## Engage NY Eureka Math Algebra 2 Module 3 Lesson 5 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 5 Exercise Answer Key

Exercise 1.

a. Write the following finite decimals as fractions (you do not need to reduce to lowest terms).

1, 1.4, 1.41, 1.414, 1.4142, 1.41421

Answer:

1.4 = \(\frac{14}{10}\)

1.41 = \(\frac{141}{100}\)

1.414 = \(\frac{1414}{1000}\)

1.4142 = \(\frac{14142}{10000}\)

1.41421 = \(\frac{141421}{100000}\)

b. Write 2^{1.4}, 2^{1.41} 2^{1.414}, and 2^{1.4142} in radical form.

Answer:

c. Use a calculator to compute decimal approximations of the radical expressions you found in part (b) to 5 decimal places. For each approximation, underline the digits that are also In the previous approximation, starting with 2.00000 done for you below. What do you notice?

2^{1} = 2 = 2.00000

Answer:

Exercise 2:

a. Write six terms of a sequence that a calculator can use to approximate 2^{π}.

(Hint: π = 3.14159…)

Answer:

{2^{3}, 2^{3.1}, 2^{3.14}, 2^{3.141}, 2^{3.1415}, 2^{3.14159}, ……}

b. Compute 2^{3.14} and 2^{π} on your calculator. In which digit do they start to differ?

Answer:

2^{3.14} = \(\sqrt[100]{2^{314}}\) ≈ 8.81524 ≈ 8.82497

They start to differ in the hundredths place.

c. How could you improve the accuracy of your estimate of 2^{π}?

Answer:

Include more digits of the decimal approximation of π in the exponent.

### Eureka Math Algebra 2 Module 3 Lesson 5 Problem Set Answer Key

Question 1.

Is it possible for a number to be both rational and irrational?

Answer:

No. Either the number can be written as \(\frac{p}{q}\) for integers p and q or it cannot. If it can, the number is rational. If it cannot, the number is irrational.

Question 2.

Use properties of exponents to rewrite the following expressions as a number or an exponential expression with only one exponent.

a. \(\left(2^{\sqrt{3}}\right)^{\sqrt{3}}\)

Answer:

8

b. \(\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}\)

Answer:

2

c. \(\left(3^{1+\sqrt{5}}\right)^{1-\sqrt{5}}\)

Answer:

\(\frac{1}{81}\)

d. \(3^{\frac{1+\sqrt{5}}{2}} \cdot 3^{\frac{1-\sqrt{5}}{2}}\)

Answer:

3

e.

\(3^{\frac{1+\sqrt{5}}{2}} \div 3^{\frac{1-\sqrt{5}}{2}}\)

Answer:

\(3^{\sqrt{5}}\)

f. \(3^{2 \cos ^{2}(x)} \cdot 3^{2 \sin ^{2}(x)}\)

Answer:

9

Question 3.

a. Between what two integer powers of 2 does 2 lie?

Answer:

2^{2} < \(2^{\sqrt{5}}\) < 2^{3}

b. Between what two integer powers of 3 does 3” lie?

Answer:

3^{3} < \(3^{\sqrt{10}}\) < 3^{4}

c. Between what two integer powers of 5 does 5’ lie?

Answer:

5^{1} < \(5^{\sqrt{3}}\) < 5^{2}

Question 4.

Use the process outlined in the lesson to approximate the number \(2^{\sqrt{5}}\). Use the approximation √5 ≈ 2.23606798.

a. Find a sequence of five intervals that contain √5 whose endpoints get successively closer to √5.

Answer:

2 < √5 < 3

2.2 < √5 < 2.3

2.23 < √5 < 2.24

2.236 < √5 < 2.237

2.2360 < √5 < 2.2361

b. Find a sequence of five intervals that contain \(2^{\sqrt{5}}\) whose endpoints get successively closer to \(2^{\sqrt{5}}\). Write your intervals in the form 2^{r} < \(2^{\sqrt{5}}\) < 2^{s} for rational numbers r and s.

Answer:

2^{2} <\(2^{\sqrt{5}}\) < 2^{3}

2^{2.2} <\(2^{\sqrt{5}}\) < 2^{2.3}

2^{2.23} <\(2^{\sqrt{5}}\) < 2^{2.24}

2^{2.236} < \(2^{\sqrt{5}}\) < 2^{2.237}

2^{2.2360} < \(2^{\sqrt{5}}\) < 2^{2.2361}

c. Use your calculator to find approximations to four decimal places of the endpoints of the intervals in part (b).

Answer:

4.0000 < \(2^{\sqrt{5}}\) < 8.0000

4.5948 < \(2^{\sqrt{5}}\) < 4.9246

4.6913 < \(2^{\sqrt{5}}\) < 4.7240

4.7109 < \(2^{\sqrt{5}}\) < 4.7142

4.7109 < \(2^{\sqrt{5}}\) < 4.7112

d. Based on your work in part (c), what is your best estimate of the value of 2?

Answer:

\(2^{\sqrt{5}}\) ≈ 4.711

e. Can we tell if 2 is rational or irrational? Why or why not?

Answer:

No. We do not have enough information to determine whether 2 has a repeated pattern in its decimal representation or not.

Question 5.

Use the process outlined in the lesson to approximate the number \(3^{\sqrt{10}}\). Use the approximation √10 ≈ 3.162 277 7.

a. Find a sequence of five intervals that contain \(3^{\sqrt{10}}\) whose endpoints get successively closer to \(3^{\sqrt{10}}\). Write your intervals in the form 3^{r} < \(3^{\sqrt{10}}\) < 3^{s} for rational numbers r and s.

Answer:

3^{3} < \(3^{\sqrt{10}}\) < 3^{4}

3^{3.1} < \(3^{\sqrt{10}}\) < 3^{3.2}

3^{3.16} < \(3^{\sqrt{10}}\) < 3^{3.17}

3^{3.162} < \(3^{\sqrt{10}}\) < 3^{3.163}

3^{3.1622} < \(3^{\sqrt{10}}\) < 3^{3.1623}

b. Use your calculator to find approximations to four decimal places of the endpoints of the intervals in part (a).

Answer:

9.0000 < \(3^{\sqrt{10}}\) < 81.0000

30.1353 < \(3^{\sqrt{10}}\) < 33.6347

32.1887 < \(3^{\sqrt{10}}\) < 32. 5443

32.2595 < \(3^{\sqrt{10}}\) < 32.2949

32.2666 < \(3^{\sqrt{10}}\) < 32.2701

c. Based on your work in part (b), what is your best estimate of the value of \(3^{\sqrt{10}}\)?

Answer:

\(3^{\sqrt{10}}\) ≈ 32.27

Question 6.

Use the process outlined in the lesson to approximate the number \(5^{\sqrt{7}}\). Use the approximation √7 ≈ 2.645 751 31.

a. Find a sequence of seven intervals that contain \(5^{\sqrt{7}}\) whose endpoints get successively closer to \(5^{\sqrt{7}}\). Write your intervals in the form 5^{r} < \(5^{\sqrt{7}}\) < 5^{s} for rational numbers r and s.

Answer:

5^{2} < \(5^{\sqrt{7}}\) < 5^{3}

5^{2.6} < \(5^{\sqrt{7}}\) < 5^{2.7}

5^{2.64} < \(5^{\sqrt{7}}\) < 5^{2.65}

5^{2.645} < \(5^{\sqrt{7}}\) < 5^{2.646}

5^{2.6457} < \(5^{\sqrt{7}}\) < 5^{2.6458}

5^{2.64575} < \(5^{\sqrt{7}}\) < 5^{2.64576}

5^{2.645751} < \(5^{\sqrt{7}}\) < 5^{2.645752}

b. Use your calculator to find approximations to four decimal places of the endpoints of the intervals in part (a).

Answer:

25.0000 < \(5^{\sqrt{7}}\) < 125.0000

65.6632 < \(5^{\sqrt{7}}\) < 77.1292

70.0295 < \(5^{\sqrt{7}}\) < 71.1657

70.5953 < \(5^{\sqrt{7}}\) < 70.7090

70.6749 < \(5^{\sqrt{7}}\) < 70.6862

70.6805 <\(5^{\sqrt{7}}\) < 70.6817

70.6807 <\(5^{\sqrt{7}}\) < 70. 6808

c. Based on your work in part (b), what is your best estimate of the value of 5?

Answer:

\(5^{\sqrt{7}}\) ≈ 70.681

Question 7.

A rational number raised to a rational power can either be rational or irrational. For example, \(4^{\frac{1}{2}}\) is rational because \(4^{\frac{1}{2}}\) = 2, and \(2^{\frac{1}{4}}\) is irrational because \(2^{\frac{1}{4}}\) = \(\sqrt[4]{2}\). In this problem, you will investigate the possibilities for an irrational number raised to an irrational power.

a. Evaluate \((\sqrt{2})^{(\sqrt{2})^{\sqrt{2}}}\).

Answer:

\((\sqrt{2})^{(\sqrt{2})^{\sqrt{2}}}\) = \((\sqrt{2})^{\sqrt{2} \cdot \sqrt{2}}\) = (√2)^{2} = 2

b. Can the value of an irrational number raised to an irrational power ever be rational?

Answer:

Yes. For instance, in part (a) above, √2 is irrational, and the number \(\sqrt{2}^{\sqrt{2}}\) is either irrational or rational. If \(\sqrt{2}^{\sqrt{2}}\) is rational, then this is an example of an irrational number raised to an irrational power that is rational. Otherwise, \(\sqrt{2}^{\sqrt{2}}\) is irrational, and part (a) is an example of an irrational number raised to an irrational power that is rational.

### Eureka Math Algebra 2 Module 3 Lesson 5 Exit Ticket Answer Key

Question 1.

Use the process outlined in the lesson to approximate the number \(2^{\sqrt{3}}\). Use the approximation √3 ≈ 1. 732 050 8.

a. Find a sequence of five intervals that contain √3 whose endpoints get successively closer to √3.

Answer:

1 < √3 < 2

1.7 < √3 < 1.8

1.73 < √3 < 1.74

1.732 < √3 < 1.733

1.7320 < √3 < 1.7321

b. Find a sequence of five intervals that contain \(2^{\sqrt{3}}\) whose endpoints get successively closer to \(2^{\sqrt{3}}\). Write your intervals in the form 2^{r} < \(2^{\sqrt{3}}\) < 2^{s} for rational numbers r and s.

Answer:

2^{1} <\(2^{\sqrt{3}}\) < 2^{2}

2^{1.7} <\(2^{\sqrt{3}}\) < 2^{1.8}

2^{1.73} <\(2^{\sqrt{3}}\) < 2^{1.74}

2^{1.732} < \(2^{\sqrt{3}}\) < 2^{1.733}

2^{1.7320} < \(2^{\sqrt{3}}\) < 2^{1.7321}

c. Use your calculator to find approximations to four decimal places of the endpoints of the intervals in part (b).

Answer:

2.0000 < \(2^{\sqrt{3}}\) < 4.0000

3.2490 < \(2^{\sqrt{3}}\) < 3.4822

3. 3173 <\(2^{\sqrt{3}}\) < 3.3404

3.3219 < \(2^{\sqrt{3}}\) < 3.3242

3.3219 < \(2^{\sqrt{3}}\) < 3.3221

d. Based on your work in part (c), what is your best estimate of the value of 2?

Answer:

\(2^{\sqrt{3}}\) ≈ 3.322