## Engage NY Eureka Math Algebra 2 Module 3 Lesson 9 Answer Key

Eureka Math Algebra 2 Module 3 Lesson 9 Opening Exercise Answer Key

Opening Exercise:

a. Evaluate WhatPower2(8). State your answer as a logarithm, and evaluate It.
log2(8) = 3

b. Evaluate WhatPower5(625). State your answer as a logarithm, and evaluate it.
log5(625) = 4

Eureka Math Algebra 2 Module 3 Lesson 9 Exploratory Challenge Answer Key

Exploratory Challenge:

Autumn is starting a new club with eight members including herself. She wants everyone to have a secret identification code made up of only A’s and B’s, known as an ID code. For example, using two characters, her ID code could be AA.

a. Using A’s and B’s, can Autumn assign each club member a unique two-character ID code using only A’s and B’s? Justify your answer. Here’s what Autumn has so far.

img 1

No, she cannot assign a unique 2-character ID code to each member. The only codes available are AA, BA, AB, and BB, and there are eight people.

b. Using A’s and B’s, how many characters would be needed to assign each club member a unique ID code? Justify your answer by showing the ID codes you would assign to each club member by completing the table above (adjust Autumn’s ID if needed).
You would need three characters in each ID code. A completed table is shown below. Students could assign one of the unique codes to any club member, so this is not the only possible solution.

img 2

When the club grew to 16 members, Autumn started noticing a pattern.
Using A’s and B’s:
i. Two people could be given a secret ID code with 1 letter: A and B.
ii. Four people could be given a secret ID code with 2 letters: AA, BA, AB, BB.
iii. Eight people could be given a secret ID code with 3 letters: AAA, BAA, ABA, BBA, AAB, BAB, ABB, BBB.

c. Complete the following statement, and list the secret ID codes for the 16 people. 16 people could be given a secret ID code with _________ letters using A’s and B’s.

img 3

16 people could be given a secret ID code with 4 characters. Notice the original members have their original three-character code with an A added to the end. Then, the newer members have the original three-character codes with a B added to the end.

img 4

d. Describe the pattern in words. What type of function could be used to model this pattern?
The number of people in the club is a power of 2. The number of characters needed to generate a unique ID code using only two characters is the exponent of the power of 2.
log2(2) = 1
log2(4) = 2
log2(8) = 3
log2(16) = 4
A logarithm function could be used to model this pattern. For 16 people, you will need a four-character ID code because log2(16) = 4.

Eureka Math Algebra 2 Module 3 Lesson 9 Exercise Answer Key

Exercises 1 – 2:

In the previous problems, the letters A and B were like the digits In a number. A four-digit ID code for Autumn’s club could be any four-letter arrangement of A’s and B’s because in her ID system, the only digits are the letters A and B.

Exercise 1.
When Autumn’s club grows to include more than 16 people, she will need five digits to assign a unique ID code to each club member. What is the maximum number of people that could be in the club before she needs to switch to a six-digit ID code? Explain your reasoning.
Since log2(32) = S and log2(64) = 6, she will need to switch to a six-digit ID code when the club grows to more than 32 members.

Exercise 2.
If Autumn has 256 members in her club, how many digits would she need to assign each club member a unique ID code using only A’s and B’s? Show how you got your answers.
She will need 8 digits because log2(256) = 8.

Example:

A thousand people are given unique identifiers made up of the digits 0, 1, 2, …, 9. How many digits would be needed for each ID number?
You would just need three digits: (000, 001, 002, …, 099, 100, 101, …, 998, 999).

Exercises 3 – 4:

Exercise 3.
There are approximately 317 million people in the United States. Compute and use log(100000000) and log(l000000000) to explain why Social Security numbers are 9 digits long.
We know that log(100000000) = 8, which is the number of digits needed to assign an ID number to loo million people. We know that log(l000000000) = 9, which is the number of digits needed to assign an ID number to 1 billion people. The United States government will not need to increase the number of digits in a Social Security number until the United States population reaches one billion.

Exercise 4.
There are many more telephones than the number of people in the United States because of people having home phones, cell phones, business phones, fax numbers, etc. Assuming we need at most 10 billion phone numbers in the United States, how many digits would be needed so that each phone number is unique? Is this reasonable? Explain.
Since log( 10000000000) = 10, you would need a ten-digit phone number in order to have ten billion unique numbers. Phone numbers in the United States are 10 digits long. If you divide 10 billion by 317 million (the number of people in the United States), that would allow for approximately 31 phone numbers per person. That is plenty of numbers for individuals to have more than one number, leaving many additional numbers for businesses and the government.

Eureka Math Algebra 2 Module 3 Lesson 9 Problem Set Answer Key

Question 1.
The student body president needs to assign each officially sanctioned club on campus a unique ID code for purposes of tracking expenses and activities. She decides to use the letters A, B, and C to create a unique three-character code for each club.

a. How many clubs can be assigned a unique ID code according to this proposal?
Since log3(27) = 3, the president could assign codes to 27 clubs according to this proposal.

b. There are actually over 500 clubs on campus. Assuming the student body president still wants to use the letters A, B, and C, how many characters would be needed to generate a unique ID code for each club?
We need to estimate log3(500). Since 35 = 243 and 36 = 729, she could use a six-character combination of letters and have enough unique IDs for up to 729 clubs.

Question 2.
Can you use the numbers 1, 2, 3, and 4 in a combination of four digits to assign a unique ID code to each of 500 people? Explain your reasoning.
log4(4) = 1
log4(16) = 2
log4(64) = 3
log4(256) =4
log4(1024) = 5
No. You would need to use a five-digit ID code using combinations of 1’s, 2’s, 3’s, and 4’s such as 11111 or 12341, or you could use the numbers 1 to 5 in four characters such as 1231, 1232, 1233, 1234, 1235, etc., because
log5(625) = 4.

Question 3.
Automobile license plates typically have a combination of letters (26) and numbers (10). Overtime, the state of New York has used different criteria to assign vehicle license plate numbers.

a. From 1973 to 1986, the state used a 3-letter and 4-number code where the three letters indicated the county where the vehicle was registered. Essex County had 13 different 3-letter codes in use. How many cars could be registered to this county?
Since log(10000) = 4, the 4-digit code could be used to register up to 10,000 vehicles. Multiply that by 13 different county codes, and up to 130,000 vehicles could be registered in Essex County.

b. Since 2001, the state has used a 3-letter and 4-number code but no longer assigns letters by county. Is this coding scheme enough to register 10 million vehicles?
Since log26(x) = 3 when x = 263 = 17,576, there are 17,576 three-letter codes. Since log(l0000) = 4, there are 10,000 four-digit codes. Multiply 17576 . 10,000 = 175,760,000, and we see that this scheme generates over 100 million license plate numbers.

Question 4.
The Richter scale uses base 10 logarithms to assign a magnitude to an earthquake based on the amount of force released at the earthquake’s source as measured by seismographs in various locations.

a. Explain the difference between an earthquake that is assigned a magnitude of 5 versus one assigned a magnitude of 7.
The difference between S and 7 is 2, so a magnitude of 7 would be 102, or 100, times greater force.

b. An earthquake with magnitude 2 can usually only be felt by people located near the earthquake’s origin, called its epicenter. The strongest earthquake on record occurred in Chile In 1960 with magnitude 9. 5. How many times stronger is the force of an earthquake with magnitude 9. 5 than the force of an earthquake with magnitude 2?
The difference between 2 and 9. 5 is 7.5, so it would be about 107.5 times the force. This is approximately 31 million times greater force.

c. What is the magnitude of an earthquake whose force is 1,000 times greater than a magnitude 4.3 quake?
Since 103 = 1000, the magnitude would be the sum of 4.3 and 3, which is 7.3.

Question 5.
Sound pressure level is measured in decibels (dB) according to the formula L = 10 log ($$\frac{I}{I0}$$), where I is the intensity of the sound and I0 is a reference intensity that corresponds to a barely perceptible sound.

a. Explain why this formula would assign 0 decibels to a barely perceptible sound.
If we let I = I0, then
L = 10 log($$\frac{I}{I0}$$)
L = 10 log(1)
L = 10 . 0
L = 0.
Therefore, the reference intensity is always 0 dB.

b. Decibel levels above 120 dB can be painful to humans. What would be the intensity that corresponds to this level?
120 = 10 log ($$\frac{I}{I0}$$)
1.2 = log($$\frac{I}{I0}$$)
$$\frac{I}{I0}$$ = 101.2
I ≈ 15.8I0
From this equation, we can see that the intensity is about 16 times greater than barely perceptible sound.

Eureka Math Algebra 2 Module 3 Lesson 9 Exit Ticket Answer Key

Question 1.
A brand-new school district needs to generate ID numbers for its student body. The district anticipates a total enrollment of 75, 000 students within the next ten years. Will a five-digit ID number using the symbols 0, 1, …, 9 be enough? Explain your reasoning.
log(10000) = 4 and log(100000) = 5, so 5 digits should be enough. However, students who enter school at the kindergarten level in the tenth year of this numbering scheme would need to keep their IDs for 13 years. Dividing 75,000 by 13 shows there would be roughly 6,000 students per grade.

Adding that many students per year would take the number of needed IDs at any one time over 100,000 in just afew more years. The district should probably use a six digit ID number.

## Engage NY Eureka Math Algebra 2 Module 31 Lesson 30 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 31 Mathematical Modeling Exercise Answer Key

Mathematical Modeling Exercise:

You have charged $1,500 for the down payment on your car to a credit card that charges 19.99% annual interest, and you plan to pay a fixed amount toward this debt each month until it is paid off. We denote the balance owed after the ne” payment has been made as b. a. What is the monthly interest rate, i? Approximate i to 5 decimal places. Answer: i = $$\frac{0.1999}{12}$$ ≈ 0.01666 b. You have been assigned to either the 50-team, the 100-team, or the 150-team, where the number indicates the size of the monthly payment R you make toward your debt. What is your value of R? Answer: Students will answer 50, 100, or 150 as appropriate. c. Remember that you can make any size payment toward a credit card debt, as long as it is at least as large as the minimum payment specified by the lender. Your lender calculates the minimum payment as the sum of 1% of the outstanding balance and the total interest that has accrued over the month, or$25, whichever is greater. Under these stipulations, what is the minimum payment? Is your monthly payment R at least as large as the minimum payment?
The minimum payment is 0.01($1500) + 0.01666($1500) = $39. 99. All given values of R are greater than the minimum payment. d. Complete the following table to show 6 months of payments. img 1 Answer: img 2 img 3 img 4 e. Write a recursive formula for the balance bn in month n in terms of the balance bn – 1. Answer: To calculate the new balance, bn, we compound interest for one month on the previous balance bn – 1 and then subtract the payment R: bn = bn – 1(1 + i) – R, with b0 = 1500. f. Write an explicit formula for the balance b in month n, leaving the expression 1 + i in symbolic form. Answer: We have the following formulas: b1 = b0(1 + i) – R b2 = b1(1 + i) – R = [b0(1 + i) – R](1 + i) – R = b0(1 + i)2 – R(1 + i) – R b3 = b2(1 + i) – R = [b0(1 + i)2 – R(1 + i) – R](1 + i) – R = b0(1 + i)3 R(1 + i)2 – R(1 + i) – R . . . . bn = b0(1 + i)n – R(1 + i)n – 1 – R(1 + i)n – 2 ….. -R(1 + i) – R g. Rewrite your formula in part (f) using r to represent the quantity (1 + i). Answer: bn = b0rn – Rrn – 1 -Rrn – 2 – Rr – R = b0rn – R(1 + r + r2 + … + rn – 1) h. What can you say about your formula in part (g)? What term do we use to describe r in this formula? Answer: The formula in part (g) contains the sum of a finite geometric series with common ratio r. i. Write your formula from part(g) in summation notation using Σ. Answer: bn = b0rn – R(1 + r + r2 + …. + rn – 1) = b0rn – R$$\left(\frac{1-r^{n}}{1-r}\right)$$ k. Find the month when your balance is paid off. Answer: The balance is paid off when bn = 0. (The final payment is less than a full payment so that the debt is not overpaid.) Students will likely do this calculation with the values of r, b0, and R substituted in. b0rn – R$$\left(\frac{1-r^{n}}{1-r}\right)$$ = 0 b0rn = R$$\left(\frac{1-r^{n}}{1-r}\right)$$ (1 – r)(b0rn) = R(1 – rn) (1 – r)(b0rn) + Rrn = R rn = $$\frac{\boldsymbol{R}}{\left(\boldsymbol{b}_{0}(\mathbf{1}-\boldsymbol{r})+\boldsymbol{R}\right)}$$ n log(r) = log$$\left(\frac{\boldsymbol{R}}{\left(\boldsymbol{b}_{0}(\mathbf{1}-\boldsymbol{r})+\boldsymbol{R}\right)}\right)$$ n = $$\frac{\log \left(\frac{\boldsymbol{R}}{\left(\boldsymbol{b}_{0}(\mathbf{1}-\boldsymbol{r})+\boldsymbol{R}\right)}\right)}{\log (\boldsymbol{r})}$$ If R = 50, then n ≈ 41.925. The debt is paid off in 42 months. If R = 100, then n ≈ 17.49. The debt is paid off in 18months. If R = 150, then n ≈ 11.0296. The debt is paid off in 12 months. l. Calculate the total amount paid over the life of the debt. How much was paid solely to interest? Answer: For R = 50: The debt is paid in 41 payments of$50, and the last payment is the amount b41 with interest:
50(41) + (1 + i)b41 = 2050 + r(b0rn – R$$\left(\frac{1-r^{n}}{1-r}\right)$$
≈ 2050 + r(45.61)
≈ 2096.37.

The total amount paid using monthly payments of $50 is$2,096.37. Of this amount, $596.37 is interest. For R = 100: The debt is paid in 17 payments of$100, and the last payment is the amount b17 with interest.

100(17) + (1 + i)b17 = 1700 + r$$\left(b_{0} r^{17}-R\left(\frac{1-r^{17}}{1-r}\right)\right)$$
≈ 1700 + r(40.52)
≈ 1740.52
The total amount paid using monthly payments of $100 is$1,740.52. Of this amount, $240.52 is interest. For R = 150: The debt is paid in 11 payments of$150, and the last payment is the amount b11 with interest.
150(11) + (1 + i)b11 = 1700 + r$$\left(b_{0} r^{n}-R\left(\frac{1-r^{n}}{1-r}\right)\right)$$
≈ 1650 +r(4.49)
≈ 1654.49
The total amount paid using monthly payments of $150 is$1,654.49. Of this amount, $154.49 is interest. ### Eureka Math Algebra 2 Module 3 Lesson 31 Problem Set Answer Key Question 1. Suppose that you have a$2, 000 balance on a credit card with a 29.99% annual interest rate, compounded monthly, and you can afford to pay $150 per month toward this debt. a. Find the amount of time it takes to pay off this debt. Give your answer in months and years. Answer: img 5 b. Calculate the total amount paid over the life of the debt. Answer: 16.419 .$150 = $2462.85 c. How much money was paid entirely to the interest on this debt? Answer:$462.85

Question 2.
Suppose that you have a $2,000 balance on a credit card with a 14.99% annual interest rate, and you can afford to pay$150 per month toward this debt.

a. Find the amount of time it takes to pay off this debt. Give your answer in months and years.
img 6

b. Calculate the total amount paid over the life of the debt.
14. 676 . $150 =$2,201.40

c. How much money was paid entirely to the interest on this debt?
$201.40 Question 3. Suppose that you have a$2,000 balance on a credit card with a 7.99% annual interest rate, and you can afford to pay $150 per month toward this debt. a. Find the amount of time it takes to pay off this debt. Give your answer in months and years. Answer: img 7 b. Calculate the total amount paid over the life of the debt. Answer: 14.009 .$150 = $2101.35 c. How much money was paid entirely to the interest on this debt? Answer:$101.35

Question 4.
Summarize the results of Problems 1, 2, and 3.
Answers will vary but should include the fact that the total interest paid in each case dropped by about half with every problem. Lower interest rates meant that the loan was paid off more quickly and that less was paid in total.

Question 5.
Brendan owes $1, 500 on a credit card with an interest rate of 12%. He is making payments of$100 every month to pay this debt off. Maggie is also making regular payments to a debt owed on a credit card, and she created the following graph of her projected balance over the next 12 months.

img 8

a. Who has the higher initial balance? Explain how you know.
Reading from the graph, Maggie’s initial balance is between $1,700 and$1,800, and we are given that Brendan’s initial balance is $1, 500, so Maggie has the larger initial balance. b. Who will pay their debt off first? Explain how you know. Answer: From the graph, it appears that Maggie will pay off her debt between months 12 and 14. Brendan’s balance in month n can be modeled by the function b = 1500(1.01)n – 100$$\left(\frac{1.01^{n}-1}{0.01}\right)$$, which is equal to zero when n ≈ 16.3. Thus, Brendan’s debt will be paid in month 17, so Maggie’s debt will be paid off first. Question 6. Alan and Emma are both making$200 monthly payments toward balances on credit cards. Alan has prepared a table to represent his projected balances, and Emma has prepared a graph.

img 9

img 10

a. What is the annual interest rate on Alan’s debt? Explain how you know.
One month’s interest on the balance of $2,000 was$41.65, so 41.65 = i(2000). Then the monthly interest rate is i = 0.020825, and the annual rate is 12i = 0. 2499, so the annual rate on Alan’s debt is 24.99%.

b. Who has the higher initial balance? Explain how you know.
From the table, we can see that Alan’s initial balance is $2,000, while Emma’s initial balance is the y-intercept of the graph, which is above$2, 000. Thus, Emma’s initial balance is higher.

c. Who will pay their debt off first? Explain how you know.
Both Alan and Emma will pay their debts off in month 12 because both of their balances in month 11 are under $100. d. What do your answers to parts (a), (b), and (c) tell you about the interest rate for Emma’s debt? Answer: Because Emma had the higher initial balance, and they made the same number of payments, Emma must have a lower interest rate on her credit card than Alan does. In fact, since the graph decreases apparently linearly, this implies that Emma has an interest rate of 0%. Question 7. Both Gary and Helena are paying regular monthly payments to a credit card balance. The balance on Gary’s credit card debt can be modeled by the recursive formula gn = gn – 1(1.01666) – 200 with g0 = 2500, and the balance on Helena’s credit card debt can be modeled by the explicit formula hn = 2000(1.01666)n – 250$$\left(\frac{1.01666^{n}-1}{0.01666}\right)$$ for n ≥ 0. a. Who has the higher Initial balance? Explain how you know. Answer: Gary has the higher initial balance. Helena’s initial balance is$2,000, and Gary’s is $2,500. b. Who has the higher monthly payment? Explain how you know. Answer: Helena has the higher monthly payment. She is paying$250 every month while Gary is paying $200. c. Who will pay their debt off first? Explain how you know. Answer: Helena will pay her debt off first since she starts at a lower balance and is paying more per month. Additionally, they appear to have the same interest rates. Question 8. In the next lesson, we will apply the mathematics we have learned to the purchase of a house. In preparation for that task, you need to come to class prepared with an idea of the type of house you would like to buy. a. Research the median housing price in the county where you live or where you wish to relocate. Answer: Answers will vary. b. Find the range of prices that are within 25% of the median price from part (a). That is, if the price from part (a) was P, then your range is 0.75P to 1.25P. Answer: Answers will vary. c. Look at online real estate websites, and find a house located in your selected county that falls into the price range specified in part (b). You will be modeling the purchase of this house in Lesson 32, so bring a printout of the real estate listing to class with you. Answer: Answers will vary. Question 9. Select a career that interests you from the following list of careers. If the career you are interested in is not on this list, check with your teacher to obtain permission to perform some independent research. Once it has been selected, use the career to answer questions in Lesson 32 and Lesson 33. img 11 Answer: ### Eureka Math Algebra 2 Module 3 Lesson 31 Exit Ticket Answer Key Question 1. Suppose that you currently have one credit card with a balance of$10,000 at an annual rate of 24. 00% interest. You have stopped adding any additional charges to this card and are determined to pay off the balance. You have worked out the formula b = b0rn – R(1 + r + r2 + …. + rn – 1), where b0 is the initial balance, bn is the balance after you have made n payments, r = 1 + i, where i is the monthly interest rate, and R is the amount you are planning to pay each month.

a. What is the monthly interest rate j? What is the growth rate, r?
The monthly interest rate ils given by i = $$\frac{0.24}{12}$$ = 0.02, and r = 1 + i = 1.02.

b. Explain why we can rewrite the given formula as bn = b0rn – R$$\left(\frac{1-r^{n}}{1-r}\right)$$.
Using summation notation and the sum formula for afinite geometric series, we have
1 + r + r2 + …. + rn – 1 = $$\sum_{k=0}^{n-1} r^{n k}$$ = $$\frac{1-r^{n}}{1-r}$$.

Then the formula becomes
bn = b0rn – R(1 + r + r2 + …. + rn – 1)
= b0rn – R$$\frac{1-r^{n}}{1-r}$$

c. How long does it take to pay off this debt if you can afford to pay a constant \$250 per month? Give the answer in years and months.
When the debt is paid off, bn ≤ 0. Then b0rn – R$$\frac{1-r^{n}}{1-r}$$ = 0, and b0rn = R$$\frac{1-r^{n}}{1-r}$$ Since b0 = 10000, R = 250, and r = 1.02, we have
10000(1.02)n ≤ 250$$\frac{1-1.02^{n}}{1-1.02}$$
10000(1.02)n ≤ -12500(1 – 1.02n)
10000(1.02)n ≤12500(1.02n – 1)
(1.02)n ≤ 1.25(1.02)n – 1.25
1.25 ≤ 0.25(1.02)n
5 ≤ 1.02n
log(5) ≤ n log(1.02)
n ≥ $$\frac{log{5}}log{1.02}$$
n ≥ 81.27
It takes 82 months to pay off this debt, which means it takes 6 years and 10 months.

## EngageNY Math Algebra 2 Module 3 Answer Key | Algebra 2 Eureka Math Module 3 Answer Key

Eureka Math Algebra 2 Module 3 Exponential and Logarithmic Functions

Eureka Math Algebra 2 Module 3 Topic A Real Numbers

Engage NY Math Algebra 2 Module 3 Topic B Logarithms

Eureka Math Algebra 2 Module 3 Mid Module Assessment Answer Key

Algebra 2 Eureka Math Module 3 Topic C Exponential and Logarithmic Functions and their Graphs

EngageNY Algebra 2 Math Module 3 Topic D Using Logarithms in Modeling Situations

Great Minds Eureka Math Algebra 2 Module 3 Topic E Geometric Series and Finance

Eureka Math Algebra 2 Module 3 End of Module Assessment Answer Key