## 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.

Answer:

\(\frac{35}{7}\) = 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.

Answer:

\(\frac{4}{33}\) = 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.

Answer:

Let x be \(0 . \overline{345}\)

10^{3}x = 10^{3}(\(0 . \overline{345}\))

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.

Answer:

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 \(0. \overline{40}\)

10^{2}y = 10^{2} (\(0. \overline{40}\))

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.

Answer:

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?

Answer:

The number \(\sqrt{33}\) is between 5 and 6 because

5^{2} < (\(\sqrt{33}\))^{2} < 6^{2}

e. For what integer x is \(\sqrt{x}\) closest to 5.25? Explain.

Answer:

(5.25)^{2} = 27.5625

Since \(\sqrt{x}\) is the square root of x, then x^{2} 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. \(\sqrt{29}\)

Answer:

Irrational because 29 is not a perfect square and \(\sqrt{29}\) has an infinite decimal expansion that does not repeat.

b. \(5 . \overline{39}\)

Answer:

Rational

c. \(\frac{12}{4}\)

Answer:

Rational

d. \(\sqrt{36}\)

Answer:

Rational

e. \(\sqrt{5}\)

Answer:

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

f. \(\sqrt [ 3 ]{ 27 }\)

Answer:

Rational

g. Ï€=3.141592â€¦

Answer:

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.

Answer:

\(\sqrt{29}\) : 5^{2} < (\(\sqrt{29}\))^{2} < 6^{2}, 5.3^{2} < (\(\sqrt{29}\))^{2} < 5.4^{2}, 5.38^{2} < (\(\sqrt{29}\))^{2} < 5.39^{2}

Question 4.

Circle the greater number in each of the pairs (a)â€“(e) below.

a. Which is greater, 8 or \(\sqrt{60}\) ?

Answer:

8

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

Answer:

\(\sqrt{26}\)

c. Which is greater, \(\sqrt [ 3 ]{ 64 }\) or \(\sqrt{16}\) ?

Answer:

The numbers are equal: \(\sqrt [ 3 ]{ 64 }\) = 4, \(\sqrt{16}\) = 4

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

Answer:

\(\sqrt{30}\)

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

Answer:

–\(\sqrt{42}\)

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.

Answer:

\(\sqrt{52}\) is between 7 and 8

\(\sqrt [ 3 ]{ 216 }\) = 6

In order from least to greatest:

\(\sqrt [ 3 ]{ 216 }\), \(\sqrt{52}\), 9

Question 5.

Answer:

a. Between which two labeled points on the number line would \(\sqrt{5}\) be located?

Answer:

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.

Answer:

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 2.2^{2} < (\(\sqrt{5}\))^{2} < 2.3^{2}, 4.84 < 5 < 5.29.

c. How could you improve the accuracy of your estimate?

Answer:

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.

\(\sqrt{5}\) : 2^{2} < (\(\sqrt{5}\))^{2} < 3^{2}, 2.2^{2} < (\(\sqrt{5}\))^{2} < 2.3^{2} 4 < 5 < 9

4.84 < 5 < 5.29

Question 6.

Determine the positive solution for each of the following equations.

a. 121 = x^{2}

Answer:

\(\sqrt{121}\) = \(\sqrt{x^{2}}\)

11 = x

11^{2} = 121

121 = 121

b. x^{3}=1000

Answer:

\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 1000 }\)

x = 10

10^{3} = 1000

1000 = 1000

c. 17 + x^{2 }= 42

Answer:

17 – 17 + x^{2} = 42 – 17

x^{2} = 25

\(\sqrt{x^{2}}\) = \(\sqrt{25}\)

x = 5

17 + 5^{2} = 42

17 + 25 = 42

42 = 42

d. x^{3 }+ 3x – 9 = x – 1 + 2x

Answer:

x^{3} + 3x – 3x – 9 = x – 1 + 2x – 3x

x^{3} – 9 = -1

x^{3} – 9 + 9 = -1 + 9

x^{3} = 8

\(\sqrt[3]{x^{3}}\) = \(\sqrt [ 3 ]{ 8 }\)

x = 2

2^{3} + (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 cm^{3}.

i. Write an equation that could be used to determine the length, l, of one side.

Answer:

V = l^{3}

216 = l^{3}

ii. Solve the equation, and explain how you solved it.

Answer:

216 = l^{3}

\(\sqrt [ 3 ]{ 216 }\) = \(\sqrt[3]{l^{3}}\)

6 = l

The length of one side is 6 cm.

To solve the equation, I had to take the cube root of both sides of the equation. The cube root of l^{3}, \(\sqrt[3]{l^{3}}\), is l. The cube root of 216, \(\sqrt [ 3 ]{ 216 }\), is 6 because 6^{3} = 216.

Therefore, the length of one side of the cube is 6 cm.