{1}\)
= \(\frac{3 × 4}{1 × 1}\)
= \(\frac{12}{1}\)
= 12
Hence, from the above,
We can conclude that we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.

Question 9.
YOU BE THE TEACHER
Our friend says she can find the area of a square given only one fractional side length. Is your friend correct? Explain.
Answer:
Yes, she can find the area of a square by only one fractional side length

Explanation:
Let the unit fractional side length of the square be: \(\frac{1}{2}\)
We know that,
The length of all the sides in a square are equal.
So,
By using this property, we can find the area of the square by taking the fractional unit side length of 1 side as the same for all the sides
So,
The area of the square = \(\frac{1}{2}\) × \(\frac{1}{2}\)
= \(\frac{1 × 1}{2 × 2}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that she can find the area of a square by only one fractional side length

Think and Grow: Modeling Real Life

Example
A zoo needs an outdoor enclosure with an area of at least \(\frac{3}{10}\) square kilometer10for a camel. Is the rectangular enclosure shown large enough for a camel?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 9
Find the area of the enclosure by multiplying the length and the width.

So, the enclosure is large enough for a camel.

Show and Grow

Question 10.
The area of a square dog kennel is 4 square yards. Will the square mat fit in the kennel?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 11
Answer:
Yes, the square mat will fit in the kennel

Explanation:
It is given that the area of a square dog kennel is 4 square yards and the side of the square mat is \(\frac{5}{3}\) yd
We know that,
The length of all the sides in a square are equal.
So,
By using this property,
The area of the square mat = \(\frac{5}{3}\) × \(\frac{5}{3}\)
= \(\frac{5 × 5}{3 × 3}\)
= \(\frac{25}{9}\) yd
Now,
The area of a square dog kennel can be written as: \(\frac{36}{9}\) yd
So,
When we compare the 2 areas,
We can get 25 < 36
Hence, from the above,
We can conclude that the square mat will fit in the kennel

Question 11.
DIG DEEPER!
The side lengths of each chalk art square are \(\frac{11}{4}\) meters. The side lengths of the square zone around each chalk square are an additional \(\frac{3}{4}\) meter. How many square meters of concrete is used to create the chalk walk shown?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 12
Answer:
The amount of the concrete used to create the chalk walk is: \(\frac{196}{16}\) square meters

Explanation:
It is given that the side lengths of each chalk art square are \(\frac{11}{4}\) meters. The side lengths of the square zone around each chalk square are an additional \(\frac{3}{4}\) meter.
So,
The total side length of the chalk walk = \(\frac{11}{4}\) + \(\frac{3}{4}\)
= \(\frac{11 + 3}{4}\)
= \(\frac{14}{4}\) meters
We know that,
The length of all sides in the square are equal.
So,
By using this property,
The amount of concrete used for the chalk walk = \(\frac{14}{4}\) × \(\frac{14}{4}\)
= \(\frac{14 × 14}{4 × 4}\)
= \(\frac{196}{16}\) square meters
Hence, from the above,
We can conclude that the amount of the concrete used to create the chalk walk is: \(\frac{196}{16}\) square meters

Find Areas of Rectangles Homework & Practice 9.6

Find the area of the shaded region.
Question 1.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 13
Answer:
The area of the shaded region is: \(\frac{1}{12}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 13
From the given figure,
The area of the shaded region = \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{1}{12}\)

Question 2.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 14
Answer:
The area of the shaded region is: \(\frac{8}{25}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 14
From the given figure,
The area of the shaded region = \(\frac{2}{5}\) × \(\frac{4}{5}\)
= \(\frac{2 × 4}{5 × 5}\)
= \(\frac{8}{25}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{8}{25}\)

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 3.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 15
Answer:
The area of the rectangle is: \(\frac{2}{24}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 15
From the given figure,
The area of the rectangle = \(\frac{2}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 2}{3 × 8}\)
= \(\frac{2}{24}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{2}{24}\)

Question 4.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 16
Answer:
The area of the rectangle is: \(\frac{35}{90}\)

Explanation:
The given figure is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 16
From the given figure,
The area of the rectangle = \(\frac{7}{10}\) × \(\frac{5}{9}\)
= \(\frac{7 × 5}{10 × 9}\)
= \(\frac{35}{90}\)
Hence, from the above,
We can conclude that the area of the shaded region is: \(\frac{35}{90}\)

Question 5.
Find the area of a rectangle with a side length of \(\frac{3}{4}\) and \(\frac{5}{12}\).
Answer:
The area of a rectangle is: \(\frac{15}{48}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{5}{12}\) and \(\frac{3}{4}\)
So,
The area of the rectangle = \(\frac{5}{12}\) × \(\frac{3}{4}\)
= \(\frac{5 × 3}{12 × 4}\)
= \(\frac{15}{48}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{15}{48}\)

Question 6.
Find the area of a square with side lengths of \(\frac{9}{16}\)
Answer:
The area of a square is: \(\frac{81}{256}\)

Explanation:
We know that,
The length of all the sides of the square is equal.
So,
The given side lengths of a square are: \(\frac{9}{16}\) and \(\frac{9}{16}\)
So,
The area of the square = \(\frac{9}{16}\) × \(\frac{9}{16}\)
= \(\frac{9 × 9}{16 × 16}\)
= \(\frac{81}{256}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{81}{256}\)

Question 7.
Open-Ended
The area of a rectangle is \(\frac{16}{24}\). What are the possible side lengths of the rectangle?
Answer:
The possible side lengths for the given area of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)

Explanation:
The given area of a rectangle is: \(\frac{16}{24}\)
So,
To find the possible side lengths of a rectangle, find the factors for the numerator and denominator of \(\frac{16}{24}\)
So,
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
So,
The possible side lengths of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)
There are so many possible side lengths of a rectangle like above
Hence, from the above,
We can conclude that the possible side lengths of a rectangle are:
A) \(\frac{4}{2}\) and \(\frac{4}{12}\)
B) \(\frac{1}{3}\) and \(\frac{4}{8}\)

Question 8.
Structure
Write an expression that represents the area of the shaded rectangle.
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 17
Answer:
The expression that represents the area of the shaded region is: \(\frac{1}{3}\) × \(\frac{1}{3}\)

Explanation:
The given shaded rectangle is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 17
From the given shaded rectangle,
The number of shaded columns is: 3
The number of shaded rows is: 3
The total number of rows is: 1
The total number of columns are: 1
So,
The expression representing the shaded region = \(\frac{The total number of rows}{The number of shaded rows}\) × \(\frac{The total number of rows}{The number of shaded rows}\)
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
Hence, from the above,
We can conclude that the expression representing the shaded region of a rectangle is: \(\frac{1}{3}\) × \(\frac{1}{3}\)

Question 9.
Modeling Real Life
The area of a square table is 9 square feet. Will the board game fit on the table?
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 18
Answer:
Yes, the board game fit on the table

Explanation:
The given area of a square table is 9 square feet
So,
The representation of the area of the square table in the fraction form is: \(\frac{36}{4}\)
Now,
The given board game is:
Big Ideas Math Answers 5th Grade Chapter 9 Multiply Fractions 9.6 18
The area of the board game can find out by multiplying its side lengths
So,
The area of the board game = \(\frac{3}{2}\) × \(\frac{5}{2}\)
= \(\frac{3 × 5}{2 × 2}\)
= \(\frac{15}{4}\)
So,
Compare the area of the square table and the area of the board game,
We get
Area of the square table > Area of the board game
Hence, from the above,
We can conclude that the board game fit on the table

Review & Refresh

Divide. Then check your answer.
Question 10.
365 ÷ 14 = _______
Answer:
365 ÷ 14 = 26 R 1

Explanation:
By using the partial quotients method,
365 ÷ 14 = ( 280 ÷ 84 ) ÷ 14
= ( 280 ÷ 14 ) + ( 84 ÷ 14 )
= 20 + 6
= 26 R 1
Hence, 365 ÷ 14 = 26 R 1

Question 11.
282 ÷ 27 = ______
Answer:
282 ÷ 27 = 10 R 12

Explanation:
By using the partial quotients method,
282 ÷ 27 = 270 ÷ 27
= 10 R 12
Hence, 270 ÷ 27 = 10 R 12

Question 12.
601 ÷ 72 = _____
Answer:
601 ÷ 72 = 8 R 25

Explanation:
By using the partial quotients method,
601 ÷ 72 = 576 ÷ 72
= 8 R 25
Hence, 601 ÷ 72 = 8 R 25

Lesson 9.7 Multiply Mixed Numbers

Explore and Grow

Find the area of the rectangle. Explain how you found your answer.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 1
Answer:
The area of the rectangle is: \(\frac{35}{4}\)

Explanation:
The given rectangle is:
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 1
From the above rectangle,
The side lengths are: 2\(\frac{1}{2}\) and 3\(\frac{1}{2}\)
The representation of the side lengths in the improper form is: \(\frac{5}{2}\) and \(\frac{7}{2}\)
So,
The area of rectangle = \(\frac{5}{2}\) × \(\frac{7}{2}\)
= \(\frac{5 × 7}{2 × 2}\)
= \(\frac{35}{4}\)
Hence,
2\(\frac{1}{2}\) × 3\(\frac{1}{2}\) = \(\frac{35}{4}\)

Structure
Find the area using a different method. Explain how you found your answer.
Answer:
The given mixed fractions of a rectangle are: 2\(\frac{1}{2}\) and 3\(\frac{1}{2}\)
To convert a mixed fraction into an improper fraction, we have to add the whole number part and the fractional part in the mixed fraction.
So,
2\(\frac{1}{2}\) = 2 + \(\frac{1}{2}\)
= \(\frac{4}{2}\) + \(\frac{1}{2}\)
= \(\frac{5}{2}\)
3\(\frac{1}{2}\) = 3 + \(\frac{1}{2}\)
= \(\frac{6}{2}\) + \(\frac{1}{2}\)
= \(\frac{7}{2}\)
So,
\(\frac{5}{2}\) × \(\frac{7}{2}\) = \(\frac{35}{4}\)

Think and Grow: Multiply Mixed Numbers

You can use a model to find the product of two mixed numbers. You can also write the mixed numbers as improper fractions and then multiply.
Example
Find 1\(\frac{1}{2}\) × 2\(\frac{3}{4}\).
One Way: Use an area model.
Step 1: Write each mixed number as a sum.
1\(\frac{1}{2}\) = 1 + \(\frac{1}{2}\) 2\(\frac{3}{4}\) = 2 + \(\frac{3}{4}\)
Step 2: Draw an area model that represents the product of the sums.
Step 3: Find the sum of the areas of the sections.

Another Way: Write each mixed number as an improper fraction, then multiply.

Show and Grow

Multiply.
Question 1.
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = ________
Answer:
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = \(\frac{15}{4}\)

Explanation:
The given fractions are: 2\(\frac{1}{2}\) and 1\(\frac{1}{2}\)
The representation of the fractions in the improper form is: \(\frac{5}{2}\) and \(\frac{3}{2}\)
So,
\(\frac{5}{2}\) × \(\frac{3}{2}\)
= \(\frac{5 × 3}{2 × 2}\)
= \(\frac{15}{4}\)
Hence,
2\(\frac{1}{2}\) × 1\(\frac{1}{2}\) = \(\frac{15}{4}\)

Question 2.
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = ______
Answer:
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{104}{12}\)

Explanation:
The given fractions are: 3\(\frac{1}{4}\) and 2\(\frac{2}{3}\)
The representation of the fractions in the improper form is: \(\frac{13}{4}\) and \(\frac{8}{3}\)
So,
\(\frac{13}{4}\) × \(\frac{8}{3}\)
= \(\frac{13 × 8}{4 × 3}\)
= \(\frac{104}{12}\)
Hence,
3\(\frac{1}{4}\) × 2\(\frac{2}{3}\) = \(\frac{104}{12}\)

Apply and Grow: Practice

Multiply.
Question 3.
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = ______
Answer:
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = \(\frac{91}{24}\)

Explanation:
The given fractions are: 1\(\frac{3}{4}\) and 2\(\frac{1}{6}\)
The representation of the fractions in the improper form is: \(\frac{7}{4}\) and \(\frac{13}{6}\)
So,
\(\frac{7}{4}\) × \(\frac{13}{6}\)
= \(\frac{13 × 7}{4 × 6}\)
= \(\frac{91}{24}\)
Hence,
1\(\frac{3}{4}\) × 2\(\frac{1}{6}\) = \(\frac{91}{24}\)

Question 4.
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = ______
Answer:
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = \(\frac{143}{18}\)

Explanation:
The given fractions are: 4\(\frac{1}{3}\) and 1\(\frac{5}{6}\)
The representation of the fractions in the improper form is: \(\frac{13}{3}\) and \(\frac{11}{6}\)
So,
\(\frac{13}{3}\) × \(\frac{11}{6}\)
= \(\frac{13 × 11}{6 × 3}\)
= \(\frac{143}{18}\)
Hence,
4\(\frac{1}{3}\) × 1\(\frac{5}{6}\) = \(\frac{143}{18}\)

Question 5.
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = ______
Answer:
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = \(\frac{323}{50}\)

Explanation:
The given fractions are: 3\(\frac{2}{5}\) and 1\(\frac{9}{10}\)
The representation of the fractions in the improper form is: \(\frac{17}{5}\) and \(\frac{19}{10}\)
So,
\(\frac{17}{5}\) × \(\frac{19}{10}\)
= \(\frac{17 × 19}{5 × 10}\)
= \(\frac{323}{50}\)
Hence,
3\(\frac{2}{5}\) × 1\(\frac{9}{10}\) = \(\frac{323}{50}\)

Question 6.
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = ______
Answer:
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = \(\frac{133}{16}\)

Explanation:
The given fractions are: 2\(\frac{3}{8}\) and 3\(\frac{1}{2}\)
The representation of the fractions in the improper form is: \(\frac{19}{8}\) and \(\frac{7}{2}\)
So,
\(\frac{19}{8}\) × \(\frac{7}{2}\)
= \(\frac{19 × 7}{8 × 2}\)
= \(\frac{133}{16}\)
Hence,
2\(\frac{3}{8}\) × 3\(\frac{1}{2}\) = \(\frac{133}{16}\)

Evaluate.
Question 7.
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = _________
Answer:
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = \(\frac{3,066}{240}\)

Explanation:
The given fractions are: 5\(\frac{1}{4}\), \(\frac{2}{5}\) and 6\(\frac{1}{12}\)
The representation of the fractions in the improper form is: \(\frac{21}{4}\) and \(\frac{73}{12}\)
So,
\(\frac{21}{4}\) × \(\frac{73}{12}\) × \(\frac{2}{5}\)
= \(\frac{21 × 73 × 2}{4 × 12 × 5}\)
= \(\frac{3,066}{240}\)
Hence,
5\(\frac{1}{4}\) × \(\frac{2}{5}\) × 6\(\frac{1}{12}\) = \(\frac{3,066}{240}\)

Question 8.
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = _________
Answer:
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = \(\frac{759}{24}\)

Explanation:
The given fractions are: 3\(\frac{2}{3}\), 10\(\frac{7}{8}\) and 2\(\frac{1}{4}\)
The representation of the fractions in the improper form is: \(\frac{11}{3}\), \(\frac{87}{8}\)  and \(\frac{9}{4}\)
So,
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\))
= \(\frac{11}{3}\) × ( \(\frac{87}{8}\) – \(\frac{9}{4}\) )
= \(\frac{11}{3}\) × ( \(\frac{87 – 18}{8}\) )
= \(\frac{11}{3}\) × \(\frac{69}{8}\)
= \(\frac{11 × 69}{8 × 3}\)
= \(\frac{759}{24}\)
Hence,
3\(\frac{2}{3}\) × (10\(\frac{7}{8}\) – 2\(\frac{1}{4}\)) = \(\frac{759}{24}\)

Question 9.
YOU BE THE TEACHER
Your friend uses the model to find 4\(\frac{1}{2}\) × 3\(\frac{2}{3}\). Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 4
Answer:
No, your friend is not correct

Explanation:
The given fractions are: 4\(\frac{1}{2}\) and 3\(\frac{2}{3}\)
The representation of the side lengths in the improper form is: \(\frac{9}{2}\) and \(\frac{11}{3}\)
So,
The area of rectangle = \(\frac{9}{2}\) × \(\frac{11}{3}\)
= \(\frac{9 × 11}{3 × 2}\)
= \(\frac{99}{6}\)
= \(\frac{33}{2}\)
= 15\(\frac{3}{2}\)
But according to your friend,
4\(\frac{1}{2}\) × 3\(\frac{2}{3}\) = 12\(\frac{1}{2}\)
Hence, from the above,
We can conclude that your friend is not correct.

Question 10.
Logic
Find the missing numbers.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 5
Answer:
The missing numbers are: 1 and \(\frac{8}{5}\)

Explanation:
According to the separation method,
To convert the mixed number into the improper fraction, we can add whole numbers and the fractions separately
Now,
Let the missing numbers be p and q
So,
p + 5 = 6
So,
p = 6 – 5 = 1
Now,
\(\frac{1}{4}\) + \(\frac{1}{q}\) = \(\frac{7}{8}\)
So,
\(\frac{1}{q}\) = \(\frac{7}{8}\) – \(\frac{1}{4}\)
= \(\frac{7}{8}\) – \(\frac{2}{8}\)
= \(\frac{7 – 2}{8}\)
= \(\frac{5}{8}\)
So,
q= \(\frac{8}{5}\)
Hence, from the above,
We can conclude that the missing numbers are: 1 and \(\frac{8}{5}\)

Think and Grow: Modeling Real Life

Example
A construction crew is paving 15 miles of a highway. The crew paves 4\(\frac{2}{10}\) miles each month. Does the crew finish paving the highway in 3\(\frac{1}{2}\) months?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 6
Find the length of the highway the crew paves by multiplying the number of months by the number of miles they pave each month. Write each mixed number as an improper fraction, then multiply.

Compare the length the crew paves to the amount that needs to be paved.
The Crew can finish paving the highway in 3\(\frac{1}{2}\) months.

Show and Grow

Question 11.
You have 3 cups of strawberries. You want to make 1\(\frac{1}{2}\) batches of the recipe. Do you have enough strawberries?
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 8
Answer:
Yes, you have enough strawberries

Explanation:
The given recipe for a strawberry smoothie is:
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 8
It is given that you want to make 1\(\frac{1}{2}\) batches of the recipe
From the table,
The number of cups of strawberry is: 1\(\frac{3}{4}\)
Now,
The representation of 1\(\frac{1}{2}\) in the iproper fraction is: \(\frac{3}{2}\)
The representation of 1\(\frac{3}{4}\) in the mixed form is: \(\frac{7}{4}\)
So,
The number of strawberries for the recipe = \(\frac{3}{2}\) × \(\frac{7}{4}\)
= \(\frac{3 × 7}{4 × 2}\)
= \(\frac{21}{8}\)
It is also given that you have 3 cups of strawberries
So,
We can write 3 in the fraction form and as the multiples of 8 is: \(\frac{24}{8}\)
So,
When we compare the number of cups we obtained and given,
We can say that we have enough strawberries to make the recipe
Hence, from the above,
We can conclude that we have enough strawberries

Question 12.
On Monday,you roller-skate 6\(\frac{1}{4}\) miles. On Tuesday, you skate 1\(\frac{2}{5}\) times as far as you did on Monday. How many total miles do you roller-skate on Monday and Tuesday combined?
Answer:
The number of miles you do roller-skate on Monday and Tuesday combined is: 15 miles

Explanation:
It is given that on Monday,you roller-skate 6\(\frac{1}{4}\) miles. On Tuesday, you skate 1\(\frac{2}{5}\) times as far as you did on Monday.
So,
The number of miles you do roller-skate on Tuesday = ( The number of miles you do roller-skate on Monday ) × 1\(\frac{2}{5}\)
= 6\(\frac{1}{4}\) × 1\(\frac{2}{5}\)
= \(\frac{25}{4}\) × \(\frac{7}{5}\)
= \(\frac{25 × 7}{5 × 4}\)
= \(\frac{175}{20}\)
Now,
The number of miles you do roller-skate on Monday and Tuesday combined = 6\(\frac{1}{4}\) + \(\frac{175}{20}\)
= \(\frac{25}{4}\) + \(\frac{175}{20}\)
= \(\frac{125}{20}\) + \(\frac{175}{20}\)
= \(\frac{175 + 125}{20}\)
= \(\frac{300}{20}\)
= 15 miles
Hence, from the above,
We can conclude that the number of miles you do roller- skate on Monday and Tuesday combined is: 15 miles

Question 13.
DIG DEEPER!
An artist paints a rectangular mural. The mural is 4\(\frac{1}{3}\) feet wide. The length is 2\(\frac{1}{4}\) times the width. What is the area of the mural?
Answer:
The area of the mural is: \(\frac{117}{12}\) square feet

Explanation:
It is given that an artist paints a rectangular mural. The mural is 4\(\frac{1}{3}\) feet wide. The length is 2\(\frac{1}{4}\) times the width.
So,
The area of the mural = ( The width of the mural ) × ( The length of the mural )
= 4\(\frac{1}{3}\) × 2\(\frac{1}{4}\)
= \(\frac{13}{3}\) × \(\frac{9}{4}\)
= \(\frac{13 × 9}{4 × 3}\)
= \(\frac{117}{12}\)
Hence, from the above,
We can conclude that the area of the mural is: \(\frac{117}{12}\) square feet

Multiply Mixed Numbers Homework & Practice 9.7

Multiply.
Question 1.
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = ______
Answer:
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{18}{16}\)

Explanation:
The given fractions are: 1\(\frac{1}{2}\) and 1\(\frac{1}{8}\)
The representation of the fractions in the improper form is: \(\frac{3}{2}\) and \(\frac{9}{8}\)
So,
\(\frac{9}{8}\) × \(\frac{3}{2}\)
= \(\frac{9 × 3}{8 × 2}\)
= \(\frac{18}{16}\)
Hence,
1\(\frac{1}{2}\) × 1\(\frac{1}{8}\) = \(\frac{18}{16}\)

Question 2.
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = ______
Answer:
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = \(\frac{99}{24}\)

Explanation:
The given fractions are: 1\(\frac{5}{6}\) and 2\(\frac{1}{4}\)
The representation of the fractions in the improper form is: \(\frac{11}{6}\) and \(\frac{9}{4}\)
So,
\(\frac{9}{4}\) × \(\frac{11}{6}\)
= \(\frac{9 × 11}{4 × 6}\)
= \(\frac{99}{24}\)
Hence,
1\(\frac{5}{6}\) × 2\(\frac{1}{4}\) = \(\frac{99}{24}\)

Evaluate.

Question 3.
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = ______
Answer:
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = \(\frac{209}{32}\)

Explanation:
The given fractions are: 2\(\frac{3}{8}\) and 2\(\frac{3}{4}\)
The representation of the fractions in the improper form is: \(\frac{19}{8}\) and \(\frac{11}{4}\)
So,
\(\frac{19}{8}\) × \(\frac{11}{4}\)
= \(\frac{19 × 11}{4 × 8}\)
= \(\frac{209}{32}\)
Hence,
2\(\frac{3}{8}\) × 2\(\frac{3}{4}\) = \(\frac{209}{32}\)

Question 4.
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = ______
Answer:
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = \(\frac{575}{42}\)

Explanation:
The given fractions are: 4\(\frac{1}{6}\) and 3\(\frac{2}{7}\)
The representation of the fractions in the improper form is: \(\frac{25}{6}\) and \(\frac{23}{7}\)
So,
\(\frac{23}{7}\) × \(\frac{25}{6}\)
= \(\frac{23 × 25}{7 × 6}\)
= \(\frac{575}{42}\)
Hence,
4\(\frac{1}{6}\) × 3\(\frac{2}{7}\) = \(\frac{575}{42}\)

Question 5.
2\(\frac{1}{3}\) × 3\(\frac{9}{10}\) × 5\(\frac{1}{5}\) = ______
Answer:
2\(\frac{1}{3}\)× 3\(\frac{9}{10}\) ×5\(\frac{1}{5}\) = \(\frac{7,098}{150}\)

Explanation:
The given fractions are: 2\(\frac{1}{3}\), 3\(\frac{9}{10}\) and 5\(\frac{1}{5}\)
The representation of the fractions in the improper form is: \(\frac{7}{3}\), \(\frac{39}{10}\)  and \(\frac{26}{5}\)
So,
\(\frac{7}{3}\) × \(\frac{39}{10}\) ×\(\frac{26}{5}\)
= \(\frac{7 × 39 × 26}{3 × 10 × 5}\)
= \(\frac{7,098}{150}\)
Hence,
2\(\frac{1}{3}\)× 3\(\frac{9}{10}\) ×5\(\frac{1}{5}\) = \(\frac{7,098}{150}\)

Question 6.
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = _______
Answer:
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = \(\frac{6,675}{480}\)

Explanation:
The given fractions are: 1\(\frac{7}{8}\), 4\(\frac{4}{5}\) and 2\(\frac{1}{12}\)
The representation of the fractions in the improper form is: \(\frac{15}{8}\), \(\frac{24}{5}\) and \(\frac{25}{12}\)
So,
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × \(\frac{25}{12}\)
= ( \(\frac{15}{8}\) + \(\frac{24}{5}\) ) × \(\frac{25}{12}\)
= ( \(\frac{75}{40}\) + \(\frac{192}{40}\) ) × \(\frac{25}{12}\)
= \(\frac{75 + 192}{40}\) × \(\frac{25}{12}\)
= \(\frac{267}{40}\) × \(\frac{25}{12}\)
= \(\frac{267 × 25}{40× 12 }\)
= \(\frac{6,675}{480}\)
Hence,
(1\(\frac{7}{8}\) + 4\(\frac{4}{5}\)) × 2\(\frac{1}{12}\) = \(\frac{6,675}{480}\)

Question 7.
Structure
Find the missing numbers.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 9
Answer:
The missing numbers are: 5 and \(\frac{3}{10}\)

Explanation:
Let the missing numbers be p and q
By using the partial products method,
( p × 4 ) + ( p × \(\frac{1}{7}\) ) = 20 + \(\frac{5}{7}\)
p × ( 4 + \(\frac{1}{7}\) ) = 20 + \(\frac{5}{7}\)
p × \(\frac{29}{7}\) = \(\frac{145}{7}\)
So,
p = \(\frac{145}{7}\) ÷ \(\frac{29}{7}\)
= \(\frac{145}{7}\) × \(\frac{7}{29}\)
= \(\frac{145 × 7}{7 × 29}\)
= 5
Now,
(q × 4 ) + ( q × \(\frac{1}{7}\) ) = \(\frac{12}{10}\) + \(\frac{3}{70}\)
q × ( 4 + \(\frac{1}{7}\) ) = \(\frac{12}{10}\) + \(\frac{3}{70}\)
q × \(\frac{29}{7}\) = \(\frac{84 + 3}{70}\)
q × \(\frac{29}{7}\) = \(\frac{87}{70}\)
So,
q = \(\frac{87}{70}\) ÷ \(\frac{29}{7}\)
= \(\frac{87}{70}\) × \(\frac{7}{29}\)
= \(\frac{87 × 7}{70 × 29}\)
= \(\frac{3}{10}\)
Hence, from the above,
We can conclude that the missing numbers are: 5 and \(\frac{3}{10}\)

Question 8.
YOU BE THE TEACHER
Your friend finds 1\(\frac{11}{12}\) × 2\(\frac{3}{8}\) . Is your friend correct? Explain.
Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions 9.7 10
Answer:
Yes your friend is correct

Explanation:
The given mixed numbers are: 1\(\frac{11}{12}\) and 2\(\frac{3}{8}\)
The representation of the mixed numbers in the fraction form is: \(\frac{23}{12}\) and \(\frac{19}{8}\)
So,
\(\frac{23}{12}\) × \(\frac{19}{8}\)
= \(\frac{23 × 19}{8 × 12}\)
= \(\frac{437}{96}\)
= 4\(\frac{53}{96}\)
Hence, from the above,
We can conclude that your friend is correct

Question 9.
Modeling Real Life
Your friend earns 7\(\frac{1}{2}\) dollars each hour. Will she earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours?
Answer:
she can’t earn enough money to buy $35 toy after working 4\(\frac{3}{4}\) hours

Explanation:
It is given that your friend earns 7\(\frac{1}{2}\) dollars each hour.
It is given that will she earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours
So,
The number of dollars she can earn = 7\(\frac{1}{2}\) × 4\(\frac{3}{4}\)
= \(\frac{15}{2}\) × \(\frac{19}{4}\)
= \(\frac{15 × 19}{4 × 2}\)
= \(\frac{285}{8}\)
Now,
The representation of $35 as the multiple of 8 is: \(\frac{280}{8}\)
So,
By comparing the given money with he money she earned,
We can say that she can’t buy the toy.
Hence, from the above,
We can conclude that she can’t earn enough money to buy a $35 toy after working 4\(\frac{3}{4}\) hours

Question 10.
Modeling Real Life
One class collects 8\(\frac{1}{4}\) pounds of recyclable materials. Another class collects 1\(\frac{1}{2}\) times as many pounds as the first class. How many pounds of recyclable materials do the two classes collect altogether?
Answer:
The number of pounds of recyclable materials the two classes collects together is: \(\frac{231}{8}\) pounds

Explanation:
It is given that one class collects 8\(\frac{1}{4}\) pounds of recyclable materials. Another class collects 1\(\frac{1}{2}\) times as many pounds as the first class.
So,
The number of pounds of recyclable materials collected by another class = 8\(\frac{1}{4}\) × 1\(\frac{1}{2}\)
= \(\frac{33}{4}\) × \(\frac{3}{2}\)
= \(\frac{99}{8}\) pounds
So,
The number of pouds of recyclable materials collected by two classes together = 8\(\frac{1}{4}\) + \(\frac{99}{8}\)
= \(\frac{33}{4}\) + \(\frac{99}{8}\)
= \(\frac{132 + 99}{8}\)
= \(\frac{231}{8}\) pounds
Hence, from the above,
We can conclude that the number of pounds of recyclable materials the two classes collects together is: \(\frac{231}{8}\) pounds

Review & Refresh

Find the product.
Question 11.
6 × 5.7 = ______
Answer:
6 × 5.7 = 34.2

Explanation:
By using the partial products method,
6 × 5.7 = 6 × ( 5 + 0.7 )
= ( 6 × 5 ) + ( 6 × 0.7 )
= 30 + 4.2
= 34.2
Hence, 6 × 5.7 = 34.2

Question 12.
0.84 × 9 = ______
Answer:
0.84 × 9 = 73.17

Explanation:
By using the partial quotients method,
0.84 × 9 = ( 0.81 + 0.03 ) × 9
= ( 0.81 × 9 ) + ( 0.03 × 9 )
= 72.9 + 0.27
= 73.17
Hence, 0.84 × 9 = 73.17

Lesson 9.8 Compare Factors and Products

Explore and Grow

Without calculating, order the rectangles by area from least to greatest. Explain your reasoning.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 1
Answer:
Let the given rectangles be named A), B), C), and D)
So,
The side lengths of A are: 1 and 1\(\frac{1}{2}\)
The side lengths of B are: 1\(\frac{1}{3}\) and \(\frac{11}{12}\)
The side lengths of C are: \(\frac{3}{2}\) and \(\frac{11}{12}\)
The side lengths of D are: \(\frac{11}{10}\) and \(\frac{3}{2}\)
Now,
The area of A is: \(\frac{1}{2}\)
The area of B is: \(\frac{44}{36}\)
The area of C is: \(\frac{33}{24}\)
The area of D is: \(\frac{33}{20}\)
For comparison make the denominators of the areas of the four rectangles equal.
So,
The area of A is: \(\frac{18}{36}\)
The area of B is: \(\frac{44}{36}\)
So, by comparing these 2 areas,
We can say that B is greater
Now,
The area of C is: \(\frac{165}{120}\)
The area of D is: \(\frac{198}{120}\)
So, by comparing these 2 areas,
We can say that D is greater
Hence, from the above,
We can conclude that the order of rectangles by areas from the greatest to the least is: B > D> A > C

Construct Arguments
Explain your strategy to your partner. Compare your strategies.
Answer:
The strategy you followed is:
A) Write the side lengths of the rectangles
B) Find the areas of the four rectangles
C) Make the denominators of all the four areas of rectangles equal
D) Compare the numerators of all the four areas of the four rectangles

Think and Grow: Compare Factors and Products

Key Idea
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
Example
Without calculating, tell whether the product 3\(\frac{1}{8}\) × \(\frac{5}{6}\) is less than, greater than, or equal to or each of its factors.

Show and Grow

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
8 × \(\frac{3}{10}\)
Answer:
The value of 8 × \(\frac{3}{10}\) is less than 8

Explanation:
The given numbers are: 8 and \(\frac{3}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{3}{10}\) is less than 1
So,
8 × \(\frac{3}{10}\) is less than 8
Hence, from the above,
We can conclude that the value of 8 × \(\frac{3}{10}\) is less than 8

Question 2.
\(\frac{4}{4}\) × 5\(\frac{2}{3}\)
Answer:
The value of \(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than \(\frac{4}{4}\)

Explanation:
The given fractions are: \(\frac{4}{4}\) and 5\(\frac{2}{3}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
We know that,
\(\frac{4}{4}\) is 1
Now,
The representation of 5\(\frac{2}{3}\) in the improper fraction form is: \(\frac{17}{3}\)
So,
\(\frac{17}{3}\) is greater than 1
So,
\(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{4}{4}\) × 5\(\frac{2}{3}\) is greater than \(\frac{4}{4}\)

Question 3.
\(\frac{4}{3}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{4}{3}\) × \(\frac{1}{6}\) is greater than \(\frac{1}{6}\) or less than \(\frac{4}{3}\)

Explanation:
The given fractions are: \(\frac{4}{3}\) and \(\frac{1}{6}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
Now,
\(\frac{4}{3}\) is greater than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{4}{3}\) × \(\frac{1}{6}\) is less than \(\frac{4}{3}\)
Hence, from the above,
We can conclude that the value of \(\frac{4}{3}\) × \(\frac{1}{6}\) is greater than \(\frac{1}{6}\) or less than \(\frac{4}{3}\)

Apply and Grow

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 4.
\(\frac{1}{4}\) × \(\frac{1}{12}\)
Answer:
The value of \(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1

Explanation:
The given fractions are: latex]\frac{1}{4}[/latex] and \(\frac{1}{12}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{1}{4}\) is less than 1
\(\frac{1}{12}\) is less than 1
So,
\(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{4}\) × \(\frac{1}{12}\) is less than 1

Question 5.
3\(\frac{4}{5}\) × 6\(\frac{7}{8}\)
Answer:
The value of 3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1

Explanation:
The given fractions are: 3\(\frac{4}{5}\) and 6\(\frac{7}{8}\)
The representation of 3\(\frac{4}{5}\) and 6\(\frac{7}{8}\) in the improper fractions form is: \(\frac{19}{5}\) and \(\frac{55}{8}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{19}{5}\) is greater than 1
\(\frac{55}{8}\) is greater than 1
So,
3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{4}{5}\) × 6\(\frac{7}{8}\) is greater than 1

Question 6.
\(\frac{1}{6}\) × \(\frac{10}{10}\)
Answer:
The value of \(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1

Explanation:
The given fractions are: \(\frac{1}{6}\) and \(\frac{10}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
We know that,
\(\frac{10}{10}\) is 1
So,
\(\frac{1}{6}\) is less than 1
So,
\(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{6}\) × \(\frac{10}{10}\) is less than 1

Question 7.
\(\frac{2}{3}\) × 5
Answer:
The value of \(\frac{2}{3}\) × 5 is less than 5

Explanation:
The given fractions are: \(\frac{2}{3}\) and 5
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{3}\) is less than 1
So,
\(\frac{2}{3}\) × 5 is less than 5
Hence, from the above,
We can conclude that the value of \(\frac{2}{3}\) × 5 is less than 5

Question 8.
\(\frac{7}{10}\) × 4\(\frac{8}{9}\)
Answer:
The value of \(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1

Explanation:
The given fractions are: \(\frac{7}{10}\) and 4\(\frac{8}{9}\)
The representation of 4\(\frac{8}{9}\) in the improper fraction form is: \(\frac{44}{9}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{7}{10}\) is less than 1
\(\frac{44}{9}\) is greater than 1
So,
\(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{7}{10}\) × 4\(\frac{8}{9}\) is greater than 1

Question 9.
\(\frac{9}{2}\) × 1\(\frac{3}{4}\)
Answer:
The value of \(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1

Explanation:
The given fractions are: \(\frac{9}{2}\) × 1\(\frac{3}{4}\)
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{9}{2}\) is greater than 1
\(\frac{7}{4}\) is greater than 1
So,
\(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{9}{2}\) × 1\(\frac{3}{4}\) is greater than 1

Without calculating, order the products from least to greatest.
Question 10.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 3
Answer:
The order of products from the least to the greatest is:
\(\frac{5}{6}\) × \(\frac{1}{3}\) < \(\frac{5}{6}\) × \(\frac{7}{7}\) < \(\frac{5}{6}\) × 1\(\frac{8}{9}\)

Explanation:
The given products are:
A) \(\frac{5}{6}\) × \(\frac{1}{3}\)
B) \(\frac{5}{6}\) × \(\frac{7}{7}\)
C) \(\frac{5}{6}\) × 1\(\frac{8}{9}\)
So,
In A),
\(\frac{5}{6}\) and \(\frac{1}{3}\) are less than 1
So,
\(\frac{5}{6}\) × \(\frac{1}{3}\) is less than 1
In B),
\(\frac{5}{6}\) is less than 1
\(\frac{7}{7}\) is 1
So,
\(\frac{5}{6}\) × \(\frac{7}{7}\) is less than 1
In C),
\(\frac{5}{6}\) is less than1
The representation of  1\(\frac{8}{9}\) in the improper fraction form is: \(\frac{17}{9}\)
So,
\(\frac{17}{9}\) is greater than 1
So,
\(\frac{5}{6}\) × 1\(\frac{8}{9}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{5}{6}\) × \(\frac{1}{3}\) < \(\frac{5}{6}\) × \(\frac{7}{7}\) < \(\frac{5}{6}\) × 1\(\frac{8}{9}\)

Question 11.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 4
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{6}\) × \(\frac{1}{4}\) < \(\frac{1}{10}\) × \(\frac{1}{4}\) < 5\(\frac{7}{10}\) × \(\frac{1}{4}\)

Explanation:
The given products are:
A) \(\frac{1}{6}\) × \(\frac{1}{4}\)
B) \(\frac{1}{10}\) × \(\frac{1}{4}\)
C) \(\frac{1}{4}\) × 5\(\frac{7}{10}\)
So,
In A),
\(\frac{1}{6}\) and \(\frac{1}{4}\) are less than 1
So,
\(\frac{1}{6}\) × \(\frac{1}{4}\) is less than 1
In B),
\(\frac{1}{10}\) is less than 1
\(\frac{1}{4}\) is less than 1
So,
\(\frac{1}{10}\) × \(\frac{1}{4}\) is less than 1
In C),
\(\frac{1}{4}\) is less than1
The representation of  5\(\frac{7}{10}\) in the improper fraction form is: \(\frac{57}{10}\)
So,
\(\frac{57}{10}\) is greater than 1
So,
\(\frac{1}{4}\) × 5\(\frac{7}{10}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{6}\) × \(\frac{1}{4}\) < \(\frac{1}{10}\) × \(\frac{1}{4}\) < 5\(\frac{7}{10}\) × \(\frac{1}{4}\)

Question 12.
YOU BE THE TEACHER
Your friend says that \(\frac{1}{2}\) × 8 is half as much as 8. Is your friend correct? Explain.
Answer:
Yes, your friend is correct

Explanation:
The given numbers are: \(\frac{1}{2}\) and 8
So,
\(\frac{1}{2}\) × 8 = \(\frac{1}{2}\) × \(\frac{8}{1}\)
= \(\frac{1 × 8}{2 × 1}\)
= 4
It is also given that according to your friend,
\(\frac{1}{2}\) × 8 is half as much as 8.
So,
8 ÷ 2 = 4
Hence, from the above,
We can conclude that your friend is correct.

Question 13.
DIG DEEPER!
Without calculating, tell whether the product is less than, greater than, or equal to 3\(\frac{3}{4}\). Explain.
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\)
Answer:
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\) is less than  3\(\frac{3}{4}\)

Explanation:
The representation of 3\(\frac{3}{4}\) in the improper fraction form is: \(\frac{15}{4}\)
Now,
\(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\)
= ( \(\frac{1}{2}\) × \(\frac{15}{4}\) ) × \(\frac{2}{7}\)
= \(\frac{15}{8}\) × \(\frac{2}{7}\)
= \(\frac{2 × 15}{7 × 8}\)
= \(\frac{30}{56}\)
= \(\frac{15}{28}\)
Now,
\(\frac{15}{4}\) is multiplied by \(\frac{7}{7}\)
So,
\(\frac{15}{4}\) = \(\frac{105}{28}\)
So,
When we compare \(\frac{105}{28}\) and \(\frac{15}{28}\)
We will get \(\frac{105}{28}\) is greater
Hence, from the above,
We can conclude that \(\left(\frac{1}{2} \times 3 \frac{3}{4}\right)\) × \(\frac{2}{7}\) is less than  3\(\frac{3}{4}\)

Think and Grow: Modeling Real Life

Example
Men’s shot put competitions use a shot with a mass of 7\(\frac{1}{4}\) kilograms. The mass of a bowling ball is \(\frac{7}{8}\) as much as the mass of the shot. Is the mass of the bowling ball less than, greater than, or equal to the mass of the shot?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 5
Write an expression to represent the mass of the bowling ball.

So, the mass of the bowling ball is greater than the mass of the shot.

Show and Grow

Question 14.
You practice playing the keyboard for 3\(\frac{1}{2}\) hours. Your friend practices playing the keyboard for \(\frac{5}{4}\) as many hours as you. Does your friend practice for fewer hours, more hours, or the same number of hours as you?
Answer:
You practice more hours than your friend

Explanation:
It is given that you practice playing the keyboard for 3\(\frac{1}{2}\) hours. Your friend practices playing the keyboard for \(\frac{5}{4}\) as many hours as you.
So,
Now,
The representation of 3\(\frac{1}{2}\) in the improper fraction form is: \(\frac{7}{2}\)
So,
The number of hours you practice playing the keyboard is: \(\frac{7}{2}\) hours
The number of hours your friend practice playing the keyboard is: \(\frac{5}{4}\) hours
So,
For comparison, make the denominators equal.
So,
Multiply \(\frac{7}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{7}{2}\) = \(\frac{14}{4}\)
By comparing the timings ,
We can say that
\(\frac{14}{4}\) > \(\frac{5}{4}\)
Hence, from the above,
We can conclude that you practice more hours than your friend

Question 15.
The original price of a telescope is $99. The sale price is \(\frac{4}{5}\) of the original price. An astrologist buys the telescope at its sale price and uses a half-off-coupon. What fraction of the original price does the astrologist pay for the telescope?
Answer:
The fraction of the original price the astrologist pays for the telescope is: \(\frac{4}{10}\)

Explanation:
It is given that the original price of a telescope is $99. The sale price is \(\frac{4}{5}\) of the original price. and an astrologist buys the telescope at its sale price and uses a half-off-coupon.
So,
The cost of the telescope that an astrologist bought = ( The \(\frac{4}{5}\)th of the original price ) × ( Half-off -coupon on the \(\frac{4}{5}\)th of the original price )
= \(\frac{4}{5}\) × \(\frac{1}{2}\)
= \(\frac{4 × 1}{5 × 2}\)
= \(\frac{4}{10}\)
Hence, from the above,
We can conclude that the fraction of the original price the astrologist pays for the telescope is: \(\frac{4}{10}\)

Question 16.
The Abraj Al-Bait Clock Tower is \(\frac{6}{10}\) kilometer tall. Zifeng Tower is \(\frac{3}{4}\) as tall as the clock tower. Is Zifeng Tower shorter than, taller than, or the same height as the Abraj Al-Bait Clock Tower? What is the height of each tower in meters?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 7
Answer:
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: \(\frac{24}{40}\)
The height of Zifeng tower is: \(\frac{18}{40}\)

Explanation:
It is given that the Abraj Al-Bait Clock Tower is \(\frac{6}{10}\) kilometer tall and Zifeng Tower is \(\frac{3}{4}\) as tall as the clock tower.
So,
The height of Abraj al-Bait Clock tower is: \(\frac{6}{10}\) kilometer
Now,
The height of Zifeng tower is = \(\frac{6}{10}\) × \(\frac{3}{4}\)
= \(\frac{6 × 3}{10 × 4}\)
= \(\frac{18}{40}\)
Now, for comparison, make the denomonators of both the towers equal.
So,
\(\frac{6}{10}\) is multiplied by \(\frac{4}{4}\)
So,
\(\frac{6}{10}\) = \(\frac{24}{40}\)
By comparing the heights of the two towers,
We can say that
\(\frac{24}{40}\) > \(\frac{18}{40}\)
Hence, from the above,
We can conclude that
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: \(\frac{24}{40}\)
The height of Zifeng tower is: \(\frac{18}{40}\)

Compare Factors and Products Homework & Practice 9.8

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
1\(\frac{3}{4}\) × 6
Answer:
The value of 1\(\frac{3}{4}\) × 6 is greater than 6

Explanation:
The given numbers are: 1\(\frac{3}{4}\) and 6
The representation of 1\(\frac{3}{4}\) in the improper fraction form is: \(\frac{7}{4}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{7}{4}\) is greater than 1
So,
1\(\frac{3}{4}\) × 6 is greater than 6
Hence, from the above,
We can conclude that the value of 1\(\frac{3}{4}\) × 6 is greater than 6

Question 2.
\(\frac{5}{12}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1

Explanation;
The given fractions are: \(\frac{5}{12}\) and \(\frac{1}{6}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{5}{12}\) is less than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{5}{12}\) × \(\frac{1}{6}\) is less than 1

Question 3.
\(\frac{2}{7}\) × \(\frac{5}{5}\)
Answer:
The value of \(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1

Explanation:
The given fractions are: \(\frac{2}{7}\) and \(\frac{5}{5}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{7}\) is less than 1
We know that
\(\frac{5}{5}\) is 1
So,
\(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{2}{7}\) × \(\frac{5}{5}\) is less than 1

Question 4.
3\(\frac{4}{5}\) × 2\(\frac{9}{10}\)
Answer:
The value of 3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1

Explanation:
the given mixed fractions are: 3\(\frac{4}{5}\) and 2\(\frac{9}{10}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
The representation of the mixed numbers in the fraction form is: \(\frac{19}{5}\) and \(\frac{29}{10}\)
So,
\(\frac{19}{5}\) is greater than 1
\(\frac{29}{10}\) is greater than
So,
3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{4}{5}\) × 2\(\frac{9}{10}\) is greater than 1

Question 5.
8 × \(\frac{2}{3}\)
Answer:
The value of  8 is greater than \(\frac{2}{3}\)

Explanation:
The given numbers are: 8 and \(\frac{2}{3}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{2}{3}\) is less than 1
So,
8 is greater than \(\frac{2}{3}\)
Hence, from the above,
We can conclude that the value of 8 is greater than \(\frac{2}{3}\)

Question 6.
1\(\frac{7}{8}\) × \(\frac{1}{4}\)
Answer:
The value of 1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1

Explanation:
The given fractions are: 1\(\frac{7}{8}\) and \(\frac{1}{4}\)
The representation of 1\(\frac{7}{8}\) in the fraction form is: \(\frac{15}{8}\)
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
\(\frac{15}{8}\) is greater than 1
\(\frac{1}{4}\) is less than 1
So,
1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1
Hence, from the above,
We can conclude that the value of 1\(\frac{7}{8}\) × \(\frac{1}{4}\) is less than 1

Without calculating, order the products from least to greatest.
Question 7.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 8
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{3}\) × \(\frac{4}{5}\) < \(\frac{1}{3}\) × \(\frac{6}{6}\) < 8\(\frac{2}{9}\) × \(\frac{1}{3}\)

Explanation:
The given products are:
A) \(\frac{1}{3}\) × \(\frac{4}{5}\)
B) \(\frac{1}{3}\) × \(\frac{6}{6}\)
C) 8\(\frac{2}{9}\) × \(\frac{1}{3}\)
So,
In A),
\(\frac{1}{3}\) and \(\frac{4}{5}\) are less than 1
So,
\(\frac{1}{3}\) × \(\frac{4}{5}\) is less than 1
In B),
\(\frac{1}{3}\) is less than 1
\(\frac{6}{6}\) is 1
So,
\(\frac{1}{3}\) × \(\frac{6}{6}\) is less than 1
In C),
\(\frac{1}{3}\) is less than1
The representation of  8\(\frac{2}{9}\) in the improper fraction form is: \(\frac{74}{9}\)
So,
\(\frac{74}{9}\) is greater than 1
So,
\(\frac{1}{3}\) × 8\(\frac{2}{9}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{3}\) × \(\frac{4}{5}\) < \(\frac{1}{3}\) × \(\frac{6}{6}\) < 8\(\frac{2}{9}\) × \(\frac{1}{3}\)

Question 8.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 9
Answer:
The order of products from the least to the greatest is:
\(\frac{1}{12}\) × \(\frac{3}{5}\) < 4 × \(\frac{3}{5}\) < 2\(\frac{1}{2}\) × \(\frac{3}{5}\)

Explanation:
The given products are:
A) \(\frac{1}{12}\) × \(\frac{3}{5}\)
B) 4 × \(\frac{3}{5}\)
C) 2\(\frac{1}{2}\) × \(\frac{3}{5}\)
So,
In A),
\(\frac{1}{12}\) and \(\frac{3}{5}\) are less than 1
So,
\(\frac{1}{12}\) × \(\frac{3}{5}\) is less than 1
In B),
\(\frac{3}{5}\) is less than 1
So,
4 × \(\frac{3}{5}\) is less than 4
In C),
\(\frac{3}{5}\) is less than1
The representation of  2\(\frac{1}{2}\) in the improper fraction form is: \(\frac{5}{2}\)
So,
\(\frac{5}{2}\) is greater than 1
So,
\(\frac{3}{5}\) × 2\(\frac{1}{2}\) is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
\(\frac{1}{12}\) × \(\frac{3}{5}\) < 4 × \(\frac{3}{5}\) < 2\(\frac{1}{2}\) × \(\frac{3}{5}\)

Question 9.
Logic
Without calculating, use >, <, or = to make the statement true. Explain.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 10
Answer:
The value of 3\(\frac{1}{3}\) × \(\frac{1}{6}\) is less than 3\(\frac{1}{2}\)

Explanation:
The given fractions are: 3\(\frac{1}{3}\), \(\frac{1}{6}\) and 3\(\frac{1}{2}\)
The representations of the 3\(\frac{1}{3}\) and  3\(\frac{1}{2}\)  in the fraction form is: \(\frac{7}{2}\) and \(\frac{10}{3}\)
So,
3\(\frac{1}{3}\) × \(\frac{1}{6}\)
= \(\frac{10}{3}\)  × \(\frac{1}{6}\)
= \(\frac{10}{18}\)
To compare the fractions, equate the denominators
So,
Multiply \(\frac{7}{2}\) by \(\frac{9}{9}\)
= \(\frac{63}{18}\)
So,
By comparison,
We can say that
\(\frac{10}{18}\) < \(\frac{63}{18}\)
Hence, from the above,
We can conclude that the value of 3\(\frac{1}{3}\) × \(\frac{1}{6}\) is less than 3\(\frac{1}{2}\)

Question 10.
Logic
Without calculating, determine which number makes the statement true.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 11
Answer:
Let the missing number be x.
The options for x are given as:
A) 1 B) \(\frac{1}{2}\) and C) 1\(\frac{4}{5}\)
So,
From the given three fractions,
The value of x is: 1\(\frac{4}{5}\)

Explanation:
The given multiplication equation is:
x × 1\(\frac{7}{8}\) is greater than 1\(\frac{7}{8}\)
So,
To find the value of x, there are 3 options. They are:
A) 1 B) \(\frac{1}{2}\) and C) 1\(\frac{4}{5}\)
Now,
Let x = 1\(\frac{4}{5}\)
So,
1\(\frac{4}{5}\) × 1\(\frac{7}{8}\)
= \(\frac{9}{5}\) × \(\frac{15}{8}\)
= \(\frac{27}{8}\)
So, by comparison,
We will get,
\(\frac{27}{8}\) is greater than \(\frac{15}{8}\)
Hence, from the above,
We can conclude that the missing number is: 1\(\frac{4}{5}\)

Question 11.
Reasoning
Why does multiplying by a fraction greater than one result in a product greater than the original number?
Answer:
In the given fraction, if the numerator is greater than the denominator, then the given fraction is greater than 1
So,
When the whole number is multiplied by the fraction which is greater than 1, the result in a product will be greater than the original number i.e, the whole number

Question 12.
Modeling Real Life
You snowboard 1\(\frac{7}{8}\) miles. Your friend snowboards 3\(\frac{2}{3}\) times as far as you. Does your friend snowboard fewer miles, more miles, or the same number of miles as you?
Answer:
The number of miles you snowboarded is less than your friend

Explanation:
It is given that you snowboard 1\(\frac{7}{8}\) miles. Your friend snowboards 3\(\frac{2}{3}\) times as far as you.
So,
The number of miles you snowboard is: 1\(\frac{7}{8}\) miles
So,
The number of miles your friend snowboard is = 1\(\frac{7}{8}\) × 3\(\frac{2}{3}\)
= \(\frac{15}{8}\) × \(\frac{11}{3}\)
= \(\frac{15 × 11}{3 × 8}\)
= \(\frac{165}{24}\) miles
So,
For comparison we have to make the denominators equal.
So,
Multiply \(\frac{15}{8}\) by \(\frac{3}{3}\)
So,
\(\frac{15}{8}\) = \(\frac{45}{24}\)
So,
By comparing, we will get
\(\frac{45}{24}\) is less than \(\frac{165}{24}\) miles
Hence, from the above,
We can conclude that the number of miles you snowboarded is less than your friend

Question 13.
Modeling Real Life
A pet owner has three dogs. The youngest dog weighs \(\frac{1}{4}\) as much as the second oldest dog. The oldest dog weighs 1\(\frac{1}{4}\) as much as the second oldest. The second oldest weighs 20 pounds. Which dog weighs the most? the least?
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 12

Answer:
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog

Explanation:
It is given that a pet owner has three dogs. The youngest dog weighs \(\frac{1}{4}\) as much as the second oldest dog. The oldest dog weighs 1\(\frac{1}{4}\) as much as the second oldest. The second oldest weighs 20 pounds.
So,
The weight of the second oldest dog is: 20 pounds
Now,
The weight of the oldest dog = 1\(\frac{1}{4}\) × ( The weight of the oldest second dog )
= 1\(\frac{1}{4}\) × 20
= \(\frac{5}{4}\) × 20
= \(\frac{5 × 20}{4 × 1}\)
= 25 pounds
Now,
The weight of the youngest dog = \(\frac{1}{4}\) × ( The weight of the second oldest dog )
= \(\frac{1}{4}\) × 20
= \(\frac{1 × 20}{4}\)
= 5 pounds
Hence, from the above,
We can conclude that
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog

Review & Refresh

Compare.
Question 14.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 13
Answer:
40.5 is greater than 40.13

Explanation:
The given decimal numbers are: 40.5 and 40.13
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 40.5 is greater than 40.13

Question 15.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 14
Answer:
13.90 is equal to 13.9

Explanation:
The given decimal numbers are: 13.90 and 13.9
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 13.90 is equal to 13.9

Question 16.
Big Ideas Math Solutions Grade 5 Chapter 9 Multiply Fractions 9.8 15
Answer:
32.006 is less than 32.06

Explanation:
The given decimal numbers are: 32.006 and 32.06
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 32.006 is less than 32.06

Multiply Fractions Performance Task

Question 1.
You see a rock formation at a national park. The formation has layers that formed millions of years ago when particles settled in the water and became rock. You make a model of the rock formation using \(\frac{3}{16}\) -inch foam sheets.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 1
a. The three types of sedimentary rocks are limestone, sandstone, and shale. Use the number of foam sheets to find the height of each sedimentary rock layer.
Answer:
The height of limestone is: \(\frac{9}{4}\) inches
The height of sandstone is: \(\frac{3}{2}\) inches
The height of shale is: \(\frac{15}{4}\) inches

Explanation:
It is given that the height of each foam sheet is: \(\frac{3}{16}\) -inches
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 1
From the table,
The number of foam sheets for Limestone is: 12
The number of foam sheets for sandstone is: 8
The number of foam sheets for shale is: 20
So,
The height of limestone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 12
= \(\frac{3}{16}\) × \(\frac{12}{1}\)
= \(\frac{9}{4}\) inches
The height of sandstone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 8
= \(\frac{3}{16}\) × \(\frac{8}{1}\)
= \(\frac{3}{2}\) inches
The height of shale is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= \(\frac{3}{16}\) × 20
= \(\frac{3}{16}\) × \(\frac{20}{1}\)
= \(\frac{15}{4}\) inches
Hence, from the above,
We can conclude that
The height of limestone is: \(\frac{9}{4}\) inches
The height of sandstone is: \(\frac{3}{2}\) inches
The height of shale is: \(\frac{15}{4}\) inches

b. What is the combined height of the sedimentary rock layers?
Answer:
The combined height of the sedimentary layers is: \(\frac{15}{2}\) inches

Explanation:
The combined height of the sedimentary rock layers = The height of limestone + The height of sandstone + The height of the shale
= \(\frac{9}{4}\) + \(\frac{3}{2}\) + \(\frac{15}{4}\)
= \(\frac{24}{4}\) + \(\frac{3}{2}\)
= \(\frac{12}{2}\) + \(\frac{3}{2}\)
= \(\frac{15}{2}\) inches
Hence, from the above,
We can conclude that the combined height of the sedimentary layers is: \(\frac{15}{2}\) inches

c. Will you use more foam sheets for the granite layers of the shale layers? Explain.
Answer:
We use  more foam sheets for the granite layers

Explanation:
From the above,
The height of the shale layers is: \(\frac{15}{4}\) inches
From the table,
The height of the granite layers is: 3 inches
When we compare the two values,
We can observe that the height of shale layers is greater than the granite layers
Hence, from the above,
We can conclude that we will use more foam sheets for granite layers

d. The height of the topsoil layer is 1\(\frac{1}{4}\) times the height of the sandstone layer. How many foam sheets do you use in the topsoil layer?
Answer:
The height of the topsoil layer is: \(\frac{15}{8}\) inches

Explanation:
From the above,
The height of the sandstone layer is: \(\frac{3}{2}\) inches
So,
The height of topsoil layer = 1\(\frac{1}{4}\) × \(\frac{3}{2}\)
= \(\frac{5}{4}\) × \(\frac{3}{2}\)
= \(\frac{15}{8}\) inches
Hence, from the above,
We can conclude that the height of the topsoil layer is: \(\frac{15}{8}\) inches

e. On your model, 1 inch represents 40 feet. What is the actual height of the rock formation?
Answer:
The actual height of the rock formation is: 495 feet

Explanation:
From the above,
The height of the sedimentary layers is: \(\frac{15}{2}\) inches
The height of the topsoil layer is: \(\frac{15}{8}\) inches
The height of the granite layer is: 3 inches
So,
The combined height of the rock formation = \(\frac{15}{2}\) + \(\frac{15}{8}\) + 3
= \(\frac{99}{8}\) inches
But, it is given that,
1 inch = 40 feet
So,
The combined height of the rock formation in feet = \(\frac{99}{8}\) inches × 40
= 495 feet
Hence, from the above,
We can conclude that the height of the rock formation in feet is: 495 feet

f. Why do you think the rock formation has layers?
Answer:
The rock formation has layers because of the tectonic plates.

Multiply Fractions Activity

Fraction Connection: Multiplication
Directions:
1. Players take turns rolling three dice.
2. On your turn, evaluate the expression indicated by your roll and cover the answer.
3. The first player to get four in a row, horizontally, vertically, or diagonally, wins?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 2
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions 3

Multiply Fractions Chapter Practice

9.1 Multiply Whole Numbers by Fractions

Multiply.
Question 1.
5 × \(\frac{1}{2}\) = _______
Answer:
5 × \(\frac{1}{2}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 5 and \(\frac{1}{2}\)
So,
5 × \(\frac{1}{2}\)
= \(\frac{5}{1}\) × \(\frac{1}{2}\)
= \(\frac{5 × 1}{1 × 2}\)
= \(\frac{5}{2}\)
Hence,
5 × \(\frac{1}{2}\) = \(\frac{5}{2}\)

Question 2.
2 × \(\frac{7}{10}\) = _______
Answer:
2 × \(\frac{7}{10}\) = \(\frac{14}{10}\)

Explanation:
The given numbers are: 2 and \(\frac{7}{10}\)
So,
2 × \(\frac{7}{10}\)
= \(\frac{2}{1}\) × \(\frac{7}{10}\)
= \(\frac{2 × 7}{1 × 10}\)
= \(\frac{14}{10}\)
Hence,
2 × \(\frac{7}{10}\) = \(\frac{14}{10}\)

Question 3.
9 × \(\frac{5}{8}\) = _______
Answer:
9 × \(\frac{5}{8}\) = \(\frac{45}{8}\)

Explanation:
The given numbers are: 9 and \(\frac{5}{8}\)
So,
9 × \(\frac{5}{8}\)
= \(\frac{9}{1}\) × \(\frac{5}{8}\)
= \(\frac{5 × 9}{1 × 8}\)
= \(\frac{45}{8}\)
Hence,
9 × \(\frac{5}{8}\) = \(\frac{45}{8}\)

Question 4.
6 × \(\frac{71}{100}\) = _______
Answer:
6 × \(\frac{71}{100}\) = \(\frac{426}{100}\)

Explanation:
The given numbers are: 6 and \(\frac{71}{100}\)
So,
6 × \(\frac{71}{100}\)
= \(\frac{6}{1}\) × \(\frac{71}{100}\)
= \(\frac{6 × 71}{1 × 100}\)
= \(\frac{426}{100}\)
Hence,
6 × \(\frac{71}{100}\) = \(\frac{426}{100}\)

Question 5.
4 × \(\frac{8}{5}\) = _______
Answer:
4 × \(\frac{8}{5}\) = \(\frac{32}{5}\)

Explanation:
The given numbers are: 4 and \(\frac{8}{5}\)
So,
4 × \(\frac{8}{5}\)
= \(\frac{4}{1}\) × \(\frac{8}{5}\)
= \(\frac{4 × 8}{1 × 5}\)
= \(\frac{32}{5}\)
Hence,
4 × \(\frac{8}{5}\) = \(\frac{32}{5}\)

Question 6.
7 × \(\frac{5}{3}\) = _______
Answer:
7 × \(\frac{5}{3}\) = \(\frac{35}{3}\)

Explanation:
The given numbers are: 7 and \(\frac{5}{3}\)
So,
7 × \(\frac{5}{3}\)
= \(\frac{7}{1}\) × \(\frac{5}{3}\)
= \(\frac{5 × 7}{1 × 3}\)
= \(\frac{35}{3}\)
Hence,
7 × \(\frac{5}{3}\) = \(\frac{35}{3}\)

9.2 Use Models to Multiply Fractions by Whole Numbers

Multiply. Use a model to help.
Question 7.
\(\frac{2}{5}\) of 20
Answer:
20 × \(\frac{2}{5}\) = 8

Explanation:
The given numbers are: 20 and \(\frac{2}{5}\)
So,
20 × \(\frac{2}{5}\)
= \(\frac{20}{1}\) × \(\frac{2}{5}\)
= \(\frac{20 × 2}{1 × 5}\)
= \(\frac{8}{1}\)
= 8
Hence,
20 × \(\frac{2}{5}\) = 8

Question 8.
\(\frac{1}{6}\) × 12
Answer:
12 × \(\frac{1}{6}\) = 2

Explanation:
The given numbers are: 12 and \(\frac{1}{6}\)
So,
12 × \(\frac{1}{6}\)
= \(\frac{12}{1}\) × \(\frac{1}{6}\)
= \(\frac{12 × 1}{1 × 6}\)
= \(\frac{2}{1}\)
= 2
Hence,
12 × \(\frac{1}{6}\) = 2

Question 9.
\(\frac{1}{3}\) × 6
Answer:
6 × \(\frac{1}{3}\) = 2

Explanation:
The given numbers are: 6 and \(\frac{1}{3}\)
So,
6 × \(\frac{1}{3}\)
= \(\frac{6}{1}\) × \(\frac{1}{3}\)
= \(\frac{6 × 1}{1 × 3}\)
= \(\frac{2}{1}\)
= 2
Hence,
6 × \(\frac{1}{3}\) = 2

Question 10.
\(\frac{5}{16}\) of 8
Answer:
8 × \(\frac{5}{16}\) = \(\frac{5}{2}\)

Explanation:
The given numbers are: 8 and \(\frac{5}{16}\)
So,
8 × \(\frac{5}{16}\)
= \(\frac{8}{1}\) × \(\frac{5}{16}\)
= \(\frac{5 × 8}{1 × 16}\)
= \(\frac{5}{2}\)
Hence,
8 × \(\frac{5}{16}\) = \(\frac{5}{2}\)

Question 11.
Modeling Real Life
You have 24 apples. You use \(\frac{1}{4}\) of them to make a single serving of applesauce. How many apples do you not use?
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 11
Answer:
The number of apples you do not use is: 18 apples

Explanation:
It is given that you have 24 apples and you use \(\frac{1}{4}\) of them to make a single serving of applesauce.
So,
The number of apples you used = ( The total number of apples ) × ( The fraction of apples you used to make a single serving of applesauce )
= 24 × \(\frac{1}{4}\)
= \(\frac{24}{1}\) × \(\frac{1}{4}\)
= \(\frac{6}{1}\)
= 6 apples
Now,
The number of apples that do not use = ( The total number of apples ) – ( The number of apples you used )
= 24 – 6
= 18 apples
Hence, from the above,
We can conclude that the number of apples that do not use is: 18 apples

9.3 Multiply Fractions and Whole Numbers

Multiply.
Question 12.
\(\frac{3}{5}\) × 15 = _______
Answer:
15 × \(\frac{3}{5}\) = 9

Explanation:
The given numbers are: 15 and \(\frac{3}{5}\)
So,
15 × \(\frac{3}{5}\)
= \(\frac{15}{1}\) × \(\frac{3}{5}\)
= \(\frac{15 × 3}{1 × 5}\)
= \(\frac{9}{1}\)
= 9
Hence,
15 × \(\frac{3}{5}\) = 9

Question 13.
\(\frac{9}{10}\) × 30 = _______
Answer:
30 × \(\frac{9}{10}\) = 27

Explanation:
The given numbers are: 30 and \(\frac{9}{10}\)
So,
30 × \(\frac{9}{10}\)
= \(\frac{30}{1}\) × \(\frac{9}{10}\)
= \(\frac{30 × 9}{1 × 10}\)
= \(\frac{27}{1}\)
= 27
Hence,
30 × \(\frac{9}{10}\) = 27

Question 14.
48 × \(\frac{3}{4}\) = _______
Answer:
48 × \(\frac{3}{4}\) = 36

Explanation:
The given numbers are: 48 and \(\frac{3}{4}\)
So,
48 × \(\frac{3}{4}\)
= \(\frac{48}{1}\) × \(\frac{3}{4}\)
= \(\frac{48 × 3}{1 × 4}\)
= \(\frac{36}{1}\)
= 36
Hence,
48 × \(\frac{3}{4}\) = 36

Question 15.
11 × \(\frac{5}{9}\) = _______
Answer:
11 × \(\frac{5}{9}\) = \(\frac{55}{9}\)

Explanation:
The given numbers are: 11 and \(\frac{5}{9}\)
So,
11 × \(\frac{5}{9}\)
= \(\frac{11}{1}\) × \(\frac{5}{9}\)
= \(\frac{5 × 11}{1 × 9}\)
= \(\frac{55}{9}\)
Hence,
11 × \(\frac{5}{9}\) = \(\frac{55}{9}\)

Question 16.
\(\frac{1}{6}\) × 19 = _______
Answer:
19 × \(\frac{1}{6}\) = \(\frac{19}{6}\)

Explanation:
The given numbers are: 19 and \(\frac{1}{6}\)
So,
19 × \(\frac{1}{6}\)
= \(\frac{19}{1}\) × \(\frac{1}{6}\)
= \(\frac{19 × 1}{1 × 6}\)
= \(\frac{19}{6}\)
Hence,
19 × \(\frac{1}{6}\) = \(\frac{19}{6}\)

Question 17.
7 × \(\frac{13}{50}\) = _______
Answer:
7 × \(\frac{13}{50}\) = \(\frac{91}{50}\)

Explanation:
The given numbers are: 7 and \(\frac{13}{50}\)
So,
7 × \(\frac{13}{50}\)
= \(\frac{7}{1}\) × \(\frac{13}{50}\)
= \(\frac{7 × 13}{1 × 50}\)
= \(\frac{91}{50}\)
Hence,
7 × \(\frac{13}{50}\) = \(\frac{91}{50}\)

9.4 Use Models to Multiply Fractions

Multiply. Use a model to help.
Question 18.
\(\frac{1}{2}\) × \(\frac{1}{10}\) = _______
Answer:
\(\frac{1}{2}\) × \(\frac{1}{10}\) = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and \(\frac{1}{10}\)
So,
\(\frac{1}{2}\) × \(\frac{1}{10}\)
= \(\frac{1 × 1}{2 × 10}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{2}\) × \(\frac{1}{10}\) = \(\frac{1}{20}\)

Question 19.
\(\frac{1}{5}\) × \(\frac{1}{9}\) = _______
Answer:
\(\frac{1}{5}\) × \(\frac{1}{9}\) = \(\frac{1}{45}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and \(\frac{1}{9}\)
So,
\(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{1 × 1}{5 × 9}\)
= \(\frac{1}{45}\)
Hence,
\(\frac{1}{5}\) × \(\frac{1}{9}\) = \(\frac{1}{45}\)

Question 20.
\(\frac{1}{6}\) × \(\frac{1}{7}\) = _______
Answer:
\(\frac{1}{6}\) × \(\frac{1}{7}\) = \(\frac{1}{42}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and \(\frac{1}{7}\)
So,
\(\frac{1}{6}\) × \(\frac{1}{7}\)
= \(\frac{1 × 1}{6 × 7}\)
= \(\frac{1}{42}\)
Hence,
\(\frac{1}{6}\) × \(\frac{1}{7}\) = \(\frac{1}{42}\)

Question 21.
\(\frac{1}{3}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and \(\frac{1}{8}\)
So,
\(\frac{1}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{3 × 8}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Question 22.
\(\frac{2}{5}\) × \(\frac{1}{3}\) = _______
Answer:
\(\frac{2}{5}\) × \(\frac{1}{3}\) = \(\frac{2}{15}\)

Explanation:
The given numbers are: \(\frac{2}{5}\) and \(\frac{1}{3}\)
So,
\(\frac{2}{5}\) × \(\frac{1}{3}\)
= \(\frac{2 × 1}{5 × 3}\)
= \(\frac{2}{15}\)
Hence,
\(\frac{2}{5}\) × \(\frac{1}{3}\) = \(\frac{2}{15}\)

Question 23.
\(\frac{2}{3}\) × \(\frac{3}{5}\) = _______
Answer:
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\)

Explanation:
The given numbers are: \(\frac{2}{3}\) and \(\frac{3}{5}\)
So,
\(\frac{2}{3}\) × \(\frac{3}{5}\)
= \(\frac{2 × 3}{3 × 5}\)
= \(\frac{2}{5}\)
Hence,
\(\frac{2}{3}\) × \(\frac{3}{5}\) = \(\frac{2}{5}\)

9.5 Multiply Fractions

Evaluate.
Question 24.
\(\frac{1}{3}\) × \(\frac{1}{8}\) = _______
Answer:
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and \(\frac{1}{8}\)
So,
\(\frac{1}{3}\) × \(\frac{1}{8}\)
= \(\frac{1 × 1}{3 × 8}\)
= \(\frac{1}{24}\)
Hence,
\(\frac{1}{3}\) × \(\frac{1}{8}\) = \(\frac{1}{24}\)

Question 25.
\(\frac{5}{6}\) × \(\frac{1}{4}\) = _______
Answer:
\(\frac{5}{6}\) × \(\frac{1}{4}\) = \(\frac{5}{24}\)

Explanation:
The given numbers are: \(\frac{5}{6}\) and \(\frac{1}{4}\)
So,
\(\frac{5}{6}\) × \(\frac{1}{4}\)
= \(\frac{5 × 1}{6 × 4}\)
= \(\frac{5}{24}\)
Hence,
\(\frac{5}{6}\) × \(\frac{1}{4}\) = \(\frac{5}{24}\)

Question 26.
\(\frac{7}{2}\) × \(\frac{2}{5}\) = _______
Answer:
\(\frac{7}{2}\) × \(\frac{2}{5}\) = \(\frac{7}{5}\)

Explanation:
The given numbers are: \(\frac{7}{2}\) and \(\frac{2}{5}\)
So,
\(\frac{7}{2}\) × \(\frac{2}{5}\)
= \(\frac{7 × 2}{2 × 5}\)
= \(\frac{7}{5}\)
Hence,
\(\frac{7}{2}\) × \(\frac{2}{5}\) = \(\frac{7}{5}\)

Question 27.
\(\frac{9}{10}\) × \(\frac{3}{7}\) = _______
Answer:
\(\frac{9}{10}\) × \(\frac{3}{7}\) = \(\frac{27}{70}\)

Explanation:
The given numbers are: \(\frac{9}{10}\) and \(\frac{3}{7}\)
So,
\(\frac{9}{10}\) × \(\frac{3}{7}\)
= \(\frac{9 × 3}{7 × 10}\)
= \(\frac{27}{70}\)
Hence,
\(\frac{9}{10}\) × \(\frac{3}{7}\) = \(\frac{27}{70}\)

Question 28.
\(\frac{4}{5}\) × \(\frac{13}{100}\) = _______
Answer:
\(\frac{4}{5}\) × \(\frac{13}{100}\) = \(\frac{52}{500}\)

Explanation:
The given numbers are: \(\frac{4}{5}\) and \(\frac{13}{100}\)
So,
\(\frac{4}{5}\) × \(\frac{13}{100}\)
= \(\frac{4 × 13}{5 × 100}\)
= \(\frac{52}{500}\)
Hence,
\(\frac{4}{5}\) × \(\frac{13}{100}\) = \(\frac{52}{500}\)

Question 29.
\(\frac{11}{25}\) × \(\frac{3}{4}\) = _______
Answer:
\(\frac{11}{25}\) × \(\frac{3}{4}\) = \(\frac{33}{100}\)

Explanation:
The given numbers are: \(\frac{11}{25}\) and \(\frac{3}{4}\)
So,
\(\frac{11}{25}\) × \(\frac{3}{4}\)
= \(\frac{11 × 3}{25 × 4}\)
= \(\frac{33}{100}\)
Hence,
\(\frac{11}{25}\) × \(\frac{3}{4}\) = \(\frac{33}{100}\)

Question 30.
9 × \(\left(\frac{2}{9} \times \frac{1}{2}\right)\) = _______
Answer:
9 × ( \(\frac{1}{2}\) × \(\frac{2}{9}\) ) = 1

Explanation:
The given numbers are: 9, \(\frac{1}{2}\) and \(\frac{2}{9}\)
So,
9 × (\(\frac{1}{2}\) × \(\frac{2}{9}\) )
= 9 × (\(\frac{1 × 2}{2 × 9}\) )
=  9 × \(\frac{1}{9}\)
= 1
Hence,
9 × ( \(\frac{1}{2}\) × \(\frac{2}{9}\) ) = 1

Question 31.
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = _______
Answer:
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{5}\)

Explanation:
The given fractions are: \(\frac{1}{10}\), \(\frac{7}{10}\) and \(\frac{3}{4}\)
So,
\(\frac{3}{4}\) × (\(\frac{1}{10}\) + \(\frac{7}{10}\) )
= \(\frac{3}{4}\) × (\(\frac{7 × 1}{10}\) )
=  \(\frac{3}{4}\) × \(\frac{8}{10}\)
= \(\frac{3 × 8}{4 × 10}\)
= \(\frac{24}{40}\)
= \(\frac{3}{5}\)
Hence,
\(\left(\frac{1}{10}+\frac{7}{10}\right)\) × \(\frac{3}{4}\) = \(\frac{3}{5}\)

Question 32.
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = _______
Answer:
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = \(\frac{3}{14}\)

Explanation:
The given fractions are: \(\frac{4}{7}\), \(\frac{5}{8}\) and \(\frac{1}{4}\)
So,
\(\frac{4}{7}\) × (\(\frac{5}{8}\) – \(\frac{1}{4}\) )
= \(\frac{4}{7}\) × (\(\frac{5 – 2}{8}\) )
=  \(\frac{4}{7}\) × \(\frac{3}{8}\)
= \(\frac{3 × 4}{7 × 8}\)
= \(\frac{12}{56}\)
= \(\frac{3}{14}\)
Hence,
\(\frac{4}{7}\) × \(\left(\frac{5}{8}-\frac{1}{4}\right)\) = \(\frac{3}{14}\)

9.6 Find Areas of Rectangles

Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 33.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 33
Answer:
The area of the rectangle is: \(\frac{1}{4}\)

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 33
From the given figure,
The side lengths of the rectangle are: \(\frac{3}{8}\) and \(\frac{2}{3}\)
So,
The area of the rectangle = \(\frac{3}{8}\) × \(\frac{2}{3}\)
= \(\frac{2 × 3}{3 × 8}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{1}{4}\)

Question 34.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 34
Answer:
The area of the rectangle is: \(\frac{35}{48}\)

Explanation:
The given figure is:
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 34
From the given figure,
The side lengths of the rectangle are: \(\frac{7}{12}\) and \(\frac{5}{4}\)
So,
The area of the rectangle = \(\frac{7}{12}\) × \(\frac{5}{4}\)
= \(\frac{7 × 5}{12 × 4}\)
= \(\frac{35}{48}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{35}{48}\)

Question 35.
Find the area of a rectangle with side lengths of \(\frac{2}{9}\) and \(\frac{1}{10}\).
Answer:
The area of the rectangle is: \(\frac{1}{45}\)

Explanation:
The given side lengths of a rectangle are: \(\frac{2}{9}\) and \(\frac{1}{10}\)
So,
The area of the rectangle = \(\frac{2}{9}\) × \(\frac{1}{10}\)
= \(\frac{2 × 1}{10 × 9}\)
= \(\frac{1}{45}\)
Hence, from the above,
We can conclude that the area of the rectangle is: \(\frac{1}{45}\)

Question 36.
Find the area of a with a side length of \(\frac{3}{4}\).
Answer:
The area of the square is: \(\frac{9}{16}\)

Explanation:
The given side length of a square is: \(\frac{3}{4}\)
We know that,
The length of all sides of a square are equal
So,
the area of the square = \(\frac{3}{4}\) × \(\frac{3}{4}\)
= \(\frac{3 × 3}{4 × 4}\)
= \(\frac{9}{16}\)
Hence, from the above,
We can conclude that the area of the square is: \(\frac{9}{16}\)

9.7 Multiply Mixed Numbers

Multiply.
Question 37.
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) = _______
Answer:
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) =\(\frac{45}{16}\)

Explanation:
The given mixed fractions are: 1\(\frac{1}{4}\) and 2\(\frac{1}{4}\)
The representation of 1\(\frac{1}{4}\) and 2\(\frac{1}{4}\) in the improper fraction form is: \(\frac{5}{4}\) and \(\frac{9}{4}\)
So,
\(\frac{5}{4}\) × \(\frac{9}{4}\)
= \(\frac{5 × 9}{4 × 4}\)
= \(\frac{45}{16}\)
Hence,
1\(\frac{1}{4}\) × 2\(\frac{1}{4}\) =\(\frac{45}{16}\)

Question 38.
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = _______
Answer:
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Explanation:
The given mixed fractions are: 3\(\frac{4}{5}\) and 1\(\frac{1}{2}\)
The representation of 3\(\frac{4}{5}\) and 1\(\frac{1}{2}\) in the improper fraction form is: \(\frac{19}{5}\)  and \(\frac{3}{2}\)
So,
\(\frac{19}{5}\) × \(\frac{3}{2}\)
= \(\frac{19 × 3}{5 × 2}\)
= \(\frac{57}{10}\)
Hence,
3\(\frac{4}{5}\) × 1\(\frac{1}{2}\) = \(\frac{57}{10}\)

Question 39.
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = _______
Answer:
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = \(\frac{624}{24}\)

Explanation:
The given mixed fractions are: 5\(\frac{1}{3}\) and 4\(\frac{7}{8}\)
The representation of 5\(\frac{1}{3}\) and 4\(\frac{7}{8}\) in the improper fraction form is: \(\frac{16}{3}\)  and \(\frac{39}{8}\)
So,
\(\frac{16}{3}\) × \(\frac{39}{8}\)
= \(\frac{39 × 16}{3 × 8}\)
= \(\frac{624}{24}\)
Hence,
5\(\frac{1}{3}\) × 4\(\frac{7}{8}\) = \(\frac{624}{24}\)

Question 40.
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = _______
Answer:
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = \(\frac{841}{480}\)

Explanation:
The given mixed fractions are: 4\(\frac{5}{6}\), 2\(\frac{9}{10}\) and \(\frac{1}{8}\)
The representation of 4\(\frac{5}{6}\) and 2\(\frac{9}{10}\) in the improper fraction form is: \(\frac{29}{6}\) and \(\frac{29}{10}\)
So,
\(\frac{29}{6}\) × \(\frac{29}{10}\)  × \(\frac{1}{8}\)
= \(\frac{29 × 29 × 1}{8 × 10 × 6}\)
= \(\frac{841}{480}\)
Hence,
4\(\frac{5}{6}\) × 2\(\frac{9}{10}\) × \(\frac{1}{8}\) = \(\frac{841}{480}\)

Question 41.
Logic
Find the missing numbers.
Big Ideas Math Answer Key Grade 5 Chapter 9 Multiply Fractions chp 41
Answer:
The missing numbers are: 2 and 11

Explanation:
Let the missing numbers be: p and q
So,
4\(\frac{1}{p}\) × 2\(\frac{1}{5}\) = 9\(\frac{q}{10}\)
So,
To solve the mixed numbers, we can separate the whole numbers and the fraction numbers
So,
\(\frac{1}{p}\) × 2\(\frac{1}{5}\) = \(\frac{q}{10}\)
\(\frac{1}{p}\) × \(\frac{11}{5}\) = \(\frac{q}{10}\)
\(\frac{1 × 11}{p × 5}\) = \(\frac{q}{10}\)
\(\frac{11}{5p}\) = \(\frac{q}{10}\)
By comparing LHS and RHS
q = 11 and p = 2
Hence, from the above,
We can conclude that the missing numbers are: 2 and 11

9.8 Compare Factors and Products

Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 42.
1\(\frac{1}{2}\) × 3
Answer:
The value of 1\(\frac{1}{2}\) × 3 is greater than 3

Explanation:
The given numbers are: 3 and 1\(\frac{1}{2}\)
The representation of 1\(\frac{1}{2}\) in the mixed form is: \(\frac{3}{2}\)
So,
\(\frac{3}{2}\) is greater than 1
So,
\(\frac{3}{2}\) × 3 is greater than 3
Hence, from the above,
We can conclude that the value of 1\(\frac{1}{2}\) × 3 is greater than 3

Question 43.
\(\frac{1}{8}\) × \(\frac{1}{6}\)
Answer:
The value of \(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1

Explanation:
The given numbers are: \(\frac{1}{8}\) and \(\frac{1}{6}\)
So,
\(\frac{1}{8}\) is less than 1
\(\frac{1}{6}\) is less than 1
So,
\(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1
Hence, from the above,
We can conclude that the value of \(\frac{1}{8}\) × \(\frac{1}{6}\) is less than 1

Question 44.
\(\frac{5}{5}\) × 2\(\frac{4}{7}\)
Answer:
The value of \(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1

Explanation:
The given numbers are: \(\frac{5}{5}\) and 2\(\frac{4}{7}\)
The representation of 2\(\frac{4}{7}\) in the improper fraction form is: \(\frac{18}{7}\)
So,
\(\frac{18}{7}\) is greater than 1
\(\frac{5}{5}\) is 1
So,
\(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{5}{5}\) × 2\(\frac{4}{7}\) is greater than 1

Question 45.
\(\frac{3}{4}\) × 2\(\frac{11}{12}\)
Answer:
The value of \(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1

Explanation:
The given numbers are: \(\frac{3}{4}\) and 2\(\frac{11}{12}\)
The representation of 2\(\frac{11}{12}\) in the improper fraction form is: \(\frac{35}{12}\)
So,
\(\frac{35}{12}\) is greater than 1
\(\frac{3}{4}\) is less than 1
So,
\(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{3}{4}\) × 2\(\frac{11}{12}\) is greater than 1

Question 46.
3\(\frac{1}{9}\) × 2\(\frac{7}{8}\)
Answer:
The value of 3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1

Explanation:
The given mixed fractions are: 3\(\frac{1}{9}\) and 2\(\frac{7}{8}\)
The representation of 3\(\frac{1}{9}\) and 2\(\frac{7}{8}\) in the improper fractions form is: \(\frac{28}{9}\) and \(\frac{23}{8}\)
So,
\(\frac{28}{9}\) is greater than 1
\(\frac{23}{8}\) is greater than 1
So,
3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1
Hence, from the above,
We can conclude that the value of 3\(\frac{1}{9}\) × 2\(\frac{7}{8}\) is greater than 1

Question 47.
\(\frac{9}{8}\) × \(\frac{5}{2}\)
Answer:
The value of \(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1

Explanation:
The given fractions are: \(\frac{9}{8}\) and \(\frac{5}{2}\)
So,
\(\frac{9}{8}\) is greater than 1
So,
\(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1
Hence, from the above,
We can conclude that the value of \(\frac{9}{8}\) × \(\frac{5}{2}\) is greater than 1

Conclusion:

Hope the information provided in the above answer key is beneficial for all the students of grade 5. Follow our Big Ideas Math Answers and make your learning fun and interesting.

Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions

Download the Big Ideas Math Book Answer Key Grade 5 Chapter 10 Divide Fractions free of cost and kick start your preparation immediately. You will get the necessary skill set needed to solve the problems related to fractions here. Access Detailed Solutions to all the problems and learn how to solve related problems when you encounter them during your exams. Seek Homework Help needed by accessing the Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions.

Big Ideas Math Book Answer Key Grade 5 Chapter 10 Divide Fractions

Cross Check the Solutions from our Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions and understand the areas you are facing difficulty. Score higher grades in your exams and refer to Big Ideas Math Book Solution Key Grade 5 Chapter 10 Divide Fractions to have strong command over fractions.

Lesson: 1 Interpret Fractions as Division

Lesson: 2 Mixed Numbers as Quotients

Lesson: 3 Divide Whole Numbers by Unit Fractions

Lesson: 4 Divide Unit Fractions by Whole Numbers

Lesson: 5 Problem Solving: Fraction Division

Chapter: 10 – Divide Fractions

Lesson 10.1 Interpret Fractions as Division

Explore and Grow`

You share 4 sheets of construction paper equally among 8 people. Write a division expression that represents the situation. What fraction of a sheet of paper does each person get? Use a model to support your answer?
Answer:
The division expression that represents the fraction of a sheet of paper does each person get is:
4 ÷ 8 = \(\frac{1}{2}\)

Explanation:
It is given that you have 4 sheets of construction paper equally among 8 people.
Hence,
The division expression that represents the fraction of a sheet of paper is:
( The number of sheets of construction paper ) ÷ ( The number of people )
= 4 ÷ 8
= \(\frac{1}{2}\)
Hence, from the above,
We can conclude that the fraction of a sheet of paper does each person get is: \(\frac{1}{2}\)

Structure
How can you check your answer using multiplication?
Answer: We can check the answer by using the partial products method or by using the simplification method.

Think and Grow: Divide Whole Numbers
You can use models to divide whole numbers that have a fraction as the quotient.
Answer: 
From the above model,
The number of colored parts is: 4
The total number of parts are: 8
So,
The fraction of the colored part out of the total number of parts = 4 ÷ 8
= \(\frac{4}{8}\) = \(\frac{1}{2}\)
In \(\frac{1}{2}\),
1 represents the quotient
2 represents the remainder
Example
Find 2 ÷ 3.
One Way: Use a tape diagram. Show 2 wholes. Divide each whole into 3 equal parts.

Another Way: Use an area model. Show 2 wholes. Divide each whole into 3 equal parts. Then separate the parts into 3 equal groups.

Show and Grow

Divide. Use a model to help

Question 1.
2 ÷ 4 =0.5
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 4
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.5

Question 2.
1 ÷ 3 = 0.33
Answer:

From the above model,
The number of colored parts is: 1
The number of total parts is: 3
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 3
= \(\frac{1}{3}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.33

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
1 ÷ 8 =0.018
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 8
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 8
= \(\frac{1}{8}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.018

Question 4.
1 ÷ 4 =0.25
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 4
= \(\frac{1}{4}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.25

Question 5.
2 ÷ 6 =0.33
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 6
= \(\frac{2}{6}\)
= \(\frac{1}{3}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.33

Question 6.
2 ÷ 5 = 0.4
Answer:
From the above model,
The number of colored parts is: 2
The number of total parts is: 5
So,
The fraction of the colored parts out of the total number of parts = 2 ÷ 5
= \(\frac{2}{5}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.4

Question 7.
3 ÷ 7 = 0.42
Answer: 
From the above model,
The number of colored parts is: 3
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 3 ÷ 7
= \(\frac{3}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.42

Question 8.
5 ÷ 6 = 0.83
Answer:
From the above model,
The number of colored parts is: 5
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 5 ÷ 6
= \(\frac{5}{6}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.83

Question 9.
How many 6s are in 1?
Answer: There are six \(\frac{1}{6}\)s in 1

Explanation:
The number of 6s in 1 can be obtained by dividing 1 into 6 equal parts.
So,
The figure obtained will be like;

From the above model,
The number of colored parts is: 1
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 6
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that there are six 6s in 1

Question 10.
How many 10s are in 9?
Answer: There are 9 \(\frac{9}{10}\)s in 9

Explanation:
The model for the number of 10s in 9 are:

From the above model,
The number of colored parts is: 9
The number of total parts is: 10
So,
The fraction of the colored parts out of the total number of parts = 9 ÷ 10
= \(\frac{9}{10}\)
Hence, from the above,
We can conclude that there are nine 9s in 10

Question 11.
Number Sense
For which equations does k = 8?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 3
Answer: Let the equations named A), B), C), and D)
So,
The four equations are:
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 3
From the above equations,
The value ‘k’ must be in the numerator.
So,
In A), the value of the numerator is: 3
In B), the value of the numerator is: k
In C), the value of the numerator is: 2
In D) the value of the numerator is: 8
So,
From the above numerator values,
We can say that “k=8” holds good for Equation B)

Question 12.
Writing
Write and solve a real-life problem for 7 ÷ 12.
Answer:
From the above model,
The number of colored parts is: 7
The number of total parts is: 12
So,
The fraction of the colored parts out of the total number of parts = 7 ÷ 12
= \(\frac{7}{12}\)
Hence,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.58

Think and Grow: Modeling Real Life

Example
Three fruit bars are shared equally among 4 friends. What fraction of a fruit bar does each friend get?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 4
Divide 3 by 4 to find what fraction of a fruit bar each friend gets.
Use an area model to find 3 ÷ 4. Show 3 whole fruit bars. Divide each fruit bar into 4 equal parts. Then separate the parts into 4 equal groups.

Show and Grow

Question 13.
You cut a 5-foot streamer into 6 pieces of equal size. What is the length of each piece in feet? in inches?
Answer: The length of each piece in feet is: \(\frac{5}{6}\)

Explanation:
It is given that you cut a 5-foot streamer into 6 equal pieces of equal size.
So,
The model representing the 6 equal pieces of the 5-foot streamer is:

From the above model,
We can see that each part in the model represents \(\frac{5}{6}\) of each part.
Hence, from the above,
We can conclude that the length of each piece of a 5-foot streamer in feet is: \(\frac{5}{6}\)

Question 14.
Four circular lemon slices are shared equally among 8 glasses of water. What fraction of a lemon slice does each glass get?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 7
Answer: The fraction of a lemon slice does each glass get is: \(\frac{1}{2}\)

Explanation:
It is given that the four circular lemon slices are shared equally among 8 glasses of water.
So,
The model representing the portion that each glass get is:

From the above model,
We can say that each part represents \(\frac{1}{2}\) part
Hence, from the above,
We can conclude that the portion of a lemon slice does glass get is: \(\frac{1}{2}\)

Question 15.
You cut a 5-foot streamer into 6 pieces of equal size. What is the length of each piece in feet? in inches?
Answer: The length of each piece in feet is: \(\frac{5}{6}\)

Explanation:
It is given that you cut a 5-foot streamer into 6 equal pieces of equal size.
So,
The model representing the 6 equal pieces of the 5-foot streamer is:

From the above model,
We can see that each part in the model represents \(\frac{5}{6}\) of each part.
Hence, from the above,
We can conclude that the length of each piece of a 5-foot streamer in feet is: \(\frac{5}{6}\)

Question 16.
DIG DEEPER!
A fruit drink is made using \(\frac{7}{4}\) quarts of orange juice and \(\frac{5}{4}\) quarts of pineapple juice. The drink is shared equally among 12 guests. What fraction of a quart does each guest get?
Answer: The fraction of a quart does each guest get is: \(\frac{1}{4}\)

Explanation:
It is given that a fruit drink is made using \(\frac{7}{4}\) quarts of orange juice and \(\frac{5}{4}\) quarts of pineapple juice.
So,
The total amount of fruit juice= \(\frac{7}{4}\) + \(\frac{5}{4}\)
= \(\frac{ 7 + 5}{4}\)
= \(\frac{12}{4}\)
It is also given that the drink is shared equally among 12 guests
So,
The fraction of a quart does each gust get = \(\frac{12}{4}\) ÷ 12
= \(\frac{12}{4}\) ÷ \(\frac{12}{1}\)
= \(\frac{12}{4}\) × \(\frac{1}{12}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that the fraction of a quart does each person get is: \(\frac{1}{4}\)

Interpret Fractions as Division Homework & Practice 10.1

Divide. Use a model to help.

Question 1.
1 ÷ 6 =0.16
Answer:


From the above model,
The number of colored parts is: 1
The number of total parts is: 6
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 6
= \(\frac{1}{6}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.16

Question 2.
1 ÷ 7 =0.14
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 7
= \(\frac{1}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.14

Question 3.
1 ÷ 5 = 0.20
Answer:
From the above model,
The number of colored parts is: 1
The number of total parts is: 5
So,
The fraction of the colored parts out of the total number of parts = 1 ÷ 5
= \(\frac{1}{5}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.20

Question 4.
3 ÷ 4 = 0.75
Answer:
From the above model,
The number of colored parts is: 3
The number of total parts is: 4
So,
The fraction of the colored parts out of the total number of parts = 3 ÷ 4
= \(\frac{3}{4}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.75

Question 5.
6 ÷ 7 = 0.85
Answer:
From the above model,
The number of colored parts is: 6
The number of total parts is: 7
So,
The fraction of the colored parts out of the total number of parts = 6 ÷ 7
= \(\frac{6}{7}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.85

Question 6.
5 ÷ 9 = 0.55
Answer:
From the above model,
The number of colored parts is: 5
The number of total parts is: 9
So,
The fraction of the colored parts out of the total number of parts = 5 ÷ 9
= \(\frac{5}{9}\)
So,
The fraction of the colored parts out of the total number of parts in the decimal form is: 0.55

Question 7.
YOU BE THE TEACHER
Your friend says \(\frac{5}{12}\) is equivalent to 12 ÷ 5. Is your friend correct? Explain.
Answer: No, your friend s not correct.

Explanation:
The given fraction is: \(\frac{5}{12}\)
From the given fraction,
The numerator is: 5
The denominator is: 12
We can write a fraction in the following form:
Fraction = \(\frac{Numerator}{Denominator}\)
So,
\(\frac{5}{12}\) is equivalent to 5 ÷ 12
But, according to your friend,
\(\frac{5}{12}\) is equivalent to 12 ÷ 5
Hence, from the above,
we can conclude that your friend is not correct.

Question 8.
Writing
Explain how fractions and division are related.

Question 9.
Structure
Write a division equation represented by the model.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8
Answer:
The division equation represented by the model is: 1 ÷ 4

Explanation:
The given model is:
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 8
From the given model,
The number of shaded parts is: 1
The total number of parts are: 4
So,
The division equation can be represented as:
Division equation = (The number of shaded parts) ÷ ( The total number of parts )
= 1 ÷ 4
= \(\frac{1}{4}\)

Question 10.
Number Sense
Eight friends share multiple vegetable pizzas, and each gets \(\frac{3}{8}\) of a pizza. How many pizzas do they share?
Answer: The total number of pizzas the eight friends shared are: 3 pizzas

Explanation:
It is given that the eight friends share multiple vegetable pizzas and each gets \(\frac{3}{8}\) of a pizza.
So,
The total number of pizzas shared by the eight friends = \(\frac{3}{8}\) × 8
= \(\frac{3}{8}\) × \(\frac{8}{1}\)
= \(\frac{3 × 8}{8 × 1}\)
= \(\frac{3}{1}\)
= 3
Hence, from the above,
We can conclude that the total number of pizzas shared by the eight friends is: 3 pizzas

Question 11.
Modeling Real Life
Seven friends each run an equal part of a 5-kilometer relay race. What fraction of a kilometer does each friend complete?
Answer: The fraction of a kilometer does each friend complete is: \(\frac{5}{7}\) kilometer

Explanation:
It is given that there are seven friends each run an equal part of a 5-kilometer relay race.
So,
The fraction that each friend run = \(\frac{The total distance} {The number of friends}\)
= \(\frac{5}{7}\)
Hence, from the above,
We can conclude that the fraction of a kilometer does each friend complete is: \(\frac{5}{7}\) kilometer

Question 12.
Modeling Real Life
A group of friends equally share 3 bags of pretzels. Each friend gets \(\frac{3}{5}\) of a bag of pretzels. How many friends are in the group?
Answer: The total number of friends in the group are: 5

Explanation:
It is given that a group of friends equally share 3 bags of pretzels and each friend gets \(\frac{3}{5}\) of a bag of pretzels.
So,
The total number of friends = \(\frac{The total number of bags}{The amount each friend gets}\)
= \(\frac{3}{1}\) × \(\frac{5}{3}\)
= \(\frac{5}{1}\)
= 5
Hence, from the above,
We can conclude that the total number of friends are: 5

Review & Refresh

Multiply.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 9
Answer: 9 × \(\frac{2}{3}\) = 6

Explanation:
The given fractions are: \(\frac{9}{1}\) and \(\frac{2}{3}\)
So,
\(\frac{9}{1}\) × \(\frac{2}{3}\)
= \(\frac{9 × 2}{1 × 3}\)
= \(\frac{6}{1}\)
= 6
Hence,
9 × \(\frac{2}{3}\) = 6

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 10
Answer: 5 × \(\frac{7}{10}\) = \(\frac{7}{2}\)

Explanation:
The given fractions are: \(\frac{5}{1}\) and \(\frac{7}{10}\)
So,
\(\frac{5}{1}\) × \(\frac{7}{10}\)
= \(\frac{5 × 7}{1 × 10}\)
= \(\frac{7}{2}\)
Hence,
5 × \(\frac{7}{10}\) = \(\frac{7}{2}\)

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 11
Answer: 3 × \(\frac{5}{12}\) = \(\frac{5}{4}\)

Explanation:
The given fractions are: \(\frac{3}{1}\) and \(\frac{5}{12}\)
So,
\(\frac{3}{1}\) × \(\frac{5}{12}\)
= \(\frac{3 × 5}{1 × 12}\)
= \(\frac{5}{4}\)
Hence,
3 × \(\frac{5}{12}\) = \(\frac{5}{4}\)

Lesson 10.2 Mixed Numbers as Quotients

Explore and Grow

You share 6 sheets of construction paper equally among 4 people. Write a division expression that represents the situation. How much paper does each person get? Use a model to support your answer.
Answer:
The division expression representing the situation is: 6 ÷ 4

Explanation:
It s given that you have shared 6 sheets of construction paper equally among 4 people
So,
The division equation representing the sharing of construction papers is: 6 ÷ 4
Now,
6 ÷ 4 = \(\frac{6}{4}\)
So,
The above equation represents that 4 is divided into 6 parts.
So,
The model representing the situation is:

From the above model,
We can say that the amount of does each person get is: 1\(\frac{1}{2}\) or 1.5 or \(\frac{3}{2}\)

Precision
Does each person get less than or more than 1 sheet of paper? Use the dividend and divisor to explain why your answer makes sense.
Answer:
From the above problem,
We can say that each person gets more than 1 paper.
So,
The division equation of the above problem is: 6 ÷ 4
The equivalent form of 6 ÷ 4 is: \(\frac{6}{4}\)
Now,
The simplest form of \(\frac{6}{4}\) is: \(\frac{3}{2}\) ( The simplest form is the division of the numerator and the denominator with the common multiple if we can divide)
The mixed form of \(\frac{3}{2}\) is: 1\(\frac{1}{2}\)

Think and Grow: Divide Whole Numbers

You can use models to divide whole numbers that have a mixed number as the quotient.
Example
Find 3 ÷ 2.
One Way:
Use a tape diagram. Show 3 wholes. Divide each whole into 2 equal parts.

Another Way: Use an area model. Show 3 wholes. Divide each whole into 2 equal parts. Then separate the parts into 2 equal groups.

Show and Grow

Divide. Use a model to help

Question 1.
5 ÷ 3 = ___
Answer: 5 ÷ 3 = 1\(\frac{2}{3}\)

Explanation;
The given division equation is: 5 ÷ 3
The model representing the division equation is:

From the above model,
5 ÷ 3 = 3 ÷ 3
= 1 R 2
Hence,
We can say that each part is divided into 1\(\frac{2}{3}\) or \(\frac{5}{3}\)

Question 2.
7 ÷ 2 = ___

Answer: 7 ÷ 2 = 3\(\frac{1}{2}\)

Explanation;
The given division equation is: 7 ÷ 2
The model representing the division equation is:

From the above model,
7 ÷ 2 = 6 ÷ 2
= 3 R 1
Hence,
We can say that each part is divided into 3\(\frac{1}{2}\) or \(\frac{7}{2}\) or 3.5

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
12 ÷ 7 = ___

Answer: 12 ÷ 7 = 1\(\frac{5}{7}\)

Explanation;
The given division equation is: 12 ÷ 7
The model representing the division equation is:

From the above model,
12 ÷ 7 = 7 ÷ 7
= 1 R 5
Hence,
We can say that each part is divided into 1\(\frac{5}{7}\) or \(\frac{12}{7}\)

Question 4.
25 ÷ 20 = ___

Answer: 25 ÷ 20 = 1\(\frac{5}{20}\) = \(\frac{5}{4}\)

Explanation;
The given division equation is: 25 ÷ 20
The model representing the division equation is:

From the above model,
25 ÷ 20 = 20 ÷ 20
= 1 R 5
Hence,
We can say that each part is divided into 1\(\frac{5}{20}\) or \(\frac{5}{4}\)

Question 5.
15 ÷ 4 = ___

Answer: 15 ÷ 4 = 3\(\frac{3}{4}\)

Explanation;
The given division equation is: 15 ÷ 4
The model representing the division equation is:

From the above model,
15 ÷ 4 = 12 ÷ 4
= 3 R 3
Hence,
We can say that each part is divided into 3\(\frac{3}{4}\) or \(\frac{15}{4}\)

Question 6.
13 ÷ 6 = ___

Answer: 13 ÷ 6 = 2\(\frac{1}{6}\)

Explanation;
The given division equation is: 13÷ 6
The model representing the division equation is:

From the above model,
13 ÷ 6 = 12 ÷ 6
= 2 R 1
Hence,
We can say that each part is divided into 2\(\frac{1}{6}\) or \(\frac{13}{6}\)

Question 7.
16 ÷ 8 = ___

Answer: 16 ÷ 8 = 2

Explanation;
The given division equation is: 16÷ 8
The model representing the division equation is:

From the above model,
16 ÷ 8
= 2 R 0
Hence,
We can say that each part is divided into 2 equal parts

Question 8.
92 ÷ 50 = ___

Answer: 92 ÷ 50 = 1\(\frac{21}{25}\)

Explanation;
The given division equation is: 92÷ 50
So,
92 ÷ 50 = 50 ÷ 50
= 1 R 42
Hence,
We can say that each part is divided into 1\(\frac{42}{50}\) or 1\(\frac{21}{25}\)

Question 9.
How many 3s are in 7?
Answer: The number of 3 in 7 are: \(\frac{7}{3}\) or 2\(\frac{1}{3}\)

Explanation:
The division equation is: 7 ÷ 3
So,
The model for the given division equation is:

From the above model,
7 ÷ 3 = 6 ÷ 3
= 2 R 1
Hence, from the above,
We can conclude that there are 2\(\frac{1}{3}\) 3s in 7

Question 10.
How many 6s are in 21?
Answer: The number of 6s in 21 are: \(\frac{21}{6}\) or 3\(\frac{3}{6}\)

Explanation:
The division equation is: 21 ÷ 6
So,
The model for the given division equation is:

From the above model,
21 ÷ 6 = 18 ÷ 6
= 3 R 3
Hence, from the above,
We can conclude that there are 3\(\frac{3}{6}\) 3s in 21

Question 11.
YOU BE THE TEACHER
Your friend says that \(\frac{35}{6}\) is equivalent to 35 ÷ 6. Is your friend correct? Explain.
Answer: Yes, your friend is correct

Explanation:
It is given that \(\frac{35}{6}\)
We know that,
The decimal equation can be converted into a fraction as \(\frac{Numerator}{Denominator}\)
So,
\(\frac{35}{6}\) = 35 ÷ 6
Hence, from the above,
We can conclude that your friend is correct

Question 12.
Writing
Write and solve a real-life problem for 24 ÷ 5.
Answer: 24 ÷ 5 = 4\(\frac{4}{5}\)

Explanation;
The given division equation is: 24÷ 5
The model for the above division equation is:

From the above model,
24 ÷ 5 = 20 ÷ 5
= 4 R 4
Hence,
We can say that each part is divided into 4\(\frac{4}{5}\)

Think and Grow: Modeling Real Life

Example
You share 7 bales of hay equally among 3 horse stalls. How many whole bales are in each stall? What fractional amount of a bale is in each stall?
Divide 7 by 3 to find how many bales of hay are in each stall. Use an area model to help.

Show and Grow

Question 13.
Six muffins are shared equally among 4 friends. How many whole muffins does each friend get? What fractional amount of a muffin does each friend get?
Answer: Each friend will get 1 muffin and 2 muffins are leftovers
The fractional part of a muffin does each friend get is: \(\frac{1}{2}\)

Explanation:
It is given that there are six muffins are shared equally among 4 friends.
So,
The number of muffins each friend get = 6 ÷ 4
= 4 ÷ 4
= 1 R 2
Hence, from the above,
We can conclude that each friend gets 1 muffin each and the fraction of each muffin get is: \(\frac{1}{2}\)

Question 14.
A cyclist bikes 44 miles in 5 days. She bikes the same distance each day. Does she bike more than 8\(\frac{1}{2}\) miles each day? Explain.
Answer: She bikes more than 8\(\frac{1}{2}\) miles each day.

Explanation:
It is given that a cyclist bikes 44 miles in 5 days.
So,
The distance that she bikes each day = 44 ÷ 5
So,
44 ÷ 5 = 40 ÷ 5
= 8 R 4
= 8\(\frac{4}{5}\) miles
But, it is given that she bikes 8\(\frac{1}{2}\) miles each day
Hence, from the above,
We can conclude that she bikes more than 8\(\frac{1}{2}\) miles each day.

Question 15.
DIG DEEPER!
At Table A, 4 students share 7 packs of clay equally. At Table B, 5 students share 8 packs of clay equally. At which table does each student get a greater amount of clay? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 15
Answer: At Table A, each student gets a greater amount of clay.

Explanation:
It is given that at Table A, 4 students share 7 packs of clay equally.
So,
The representation of clay at table A is: \(\frac{7}{4}\)
It is also given that at Table B, 5 students share 8 packs of clay equally.
So,
The representation of clay at table B is: \(\frac{8}{5}\)
So,
For comparison, equate the denominators.
So,
Multiply the first fraction at table A by \(\frac{5}{5}\) and the fraction at table B by \(\frac{4}{4}\)
So,
\(\frac{7}{4}\) × \(\frac{5}{5}\)
= \(\frac{35}{20}\)
So,
\(\frac{8}{5}\) × \(\frac{4}{4}\)
= \(\frac{32}{20}\)
Hence, from the above,
We can conclude that at table A, the students will get more amount of clay.

Mixed Numbers as Quotients Homework & Practice 10.2

Divide. Use a model to help.

Question 1.
5 ÷ 2 = ___
Answer: 5 ÷ 2 = 2\(\frac{1}{2}\)

Explanation;
The given division equation is: 5 ÷ 2
The model representing the division equation is:

From the above model,
5 ÷ 2 = 4 ÷ 2
= 2 R 1
Hence,
We can say that 5 ÷ 2 = 2\(\frac{1}{2}\) or 2.5 or \(\frac{5}{2}\)

Question 2.
10 ÷ 7 = ___
Answer: 10 ÷ 7 = 1\(\frac{3}{7}\) = \(\frac{10}{7}\)

Explanation;
The given division equation is: 10 ÷ 7
The model representing the division equation is:

From the above model,
10 ÷ 7 = 7 ÷ 7
= 1 R 3
Hence,
We can say that 10 ÷ 7 = 1\(\frac{3}{7}\) or \(\frac{10}{7}\)

Question 3.
3 ÷ 9 = ___
Answer: 3 ÷ 9 = \(\frac{1}{3}\)

Explanation;
The given division equation is: 3 ÷ 9
The model representing the division equation is:

From the above model,
3 and 9 are the multiples of 3.
So,
3 ÷ 9 = \(\frac{1}{3}\)
Hence,
We can say that 3 ÷ 9 = \(\frac{1}{3}\)

Question 4.
11 ÷ 4 = ___
Answer: 11 ÷ 4 = 2\(\frac{3}{4}\)

Explanation;
The given division equation is: 11 ÷ 4
The model representing the division equation is:

From the above model,
11 ÷ 4 = 8 ÷ 4
= 2 R 3
Hence,
We can say that 11 ÷ 4 = \(\frac{11}{4}\) or 2\(\frac{3}{4}\)

Question 5.
13 ÷ 6 = ___
Answer: 13 ÷ 6 = 2\(\frac{1}{6}\)

Explanation;
The given division equation is: 13 ÷ 6
The model representing the division equation is:

From the above model,
13 ÷ 6 = 12 ÷ 6
= 2 R 1
Hence,
We can say that 13 ÷ 6 = \(\frac{13}{6}\) or 2\(\frac{1}{6}\)

Question 6.
45 ÷ 8 = ___
Answer: 45 ÷ 8 = 5\(\frac{5}{8}\)

Explanation;
The given division equation is: 45 ÷ 8
The model representing the division equation is:

From the above model,
45 ÷ 8 = 40 ÷ 8
= 5 R 5
Hence,
We can say that 45 ÷ 8 = \(\frac{45}{8}\) or 5\(\frac{5}{8}\)

Question 7.
Number Sense
Between which two whole numbers is the quotient of 74 and 9?
Answer: The quotient of 74 and 9 is between 8 and 9

Explanation:
The given two numbers are 7 and 9
So,
By using the partial quotients method,
74 ÷ 9= 72 ÷ 9
= 8 R 2
So,
74 ÷ 9 = \(\frac{74}{9}\) or 8\(\frac{2}{9}\) or 8.3
Hence, from the above,
We can conclude that the quotient of 74 and 9 is between 8 and 9

Question 8.
Reasoning
Three friends want to share 22 baseball cards. For this situation, why does the quotient 7 R1 make more sense than the quotient 7\(\frac{1}{3}\)?
Answer:
It is given that three friends want to share 22 baseball cards.
So,
We have to find the number of baseball cards each friend possesses.
So,
It is sufficient to write the number of baseball cards possessed by each friend in the remainder form rather than the fraction form.
So,
The number of baseball cards possessed by each friend = \(\frac{The total number of baseball cards}{The number of friends}\)
= 22 ÷ 3
= 21 ÷ 3
= 7 R 1
Hence, from the above,
We can conclude that the remainder form is sufficient to find the number of baseball cars possessed by each friend rather than the fraction form.

Question 9.
DIG DEEPER!
Is \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5? Explain.
Answer: Yes, \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5

Explanation:
The given fraction and the number is: \(\frac{2}{5}\) and 3
So,
\(\frac{2}{5}\) × 3 = \(\frac{2}{5}\) × \(\frac{3}{1}\)
= \(\frac{2 × 3}{5}\)
= 2 × 3 ÷ 5
Hence, from the above,
We can conclude that \(\frac{2}{5}\) × 3 equivalent to 2 × 3 ÷ 5

Question 10.
Modeling Real Life
A bag of 4 balls weighs 6 pounds. Each ball weighs the same amount. What is the weight of each ball?
Answer: The weight of each ball is: \(\frac{3}{2}\) pounds or 1.5 pounds

Explanation:
It is given that a bag of 4 balls weighs 6 pounds
So,
The weight of each ball = \(\frac{The total weight of the balls}{The number of balls}\)
= 6 ÷ 4
Since 6 and 4 are the multiples of 2, divide the two numbers by 2
So,
6 ÷ 4 = 3 ÷ 2
So,
3 ÷ 2 = 2 ÷ 2
= 1 R 1
= 1\(\frac{1}{2}\) pounds
Hence, from the above,
We can conclude that the weight of each ball is: 1\(\frac{1}{2}\) pounds or 1.5 pounds

Question 11.
Modeling Real Life
Zookeepers order 600 pounds of bamboo for the pandas. The bamboo lasts 7 days. How many whole pounds of bamboo do the pandas eat each day? What fractional amount of a pound do the pandas eat each day?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 17
Answer:
The amount of bamboos the pandas eat each day is around 85 pounds
The amount of bamboos the pandas eat each day in the fraction form is: 85\(\frac{5}{7}\)

Explanation:
It is given that zookeepers order 600 pounds of bamboo for the pandas and the bamboos last 7 days for the pandas
So,
The number of bamboos the pandas eat each day = 600 ÷ 7
So,
By using the partial quotients method,
600 ÷ 7 = ( 560 + 35 ) ÷ 7
= ( 560 ÷ 7 ) + ( 35 ÷ 7 )
= 80 + 5
= 85 R 5
Hence, from the above,
We can conclude that
The amount of bamboos the pandas eat each day is around 85 pounds
The amount of bamboos the pandas eat each day in the fraction form is: 85\(\frac{5}{7}\)

Question 12.
Modeling Real Life
A plumber has 20 feet of piping. He cuts the piping into 6 equal pieces. Is each piece greater than, less than, or equal to 3\(\frac{1}{2}\) feet?
Answer: Each piece is less than 3\(\frac{1}{2}\) feet

Explanation:
It is given that a plumber has 20 feet of piping and he cuts the piping into 6 equal pieces.
So,
The length of each piece = 20 ÷ 6
By using the partial quotients method,
20 ÷ 6 = 18 ÷ 6
= 3 R 2
So,
20 ÷ 6 = 3\(\frac{2}{6}\)
Now,
3\(\frac{1}{2}\) = \(\frac{7}{2}\)
3\(\frac{2}{6}\) = \(\frac{20}{6}\)
For comparison, we have to equate whether the denominators or the numerators.
So,
Multiply 3\(\frac{1}{2}\) with \(\frac{3}{3}\)
So,
3\(\frac{1}{2}\) = \(\frac{21}{6}\)
Hence, from the above,
We can conclude that each piece is less than 3\(\frac{1}{2}\) feet

Review & Refresh

Add.

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 18
Answer: \(\frac{2}{9}\) + \(\frac{2}{3}\) = \(\frac{8}{9}\)

Explanation:
The two given fractions are: \(\frac{2}{9}\) and \(\frac{2}{3}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{2}{3}\)  with \(\frac{3}{3}\)
So,
\(\frac{2}{3}\)  = \(\frac{6}{9}\)
Hence, from the above,
We can conclude that \(\frac{2}{9}\) + \(\frac{2}{3}\) = \(\frac{8}{9}\)

Question 14.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 19
Answer: \(\frac{1}{10}\) + \(\frac{3}{4}\) = \(\frac{34}{40}\)

Explanation:
The two given fractions are: \(\frac{1}{10}\) and \(\frac{3}{4}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{1}{10}\)  with \(\frac{4}{4}\)
Multiply \(\frac{3}{4}\)  with \(\frac{10}{10}\)
So,
\(\frac{1}{10}\)  = \(\frac{4}{40}\)
\(\frac{3}{4}\)  = \(\frac{30}{40}\)
Hence, from the above,
We can conclude that \(\frac{1}{10}\) + \(\frac{3}{4}\) = \(\frac{34}{40}\)

Question 15.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 20
Answer: \(\frac{3}{5}\) + \(\frac{5}{6}\) + \(\frac{1}{5}\) = \(\frac{49}{30}\)

Explanation:
The three given fractions are: \(\frac{3}{5}\) , \(\frac{5}{6}\) and \(\frac{1}{5}\)
So, in addition, we have to make either the numerators or the denominators equal
So,
Multiply \(\frac{3}{5}\)  with \(\frac{6}{6}\)
Multiply \(\frac{5}{6}\)  with \(\frac{5}{5}\)
Multiply \(\frac{1}{5}\)  with \(\frac{6}{6}\)
So,
\(\frac{3}{5}\)  = \(\frac{18}{30}\)
\(\frac{5}{6}\)  = \(\frac{25}{30}\)
\(\frac{1}{5}\)  = \(\frac{6}{30}\)
Hence, from the above,
We can conclude that \(\frac{3}{5}\) + \(\frac{5}{6}\) +\(\frac{1}{5}\)  = \(\frac{49}{30}\)

Lesson 10.3 Divide Whole Numbers by Unit Fractions

Explore and Grow

Write a real-life problem that can be represented by 6 ÷ \(\frac{1}{2}\)?
Answer:
Suppose, we have an apple and there are 6 children and we are giving each child half of the piece.
So,
Each child receives 6 ÷ \(\frac{1}{2}\) piece of the apple

What is the solution to the problem? Use a model to support your answer?
Answer:
The above problem is the division of an apple among the six children
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
The amount each child receive from an apple = 6 ÷ \(\frac{1}{2}\)
= 6 × \(\frac{2}{1}\)
= \(\frac{6}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 6 × 2}{1 × 1}\)
= 12

Structure
How can you check your answer using multiplication?
Answer:
We can check the answer using multiplication by the two rules regarding division and multiplication. They are:
A) a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
B) a= \(\frac{a}{1}\)

Think and Grow: Divide Whole Numbers by Unit Fractions

You can use models to divide whole numbers by unit fractions.
Example
Find 4 ÷ \(\frac{1}{3}\)
One Way:
Use a tape diagram to find how many \(\frac{1}{3}\)s are in 4. There are 4 wholes.
Divide each whole into 3 equal parts. Each part is \(\frac{1}{3}\).
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 21
Because there are 3 one-thirds in 1 whole, there are
4 × 3 equal parts = 12 one-thirds in 4 wholes.

Show and Grow

Divide. Use a model to help

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 23
Answer: 3 ÷ \(\frac{1}{2}\) = 6

Explanation:
The given numbers are: 3 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
3 ÷ \(\frac{1}{2}\)  = 3 × \(\frac{2}{1}\)
= \(\frac{3}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 3 × 2}{1 × 1}\)
= 6
Hence,
3÷ \(\frac{1}{2}\) = 6

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 24
Answer: 2 ÷ \(\frac{1}{5}\) = 10

Explanation:
The given numbers are: 2 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{5}\)  = 2 × \(\frac{5}{1}\)
= \(\frac{5}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 5 × 2}{1 × 1}\)
= 10
Hence,
2÷ \(\frac{1}{5}\) = 10

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.

Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 25
Answer: 1 ÷ \(\frac{1}{3}\) = 3

Explanation:
The given numbers are: 1 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
1 ÷ \(\frac{1}{3}\)  = 1 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{1}{1}\)
= \(\frac{ 3 × 1}{1 × 1}\)
= 3
Hence,
1÷ \(\frac{1}{3}\) = 3

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 26
Answer: 3 ÷ \(\frac{1}{5}\) = 15

Explanation:
The given numbers are: 3 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
3 ÷ \(\frac{1}{5}\)  = 3 × \(\frac{5}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 3 × 5}{1 × 1}\)
= 15
Hence,
3÷ \(\frac{1}{5}\) = 15

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 27
Answer: 5 ÷ \(\frac{1}{3}\) = 15

Explanation:
The given numbers are: 5 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
5 ÷ \(\frac{1}{3}\)  = 5 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 3 × 5}{1 × 1}\)
= 15
Hence,
5÷ \(\frac{1}{3}\) = 15

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 28
Answer: 4 ÷ \(\frac{1}{4}\) = 16

Explanation:
The given numbers are: 4 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{4}\)  = 4 × \(\frac{4}{1}\)
= \(\frac{4}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 4 × 4}{1 × 1}\)
= 16
Hence,
4÷ \(\frac{1}{4}\) = 16

Question 7.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 29
Answer: 7 ÷ \(\frac{1}{2}\) = 14

Explanation:
The given numbers are: 7 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{2}\)  = 7 × \(\frac{2}{1}\)
= \(\frac{7}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 7 × 2}{1 × 1}\)
= 14
Hence,
7÷ \(\frac{1}{2}\) = 14

Question 8.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 30
Answer: 2 ÷ \(\frac{1}{7}\) = 14

Explanation:
The given numbers are: 2 and \(\frac{1}{7}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{7}\)  = 2 × \(\frac{7}{1}\)
= \(\frac{2}{1}\) × \(\frac{7}{1}\)
= \(\frac{ 7 × 2}{1 × 1}\)
= 14
Hence,
2÷ \(\frac{1}{7}\) = 14

Question 9.
How many \(\frac{1}{4}\)s are in 5?
Answer: There are 20 \(\frac{1}{4}\)s in 5

Explanation:
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of \(\frac{1}{4}\)s in 5
So,
5 ÷ \(\frac{1}{4}\)  = 5 × \(\frac{4}{1}\)
= \(\frac{5}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 5 × 4}{1 × 1}\)
= 20
Hence, from the above,
We can conclude that there are 20 \(\frac{1}{4}\)s in 5.

Question 10.
How many \(\frac{1}{6}\)s are in 2?
Answer: There are 12 \(\frac{1}{6}\)s in 2

Explanation:
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of \(\frac{1}{6}\)s in 2
So,
2 ÷ \(\frac{1}{6}\)  = 2 × \(\frac{6}{1}\)
= \(\frac{2}{1}\) × \(\frac{6}{1}\)
= \(\frac{ 2 × 6}{1 × 1}\)
= 12
Hence, from the above,
We can conclude that there are 12 \(\frac{1}{6}\)s in 2.

Question 11.
YOU BE THE TEACHER
Newton finds 6 ÷ \(\frac{1}{3}\). Is he correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 31
Answer: No, Newton is not correct

Explanation:
The given division equation is: 6 ÷ \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
6 ÷ \(\frac{1}{3}\)  = 6 × \(\frac{3}{1}\)
= \(\frac{6}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 3 × 6}{1 × 1}\)
= 18
But, according to Newton,
6 ÷ \(\frac{1}{3}\) = 2
Hence, from the above,
We can conclude that Newton is not correct.

Question 12.
Writing
Write and solve a real-life problem for 4 ÷ \(\frac{1}{2}\).
Answer:
Suppose we have 4 bags of wheat and we have to distribute the 4 bags by dividing each bag of wheat in half
So,
Each person receives 4 ÷ \(\frac{1}{2}\) bag of wheat
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{2}\)  = 4 × \(\frac{2}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 4 × 2}{1 × 1}\)
= 8
Hence, from the above,
We can conclude that there are 8 bags of wheat when divide the 4 bags of wheat in half.

Think and Grow: Modeling Real Life

Example
A chef makes 3 cups of salsa. A serving of salsa is \(\frac{1}{8}\) cup. How many servings does the chef make?
To find the number of servings, find the number of \(\frac{1}{8}\) cups in 3 cups.
Use an area model to find 3 ÷ \(\frac{1}{8}\). Divide each cup into 8 equal parts.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 32

Show and Grow

Question 13.
A litter of kittens weighs a total of 2 pounds. Each newborn kitten weighs \(\frac{1}{4}\) pound. How many kittens are in the litter?
Answer: The number of kittens in the litter are: 8 kittens

Explanation:
It is given that a litter of kittens weighs a total of 2 pounds and each newborn kitten weighs \(\frac{1}{4}\) pound.
So,
The number of kittens in the litter = \(\frac{The total weight of litter}{The weight of each newborn kitten}\)
= 2 ÷ \(\frac{1}{4}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{4}\)  = 2 × \(\frac{4}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 4 × 2}{1 × 1}\)
= 8
Hence, from the above,
We can conclude that the number of kittens in the litter are: 8 kittens

Question 14.
You put signs on a walking trail that is 7 miles long. You put a sign at the start and at the end of the trail. You also put a sign every \(\frac{1}{10}\) mile. How many signs do you put on the trail?
Answer: The total number of signs you put on the trail is: 72

Explanation:
It is given that you put signs on a walking trail that is 7 miles long and you put a sign at the start and at the end of the trail.
It is also given that you put a sign every \(\frac{1}{10}\) mile.
So,
The total number of signs you put on the trail = The sign at the start of the trail + The sign at the end of the trail + The total number of signs for \(\frac{1}{10}\) mile
Now,
The total number of signs for \(\frac{1}{10}\) mile = 7 ÷ \(\frac{1}{10}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{10}\)  = 7 × \(\frac{10}{1}\)
= \(\frac{7}{1}\) × \(\frac{10}{1}\)
= \(\frac{ 7 × 10}{1 × 1}\)
= 70
So,
The total number of signs you put on the trail = 1 + 1 + 70
= 72
hence, from the above,
We can conclude that there are 72 signs that you put on the trail

Question 15.
DIG DEEPER!
You have 2 boards that are each 8 feet long. You cut \(\frac{1}{2}\)– foot pieces to make square picture frames. How many picture frames can you make?
Answer: The number of picture frames you can make is: 32

Explanation:
It is given that you have 2 boards that are each 8 feet long.
So,
The total length of 2 boards = 2 × 8 = 16 feet
It is also given that you cut \(\frac{1}{2}\)– foot pieces to make square picture frames.
So,
The total number of picture frames = \(\frac{The total length of 2 boards}{The length of each square frame}\)
= 16 ÷ \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
16 ÷ \(\frac{1}{2}\)  = 16 × \(\frac{2}{1}\)
= \(\frac{16}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 16 × 2}{1 × 1}\)
= 32
Hence, from the above,
We can conclude that we can make 32 picture frames.

Divide Whole Numbers by Unit Fractions Homework & Practice 10.3

Divide. Use a model to help.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 33
Answer: 1 ÷ \(\frac{1}{9}\) = 9

Explanation:
The given numbers are: 1 and \(\frac{1}{9}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
1 ÷ \(\frac{1}{9}\)  = 1 × \(\frac{9}{1}\)
= \(\frac{1}{1}\) × \(\frac{9}{1}\)
= \(\frac{ 1 × 9}{1 × 1}\)
= 9
Hence,
1÷ \(\frac{1}{9}\) = 9

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 34
Answer: 2 ÷ \(\frac{1}{3}\) = 6

Explanation:
The given numbers are: 2 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{3}\)  = 2 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 3 × 2}{1 × 1}\)
= 6
Hence,
2÷ \(\frac{1}{3}\) = 6

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 35
Answer: 5 ÷ \(\frac{1}{2}\) = 10

Explanation:
The given numbers are: 5 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
5 ÷ \(\frac{1}{2}\)  = 5 × \(\frac{2}{1}\)
= \(\frac{5}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 5 × 2}{1 × 1}\)
= 10
Hence,
5÷ \(\frac{1}{2}\) = 10

Divide. Use a model to help.

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 36
Answer: 9 ÷ \(\frac{1}{4}\) = 36

Explanation:
The given numbers are: 9 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{4}\)  = 9 × \(\frac{4}{1}\)
= \(\frac{9}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 9 × 4}{1 × 1}\)
= 36
Hence,
9÷ \(\frac{1}{4}\) = 36

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 37
Answer: 7 ÷ \(\frac{1}{3}\) = 21

Explanation:
The given numbers are: 7 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{3}\)  = 7 × \(\frac{3}{1}\)
= \(\frac{3}{1}\) × \(\frac{7}{1}\)
= \(\frac{ 3 × 7}{1 × 1}\)
= 21
Hence,
7÷ \(\frac{1}{3}\) = 21

Question 6.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 38
Answer: 8 ÷ \(\frac{1}{5}\) = 40

Explanation:
The given numbers are: 8 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
8 ÷ \(\frac{1}{5}\)  = 8 × \(\frac{5}{1}\)
= \(\frac{8}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 8 × 5}{1 × 1}\)
= 40
Hence,
8÷ \(\frac{1}{5}\) = 40

Question 7.
Number Sense
Explain how you can check your answer for Exercise 6.
Answer:
We can check the answer for exercise 6 by using the below model:

From the above model,
Each part represents \(\frac{8}{5}\)
So,
The total value of the 5 parts is: \(\frac{40}{5}\)
Hence,
In the above way, we can say that we check the answer

Question 8.
YOU BE THE TEACHER
Descartes finds 5 ÷ \(\frac{1}{4}\). Is he correct? Explain.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 38.1
Answer: Yes, he is correct

Explanation:
We can write 5 as \(\frac{20}{4}\) or \(\frac{5}{1}\)
But, we only take \(\frac{20}{4}\) because the divided number given is 4
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{20}{4}\) ÷ \(\frac{1}{4}\)
= \(\frac{20}{4}\) × \(\frac{4}{1}\)
= \(\frac{ 20 × 4}{4 × 1}\)
= 20
Hence, from the above,
We can conclude that Descartes is correct.

Question 9.
Modeling Real Life
You need \(\frac{1}{2}\) pound of clay to make a pinch pot. How many pinch pots can you make with 12 pounds of clay?
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 42
Answer: You can make 24 pinch pots with 12 pounds of clay

Explanation:
It is given that you need \(\frac{1}{2}\) pound of clay to make a pinch pot.
It is also given that you have 12 pounds of clay
So,
The number of pinch pots you can make by using 12 pounds of clay = \(\frac{The total amount of clay}{The amount of clay used to make each pinch pot}\)
= 12 ÷ \(\frac{1}{2}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
12 ÷ \(\frac{1}{2}\)
= \(\frac{12}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 12 × 2}{4 × 1}\)
= 24
Hence, from the above,
We can conclude that we can make 24 pinch pots by using 12 pounds of clay.

Question 10.
Modeling Real Life
Your art teacher has 5 yards of yellow string and 4 yards of green string. She cuts both colors \(\frac{1}{3}\)-yard pieces to hang of string into student artwork. How many pieces of student artwork can she hang?
Answer: The number of pieces of student artwork she can hang is: 27

Explanation:
It is given that your art teacher has 5 yards of yellow string and 4 yards of green string.
So,
The total number of yards of string = 5 + 4 = 9 yards of string
It is also given that she cuts both colors \(\frac{1}{3}\)-yard pieces to hang of string into student artwork.
So,
The number of pieces of student artwork she can hang = \(\frac{The total number of yards of strings}{The length of each yard f string}\)
= 9 ÷ \(\frac{1}{3}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{3}\)
= \(\frac{9}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 9 × 3}{1 × 1}\)
= 27
Hence, from the above,
We can conclude that there are 27 pieces of student artwork that she can hang.

Review & Refresh

Question 11.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 43
Answer: \(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{6}{20}\)

Explanation:
The given fractions are: \(\frac{3}{4}\) and \(\frac{2}{5}\)
So,
\(\frac{2}{5}\) × \(\frac{3}{4}\)
= \(\frac{2 × 3}{5 × 4}\)
= \(\frac{6}{20}\)
Hence,
\(\frac{2}{5}\) × \(\frac{3}{4}\) = \(\frac{6}{20}\)

Question 12.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 44
Answer: \(\frac{1}{8}\) × \(\frac{5}{8}\) = \(\frac{5}{64}\)

Explanation:
The given fractions are: \(\frac{1}{8}\) and \(\frac{5}{8}\)
So,
\(\frac{1}{8}\) × \(\frac{5}{8}\)
= \(\frac{1 × 5}{8 × 8}\)
= \(\frac{5}{64}\)
Hence,
\(\frac{1}{8}\) × \(\frac{5}{8}\) = \(\frac{5}{64}\)

Question 13.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 45
Answer: \(\frac{4}{9}\) × \(\frac{2}{7}\) = \(\frac{8}{63}\)

Explanation:
The given fractions are: \(\frac{4}{9}\) and \(\frac{2}{7}\)
So,
\(\frac{4}{9}\) × \(\frac{2}{7}\)
= \(\frac{2 × 4}{7 × 9}\)
= \(\frac{8}{63}\)
Hence,
\(\frac{4}{9}\) × \(\frac{2}{7}\) = \(\frac{8}{63}\)

Lesson 10.4 Divide Unit Fractions by Whole Numbers

Write a real-life problem that can be represented by \(\frac{1}{2}\) ÷ 3?
Answer:
Suppose we have 3 people and those 3 people each has to share \(\frac{1}{2}\) of the apple

What is the solution to the problem? Use a model to support your answer?
Answer:
The above problem is: We have to share \(\frac{1}{2}\) each for the 3 people
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 3
= \(\frac{1}{2}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{2 × 3}\)
= \(\frac{1}{6}\)
Hence,
\(\frac{1}{6}\) is the solution to the above problem.

Precision
Is the answer greater than or less than 1? Explain?
Answer: The answer is less than 1

Explanation:
The answer for the problem is: \(\frac{1}{6}\)
So,
For the comparison of \(\frac{1}{6}\) with 1, we have to see whether the numerators or the denominators are equal or not
So, in this case, the numerators are equal
So, compare the denominators
So,
1 < 6
Hence, from the above,
We can conclude that \(\frac{1}{6}\) is less than 1

Think and Grow: Divide Unit Fractions by Whole Numbers

You can use models to divide unit fractions by whole numbers.

Show and Grow

Divide. Use a model to help.

Question 1.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 48
Answer: \(\frac{1}{4}\) ÷ 2 = \(\frac{1}{8}\)

Explanation:
The given numbers are: \(\frac{1}{4}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) ÷ 2
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 4}\)
= \(\frac{1}{8}\)
Hence,
\(\frac{1}{4}\) ÷ 2 = \(\frac{1}{8}\)

Question 2.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 49
Answer: \(\frac{1}{2}\) ÷ 5 = \(\frac{1}{10}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 5
= \(\frac{1}{2}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{2 × 5}\)
= \(\frac{1}{10}\)
Hence,
\(\frac{1}{2}\) ÷ 5 = \(\frac{1}{10}\)

Apply and Grow: Practice

Divide. Use a model to help.

Question 3.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 50
Answer: \(\frac{1}{5}\) ÷ 3 = \(\frac{1}{15}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 3
= \(\frac{1}{5}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{5 × 3}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{1}{5}\) ÷ 3 = \(\frac{1}{15}\)

Question 4.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 51
Answer: \(\frac{1}{6}\) ÷ 2 = \(\frac{1}{12}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 2
= \(\frac{1}{6}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{6}\) ÷ 2 = \(\frac{1}{12}\)

Question 5.
Big Ideas Math Answer Key Grade 5 Chapter 10 Divide Fractions 52
Answer: \(\frac{1}{3}\) ÷ 5 = \(\frac{1}{15}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 5
= \(\frac{1}{3}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{3 × 5}\)
= \(\frac{1}{15}\)
Hence,
\(\frac{1}{3}\) ÷ 5 = \(\frac{1}{15}\)

Question 6.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 53
Answer: \(\frac{1}{5}\) ÷ 4 = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 4
= \(\frac{1}{5}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{5 × 4}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{5}\) ÷ 4 = \(\frac{1}{20}\)

Question 7.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 54
Answer: \(\frac{1}{3}\) ÷ 3 = \(\frac{1}{9}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 3
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{3 × 3}\)
= \(\frac{1}{9}\)
Hence,
\(\frac{1}{3}\) ÷ 3 = \(\frac{1}{9}\)

Question 8.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 55
Answer: \(\frac{1}{8}\) ÷ 2 = \(\frac{1}{16}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 2
= \(\frac{1}{8}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 8}\)
= \(\frac{1}{16}\)
Hence,
\(\frac{1}{8}\) ÷ 2 = \(\frac{1}{16}\)

Question 9.
How many 6s are in \(\frac{1}{2}\)?
Answer: There are \(\frac{1}{12}\) 6s in \(\frac{1}{2}\)

Explanation:
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of 6s in \(\frac{1}{2}\)
So,
\(\frac{1}{2}\) ÷ 6
= \(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{ 1 × 1}{2 × 6}\)
= \(\frac{1}{12}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{12}\) 6s in \(\frac{1}{2}\)

Question 10.
How many 2s are in \(\frac{1}{3}\) ?
Answer: There are \(\frac{1}{6}\) 2s in \(\frac{1}{3}\)

Explanation:
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
Now,
We have to find the number of 2s in \(\frac{1}{3}\)
So,
\(\frac{1}{3}\) ÷ 2
= \(\frac{1}{3}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 3}\)
= \(\frac{1}{6}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{6}\) 2s in \(\frac{1}{2}\)

Question 11.
Writing
Write and solve a real-life problem for
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 56
Answer:
Suppose a box has 7 chocolates. We have to divide these seven chocolates into further \(\frac{1}{2}\) parts so that the chocolates can be distributed to more people
So,
The each part of chocolate we can get = \(\frac{1}{2}\) ÷ 7
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 7
= \(\frac{1}{2}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{2 × 7}\)
= \(\frac{1}{14}\)
Hence, from the above,
We can conclude that we can get \(\frac{1}{14}\) part of each chocolate.

Question 12.
Reasoning
Complete the statements.

Think and Grow: Modeling Real Life

You melt \(\frac{1}{4}\) quart of soap. You pour the soap into 4 of the same-sized molds. What fraction of a quart of soap does each mold hold?
You are dividing \(\frac{1}{4}\) quart into 4 equal parts, so you need to find \(\frac{1}{4}\) ÷ 4.

Show and Grow

Question 13.
You buy \(\frac{1}{2}\) pound of grapes. You equally divide the grapes into 2 bags. What fraction of a pound of grapes do you put into each bag?
Answer: The fraction of a pound of grapes you put into each bag is: \(\frac{1}{8}\) pound

Explanation:
It is given that you buy \(\frac{1}{2}\) pound of grapes.
It is also given that you equally divide the grapes into 2 bags.
So,
The number of grapes in each bag = \(\frac{1}{2}\) ÷ 2
Now,
The fraction of pound of grapes you put into each bag = \(\frac{The number of grapes in each bag}{2}\)
=  ( \(\frac{1}{2}\) ÷ 2 ) ÷ 2
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
( \(\frac{1}{2}\) ÷ 2 ) ÷ 2
= ( \(\frac{1}{2}\) × \(\frac{1}{2}\) ) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 2}\) × \(\frac{1}{2}\)
= \(\frac{1}{4}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{2 × 4}\)
= \(\frac{1}{8}\)
Hence, from the above
We can conclude that the fraction of pound of grapes in each bag is: \(\frac{1}{8}\) pound

Question 14.
You have \(\frac{1}{8}\) cup of red sand, \(\frac{1}{4}\) cup of blue sand, and \(\frac{1}{2}\) cup of white sand. You equally divide the sand into 3 containers. What fraction of a cup of sand do you pour into each container?
Answer: The fraction of a cup of sand you pour into each container is: \(\frac{7}{24}\)

Explanation:
It is given that you have \(\frac{1}{8}\) cup of red sand, \(\frac{1}{4}\) cup of blue sand, and \(\frac{1}{2}\) cup of white sand.
So,
The total amount of sand = \(\frac{1}{8}\) cup of red sand + \(\frac{1}{4}\) cup of blue sand + \(\frac{1}{2}\) cup of white sand
In addition, we have to see either the numerators are equal or the denominators are equal.
If the numerators are equal we have to ake the denominators also equal.
So,
\(\frac{1}{4}\) is multplied by \(\frac{2}{2}\)
\(\frac{1}{2}\) is multiplied by \(\frac{4}{4}\)
So,
\(\frac{1}{4}\) = \(\frac{2}{8}\)
\(\frac{1}{2}\) = \(\frac{4}{8}\)
So,
\(\frac{1}{8}\) + \(\frac{2}{8}\) + \(\frac{4}{8}\) = \(\frac{7}{8}\)
It is also given that all the sand is equally distributed into 3 containers
So,
The amount of sand in each container = \(\frac{7}{8}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{7}{8}\) ÷ 3
= \(\frac{7}{8}\) × \(\frac{1}{3}\)
= \(\frac{ 7 × 1}{8 × 3}\)
= \(\frac{7}{24}\)
Hence, from the above,
We can conclude that the amount of sand in each container is: \(\frac{7}{24}\) cup.

Question 15.
DIG DEEPER!
You, your friend, and your cousin share \(\frac{1}{2}\) of a vegetable pizza and \(\frac{1}{4}\) of a cheese share pizza. The pizzas are the same size. What fraction of a pizza do you get in all?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 58

Divide. Use a model to help
Answer: The fraction of a pizza you got is: \(\frac{3}{12}\)

Explanation:
It is given that you, your friend, and your cousin share \(\frac{1}{2}\) of a vegetable pizza and \(\frac{1}{4}\) of a cheese share pizza.
So,
The total amount of pizza = \(\frac{1}{2}\) of a vegetable pizza + \(\frac{1}{4}\) of a cheese share pizza
In addition, we have to see either the numerators are equal or the denominators are equal.
If the numerators are equal we have to ake the denominators also equal.
So,
\(\frac{1}{2}\) is multplied by \(\frac{2}{2}\)
So,
\(\frac{1}{2}\) = \(\frac{2}{4}\)
So,
\(\frac{2}{4}\) + \(\frac{1}{4}\) = \(\frac{3}{4}\)
So,
The fraction of pizza each get = \(\frac{The total amount of pizza}{3}\)
= \(\frac{3}{4}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{3}{4}\) ÷ 3
= \(\frac{3}{4}\) × \(\frac{1}{3}\)
= \(\frac{ 3 × 1}{4 × 3}\)
= \(\frac{3}{12}\)
Hence, from the above,
We can conclude that the fraction of pizza each get is: \(\frac{3}{12}\)

Divide Unit Fractions by Whole Numbers Homework & Practice 10.4

Question 1.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 59
Answer: \(\frac{1}{3}\) ÷ 4 = \(\frac{1}{12}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 4
= \(\frac{1}{3}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{3 × 4}\)
= \(\frac{1}{12}\)
Hence,
\(\frac{1}{3}\) ÷ 4 = \(\frac{1}{12}\)

Question 2.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 60
Answer: \(\frac{1}{6}\) ÷ 3 = \(\frac{1}{18}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 3
= \(\frac{1}{6}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{6 × 3}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{6}\) ÷ 3 = \(\frac{1}{18}\)

Question 3.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 61
Answer: \(\frac{1}{4}\) ÷ 5 = \(\frac{1}{20}\)

Explanation:
The given numbers are: \(\frac{1}{4}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{4}\) ÷ 5
= \(\frac{1}{4}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{5 × 4}\)
= \(\frac{1}{20}\)
Hence,
\(\frac{1}{4}\) ÷ 5 = \(\frac{1}{20}\)

Divide. Use a model to help.

Question 4.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 62
Answer: \(\frac{1}{5}\) ÷ 9 = \(\frac{1}{45}\)

Explanation:
The given numbers are: \(\frac{1}{5}\) and 9
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 9
= \(\frac{1}{5}\) × \(\frac{1}{9}\)
= \(\frac{ 1 × 1}{5 × 9}\)
= \(\frac{1}{45}\)
Hence,
\(\frac{1}{5}\) ÷ 9 = \(\frac{1}{45}\)

Question 5.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 63
Answer: \(\frac{1}{8}\) ÷ 6 = \(\frac{1}{48}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 6
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 6
= \(\frac{1}{8}\) × \(\frac{1}{6}\)
= \(\frac{ 1 × 1}{8 × 6}\)
= \(\frac{1}{48}\)
Hence,
\(\frac{1}{8}\) ÷ 6 = \(\frac{1}{48}\)

Question 6.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 64
Answer: \(\frac{1}{7}\) ÷ 4 = \(\frac{1}{28}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 4
= \(\frac{1}{7}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{7 × 4}\)
= \(\frac{1}{28}\)
Hence,
\(\frac{1}{7}\) ÷ 4 = \(\frac{1}{28}\)

Question 7.
YOU BE THE TEACHER
Your friend divides \(\frac{1}{3}\) by 7 to get \(\frac{1}{21}\). He checks his answer by multiplying \(\frac{1}{21}\) × \(\frac{1}{3}\). Does your friend check his answer correctly? Explain.
Answer: No, your friend does not check his answer correctly

Explanation:
It is given that your friend divides \(\frac{1}{3}\) by 7 to get \(\frac{1}{21}\).
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 7
= \(\frac{1}{3}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
It is also given that your friend checks his answer by multiplying \(\frac{1}{21}\) × \(\frac{1}{3}\).
Now,
\(\frac{1}{21}\) × \(\frac{1}{3}\)
= \(\frac{1 × 1}{21 × 3}\)
= \(\frac{1}{63}\)
But, your friend wanted to check whether \(\frac{1}{21}\) × \(\frac{1}{3}\) = \(\frac{1}{7}\)
But, the value becomes \(\frac{1}{63}\)
Hence, from the above,
We can conclude that your friend does not check the answer correctly.

Question 8.
Logic
Find the missing numbers.

Question 9.
Modeling Real Life
You win tickets that you can exchange for prizes. You exchange \(\frac{1}{5}\) of your tickets and then divide them equally among 3 prizes. What fraction of your tickets do you spend on each prize?
Answer: The fraction of your tickets you spend on each prize is: \(\frac{1}{15}\)

Explanation:
It is given that you win tickets that you can exchange for prizes.
It is also given that you exchange \(\frac{1}{5}\) of your tickets and then divide them equally among 3 prizes
So,
The fraction of the tickets spent on each prize = \(\frac{The value of Exchange}{The number of prizes}\)
= \(\frac{1}{5}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{5}\) ÷ 3
= \(\frac{1}{5}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{5 × 3}\)
= \(\frac{1}{15}\)
Hence, from the above,
We can conlude that the fraction of tickets you spend on each prize is: \(\frac{1}{15}\)

Question 10.
DIG DEEPER!
You have \(\frac{1}{8}\) gallon of melted crayon wax. You pour the wax equally into 8 different molds to make new crayons. What fraction of a cup of melted wax is in each mold? Think: 1 gallon is 16 cups.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 66
Answer: The fraction of a cup of melted wax in each mold is: \(\frac{1}{4}\)

Explanation:
It is given that you have \(\frac{1}{8}\) gallon of melted crayon wax.
It is also given that you pour the wax equally into 8 different molds to make new crayons.
So,
The fraction of melted crayon wax in each mold in gallons = \(\frac{The total amount of melted crayon wax }{The number of molds}\)
= \(\frac{1}{8}\) ÷ 8
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 8
= \(\frac{1}{8}\) × \(\frac{1}{8}\)
= \(\frac{ 1 × 1}{8 × 8}\)
= \(\frac{1}{64}\) gallons
But, it is given that
1 gallon = 16 cups
So,
The total number of cups that the melted crayon wax contained = \(\frac{1}{64}\) × \(\frac{16}{1}\)
= \(\frac{1 × 16 }{64 × 1}\)
= \(\frac{1}{4}\)
Hence, from the above,
We can conclude that there are \(\frac{1}{4}\) cups of melted crayon wax in each mold.

Review & Refresh

Question 11.
0.9 ÷ 0.1 = ___
Answer: 0.9 ÷ 0.1 = 9

Explanation:
The given decimal numbers are: 0.9 and 0.1
The representation of the decimal numbers in the fraction form is: \(\frac{9}{10}\) and \(\frac{1}{10}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{9}{10}\) ÷ \(\frac{1}{10}\)  = \(\frac{9}{10}\) × \(\frac{10}{1}\)
= \(\frac{ 9 × 10}{10 × 1}\)
= 9
Hence, 0.9 ÷ 0.1 = 9

Question 12.
38.6 ÷ 100 = ___

Answer: 38.6 ÷ 100 = 0.386

Explanation:
The given numbers are: 38.6 and 100
The representation of the numbers in the fraction form is: \(\frac{386}{10}\) and \(\frac{100}{1}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{386}{10}\) ÷ \(\frac{100}{1}\)  = \(\frac{386}{10}\) × \(\frac{1}{100}\)
= \(\frac{ 386 × 1}{100 × 10}\)
= \(\frac{386}{1000}\)
= 0.386
Hence, 38.6 ÷ 100 = 0.386

Question 13.
2.57 ÷ 0.01 = ___
Answer: 2.57 ÷ 0.01 = 257

Explanation:
The given decimal numbers are: 2.57 and 0.01
The representation of the decimal numbers in the fraction form is: \(\frac{257}{100}\) and \(\frac{1}{100}\)
Now,
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{257}{100}\) ÷ \(\frac{1}{100}\)  = \(\frac{257}{100}\) × \(\frac{100}{1}\)
= \(\frac{ 257 × 100}{100 × 1}\)
= 257
Hence, 2.57 ÷ 0.01 = 257

Lesson 10.5 Problem Solving: Fraction Division

Explore and Grow

You want to make a \(\frac{1}{3}\) batch of the recipe. How you can use division to find the amount of each ingredient you need?
Answer:
It is given that you want to make a \(\frac{1}{3}\) batch of the recipe.
So,
From \(\frac{1}{3}\),
1 represents a batch of the recipe
3 represents the total number of ingredients in a batch
So,
The amount of each ingredient you need = \(\frac{The amount of the batch of the recipe }{The total number of ingredients}\)
= \(\frac{1}{3}\) ÷ 3
Now,
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 3
= \(\frac{1}{3}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{3 × 3}\)
= \(\frac{1}{9}\)
Hence, from the above,
We can conclude that the amount of each ingredient you need is: \(\frac{1}{9}\)

Reasoning
Without calculating, explain how you can tell whether you need more than or less than 1 tablespoon of olive oil?
Answer: You need less than 1 tablespoon of olive oil

Explanation:
From the above problem,
The amount of each ingredient is: \(\frac{1}{9}\)
Since the amount of each ingredient is less than 1, you need less than 1 tablespoon of olive oil

Think and Grow: Problem Solving: Fraction Division

Example
You have 4 cups of yellow paint and 3 cups of blue paint. How many batches of green paint can you make?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 67

Understand the Problem

What do you know?

  • You have 4 cups of yellow paint and 3 cups of blue paint.
  • One batch of green paint is made of \(\frac{1}{2}\) cup of yellow and \(\frac{1}{3}\) cup of blue.

What do you need to find?

  • You need to find how many batches of green paint you can make.

Make a Plan
How will you solve?

  • Find how many batches are possible from yellow, and how many from blue.
  • Choose the lesser number of batches.

Solve

So, you can make 8 batches of green paint.

Show and Grow

Question 1.
In the example, explain why you choose the fewer number of batches.
Answer: In the above example, the yellow paint has the less number of batches as the amount of each batch of yellow paint-filled is more than the batch of green paint
Hence,
We choose the fewer number of batches of yellow paint

Apply and Grow: Practice

Understand the problem. What do you know? What do you need to find? Explain.

Question 2.
A landowner donates 3 acres of land to a city. The mayor of the city uses 1 acre of the land for a playground and the rest of the land for community garden plots. Each garden plot is \(\frac{1}{3}\) acre. How many plots are there?
Understand the problem. Then make a plan. How will you solve it? Explain?
Answer: The number of plots in the community is: 6

Explanation:
It is given that a landowner donates 3 acres of land to a city and the mayor of the city uses 1 acre of the land for a playground and the rest of the land for community garden plots.
So,
The portion of the land used for community garden plots is: 2 acres
It is also given that each garden plot is \(\frac{1}{3}\) acre.
So,
The number of plots = \(\frac{The portion of the land used for community garden plots}{The area of each garden plot}\)
= 2 ÷ \(\frac{1}{3}\)
= 2 × \(\frac{3}{1}\)
= \(\frac{2}{1}\) × \(\frac{3}{1}\)
= 6
Hence, from the above,
We can conclude that there are 6 plots

Question 3.
A craftsman uses \(\frac{3}{4}\) gallon of paint to paint 4 identical dressers. He uses the same amount of paint on each dresser. How much paint does he use to paint 7 of the same dressers?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 69
Answer: The paint used by the craftsman to paint 7 of the same dressers is: \(\frac{21}{16}\)

Explanation:
It is given that a craftsman uses \(\frac{3}{4}\) gallon of paint to paint 4 identical dressers.
So,
The paint used to paint each dresser = \(\frac{3}{4}\) ÷ 4
= \(\frac{3}{4}\) × \(\frac{1}{4}\)
= \(\frac{3}{16}\) gallon
So,
The amount of paint used to paint the 7 identical dressers = \(\frac{The paint used to paint each dresser}{1}\) × 7
= \(\frac{3}{16}\) × \(\frac{7}{1}\)
= \(\frac{3 × 7}{16 × 1}\)
= \(\frac{21}{16}\) gallon
Hence, from the above,
We can conclude that the paint used to paint 7 identical dressers is: \(\frac{21}{16}\) gallon

Question 4.
An airplane travels 125 miles in \(\frac{1}{4}\) hour. It travels the same number of miles each hour. How many miles does the plane travel in 5 hours?
Answer: The number of miles the plane travel in 5 hours is: 2,500 miles

Explanation:
It is given that an airplane travels 125 miles in \(\frac{1}{4}\) hour
So,
The number of miles traveled by plane in 1 hour = 125 ÷ \(\frac{1}{4}\)
= 125 × \(\frac{4}{1}\)
= 125 × 4
= 500 miles
So,
The number of miles traveled by plane in 5 hours = ( The number of miles traveled by plane in 1 hour ) × 5
= 500 × 5
= 2,500 miles
Hence, from the above,
We can conclude that the number of miles traveled by plane in 5 hours is: 2,500 miles

Question 5.
You make bows for gifts using \(\frac{2}{3}\) yard of ribbon for each bow. You have 4 feet of red ribbon and 5 feet of green ribbon. How many bows can you make?
Answer: The number of bows you can make is: 2 bows

Explanation:
It is given that you make bows for gifts using \(\frac{2}{3}\) yard of ribbon for each bow.
It is also given that you have 4 feet of red ribbon and 5 feet of green ribbon
So,
The total length of ribbon = 5 + 4 = 9 feet
we know that,
1 foot = \(\frac{1}{3}\) yards
So,
9 feet = 9 × \(\frac{1}{3}\) yards
= \(\frac{9}{1}\) yards × \(\frac{1}{3}\) yards
= 3 yards
So,
The number of bows you can make = \(\frac{2}{3}\) yards × 3
= 2 bows
Hence, from the above,
We can conclude that the number of bows we can make is: 2

Question 6.
A landscaper buys 1 gallon of plant fertilizer. He uses \(\frac{1}{5}\) of the fertilizer, and then divides the rest into 3 smaller bottles. How many gallons does he put into each bottle?
Answer: The number of gallons he put into each bottle is: \(\frac{4}{15}\)

Explanation:
It is given that a landscaper buys 1 gallon of plant fertilizer and he uses \(\frac{1}{5}\) of the fertilizer
So,
The remaining amount of the fertilizer = 1 – \(\frac{1}{5}\)
= \(\frac{4}{5}\) gallons
It is also given that he divided the remaining amount of fertilizer into 3 smaller bottles.
So,
The amount of fertilizer put into each bottle = \(\frac{The remaining amount of the fertilizer}{The total number of bottles}\)
= \(\frac{4}{5}\) ÷ 3
= \(\frac{4}{5}\) × \(\frac{1}{3}\)
= \(\frac{4 × 1}{5 × 3}\)
= \(\frac{4}{15}\) gallons
hence, from the above,
We can conclude that the amount of remaining fertilizer put into each bottle is: \(\frac{4}{15}\) gallons

Think and Grow: Modeling Real Life

Example
A sponsor donates $0.10 to a charity for every \(\frac{1}{4}\) kilometer of the triathlon an athlete completes. The athlete completes the entire triathlon. How much money does the sponsor donate?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 70
Think: What do you know? What do you need to find? How will you solve?
Write and solve an equation.
Add 1.9, 90, and 21.1 to find how many kilometers the athlete completes.
Divide the sum by \(\frac{1}{4}\) to find how many \(\frac{1}{4}\) kilometers the athlete completes.
Multiply the quotient by $0.10 to find how much money the sponsor donates.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 71
Let m represent the total amount of money donated.

Show and Grow

Question 7.
You earn $5 for every \(\frac{1}{2}\) hour you do yard work. How much money do you earn in 1 week?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 73
Answer: The amount you earn in 1 week is: $700

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 73
From the above table,
The total amount of time = 5\(\frac{1}{2}\) + 3 + 1\(\frac{1}{2}\)
= \(\frac{11}{2}\) + 3 + \(\frac{3}{2}\)
= \(\frac{11 + 3}{2}\) + 3
= 7 + 3
= 10 hours
It is given that you earn $5 for every \(\frac{1}{2}\) hour you do yard work
So,
The amount earned in 10 hours in a day = 10 ÷\(\frac{1}{2}\) × 5 ( Since we have the time in hours but the money earned is given in half an hour basis )
= 20 × 5
= $100
We know that 1 week = 7 days
So,
The amount earned in 1 week = 100 × 7 = $700
hence, from the above,
We can conclude that we can earn $700 in a week

Problem Solving: Fraction Division Homework &  10.5

Understand the problem. Then make a plan. How will you solve? Explain.

Question 1.
A train travels 75 miles in \(\frac{1}{2}\) hour. How many miles does the train travel in 8 hours?
Answer: The number of miles the train travel in  hours is: 1,200 miles

Explanation:
It is given that a train travels 75 miles in \(\frac{1}{2}\) hour.
So,
The number of miles the train travel in 1 hour = 75 ÷ \(\frac{1}{2}\)
= 75 × 2
= 150 miles
So,
The number of miles the train travel in 8 hours = The number of miles traveled by train in 1 hour × 8
= 150 × 8
= 1,200 miles
Hence, from the above,
We can conclude that the train travels 1,200 miles in 8 hours.

Question 2.
You need \(\frac{2}{3}\) yard of fabric to create a headband. You have 12 feet of blue fabric and 4 feet of yellow fabric. How many headbands can you make with all of the fabric?
Answer: The number of headbands you can make with all of the fabric is: 8 headbands

Explanation:
It is given that you need \(\frac{2}{3}\) yard of fabric to create a headband.
It is also given that you have 12 feet of blue fabric and 4 feet of yellow fabric.
So,
The total length of the fabric = 12 + 4 = 16 feet
We know that
1 foot = \(\frac{1}{3}\) yards
So,
16 feet = \(\frac{16}{3}\) yards
So,
The number of headbands you can create with all the fabric = \(\frac{The total length of the fabric}{The length of each fabric}\)
= \(\frac{16}{3}\) ÷ \(\frac{2}{3}\)
= \(\frac{16}{3}\) × \(\frac{3}{2}\)
= \(\frac{16 × 3}{3 × 2}\)
= 8 headbands
Hence, from the above,
We can conclude that we can create 8 headbands with all the fabric.

Question 3.
An art teacher has 8 gallons of paint. Her class uses \(\frac{3}{4}\) of the paint. The teacher divides the rest of the paint into 4 bottles. How much paint is in each bottle?
Answer: The amount of paint in each bottle is: \(\frac{1}{2}\)

Explanation:
It is given that an art teacher has 8 gallons of paint and her class uses \(\frac{3}{4}\) of the paint.
So,
The remaining amount of paint = \(\frac{1}{4}\) × 8
= \(\frac{1}{4}\) × \(\frac{8}{1}\)
=\(\frac{1 × 8}{4 × 1}\)
= 2 gallons
It is also given that the remaining amount of the paint divided into 4 bottles by the teacher
So,
The amount of paint present in each bottle = 2 ÷ 4
= \(\frac{1}{2}\) gallons
Hence, from the above,
We can conclude that the amout of paint present in each bottle is: \(\frac{1}{2}\) gallons

Question 4.
You mix 3\(\frac{1}{4}\) cups of frozen strawberries and 4\(\frac{1}{2}\) cups of frozen blueberries in a bowl. A smoothie requires \(\frac{1}{2}\) cup of your berry mix. How many smoothies can you make?
Answer: The number of smoothies you can make is:

Explanation:
It is given that you mix 3\(\frac{1}{4}\) cups of frozen strawberries and 4\(\frac{1}{2}\) cups of frozen blueberries in a bowl.
So,
The amount of berry mix = 3\(\frac{1}{4}\) cups of frozen strawberries + 4\(\frac{1}{2}\) cups of frozen blueberries
= 3\(\frac{1}{4}\) + 4\(\frac{1}{2}\)
= \(\frac{13}{4}\) + \(\frac{9}{2}\)
In addition, equate the denominators
So,
Multiply \(\frac{9}{2}\) with \(\frac{2}{2}\)
So,
\(\frac{9}{2}\) = \(\frac{18}{4}\)
So,
The amount of berry mix = \(\frac{13}{4}\) + \(\frac{18}{4}\)
= \(\frac{31}{4}\)
Now,
It is also given that the smoothie requires \(\frac{1}{2}\) cup of your berry mix.
So,
The number of smoothies = \(\frac{31}{4}\) ÷ \(\frac{1}{2}\)
= \(\frac{31}{4}\) × \(\frac{2}{1}\)
= \(\frac{31 × 2}{4 × 1}\)
= \(\frac{31}{2}\)
Hence, from the above,
We can conclude that the number of smoothies we can make are: \(\frac{31}{2}\)

Question 5.
Modeling Real Life
A sponsor donates $0.10 for every \(\frac{1}{4}\) dollar donated at the locations shown. How much money does the sponsor donate?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 74
Answer: The amount of money the sponsor donates is: $40.4

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 74
From the above table,
The total amount of money collected = 25.25 + 12.50 + 63.25
= $101
It is given that a sponsor donates $0.10 for every \(\frac{1}{4}\) dollar
So,
the total amount of donated = The total amount of money collected ÷ \(\frac{1}{4}\) × 0.10
= 101 ÷ \(\frac{1}{4}\) × 0.10
= 101 × 4 × 0.10
= 04 × 0.10
= $40.4
Hence, from the above,
We can conclude that the amount of money donated by a sponsor is: $40.4

Question 6.
DIG DEEPER!
A nurse earns $16 for every \(\frac{1}{2}\) hour at work. How much money does she earn in 5 days?
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 75
Answer: The money she earns in 5 days is: $1,280

Explanation:
The given table is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 75
From the above table,
The total amount of time = 6\(\frac{3}{4}\) + 1 + \(\frac{1}{4}\)
= \(\frac{27}{4}\) + 1 + \(\frac{1}{4}\)
= \(\frac{27 + 1}{4}\) + 1
= 7 + 1
= 8 hours
It is given that a nurse earns $16 for every \(\frac{1}{2}\) hour at work.
So,
The money she earned for 1 hour = 16 ÷ \(\frac{1}{2}\)
= 16 × 2 = $32
So,
The money earned for 8 hours = The money earned in 1 hour × 8
= 32 × 8 = $256
So,
The money earned in 5 days = The money earned in 1 day × 5
= 256 × 5 = $1,280
hence, from the above,
we can conclude that she can earn $1,280 in 5 days.

Review & Refresh

Find the quotient. Then check your answer.

Question 7.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 76
Answer:  186 ÷ 12 = 1 R 4

Explanation:
Let 185.88 be rounded to 186
So,
By using the partial quotients method,
186 ÷ 12 = ( 120 + 36 + 24 ) ÷ 12
= ( 120 ÷ 12 ) + ( 36 ÷ 12 ) + ( 24 ÷ 12 )
= 10 + 3 + 2
= 17 R 4
Hence, 186 ÷ 12 = 17 R 4

Question 8.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 77
Answer: 74 ÷ 24 = 3 R 2

Explanation:
Let 74.4 be rounded to 74
So,
By using the partial quotients method,
74 ÷ 24 = 72 ÷ 24
= 3 R 2
Hence, 74 ÷ 24 = 3 R 2

Question 9.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 78
Answer: 42 ÷ 46 = 0.9

Explanation:
Let 42.32 be rounded to 42
So,
By using the partial quotients method,
42 ÷ 46 = 0.9
Hence,
42 ÷ 46 = 0.9

Divide Fractions Performance Task 10

Your city has a robotics competition. Each team makes a robot that travels through a maze. The time each robot spends in the maze is used to find the team’s score.
1. One-third of the students in your grade participate in the competition. The number of participating students is divided into 12 teams.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 79
a. What fraction of the total number of students in your grade is on each team?
Answer: The fraction of the total number of students in your grade is: \(\frac{1}{36}\)

Explanation:
It is given that there are \(\frac{1}{3}\) of the students in your grade are participating in the competition.
It is also given that the participating students are divided into 12 teams.
So,
The fraction of the total number of students in each team = \(\frac{The number of participating students}{Th total number of teams}\)
= \(\frac{1}{3}\) ÷ 12
= \(\frac{1}{3}\) ÷ \(\frac{12}{1}\)
= \(\frac{1}{3}\) × \(\frac{1}{12}\)
= \(\frac{1 × 1}{12 × 3}\)
= \(\frac{1}{36}\)
Hence, from the above,
We can conclude that the fraction of students that are in each team is: \(\frac{1}{36}\)

b. There are 3 students on each team. How many students are in your grade?
Answer: The number of students in your grade is: 36

Explanation:
It is given that the number of students is divided into 12 teams
It is also given that there are 3 students on each team
So,
The total number of students = The number of teams × The number of students in each team
= 12 × 3 = 36 students
Hence, from the above,
We can conclude that there are 36 students in your grade

Question 2.
The maze for the competition is shown.
a. Write the length of the maze in feet.
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 80
Answer:
The given maze for the competition is:
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 80
From the above maze,
The total length of the maze is: 8 feet 6 inches
We know that,
1 foot = 12 inches
Hence,
1 inch = \(\frac{1}{12}\) feet
So,
6 inches = 6 × \(\frac{1}{12}\)
= \(\frac{1}{12}\) × \(\frac{6}{1}\)
= \(\frac{1}{2}\) feet
So,
The total length of the maze in feet = 8 feet + \(\frac{1}{2}\) feet
= 8.5 feet or 8\(\frac{1}{2}\) feet

b. The length of the maze is divided into 6 equal sections. What is the length of each section of the maze?
Answer: The length of each section of the maze is: \(\frac{17}{12}\) feet

Explanation:
From the above Exercise,
The total length of the maze in feet is: 8.5 feet or 8\(\frac{1}{2}\) feet
It is given that the length of the maze is divided into 6 equal sections
So,
The length of each section of the maze = 8\(\frac{1}{2}\) ÷ 6
= 8\(\frac{1}{2}\) ÷ \(\frac{6}{1}\)
= 8\(\frac{1}{2}\) × \(\frac{1}{6}\)
= \(\frac{17}{2}\) × \(\frac{1}{6}\)
= \(\frac{17}{12}\) feet
Hence, from the above,
We can conclude that the length of each section in a maze is: \(\frac{17}{12}\) feet

Question 3.
Each team has 200 seconds to complete the maze. The rules require judges to use the expression (200 – x) ÷ \(\frac{1}{5}\), where x is the total number of seconds, to find a team’s total score.
a. Your robot completes the maze in 3 minutes 5 seconds. How many points does your team earn?
Answer: The number of points your team earn is: 75 points

Explanation:
It is given that each team has 200 seconds to complete the maze and the rules require judges to use the expression (200 – x) ÷ \(\frac{1}{5}\) where x is the total number of seconds
It is also given that your robot completes the maze in 3 minutes 5 seconds
We know that,
1 minute = 60 seconds
So,
The time taken by the robot to complete the maze in seconds = ( 3 × 60 ) + 5
= 185 seconds
So,
x= 185
So,
200 – x = 200 – 185 = 15
So,
The number of points the team earned = ( 200 – x ) ÷ \(\frac{1}{5}\)
= 15 ÷ \(\frac{1}{5}\)
= 15 × 5
= 75 points
Hence, from the above,
We can conclude that the number of points earned by the team is: 75 points

b. Do you think the team with the most points or the fewest points wins? Use an example to justify your answer.
Answer: The team with the most points wins the competition because
Reason:
Suppose team A takes 2 minutes and team B takes 3 minutes to complete the competition
So,
The time is taken by team A in seconds = 120 seconds
So,
x= 120
So,
200 – x = 200 – 120 = 80
Now,
The time is taken by team B in seconds = 180 seconds
So,
x= 180
So,
200 – x = 200 – 180 = 20
Now,
The number of points earned by team A = 80 ÷ \(\frac{1}{5}\)
= 400 points
The number of points earned by team B = 20 ÷ \(\frac{1}{5}\)
= 100 points
Hence, from the above,
We can conclude that the team with more points wins the competition

Divide Fractions Activity

Fraction Connection: Division

Directions:

  1. Players take turns rolling three dice.
  2. On your turn, evaluate the expression indicated by your roll and cover the answer.
  3. The first player to get four in a row, horizontally, vertically, or diagonally, wins!

Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 81
Big Ideas Math Answers 5th Grade Chapter 10 Divide Fractions 82

Divide Fractions Chapter Practice 10

10.1 Interpret Fractions as Division

Divide. Use a model to help.

Question 1.
1 ÷ 2 = ___
Answer: 1 ÷ 2 = \(\frac{1}{2}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
1 ÷ 2 = \(\frac{1}{2}\)

Question 2.
3 ÷ 10 = __
Answer: 3 ÷ 10 = \(\frac{3}{10}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
3 ÷ 10 = \(\frac{3}{10}\)

Question 3.
4 ÷ 7 = __
Answer: 4 ÷ 7 = \(\frac{4}{7}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
4 ÷ 7 = \(\frac{4}{7}\)

Question 4.
11 ÷ 15 = ___
Answer: 11 ÷ 15 = \(\frac{11}{15}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
11 ÷ 15 = \(\frac{11}{15}\)

Question 5.
8 ÷ 9 = ___
Answer: 8 ÷ 9 = \(\frac{8}{9}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
8 ÷ 9 = \(\frac{8}{9}\)

Question 6.
13 ÷ 20 = ___
Answer: 13 ÷ 20 = \(\frac{13}{20}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
Hence,
13 ÷ 20 = \(\frac{13}{20}\)

Question 7.
Modeling Real Life
Nine friends equally share 12 apples. What fraction of an apple does each friend get?
Answer: The fraction of an apple each friend get is: \(\frac{9}{12}\)

Explanation:
It is given that nine friends equally share 12 apples.
So,
The fraction of an apple each friend get = \(\frac{The number of friends}{The number of apples}\)
= \(\frac{9}{12}\)
Hence, from the above,
We can conclude that the fraction of an apple each friend get is: \(\frac{9}{12}\)

10.2 Mixed Numbers as Quotients

Divide. Use a model to help

Question 8.
8 ÷ 3 = ___

Answer: 8 ÷ 3 = 2\(\frac{2}{3}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
8 ÷ 3 = \(\frac{8}{3}\)
By using the partial quotients method,
8 ÷ 3 = 6 ÷ 3
= 2 R 2
Hence,
8 ÷ 3 = 2\(\frac{2}{3}\)

Question 9.
6 ÷ 5 = ___
Answer: 6 ÷ 5 = 1\(\frac{1}{5}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
6 ÷ 5 = \(\frac{6}{5}\)
By using the partial quotients method,
6 ÷ 5 = 5 ÷ 5
= 1 R 1
Hence,
6 ÷ 5 = 1\(\frac{1}{5}\)

Question 10.

10 ÷ 4 = ___
Answer: 10 ÷ 4 = 2\(\frac{2}{4}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
10 ÷ 4 = \(\frac{10}{4}\)
By using the partial quotients method,
10 ÷ 4 = 8 ÷ 4
= 2 R 2
Hence,
10 ÷ 4 = 2\(\frac{2}{4}\)

Question 11.

20 ÷ 11 = __
Answer: 20 ÷ 11 = 1\(\frac{9}{11}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
20 ÷ 11 = \(\frac{20}{11}\)
By using the partial quotients method,
20 ÷ 11 = 11 ÷ 11
= 1 R 9
Hence,
20 ÷ 11 = 1\(\frac{9}{11}\)

Question 12.

25 ÷ 2 = ___
Answer: 25 ÷ 2 = 12\(\frac{1}{2}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
25 ÷ 2 = \(\frac{25}{2}\)
By using the partial quotients method,
25 ÷ 2 = 24 ÷ 2
= 12 R 1
Hence,
25 ÷ 2 = 12\(\frac{1}{2}\)

Question 13.

64 ÷ 9 = ___
Answer: 64 ÷ 9 = 7\(\frac{1}{9}\)

Explanation:
We know that,
a ÷ b = \(\frac{a}{b}\)
a\(\frac{b}{c}\) = \(\frac{ac + b}{c}\)
\(\frac{a}{b}\) = Remainder\(\frac{Quotient}{Divisor}\)
So,
64 ÷ 9 = \(\frac{64}{9}\)
By using the partial quotients method,
64 ÷ 9 = 63 ÷ 9
= 7 R 1
Hence,
64 ÷ 9 = 7\(\frac{1}{9}\)

10.3 Divide Whole Numbers by Unit Fractions

Divide. Use a model to help.

Question 14.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 83
Answer: 4 ÷ \(\frac{1}{2}\) = 8

Explanation:
The given numbers are: 4 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
4 ÷ \(\frac{1}{2}\)  = 4 × \(\frac{2}{1}\)
= \(\frac{4}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 2 × 4}{1 × 1}\)
= 8
Hence,
4÷ \(\frac{1}{2}\) = 8

Question 15.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 84
Answer: 6 ÷ \(\frac{1}{5}\) = 30

Explanation:
The given numbers are: 6 and \(\frac{1}{5}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
6 ÷ \(\frac{1}{5}\)  = 6 × \(\frac{5}{1}\)
= \(\frac{6}{1}\) × \(\frac{5}{1}\)
= \(\frac{ 6 × 5}{1 × 1}\)
= 30
Hence,
6÷ \(\frac{1}{5}\) = 30

Question 16.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 85
Answer: 7 ÷ \(\frac{1}{4}\) = 28

Explanation:
The given numbers are: 7 and \(\frac{1}{4}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
7 ÷ \(\frac{1}{4}\)  = 7 × \(\frac{4}{1}\)
= \(\frac{7}{1}\) × \(\frac{4}{1}\)
= \(\frac{ 7 × 4}{1 × 1}\)
= 28
Hence,
7÷ \(\frac{1}{4}\) = 36

Question 17.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 86
Answer: 8 ÷ \(\frac{1}{3}\) = 24

Explanation:
The given numbers are: 8 and \(\frac{1}{3}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
8 ÷ \(\frac{1}{3}\)  = 8 × \(\frac{3}{1}\)
= \(\frac{8}{1}\) × \(\frac{3}{1}\)
= \(\frac{ 8 × 3}{1 × 1}\)
= 24
Hence,
8÷ \(\frac{1}{3}\) = 24

Question 18.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 87
Answer: 9 ÷ \(\frac{1}{2}\) = 18

Explanation:
The given numbers are: 9 and \(\frac{1}{2}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
9 ÷ \(\frac{1}{2}\)  = 9 × \(\frac{2}{1}\)
= \(\frac{9}{1}\) × \(\frac{2}{1}\)
= \(\frac{ 9 × 2}{1 × 1}\)
= 18
Hence,
9÷ \(\frac{1}{2}\) = 18

Question 19.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 88
Answer: 2 ÷ \(\frac{1}{10}\) = 20

Explanation:
The given numbers are: 2 and \(\frac{1}{10}\)
We know that,
a ÷ \(\frac{a}{b}\) = a × \(\frac{b}{a}\)
a= \(\frac{a}{1}\)
So,
2 ÷ \(\frac{1}{10}\)  = 2 × \(\frac{10}{1}\)
= \(\frac{2}{1}\) × \(\frac{10}{1}\)
= \(\frac{ 2 × 10}{1 × 1}\)
= 20
Hence,
2÷ \(\frac{1}{10}\) = 20

10.4 Divide Unit Fractions by Whole Numbers

Divide. Use a model to help.

Question 20.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 89
Answer: \(\frac{1}{7}\) ÷ 2 = \(\frac{1}{14}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 2
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 2
= \(\frac{1}{7}\) × \(\frac{1}{2}\)
= \(\frac{ 1 × 1}{7 × 2}\)
= \(\frac{1}{14}\)
Hence,
\(\frac{1}{7}\) ÷ 2 = \(\frac{1}{14}\)

Question 21.
Big Ideas Math Answers Grade 5 Chapter 10 Divide Fractions 90
Answer: \(\frac{1}{2}\) ÷ 9 = \(\frac{1}{18}\)

Explanation:
The given numbers are: \(\frac{1}{2}\) and 9
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{2}\) ÷ 9
= \(\frac{1}{2}\) × \(\frac{1}{9}\)
= \(\frac{ 1 × 1}{2 × 9}\)
= \(\frac{1}{18}\)
Hence,
\(\frac{1}{2}\) ÷ 9 = \(\frac{1}{18}\)

Question 22.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 91
Answer: \(\frac{1}{3}\) ÷ 7 = \(\frac{1}{21}\)

Explanation:
The given numbers are: \(\frac{1}{3}\) and 7
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{3}\) ÷ 7
= \(\frac{1}{3}\) × \(\frac{1}{7}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{3}\) ÷ 7 = \(\frac{1}{21}\)

Question 23.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 92
Answer: \(\frac{1}{6}\) ÷ 5 = \(\frac{1}{30}\)

Explanation:
The given numbers are: \(\frac{1}{6}\) and 5
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{6}\) ÷ 5
= \(\frac{1}{6}\) × \(\frac{1}{5}\)
= \(\frac{ 1 × 1}{6 × 5}\)
= \(\frac{1}{30}\)
Hence,
\(\frac{1}{6}\) ÷ 5 = \(\frac{1}{30}\)

Question 24.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 93
Answer: \(\frac{1}{7}\) ÷ 3 = \(\frac{1}{21}\)

Explanation:
The given numbers are: \(\frac{1}{7}\) and 3
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{7}\) ÷ 3
= \(\frac{1}{7}\) × \(\frac{1}{3}\)
= \(\frac{ 1 × 1}{7 × 3}\)
= \(\frac{1}{21}\)
Hence,
\(\frac{1}{7}\) ÷ 3 = \(\frac{1}{21}\)

Question 25.
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 94

Answer: \(\frac{1}{8}\) ÷ 4 = \(\frac{1}{32}\)

Explanation:
The given numbers are: \(\frac{1}{8}\) and 4
We know that,
\(\frac{a}{b}\) ÷ a  = \(\frac{a}{b}\) × \(\frac{1}{a}\)
a= \(\frac{a}{1}\)
So,
\(\frac{1}{8}\) ÷ 4
= \(\frac{1}{8}\) × \(\frac{1}{4}\)
= \(\frac{ 1 × 1}{8 × 4}\)
= \(\frac{1}{32}\)
Hence,
\(\frac{1}{8}\) ÷ 4 = \(\frac{1}{32}\)

10.5 Problem Solving: Fraction Division

Question 26.
A mechanic buys 1 gallon of oil. She uses \(\frac{1}{6}\) of the oil, and then divides the rest into 4 smaller bottles. How much does she put into each bottle?
Big Ideas Math Solutions Grade 5 Chapter 10 Divide Fractions 95
Answer: The amount of oil she put into each bottle is: \(\frac{5}{24}\)

Explanation:
It is given that a mechanic buys 1 gallon of oil and she uses \(\frac{1}{6}\) of the oil
So,
the remaining part of the oil = 1 –  \(\frac{1}{6}\)
= \(\frac{5}{6}\)
It is also given that she divides the rest of the oil into 4 smaller bottles.
So,
The amount of oil in each bottle = \(\frac{The remaining part of the oil}{The number of bottles}\)
= \(\frac{5}{6}\) ÷ 4
= \(\frac{5}{6}\) × \(\frac{1}{4}\)
= \(\frac{5 × 1}{6 × 4}\)
= \(\frac{5}{24}\)
Hence, from the above,
We can conclude that the amount of oil in each bottle is: \(\frac{5}{24}\)

Conclusion:

Make use of the quick links and try to solve the problems in a simple manner. Redefine your true self with the BIM Answer Key for Grade 5 curated by subject experts. Test your knowledge by solving the questions which are given at the end of the chapter.