# Eureka Math Grade 8 Module 7 Mid Module Assessment Answer Key

## Engage NY Eureka Math 8th Grade Module 7 Mid Module Assessment Answer Key

### Eureka Math Grade 8 Module 7 Mid Module Assessment Task Answer Key

Question 1.
a. What is the decimal expansion of the number $$\frac{35}{7}$$? Is the number $$\frac{35}{7}$$ rational or irrational? Explain.
$$\frac{35}{7}$$ = 5.000….
The number $$\frac{35}{7}$$ is a rational number. Rational numbers have decimal expansions that repeat. In this case, the decimal that repeats is a zero.

b. What is the decimal expansion of the number $$\frac{4}{33}$$? Is the number $$\frac{4}{33}$$ rational or irrational? Explain.
$$\frac{4}{33}$$ = 0.1212……… = $$0 . \overline{12}$$

The number $$\frac{4}{33}$$ is a rational number. Rational numbers have decimal expansions that repeat. The digits 12 repeat in the decimal expansion of $$\frac{4}{33}$$. So, $$\frac{4}{33}$$ is rational.

Question 2.
a. Write $$0 . \overline{345}$$ as a fraction.
Let x be $$0 . \overline{345}$$
103x = 103($$0 . \overline{345}$$)
1000 x = 1000($$0 . \overline{345}$$)
1000x = $$345 . \overline{345}$$
1000x = 345 + x
1000x – x = 345 + x – x
999x = 345
x = $$\frac{345}{999}$$ = $$\frac{115}{333}$$

b. Write $$2.8 \overline{40}$$ as a fraction.
Let x be $$2.8 \overline{40}$$
10x = $$2.8 \overline{40}$$
10x = 28 + $$0. \overline{40}$$
10x = $$\frac{(28)(99)+40}{99}$$
x = $$\frac{2812}{99}$$($$\frac{1}{10}$$)
x = $$\frac{2812}{990}$$ = $$\frac{1406}{495}$$

Let y be $$0. \overline{40}$$
102y = 102 ($$0. \overline{40}$$)
100y = $$40. \overline{40}$$
100y = 40 + y
100y – y = 40 + y – y
99y = 40
y = $$\frac{40}{99}$$
$$2.8 \overline{40}$$ = $$\frac{2812}{990}$$ = $$\frac{1406}{495}$$

c. Brandon stated that 0.66 and $$\frac{2}{3}$$ are equivalent. Do you agree? Explain why or why not.
No, I do not agree with brandon. The decimal 0.66 is equal to the fraction $$\frac{66}{100}$$ = $$\frac{33}{50}$$, not $$\frac{2}{3}$$. Also, the number $$\frac{2}{3}$$ is equal to the infinit decimal $$0 . \overline{6}$$. The number 0.66 is a finite decimal. Therefore, 0.66 and $$\frac{2}{3}$$ are not equivlent.

d. Between which two positive integers does $$\sqrt{33}$$ lie?
The number $$\sqrt{33}$$ is between 5 and 6 because
52 < ($$\sqrt{33}$$)2 < 62

e. For what integer x is $$\sqrt{x}$$ closest to 5.25? Explain.
(5.25)2 = 27.5625
Since $$\sqrt{x}$$ is the square root of x, then x2 will give me the integer that belongs in the square root.
(5.25)2 = 27.5625, which is closest to the integer 28.

Question 3.
Identify each of the following numbers as rational or irrational. If the number is irrational, explain how you know.
a. $$\sqrt{29}$$
Irrational because 29 is not a perfect square and $$\sqrt{29}$$ has an infinite decimal expansion that does not repeat.

b. $$5 . \overline{39}$$
Rational

c. $$\frac{12}{4}$$
Rational

d. $$\sqrt{36}$$
Rational

e. $$\sqrt{5}$$
Irrational because 5 is not a perfect square and $$\sqrt{5}$$ has an infinite decimal expansion that does not repeat.

f. $$\sqrt [ 3 ]{ 27 }$$
Rational

g. Ï€=3.141592â€¦
Irrational because pi has a decimal expansion that does not repeat.

h. Order the numbers in parts (a)â€“(g) from least to greatest, and place them on a number line.
$$\sqrt{29}$$ : 52 < ($$\sqrt{29}$$)2 < 62, 5.32 < ($$\sqrt{29}$$)2 < 5.42, 5.382 < ($$\sqrt{29}$$)2 < 5.392

Question 4.
Circle the greater number in each of the pairs (a)â€“(e) below.
a. Which is greater, 8 or $$\sqrt{60}$$ ?
8

b. Which is greater, 4 or $$\sqrt{26}$$ ?
$$\sqrt{26}$$

c. Which is greater, $$\sqrt [ 3 ]{ 64 }$$ or $$\sqrt{16}$$ ?
The numbers are equal: $$\sqrt [ 3 ]{ 64 }$$ = 4, $$\sqrt{16}$$ = 4

d. Which is greater, $$\sqrt [ 3 ]{ 125 }$$5 or $$\sqrt{30}$$ ?
$$\sqrt{30}$$

e. Which is greater, -7 or –$$\sqrt{42}$$ ?
–$$\sqrt{42}$$

f. Put the numbers 9, $$\sqrt{52}$$ , and $$\sqrt [ 3 ]{ 216 }$$ in order from least to greatest. Explain how you know which order to put them in.
$$\sqrt{52}$$ is between 7 and 8
$$\sqrt [ 3 ]{ 216 }$$ = 6
In order from least to greatest:
$$\sqrt [ 3 ]{ 216 }$$, $$\sqrt{52}$$, 9

Question 5.

a. Between which two labeled points on the number line would $$\sqrt{5}$$ be located?
The number $$\sqrt{5}$$ is between 2.2 and 2.3.

b. Explain how you know where to place $$\sqrt{5}$$ on the number line.
I knew that $$\sqrt{5}$$ was between 2 and 3 but closer to 2. So next, I checked intervals of tenths beginning with 2.0 to 2.1. The interval that $$\sqrt{5}$$ fit between was 2.2 and 2.3 because 2.22 < ($$\sqrt{5}$$)2 < 2.32, 4.84 < 5 < 5.29.

c. How could you improve the accuracy of your estimate?
To improve the estimate, I would have to continue with the method of rational approximation to determine which interval of hundreths $$\sqrt{5}$$ fits between. Once I knew the interval of hundreths, I would check the interval of thousandths, and so on.
$$\sqrt{5}$$ : 22 < ($$\sqrt{5}$$)2 < 32, 2.22 < ($$\sqrt{5}$$)2 < 2.32 4 < 5 < 9
4.84 < 5 < 5.29

Question 6.
Determine the positive solution for each of the following equations.
a. 121 = x2
$$\sqrt{121}$$ = $$\sqrt{x^{2}}$$
11 = x
112 = 121
121 = 121

b. x3=1000
$$\sqrt[3]{x^{3}}$$ = $$\sqrt [ 3 ]{ 1000 }$$
x = 10
103 = 1000
1000 = 1000

c. 17 + x2 = 42
17 – 17 + x2 = 42 – 17
x2 = 25
$$\sqrt{x^{2}}$$ = $$\sqrt{25}$$
x = 5
17 + 52 = 42
17 + 25 = 42
42 = 42

d. x3 + 3x – 9 = x – 1 + 2x
x3 + 3x – 3x – 9 = x – 1 + 2x – 3x
x3 – 9 = -1
x3 – 9 + 9 = -1 + 9
x3 = 8
$$\sqrt[3]{x^{3}}$$ = $$\sqrt [ 3 ]{ 8 }$$
x = 2
23 + (3)(2) – 9 = 2 – 1 + (2)(2)
8 + 6 – 9 = 2 – 1 + 4
14 – 9 = 1 + 4
5 = 5

e. The cube shown has a volume of 216 cm3.
i. Write an equation that could be used to determine the length, l, of one side.

V = l3
216 = l3

ii. Solve the equation, and explain how you solved it.
$$\sqrt [ 3 ]{ 216 }$$ = $$\sqrt[3]{l^{3}}$$
To solve the equation, I had to take the cube root of both sides of the equation. The cube root of l3, $$\sqrt[3]{l^{3}}$$, is l. The cube root of 216, $$\sqrt [ 3 ]{ 216 }$$, is 6 because 63 = 216.