We included HMH Into Math Grade 7 Answer Key PDF Module 5 Review to make students experts in learning maths.
HMH Into Math Grade 7 Module 5 Review Answer Key
Vocabulary
dividend
divisor
factor
product
quotient
rational number
For Problems 1-5, choose the correct term or terms from the Vocabulary box to label each boxed number.
Question 1.
Answer:
Question 6.
Complete the table to show the total changes and the average (mean) daily changes in the prices of four stocks over a 5-day period.
Answer:
Question 7.
An equation is shown, where a < 0, c > 0, and |a| > |c|.
a ∙ b = c
Which statements about the value of b are true?
A. b < 0 B. b > 0
C. b > 1
D. |b| < 1
E. |b| < 0 F. |b| > 1
Answer:
B. b > 0
F. |b| > 1
Explanation:
The statements about the value of b true are b > 0, |b| > 1.
Question 8.
Use Tools A horse’s weight decreases by \(\frac{1}{2}\) pound per week. At this rate, how many weeks will it take for the horse to lose a total of 5\(\frac{1}{2}\) pounds? Write an expression that could be used to model the situation. State what strategy and tool you will use to answer the question, explain your choice, and then find the answer.
Answer:
11 weeks.
Explanation:
A horse’s weight decreases by \(\frac{1}{2}\) pound per week.
total of 5\(\frac{1}{2}\) pounds = \(\frac{11}{2}\)
\(\frac{11}{2}\) = \(\frac{1}{2}\)
\(\frac{11}{2}\) × 2
11 weeks.
Question 9.
An equation is shown, where p <0. p ∙ q = r A. Assume r > 0. Plot a point on the number line to identify a possible location for q.
Answer:
B. Assume r < 0. Plot a point on the number line to identify a possible location for q.
Answer:
Question 10.
The total change in the outdoor temperature over 6 hours was —21 °F. What was the mean change in temperature per hour?
____ °F per hour
Answer:
3.5 °F per hour.
Explanation:
According to the given question
The total change in the outdoor temperature over 6 hours was —21 °F.
21 ÷ 6
3.5 °F per hour.
Question 11.
Which expressions are equivalent to (-\(\frac{1}{3}\))(-2)(6)? Select Equivalent or Not Equivalent for each expression.
Answer:
Question 12.
The depth of water in a swimming pool decreases by \(\frac{1}{8}\) inch per day due to evaporation. No water ¡s added to the pool for a 2-week period. What is the total change in the depth of the water during this time period?
A. -1\(\frac{3}{4}\) inches
B. –\(\frac{1}{4}\) inch
C. 1\(\frac{7}{8}\) inches
D. 16 inches
Answer:
-1\(\frac{3}{4}\) inches.
Explanation:
The depth of water in a swimming pool decreases by \(\frac{1}{8}\) inch per day.
No water is added for 2 weeks.
2 weeks = 14 days
–\(\frac{1}{8}\) × 14
–\(\frac{14}{8}\)
–\(\frac{7}{4}\)
-1\(\frac{3}{4}\) inches.
Question 13.
Which expression is equivalent to – (\(\frac{2}{5}\))?
A. \(-\frac{(-2)}{5}\)
B. \(-\frac{2}{(-5)}\)
C. \(\frac{-2}{-5}\)
D. \(\frac{2}{-5}\)
Answer:
\(\frac{2}{-5}\)
Explanation:
The expression that is equivalent to – (\(\frac{2}{5}\)) is \(\frac{2}{-5}\).
Question 14.
Which decimal is equivalent to \(\frac{5}{9}\)?
A. \(0 . \overline{5}\)
B. 0.55
C. 1.8
D. 5.9
Answer:
0.55
Explanation:
Given,
\(\frac{5}{9}\)
0.55