We included **HMH Into Math Grade 6 Answer Key PDF** **Module 8 Review **to make students experts in learning maths.

## HMH Into Math Grade 6 Module 8 Review Answer Key

**Vocabulary**

algebraic expression

base

coefficient

constant

equivalent expression

evaluate

exponent

like term

numerical expression

term

variable

**Complete the following to review your vocabulary for this module.**

Question 1.

For the expression 6^{4}:

A. The ___ is 6 and the ____ is 4.

B. A(n) ____ using repeated multiplication is 6 × 6 × 6 × 6.

Answer:

A. base, exponent,

B. Equivalent expression,

Explanation:

Given expression 6^{4}, A. The base is 6 and the exponent is 4.

The base number tells what number is being multiplied. The exponent a small number written above and to the right of the base number tells us how many times the base number is being multiplied.

B. An equivalent expression using repeated multiplication is 6 × 6 × 6 × 6.

Question 2.

A(n) ____ expression contains at least one variable, while a(n) _____ expression contains only numbers and operations.

Answer:

algebraic expression, numerical expression,

Explanation:

A(n) algebraic expression contains at least one variable while a(n) numerical expression contains only numbers and operations.

Question 3.

For the expression 4x + 7, the ____ is 4, the ____ is x, and the ___ is 7.

Answer:

Coefficient, Variable, Constant,

Explanation:

For the expression 4x + 7, the coefficient is 4 the variable is x and the constant is 7.

Coefficient is a number which is placed before another quantity and which multiplies it here it is 4,

A variable is an alphabet or term that represents an unknown number or unknown value or unknown quantity here it is x, A fixed and well-defined number or other non-changing mathematical object here it is 7.

Question 4.

To _____ an algebraic or numerical expression, find its value.

Answer:

Evaluate,

Explanation:

To evaluate an algebraic expression means to find the value of the expression when the variable is replaced by a given number.

Question 5.

_____ are terms with the same variables raised to the same exponents.

Answer:

Like terms,

Explanation:

Like terms are terms with the same variables raised to the same exponents.

**Concepts and Skills**

**For Problems 6-9. write an equivalent expression and evaluate.**

Question 6.

5 × 5 × 5 × 5 = ___ = ____

Answer:

5^{4}, 625,

Explanation

Given to write an equivalent expression for 5 X 5 X 5 X 5 so it is 5^{4 }and 4 times 5 is 625.

Question 7.

____ = 4^{3} = ____

Answer:

4 X 4 X 4 and 64,

Explanation

Given to write an equivalent expression for 4^{3 }it is 4 X 4 X 4 and 3 times 4 is 64.

Question 8.

___ = 2^{5} = ____

Answer:

2 X 2 X 2 X 2 X 2 and 32,

Explanation

Given to write an equivalent expression for 2^{5 }it is 2 X 2 X 2 X 2 X 2 and 5 times 2 is 32.

Question 9.

3 × 3 × 3 × 3 × 3 = ____

Answer:

3^{5}, 243,

Explanation

Given to write an equivalent expression for 3 X 3 X 3 X 3 X 3 so it is 3^{5 }and 5 times 3 is 243.

Question 10.

**Use Tools** Howard buys 5 pounds of apples at $2.50 per pound and 3 pounds of grapes at $1.50 per pound. What is the total cost of the fruit. State what strategy and tool you will use to answer the question, explain your choice, and then find the answer.

_____________________

_____________________

Answer:

Multiplication and Addition, The total cost of the fruit is $17.0.

Explanation:

Given Howard buys 5 pounds of apples at $2.50 per pound and 3 pounds of grapes at $1.50 per pound. The strategy I use to find the total cost of the fruits is first multiplication then addition

first we find cost for apples and then cost for grapes and last we add to find the total cost for the fruits as (5 X $2.50) + (3 X $1.50) = $12.5 + $4.5 = $17.0 therefore total cost of the fruits is $17.0.

**For Problems 11—12, identify the variable, coefficient, and constant term of the expression.**

Question 11.

12p + 47

variable: ____

coefficient: ____

constant: ____

Answer:

Variable : p, Coefficient : 12, Constant : 47,

Explanation:

In the given expression 12p + 47, variable is p, coefficient is 12 and constant is 47.

Question 12.

m + 7.5

variable: ___

coefficient: ________

constant: _____

Answer:

Variable : m, Coefficient : 1, Constant : 7.5,

Explanation:

In the given expression m + 7.5 variable is m, coefficient is 1 and constant is 7.5.

Question 13.

Write an equivalent expression using the Distributive Property.

35 + 21 = 7(____ + ____)

Answer:

7(5 + 3),

Explanation:

An equivalent expression for 35 + 21 using the Distributive Property is

as common factor for 35 and 21 is 7 because 7 X 5 = 35, 7 X 3 = 21, therfore it is 7(5 + 3).

Question 14.

Write an algebraic expression for 24 more than the product of 2 and x.

_____________________

Answer:

Algebraic expression = 2x + 24,

Explanation:

Given to write an algebraic expression for 24 more than the product of 2 and x is 2x + 24.

Question 15.

Barbara has b bags that each contain 4 pounds of rice. She has another bag that has 3 pounds of rice. Write an algebraic expression that shows how many pounds of rice Barbara has.

_____________________

Answer:

4b + 3,

Explanation:

Given Barbara has b bags that each contain 4 pounds of rice.

She has another bag that has 3 pounds of rice. An algebraic expression that shows number of

pounds of rice Barbara has is 4b +3.

Question 16.

Evaluate the expression 5(m – 2) + 10w when m = 8.4 and w = 1.25.

A. 8.75

B. 44.5

C. 80.25

D. 52.5

Answer:

B. 44.5,

Explanation:

Evaluating the expression 5(m – 2) + 10w when m = 8.4 and w = 1.25 is 5(8.4 – 2) + 10 X 1.25 =

(5 X 6.4) + 12.5 = 32 + 12.5 = 44.5 matches with B.

Question 17.

Evaluate the expression 6x + \(\frac{2}{3}\) – 4y + \(\frac{1}{2}\) when x = \(\frac{3}{4}\) and y = \(\frac{1}{6}\).

A. 2\(\frac{2}{3}\)

B. 5

C. 1\(\frac{1}{2}\)

D. 4

Answer:

B. 5,

Explanation:

Given to evaluate the expression 6x + \(\frac{2}{3}\) – 4y + \(\frac{1}{2}\)

when x = \(\frac{3}{4}\) and y = \(\frac{1}{6}\) is 6 X \(\frac{3}{4}\) + \(\frac{2}{3}\) – 4 X \(\frac{1}{6}\) + \(\frac{1}{2}\) =

\(\frac{6 X 3}{4}\) + \(\frac{2}{3}\) – \(\frac{4 X1}{6}\) + \(\frac{1}{2}\) = \(\frac{9}{2}\) + \(\frac{2}{3}\) – \(\frac{2}{3}\) + \(\frac{1}{2}\) first we make all common denominator so we multiply numerators

and denominators like this \(\frac{9 X 3}{2 X 3}\) + \(\frac{2 X 2}{3 X 2}\) – \(\frac{2 X 2}{3 X 2}\) + \(\frac{1 X 3}{2 X 3}\) = \(\frac{27}{6}\) + \(\frac{4}{6}\) – \(\frac{4}{6}\) + \(\frac{3}{6}\) now all have common denominators 6 so we add and subtract numerators as \(\frac{27 + 4 – 4 + 3}{6}\) =

\(\frac{30}{6}\) = 5, matches with bit 5.

Question 18.

Which expressions are equivalent to 8(2s + 6)? Select all that apply.

A. 16s + 6

B. 16s + 48

C. 10s + 48

D. 4(4s + 12)

E. 2(4s + 1) + 4(2s + 1)

Answer:

B. 16s + 48, D. 4(4s + 12),

Explanation:

Given expressions are equivalent to 8(2s + 6) are as 8(2s + 6) = 16s + 48,

checking with bit A. 16s + 6 which will not match, checking with bit B. 16s + 48 matches,

checking with bit C. 10s + 48 will not match, checking with bit D. 4(4s + 12) = 16s + 48 matches,

checking with bit E. 2(4s + 1) + 4(2s + 1), 8s + 2 + 4s + 1 = 12s + 3 which will not match,

therefore bits B. 16s + 48 and D. 4(4s + 12) matches.

Question 19.

Write three expressions that are equivalent to 24k + 12k.

Answer:

2k(12 + 6), 3k(8 + 4), 4k(6 + 3),

Explanation:

Given to write three expressions for 24k + 12k are we see common factors for 12 and 24,

for 12 it is 1,2,3,4,6,12 and for 24 it is 1,2,3,4,6,12,24 so the equivalent expressions for 24k + 12k are

2k(12 + 6), 3k(8 + 4), 4k(6 + 3).

Question 20.

A bag of plums costs $3 per pound and a bag of oranges costs $2 per pound. If Cammie buys x pounds of plums and y pounds of oranges, what expression could she write to find the total amount she will spend?

Answer:

Expression is 3x + 2y,

Explanation:

Given a bag of plums costs $3 per pound and a bag of oranges costs $2 per pound. If Cammie buys

x pounds of plums and y pounds of oranges, The expression could she write to find the

total amount she will spend is 3x + 2y.