Math in Focus

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 2 Lesson 2.5 Operations with Rational Numbers to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 2 Lesson 2.5 Answer Key Operations with Rational Numbers

Math in Focus Grade 7 Chapter 2 Lesson 2.5 Guided Practice Answer Key

Copy and Complete

Question 1.

Answer:
–\(\frac{1}{9}\) + \(\frac{2}{-5}\)
Write equivalent fractions using the LCD
\(\frac{(-1)(-5)}{(9)(-5)}\) + \(\frac{2(9)}{(-5)(9)}\)
= \(\frac{5}{-45}\) + \(\frac{18}{-45}\)
= \(\frac{5+18}{-45}\)
= \(\frac{23}{-45}\)

Question 2.

Answer:
-1 \(\frac{1}{6}\) + 3\(\frac{4}{9}\)
= -1 – \(\frac{1}{6}\) + 3 + \(\frac{4}{9}\)
= -1 + 3 = 2
= –\(\frac{1}{6}\) + \(\frac{4}{9}\)
LCD of \(\frac{1}{6}\) and \(\frac{4}{9}\) is 18
= –\(\frac{3}{18}\) + \(\frac{8}{18}\) = \(\frac{5}{18}\)
2 + \(\frac{5}{18}\) = 2\(\frac{5}{18}\)

Question 3.
\(\frac{1}{6}\) + \(\left(\frac{-5}{9}\right)\) + \(\left(\frac{-1}{3}\right)\)
Add two rational numbers at a time, working from left to right.

Answer:

Method 2
Use a common denominator for all three fractions.

Answer:
\(\frac{1}{6}\) + \(\left(\frac{-5}{9}\right)\) + \(\left(\frac{-1}{3}\right)\)
= \(\frac{3}{18}\) + \(\left(\frac{-10}{18}\right)\) + \(\left(\frac{-6}{18}\right)\)
= (3 -10 – 6)/18
= -13/18

Evaluate each expression.

Question 4.
\(\frac{1}{4}\) – \(\frac{3}{10}\)
Answer:
The fractions have unlike denominators.
So, find the LCD for 1/4 and 3/10 that is 20.
\(\frac{1}{4}\) × \(\frac{5}{5}\) – \(\frac{3}{10}\) × \(\frac{2}{2}\)
\(\frac{5}{20}\) – \(\frac{6}{20}\) = – \(\frac{1}{20}\)

Question 5.
\(\frac{7}{8}\) – \(\frac{9}{10}\)
Answer:
Given,
\(\frac{7}{8}\) – \(\frac{9}{10}\)
The fractions have unlike denominators.
LCD of 8 and 10 is 40
\(\frac{7}{8}\) × \(\frac{5}{5}\) – \(\frac{9}{10}\) × \(\frac{4}{4}\)
\(\frac{35}{40}\) – \(\frac{36}{40}\) = – \(\frac{1}{40}\)

Question 6.
3\(\frac{1}{4}\) – 7\(\frac{5}{6}\)
Answer:
Given
3\(\frac{1}{4}\) – 7\(\frac{5}{6}\)
Rewriting the equation
3 + \(\frac{1}{4}\) – 7 – \(\frac{5}{6}\)
3 – 7 = -4
Now subtract the fractions
\(\frac{1}{4}\) – \(\frac{5}{6}\)
LCM of 4 and 6 is 12
\(\frac{1}{4}\) × \(\frac{3}{3}\) – \(\frac{5}{6}\) × \(\frac{2}{2}\)
\(\frac{3}{12}\) – \(\frac{10}{12}\) = – \(\frac{7}{12}\)
-4 – \(\frac{7}{12}\) = -4\(\frac{7}{12}\)

Question 7.
\(\frac{3}{7}\) – \(\frac{27}{28}\) – \(\frac{3}{14}\)
Answer:
Given
\(\frac{3}{7}\) – \(\frac{27}{28}\) – \(\frac{3}{14}\)
LCM of 7, 14, 28 is 28.
\(\frac{3}{7}\) × \(\frac{4}{4}\) – \(\frac{27}{28}\) × \(\frac{1}{1}\) – \(\frac{3}{14}\) × \(\frac{2}{2}\)
\(\frac{12}{28}\) – \(\frac{27}{28}\) – \(\frac{6}{28}\)
= (12 – 27 – 6)/28
= -21/28
= -3/4

Solve.

Question 8.
Philadelphia suffered a severe snowstorm in 1996 that left 30\(\frac{7}{10}\) inches of snow on the ground. Another severe snowstorm occured in 2010, when 28\(\frac{1}{2}\) inches of snow fell.
a) Write a subtraction expression for the difference in depth of these two record snowfalls.
Answer:
30\(\frac{7}{10}\) – 28\(\frac{1}{2}\)

b) Rewrite the expression as an addition expression.
Answer: – 28\(\frac{1}{2}\) + 30\(\frac{7}{10}\)

c) Find the difference in these two record snowfalls.
Answer:
30\(\frac{7}{10}\) – 28\(\frac{1}{2}\)
Rewriting the equation
30 + \(\frac{7}{10}\) – 28 – \(\frac{1}{2}\)
30 – 28 = 2
\(\frac{7}{10}\) – \(\frac{1}{2}\)
LCD = 10
\(\frac{7}{10}\) – \(\frac{5}{10}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)
2 + \(\frac{1}{5}\) = 2 \(\frac{1}{5}\)

Copy and Complete.

Question 9.

Answer:
–\(\frac{4}{5}\) × \(\frac{20}{21}\)
Multiply the numerators and multiply the denominators
= (-4 × 20)/(5 × 21)
= –\(\frac{80}{105}\)

Question 10.

Answer:
-3\(\frac{1}{4}\) × -2\(\frac{2}{3}\)
Write as improper fractions
–\(\frac{13}{4}\) × –\(\frac{8}{3}\)
= \(\frac{104}{12}\)
= \(\frac{52}{6}\)
= \(\frac{26}{3}\)
= 8\(\frac{2}{3}\)

Evaluate each quotient.

Question 11.
\(\frac{3}{20}\) ÷ \(\left(-\frac{6}{35}\right)\)
Answer:
Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.
\(\frac{3}{20}\) × –\(\frac{35}{6}\)
= (3 × -35)/(20 × 6)
= -105/120
= -7/8

Question 12.
-3\(\frac{1}{3}\) ÷ \(\left(-1 \frac{1}{4}\right)\)
Answer:
Given,
-3\(\frac{1}{3}\) ÷ \(\left(-1 \frac{1}{4}\right)\)
\(\frac{10}{3}\) ÷ \(\frac{5}{4}\)
LCD of 3 and 4 is 12.
\(\frac{10}{3}\) × \(\frac{4}{4}\) ÷ \(\frac{5}{4}\) × \(\frac{3}{3}\)
\(\frac{40}{12}\) ÷ \(\frac{15}{12}\)
\(\frac{40}{12}\) = 2 \(\frac{2}{3}\)

Question 13.

Answer:

Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.
\(\frac{1}{4}\) ÷ \(\frac{-3}{8}\) = \(\frac{-2}{3}\)

Solve. Show your work.

Question 14.
A pancake recipe requires 1\(\frac{2}{3}\) cups of flour to make 20 pancakes and you have 9 cups of flour.
a) How many pancakes can you make with 1 cup of flour?
Answer:
Given,
A pancake recipe requires 1\(\frac{2}{3}\) cups of flour to make 20 pancakes
Let 1 cup of flour is x
1 \(\frac{2}{3}\) x = 20
x = 12

b) How many pancakes can you make with 9 cups of flour?
Answer:
1 cup = 12 pancakes
9 cups = 9 × 12 = 108
Thus you can make 81 pancakes with 9 cups of flour.

c) Do you have enough to make 100 pancakes? Explain your reasoning.
Answer:
1 cup of flour = 12 pancakes
9 cups = 9 × 12 = 108
Thus you can make 81 pancakes with 9 cups of flour.
As I can make 108 pancakes with 9 cups of flour, I can conclude that I have enough flour to make 100 pancakes

Math in Focus Course 2A Practice 2.5 Answer Key

Evaluate each expression. Give your answer in simplest form.

Question 1.
\(\frac{1}{2}\) + \(\left(-\frac{5}{6}\right)\)
Answer:
Given,
\(\frac{1}{2}\) + \(\left(-\frac{5}{6}\right)\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
\(\frac{1}{2}\) – \(\frac{5}{6}\)
LCD of 2 and 6 is 6.
\(\frac{1}{2}\) × \(\frac{3}{3}\) – \(\frac{5}{6}\) × \(\frac{1}{1}\)
\(\frac{3}{6}\) – \(\frac{5}{6}\) = – \(\frac{2}{6}\) or – \(\frac{1}{3}\)

Question 2.
–\(\frac{6}{7}\) + \(\frac{3}{14}\)
Answer:
Given,
–\(\frac{6}{7}\) + \(\frac{3}{14}\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
LCD of 7 and 14 is 14.
–\(\frac{12}{14}\) + \(\frac{3}{14}\) = –\(\frac{9}{14}\)

Question 3.
–\(\frac{1}{7}\) + \(\left(\frac{-3}{5}\right)\)
Answer:
Given,
–\(\frac{1}{7}\) + \(\left(\frac{-3}{5}\right)\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
-(\(\frac{1}{7}\) + \(\frac{3}{5}\))
LCD of 7 and 5 is 35
-(\(\frac{5}{35}\) + \(\frac{21}{35}\))
–\(\frac{26}{35}\)

Question 4.
\(\frac{1}{2}\) + \(\left(-\frac{2}{5}\right)\) + \(\frac{1}{4}\)
Answer:
Given,
\(\frac{1}{2}\) + \(\left(-\frac{2}{5}\right)\) + \(\frac{1}{4}\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
\(\frac{1}{2}\) + \(\frac{1}{4}\) – \(\frac{2}{5}\)
\(\frac{3}{4}\) – \(\frac{2}{5}\)
LCD of 4 and 5 is 20.
\(\frac{15}{20}\) – \(\frac{8}{20}\) = \(\frac{7}{20}\)

Question 5.
–\(\frac{1}{7}\) + \(\left(\frac{-5}{6}\right)\) + \(\left(\frac{-1}{3}\right)\)
Answer:
Given,
–\(\frac{1}{7}\) + \(\left(\frac{-5}{6}\right)\) + \(\left(\frac{-1}{3}\right)\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
–\(\frac{1}{7}\) – \(\frac{5}{6}\) – \(\frac{1}{3}\)
– (\(\frac{1}{7}\) + \(\frac{5}{6}\) + \(\frac{1}{3}\) )
LCD of 7, 6 and 3 is 42.
– (\(\frac{6}{42}\) + \(\frac{35}{42}\) + \(\frac{14}{42}\))
– (6 + 35 + 14)/42
– \(\frac{55}{42}\)

Question 6.
\(\frac{3}{5}\) – \(\frac{2}{3}\)
Answer:
Given,
\(\frac{3}{5}\) – \(\frac{2}{3}\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
LCD of 5 and 3 is 15.
\(\frac{9}{15}\) – \(\frac{10}{15}\) = – \(\frac{1}{15}\)

Question 7.
–\(\frac{1}{7}\) – \(\frac{3}{14}\)
Answer:
Given,
–\(\frac{1}{7}\) – \(\frac{3}{14}\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
LCD of 7 and 14 is 14.
-(\(\frac{1}{7}\) + \(\frac{3}{14}\))
-(\(\frac{2}{14}\) + \(\frac{3}{14}\))
– \(\frac{5}{14}\)

Question 8.
–\(\frac{1}{5}\) – \(\left(\frac{-2}{7}\right)\)
Answer:
Given,
–\(\frac{1}{5}\) – \(\left(\frac{-2}{7}\right)\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
– \(\frac{1}{5}\) + \(\frac{2}{7}\)
LCD of 7 and 5 is 35
-(\(\frac{7}{35}\) + \(\frac{10}{35}\))
–\(\frac{17}{35}\)

Question 9.
\(\frac{1}{3}\) – \(\left(-\frac{2}{5}\right)\) – \(\frac{3}{4}\)
Answer:
Given,
\(\frac{1}{3}\) – \(\left(-\frac{2}{5}\right)\) – \(\frac{3}{4}\)
The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.
\(\frac{1}{3}\) + \(\frac{2}{5}\) – \(\frac{3}{4}\)
LCD of 3, 5, and 7 is 60.
\(\frac{20}{60}\) + \(\frac{24}{60}\) – \(\frac{45}{60}\)
\(\frac{44}{60}\) – \(\frac{45}{60}\) = – \(\frac{1}{60}\)

Evaluate each product. Give your answer in simplest form.

Question 10.
–\(\frac{7}{25}\) • \(\frac{5}{14}\)
Answer:
Given,
–\(\frac{7}{25}\) • \(\frac{5}{14}\)
-(7 × 5)/(25 × 14)
= -1/10

Question 11.
\(\frac{5}{8}\) • \(\left(-\frac{4}{15}\right)\)
Answer:
Given,
\(\frac{5}{8}\) • \(\left(-\frac{4}{15}\right)\)
(5 × -4)/(8 × 15)
= -20/120
= -1/6

Question 12.
\(\frac{7}{30}\) • \(\left(-\frac{6}{7}\right)\)
Answer:
Given,
\(\frac{7}{30}\) • \(\left(-\frac{6}{7}\right)\)
(7 × -6)/(30 × 7)
= -1/5

Question 13.
–\(\frac{8}{27}\) • \(\left(-\frac{9}{40}\right)\)
Answer:
Given,
–\(\frac{8}{27}\) • \(\left(-\frac{9}{40}\right)\)
= (-8 × -9)/(27 × 40)
= 1/15

Question 14.
–\(\frac{11}{16}\) • \(\left(-\frac{4}{33}\right)\)
Answer:
Given,
–\(\frac{11}{16}\) • \(\left(-\frac{4}{33}\right)\)
(-11 × -4)/(16 × 33)
1/12

Question 15.
\(\frac{5}{8}\) • \(\left(-2 \frac{4}{5}\right)\)
Answer:
Given,
\(\frac{5}{8}\) • \(\left(-2 \frac{4}{5}\right)\)
\(\frac{5}{8}\) • \(\left(\frac{-6}{5}\right)\)
(5 × -6)/(8 × 5)
-3/4

Question 16.
–\(\frac{3}{22}\) • 1\(\frac{5}{6}\)
Answer:
Given,
–\(\frac{3}{22}\) • 1\(\frac{5}{6}\)
Convert from mixed fraction to the improper fraction to find the product of two fractions.
–\(\frac{3}{22}\) • \(\frac{11}{6}\)
= –\(\frac{3×11}{22×6}\)
= –\(\frac{33}{132}\)
= –\(\frac{1}{4}\)

Question 17.
3\(\frac{1}{8}\) • \(\left(-\frac{3}{10}\right)\)
Answer:
Given,
3\(\frac{1}{8}\) • \(\left(-\frac{3}{10}\right)\)
Convert from mixed fraction to the improper fraction to find the product of two fractions.
\(\frac{25}{8}\) • –\(\frac{3}{10}\)
= –\(\frac{75}{80}\)
= –\(\frac{15}{16}\)

Question 18.
-4\(\frac{1}{2}\) • \(\left(-1 \frac{8}{9}\right)\)
Answer:
Given,
-4\(\frac{1}{2}\) • \(\left(-1 \frac{8}{9}\right)\)
Convert from mixed fraction to the improper fraction to find the product of two fractions.
–\(\frac{9}{2}\) • –\(\frac{17}{9}\)
minus × minus = plus
We get
\(\frac{9}{2}\) • \(\frac{17}{9}\)
= \(\frac{51}{6}\)
= \(\frac{17}{2}\)

Evaluate each quotient. Give your answer in simplest form.

Question 19.
-10 ÷ \(\left(-\frac{5}{6}\right)\)
Answer:
Given,
-10 ÷ \(\left(-\frac{5}{6}\right)\)
First cancel the negative signs
We get
10 ÷ \(\frac{5}{6}\)
(10 × 6)/(1 × 5)
= 60/5
= 12

Question 20.
\(\frac{9}{25}\) ÷ (-18)
Answer:
Given,
\(\frac{9}{25}\) ÷ (-18)
(9 × 1)/25 × -28
=\(\frac{9}{-450}\)
= –\(\frac{1}{50}\)

Question 21.
–\(\frac{3}{8}\) ÷ \(\left(-\frac{1}{8}\right)\)
Answer:
Given,
–\(\frac{3}{8}\) ÷ \(\left(-\frac{1}{8}\right)\)
First cancel the negative signs
We get
\(\frac{3}{8}\) ÷ \(\frac{1}{8}\) = 3

Question 22.
–\(\frac{1}{4}\) ÷ \(\frac{3}{8}\)
Answer:
Given,
–\(\frac{1}{4}\) ÷ \(\frac{3}{8}\)
= -(1 × 8)/(4 × 3) = -8/12
= -2/3

Question 23.
\(\frac{5}{12}\) ÷ \(\left(-\frac{1}{6}\right)\)
Answer:
Given,
\(\frac{5}{12}\) ÷ \(\left(-\frac{1}{6}\right)\)
(5 × 6)/(-12 × 1)
= -30/12
= -5/2
= \(\left(-2 \frac{1}{2}\right)\)

Question 24.
-1\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)
Answer:
Given,
-1\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)
Convert from mixed fraction to the improper fraction
–\(\frac{5}{4}\) ÷ \(\frac{3}{4}\)
= -(5 × 4)/(4 × 3)
= –\(\frac{20}{12}\)
= -1\(\frac{2}{3}\)

Question 25.
\(\frac{8}{15}\) ÷ \(\left(-2 \frac{2}{3}\right)\)
Answer:
Given,
\(\frac{8}{15}\) ÷ \(\left(-2 \frac{2}{3}\right)\)
Convert from mixed fraction to the improper fraction
\(\frac{8}{15}\) ÷ –\(\frac{8}{3}\)
= (8 × 3)/(15 × -8)
= –\(\frac{1}{5}\)

Question 26.
3\(\frac{3}{4}\) ÷ \(\left(-\frac{1}{4}\right)\)
Answer:
Given,
3\(\frac{3}{4}\) ÷ \(\left(-\frac{1}{4}\right)\)
Convert from mixed fraction to the improper fraction
\(\frac{15}{4}\) ÷ –\(\frac{1}{4}\)
= (15 × 4)/(4 × -1)
= –\(\frac{60}{4}\)
= -15

Question 27.
2\(\frac{1}{2}\) ÷ \(\left(-1 \frac{2}{3}\right)\)
Answer:
Given,
2\(\frac{1}{2}\) ÷ \(\left(-1 \frac{2}{3}\right)\)
Convert from mixed fraction to the improper fraction
\(\frac{5}{2}\) ÷ –\(\frac{5}{3}\)
= -(5 × 3)/(2 × 5)
= –\(\frac{3}{2}\) or -1\(\frac{1}{2}\)

Question 28.
-2\(\frac{2}{7}\) ÷ \(\left(-1 \frac{3}{7}\right)\)
Answer:
Given,
-2\(\frac{2}{7}\) ÷ \(\left(-1 \frac{3}{7}\right)\)
Convert from mixed fraction to the improper fraction
–\(\frac{16}{7}\) ÷  –\(\frac{10}{7}\)
Cancel the negative signs and then perform division operation
\(\frac{16}{7}\) ÷ \(\frac{10}{7}\)
= (16 × 7)/(7 × 10)
= \(\frac{8}{5}\)
= 1 \(\frac{3}{5}\)

Question 29.

Answer:
Given,

-7 ÷ –\(\frac{7}{3}\)
Cancel the negative signs and then perform division operation
(7 × 3)/(7)
= 21/7
= 3

Question 30.

Answer:
Given,

–\(\frac{2}{3}\) ÷ 4
= -(2 × 1)/(3 × 4)
= -2/12
= –\(\frac{1}{6}\)

Question 31.

Answer:
Given,

Cancel the negative signs and then perform division operation
\(\frac{3}{4}\) ÷ \(\frac{5}{8}\)
= (3 × 8)/(4 × 5)
= 24/20
= 1\(\frac{1}{5}\)

Question 32.

Answer:
Given,

–\(\frac{1}{5}\) ÷ 1\(\frac{2}{15}\)
–\(\frac{1}{5}\) ÷ \(\frac{17}{15}\)
= -(1 × 15)/(5 × 17)
= –\(\frac{3}{17}\)

Solve. Show your work.

Question 33.
David hiked 15\(\frac{9}{10}\) miles on Saturday and 6\(\frac{7}{10}\) miles on Sunday. How much farther did David bike on Saturday than on Sunday?
Answer:
David hiked 15\(\frac{9}{10}\) miles on Saturday and 6\(\frac{7}{10}\) miles on Sunday.
15\(\frac{9}{10}\) – 6\(\frac{7}{10}\) = 9 \(\frac{1}{5}\)

Question 34.
A recipe calls for \(\frac{3}{4}\) cup of flour, but Kelli has only \(\frac{1}{4}\) cup of flour. How much more flour does she need?

Answer:
Given,
A recipe calls for \(\frac{3}{4}\) cup of flour, but Kelli has only \(\frac{1}{4}\) cup of flour.
\(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{2}\)
Thus Kelli needs \(\frac{1}{2}\) cup of flour.

Question 35.
A weather report showed that the rainfall in Janesville was 2\(\frac{2}{3}\) inches during the first half of January. At the end of January, the total rainfall was 3\(\frac{1}{4}\) inches. How much did it rain in the second half of January?
Answer:
Given,
A weather report showed that the rainfall in Janesville was 2\(\frac{2}{3}\) inches during the first half of January.
At the end of January, the total rainfall was 3\(\frac{1}{4}\) inches.
3\(\frac{1}{4}\) – 2\(\frac{2}{3}\)
Convert from mixed fraction to the improper fraction
\(\frac{13}{4}\) – \(\frac{8}{3}\)
LCD of 4 and 3 is 12
\(\frac{39}{12}\) – \(\frac{32}{12}\) = \(\frac{7}{12}\)

Question 36.
The sum of two rational numbers is 5\(\frac{1}{2}\). If one of the numbers is 6\(\frac{3}{14}\), find the other number.
Answer:
The sum of two rational numbers is 5\(\frac{1}{2}\). If one of the numbers is 6\(\frac{3}{14}\)
x + 6\(\frac{3}{14}\) = 5\(\frac{1}{2}\)
x = 5\(\frac{1}{2}\) – 6\(\frac{3}{14}\)
x = 5 + \(\frac{1}{2}\) – 6 – \(\frac{3}{14}\)
x = -1 + \(\frac{2}{7}\) = –\(\frac{5}{7}\)

Question 37.
MathJournal Peter adds \(\frac{1}{a}\) + \(\left(-\frac{1}{b}\right)\) and says the answer is \(\frac{1}{a-b}\). Give an example to show that Peter is wrong.
Answer:
\(\frac{1}{a}\) + \(\left(-\frac{1}{b}\right)\)
\(\frac{1}{a}\) – \(\frac{1}{b}\)
LCD of a and b is ab
\(\frac{b}{ab}\) – \(\frac{a}{ab}\)
\(\frac{b-a}{ab}\)
By this we can say that Peter is wrong.

Question 38.
MathJournal Jo multiplies two mixed numbers, -4\(\frac{3}{5}\) and 1\(\frac{2}{7}\) as follows:

Describe Jo’s mistakes. What is the correct answer?
Answer:
-4\(\frac{3}{5}\) × 1\(\frac{2}{7}\)
Convert from mixed fraction to the improper fraction
–\(\frac{23}{5}\) × \(\frac{9}{7}\)
–\(\frac{207}{35}\)
-5\(\frac{32}{35}\)

Question 39.
A clock’s battery is running low. Every 6 hours, it slows down by \(\frac{1}{2}\) hour. By how much does it slow down in 1 hour?
Answer:
If its running low and every six hours it slows down half an hour, we need to divide 6 from 1/2.
6 divided by 1/2 is 12
It will slow down 12 minutes every hour.

Question 40.
A weighs 5\(\frac{1}{2}\) pounds and package B weighs 1\(\frac{1}{4}\) pounds. Find the average weight of the two packages.
Answer:
Given,
A weighs 5\(\frac{1}{2}\) pounds and package B weighs 1\(\frac{1}{4}\) pounds.
5\(\frac{1}{2}\) + 1\(\frac{1}{4}\) = 6 \(\frac{3}{4}\)
Convert from mixed fraction to the improper fraction
6 \(\frac{3}{4}\) = \(\frac{27}{4}\)
Average of two packets
\(\frac{27}{4}\) ÷ 2 = \(\frac{27}{8}\) = 3 \(\frac{3}{8}\)

Question 41.
A scientist measured the weight of some damp soil. After exposing the soil to the air for 4\(\frac{1}{2}\) weeks, the scientist found that the weight had decreased by 5\(\frac{1}{8}\) ounces. Find the average weight loss per week.
Answer:
Let the weight loss per week is x
A scientist measured the weight of some damp soil. After exposing the soil to the air for 4\(\frac{1}{2}\) weeks, the scientist found that the weight had decreased by 5\(\frac{1}{8}\) ounces.
4\(\frac{1}{2}\) weeks = 5\(\frac{1}{8}\)
1 week = x
x × 4\(\frac{1}{2}\) = 5\(\frac{1}{8}\)
x = 5\(\frac{1}{8}\) ÷ 4\(\frac{1}{2}\)
x = \(\frac{41}{8}\) ÷ \(\frac{9}{2}\)
x = 1 \(\frac{5}{36}\) ounces

Question 42.
A plank measures 4\(\frac{3}{4}\) feet. Elizabeth cuts off \(\frac{2}{5}\) of the plank. How long is the plank now?
Answer:
Given,
A plank measures 4\(\frac{3}{4}\) feet.
Elizabeth cuts off \(\frac{2}{5}\) of the plank.
4\(\frac{3}{4}\) – \(\frac{2}{5}\) = 4 \(\frac{7}{20}\)
Thus the plank is 4 \(\frac{7}{20}\) feet long now.

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 2 Lesson 2.4 Operations with Integers to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 2 Lesson 2.4 Answer Key Operations with Integers

Math in Focus Grade 7 Chapter 2 Lesson 2.4 Guided Practice Answer Key

Evaluate each expression.

Question 1.
14 + 8 – 96
14 + 8 – 9 • 6 = 14 + 8 – Multiply.
= 14 + 8 + Rewrite subtraction as adding the opposite.
= 14 + () + 8 Use the commutative property of addition.
= + 8 Add.
= Add.
Answer:
14 + 8 – 9 • 6
= 14 + 8 – 54
= 14 + (-54) + 8
= -40 + 8
= -32
Explanation:
The given expression is 14 + 8 – 9 • 6. Use BODMAS rule.
First we have to perform multiplication operation. Multiply 9 with 6 the product is 54.
We have to use – symbol before 54. Now the expression is 14 + 8 – 54.
Now, use the commutative property of addition the expression is 14 + (-54) + 8.
Add 14 with -54 the sum is -40. The expression is -40 + 8.
Add -40 with 8 the sum is -32.

Question 2.
(-25 – 5) ÷ 6 – 21
(-25 – 5) ÷ 6 – 21 = ÷ 6 – 21 Subtract within the parentheses.
= – 21 Divide.
= Subtract.
Answer:
(-25 – 5) ÷ 6 – 21
= (-30) ÷ 6 – 21
= -5 – 21
= -26
Explanation:
The given expression is (-25 – 5) ÷ 6 – 21. Use BODMAS rule.
First we have to perform operation within the parentheses. Subtract -25 with -5 the difference is -30. Now the expression is  (-30) ÷ 6 – 21.
Secondly we have to perform division operation. Divide -30 by 6 the quotient is -5.
The expression is -5 -21. Subtract -5 with – 21 the result is -26.

Question 3.
-14 – (3 + 3) • 2
Answer:
-14 – (3 + 3) • 2
= -14 – (6) • 2
= -14 – 12
= – 26
Explanation:
The given expression is -14 – (3 + 3) • 2. Use BODMAS rule.
First we have to perform operation within the parentheses. Add 3 with 3 the sum is 6. Now the expression is  -14 -(6) • 2
Second we have to perform multiplication operation. Multiply 6 with 2 the product is 12. Now the expression is -14 -12.
Third we have to perform subtraction operation. Subtract -14 with -12 the result is -26.

Solve.

Question 4.
Joseph drew a hexagon on a 3-inch square piece of paper. He cut four identical right triangles from the four corners of the paper. The height of each triangle is \(\frac{1}{2}\) the length of the paper. The base of each triangle is \(\frac{1}{3}\) the length of the paper. What is the area of the paper that remained after these triangles are cut off?

Height of a triangle:
Base of a triangle:
Area of remaining paper:
Area of original paper – Area of four cut-off triangles
= 3 • 3 – 4 • (\(\frac{1}{2}\) • • ) Write an expression.
= – Multiply.
= in² Subtract.
The area of the remaining paper is square inches.
Answer:
Length of the paper = 3 inches
Height of a triangle: 1/2 • length of the paper = 1/2 • 3 = 3/2
Base of a triangle: 1/3 • length of the paper = 1/3 • 3 = 1
Area of four cut-off triangles = 4 • (1/2 • base • height)
Area of remaining paper = Area of original paper – Area of four cut-off triangles
= 3 • 3 – 4 • (1/2 • 1 • 3/2)
= 9 – 4 •(3/4)
= 9 – 3
= 6 in²
The area of the remaining paper is 6 square inches.

Math in Focus Course 2A Practice 2.4 Answer Key

Evaluate each expression.

Question 1.
-3 • 5 + 7
Answer:
-3 • 5 + 7
= – 15 + 7
= -8
Explanation:
The given expression is -3 • 5 + 7. Use BODMAS rule.
First multiply -3 with 5 the product is -15. Now the expression is -15 + 7.
Next perform addition operation. Add -15 with 7 the sum is -8.

Question 2.
50 ÷ (-5) + (-4)
Answer:
50 ÷ (-5) + (-4)
= 50 ÷ -5 – 4
= -10 – 4
= -14
Explanation:
The given expression is 50 ÷ (-5) + (-4). Use BODMAS rule.
First perform operations within the parentheses. Now the expression is 50 ÷ – 5 – 4.
Next perform division operation. Divide 50 by -5 the quotient is -10. Now the expression is -10 -4.
Subtract -10 with -4 the result is -14.

Question 3.
4 • (-6) + (-3) • 5
Answer:
4 • (-6) + (-3) • 5
= 4 • -6  – 3 • 5
= -24 – 15
= -39
Explanation:
The given expression is 4 • (-6) + (-3) • 5. Use BODMAS rule.
First perform operations within the parentheses. Now the expression is 4 • -6  – 3 • 5.
Next perform multiplication operation. Multiply 4 with -6 the product is -24. Multiply -3 with 5 the product is -15. Now the expression is -24 – 15.
Subtract -24 with -15 the result is -39.

Question 4.
11 – 2 • 8 – (-9)
Answer:
= 11 – 2 • 8 – (-9)
= 11 – 2 • 8 + 9
= 11 – 16 + 9
= 20 – 16
= 4
Explanation:
The given expression is 11 – 2 • 8 – (-9). Use BODMAS rule.
First perform operations within the parentheses. Now the expression is 11 – 2 • 8 + 9.
Next perform multiplication operation. Multiply 2 with 8 the product is 16. Now the expression is 11 – 16 + 9.
Add 11 with 9 the sum is 20. The expression is 20 -16.
Subtract 16 from 20 the result is 4.

Question 5.
-64 ÷ 4. 5 – 37
Answer:
-64 ÷ 4. 5 – 37
= -16. 5 – 37
= – 80 – 37
= – 117
Explanation:
The given expression is-64 ÷ 4. 5 – 37. Use BODMAS rule.
First perform division operation. Divide 64 by 4 the quotient is 16. Now the expression is -16. 5 – 37.
Next perform multiplication operation. Multiply 16 with 5 the product is 80. Now the expression is -80 – 37.
subtract -80 with -37 the result is -117.

Question 6.
-28 – 350 ÷ 7 + 8
Answer:
-28 – 350 ÷ 7 + 8
= -28 – 50 + 8
= -28 -42
= – 70
Explanation:
The given expression is -28 – 350 ÷ 7 + 8. Use BODMAS rule.
First perform division operation. Divide 350 by 7 the quotient is 50. Now the expression is -28 – 50 + 8.
Next perform addition operation. Add -50 with 8 the sum is -42. Now the expression is -28 -42.
subtract -28 with -42 the result is -70.

Question 7.
1oo – (8 – 15) • 9
Answer:
1oo – (8 – 15) • 9
= 100 – (-7) • 9
= 100 + 63
= 163
Explanation:
The given expression is 1oo – (8 – 15) • 9. Use BODMAS rule.
First perform operations within the parentheses. Subtract 15 from 8 the difference is -7. Now the expression is 100 – (-7) • 9.
Next perform multiplication operation. Multiply -7 with 9 the product is -63. Now the expression is 100 + 63.
Add 100 with 63 the sum is 163.

Question 8.
70 ÷ (-4 – 3) + 60
Answer:
70 ÷ (-4 – 3) + 60
= 70 ÷ (-7) + 60
= – 10 + 60
= 50
Explanation:
The given expression is 70 ÷ (-4 – 3) + 60. Use BODMAS rule.
First perform operations within the parentheses. Subtract -4 with -3 the result is -7. Now the expression is 70 ÷ (-7) + 60.
Next perform division operation. Divide 70 by – 7 the quotient is -10. Now the expression is -10 + 60.
Add -10 with 60 the result is 50.

Question 9.
(4 + 2) • 3 – 8 • (2 + 3)
Answer:
(4 + 2) • 3 – 8 • (2 + 3)
= 6 • 3 – 8 • 5
= 18 – 40
= -22
Explanation:
The given expression is (4 + 2) • 3 – 8 • (2 + 3). Use BODMAS rule.
First perform operations within the parentheses. Add 4 with 2 the sum is 6. Add 2 with 3 the sum is 5. Now the expression is 6 • 3 – 8 • 5.
Next perform multiplication operation. Multiply 6 with 3 the product is 18. Multiply 8 with 5 the product is 40. Now the expression is 18 – 40.
Subtract 40 from 18 the difference is -22.

Question 10.
70 ÷ (-4 – 3) + 60
Answer:
= 70 ÷ (-7) + 60
= – 10 + 60
= 50
Explanation:
The given expression is 70 ÷ (-4 – 3) + 60. Use BODMAS rule.
First perform operations within the parentheses. Subtract -4 with -3 the result is -7. Now the expression is 70 ÷ (-7) + 60.
Next perform division operation. Divide 70 by – 7 the quotient is -10. Now the expression is -10 + 60.
Add -10 with 60 the result is 50.

Question 11.
15 ÷ (4 + 1) – 8.3
Answer:
15 ÷ (4 + 1) – 8 • 3
= 15 ÷ 5 – 8 • 3
= 3 – 8 • 3
= 3 – 24
= -21
Explanation:
The given expression is 15 ÷ (4 + 1) – 8 • 3. Use BODMAS rule.
First perform operations within the parentheses. Add 4 with 1 the sum is 5. Now the expression is 15 ÷ 5 – 8 • 3.
Next perform division operation. Divide 15 by 5 the quotient is 3. Now the expression is 3 – 8 • 3.
Perform multiplication operation. Multiply 8 with 3 the product is 24. Now the expression is 3 – 24.
Subtract 24 from 3 the difference is -21.

Question 12.
24 ÷ 4 – (-13 + 3) • 2
Answer:
24 ÷ 4 – (-13 + 3) • 2
= 24 ÷ 4 + 10 • 2
= 6 + 10 • 2
= 6 + 20
= 26
Explanation:
The given expression is 24 ÷ 4 – (-13 + 3) • 2. Use BODMAS rule.
First perform operations within the parentheses. Add 3 with -13 the sum is -10. Now the expression is 24 ÷ 4 + 10 • 2.
Next perform division operation. Divide 24 by 4 the quotient is 6. Now the expression is 6 + 10 • 2.
Perform multiplication operation. Multiply 10 with 2 the product is 20. Now the expression is 6 + 20.
Add 6 with 20 the sum is 26.

Question 13.
-12 + 50 ÷ (-2 – 3) + 72 ÷ (4 + 2)
Answer:
-12 + 50 ÷ (-2 – 3) + 72 ÷ (4 + 2)
= -12 + 50 ÷ (-5) + 72 ÷  6
= -12 – 10 + 12
= -12 + 2
= -10
Explanation:
The given expression is -12 + 50 ÷ (-2 – 3) + 72 ÷ (4 + 2). Use BODMAS rule.
First perform operations within the parentheses. Add -2 with -3 the sum is -5.  Add 4 with 2 the sum is 6. Now the expression is-12 + 50 ÷ (-5) + 72 ÷  6.
Next perform division operation. Divide 50 by -5 the quotient is -10. Divide 72 by 6 the quotient is 12. Now the expression is -12 – 10 + 12.
Add -10 with 12 the sum is 2. Now the expression is -12 + 2.
Add -12 with 2 the sum is -10.

Question 14.
180 ÷ (4 + 16) – 8 • 3 + 7 • (2 + 3)
Answer:
180 ÷ (4 + 16) – 8 • 3 + 7 • (2 + 3)
= 180 ÷ 20 – 8 • 3 + 7 • 5
= 9 –  8 • 3 + 7 • 5
= 9 – 24 + 35
= 9 + 11
= 20
Explanation:
The given expression is 180 ÷ (4 + 16) – 8 • 3 + 7 • (2 + 3). Use BODMAS rule.
First perform operations within the parentheses. Add 4 with 16 the sum is 20. Add 2 with 3 the sum is 5. Now the expression is 180 ÷ 20 – 8 • 3 + 7 • 5.
Next perform division operation. Divide 180 by 20 the quotient is 9. Now the expression is9 –  8 • 3 + 7 • 5.
Perform multiplication operation. Multiply 8 with 3 the product is 24. Multiply 7 with 5 the product is 35. Now the expression is 9 – 24 + 35.
Add -24 with 35 the sum is 11.
Add 9 with 11 the sum is 20.

Solve. Show your work.

Question 15.
Emily made a sketch of an octagonal window on a 27-inch square piece of paper. First she cut four identical isosceles triangles from the corners of the paper. Then she cut a square from the center of the octagon. Each leg of a cut-off triangle is \(\frac{1}{3}\) the length of the paper. The side length of the cut-out square is also \(\frac{1}{3}\) the length of the paper. What ¡s the area of the sketch after she removed the triangles and the square?

Answer:
Length of the paper = 27-inch 
Area of square piece of paper = side • side = 27 in • 27 in = 729 square inches
Each leg of a cut-off triangle = 1/3 • length of the paper = 1/3 • 27 =9 inches
Side length of a cut-out square = 1/3 • length of the paper = 1/3 • 27 = 9 inches
Area of a square = side x side
Area of four cut-off triangles = 4 • (1/2 • base • height)
From the above image we can observe both base and height are same.
Area of octagonal window = Area of a square piece of paper – Area of four cut-off triangles – Area of a square
= 729 – 4 • (1/2 • 9 • 9) –  9 • 9
= 729 – 4 • (81/2) – 81
= 729 – 2 • 81 – 81
= 729 – 162 – 81
= 486 square inches
The area of the sketch after she removed the triangles and the square is 486 square inches.

Question 16.
Math Journal Suppose that Lydia shows you some of her homework:

Lydia made a common error when she used the distributive property to evaluate the expression -2(6 – 8). Evaluate the expression using the order of operations. Then explain how Lydia can correctly use the distributive property to evaluate the expression.
Answer:
Distributive property = A ( B+ C) = AB + AC
Use distributive property to evaluate the given expression -2 (6 – 8).
-2 (6 – 8) = -2 • 6 + (-2) • (-8)
= -12 + 16
= 4
– 2 (6 – 8) = 4

Question 17.
Sylvia took three turns in a video game. She scored -120 points during her first turn, 320 points during her second turn, and -80 points during her third turn. What was her average score for the three turns?
Answer:
Sylvia scored -120 points during her first turn.
Sylvia scored 320 points during her second turn.
Sylvia scored -80 points during her third turn.
Average = (-120 + 320 -80)/ 3 = 120/3 = 40 
Average score for the three turns is 40 points.

Question 18.
Math Journal Benjamin wrote: -20 + 4 • 2 + 7 – 35 = -19. Where can he place the parentheses so that the equation will be a true statement?
Answer:
-20 + 4 • 2 + 7 – 35 = -19
= -20 + 4 • (2 + 7) – 35
= -20 + 4 • 9 – 35
= -20 + 36 – 35
= -20 + 1
= -19
The equation will be a true statement by keeping parentheses to 2 + 7. The equation is – 20 + 4 • (2 + 7) – 35 = -19

This handy Math in Focus Grade 1 Workbook Answer Key Chapter 7 Numbers to 20 detailed solutions for the textbook questions.

Math in Focus Grade 1 Chapter 7 Answer Key Numbers to 20

Count how many stickers the Fill in the blanks.

Question 1.

Answer:

Draw how many stickers you have. Then fill in the blanks.

Question 2.

I have _______ stickers.
Answer: I have 11 stickers

Write about the number of stickers everyone has. Fill in the blanks with the correct names.

Question 3.
____ has more stickers than _____________
Answer:
Pete has more stickers than Ty.

Question 4.
________ has fewer stickers than _______
Answer:
Ty has fewer stickers than Pete

Question 5.
_____________ has the greatest number of stickers.
Answer:
Pete has the greatest number of stickers.

Question 6.
_____________ has the least number of stickers.
Answer:
Ty has the least number of stickers.

Put On Your Thinking Cap!

Challenging Practice

Question 1.
Class 1A of Greenfield School holds a basketball contest. Find out who won.
CLUES
Rita scores the least number of baskets. John scores 3 more baskets than Rita. Dion scores more baskets than Rachel but less than Frank.

Write the names next to the number of baskets scored.

Who won the contest? ____

Answer: Frank won the contest

Fill in the blanks.

Question 2.
10 + ___ = 15
Answer: 5
10 + 5 = 15

Question 3.
10 + ___ = 11
Answer: 1
10 + 1 = 11

Question 4.
10 + ___ = 18
Answer: 8
10 + 8 = 18

Question 5.
___ + 10 = 14
Answer: 4
4 + 10 = 14

Question 6.
___ + 10 = 17
Answer: 7
7 + 10 = 17

Write the correct names.

Question 7.
These are the numbers of 12 players on a team.

Whose names have the following numbers?

Answer:

Put On Your Thinking Cap!

Problem Solving

Use the clues on the next page. Help Tony find which numbers his counters covered.

Read what Tony’s friends said.
Circle the numbers that were covered on Tony’s card.

 

Answer:

Chapter Review/Test

Vocabulary

Unscramble the letters to spell each number.

Question 1.
15 fieetfn
_______
Answer:
Fifteen

Question 2.
11 evleen
________
Answer:
Eleven

Question 3.
18 egeethni
_______
Answer:
Eighteen

Question 4.
20 twynet
_____________
Answer:
Twenty

Fill in the blank with the correct word.

place-value chart compare

Question 5.
You can show numbers as tens and ones in a _______
Answer:
You can show numbers as tens and ones in a place-value chart

Question 6.
When you ____ 12 and 15,12 is the number that is less.
Answer:
When you Compare 12 and 15,12 is the number that is less.

Concepts and Skills

Count. Write the number.

Question 7.

Answer: 13

Question 8.

Answer: 17

Fill in the blanks.

Question 9.

Set ___ has ___ more oranges than Set ____
Set ___ has ___ fewer oranges than Set ____.
Which is the greater number? ____
Answer:

Complete the number pattern.

Question 10.
13, 14, ___, ____, 17, ___, 19
Answer: 13, 14, 15, 16, 17, 18, 19

Question 11.
19, 17, ___, 13, 11, ___
Answer:
19, 17, 15, 13, 11, 9

Write the numbers in order from least to greatest.

Question 12.

Answer:

Write the numbers in order from greatest to least.

Question 13.

Answer:

Problem Solving
Read the clues.
Then cross out the numbers to solve.

Example

I am greater than 13.
I am less than 17.
Of the numbers that are left:
I am not the least.
I am not the greatest.
What number am I? 15

Question 14.

I am less than 20.
I am more than 13.
I am less than 17.
I am 4 more than 12.
What number am I? _______
Answer: I am number 15

Question 15.

a. I am more than 10.
I am less than 20.
I am more than 12.
I am less than 15.
Of the numbers that are left: I am the greater number. What number am I? ____
Answer:

b. Draw the number in the place-value chart.

Answer:

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 2 Lesson 2.3 Multiplying and Dividing Integers to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 2 Lesson 2.3 Answer Key Multiplying and Dividing Integers

Math in Focus Grade 7 Chapter 2 Lesson 2.3 Guided Practice Answer Key

Lesson Objective

• Multiply and divide integers.

Hands-On Activity

Explore Multiplication Rules Using Repeated Addition

Work individually.

You can think of multiplying integers as repeated addition.

Step 1
Use a number line to model and complete the multiplication of integers as repeated addition.

a) Evaluate 3 – 2.

3 • 2 means three groups of 2.
3 • 2 = 2 • 3
= 2 + 2 + 2
= 6

Commutative property of multiplication:
Two or more numbers can be multiplied in any order.

b) Evaluate 3 • (-2).

3 • (- 2) means three groups of -2.
3 • (-2) = (-2) • 3
= -2 + (-2) + (-2)
=
The expression 3 • (-2) can also be written as 3(-2).
The expression (-2) • 3 can also be written as -2(3).

Step 2.
Copy and complete the table using repeated addition.

Answer:

Math Journal
Study the pattern in the table from step 2.
a) What do you observe about the sign of the product of two positive integers?
Answer:
I observed the product of two positive integers with ‘ + ‘ sign.

b) What do you observe about the sign of the product of a positive and a negative integer?
Answer:
I observed the product of a positive and a negative integer with ‘ – ‘ sign.

Step 3.
Use a number line to model and complete the multiplication as addition of the opposite.

a) Evaluate -3 • 2.

You can say that -3 • 2 is the opposite of three groups of 2, 6.
-3 • 2 = -(3)(2)
=
Answer:
I can say that -3 • 2 is the opposite of three groups of 2, 6.
-3 • 2 = -(3)(2)

b) Evaluate -3 • (-2).

You can say that -3 • (-2) is the opposite of three groups of -2, -6.
-3 • (-2) = -(3)(-2)
= -()
=
Answer:
I can say that -3 • (-2) is the opposite of three groups of -2, -6.
-3 • (-2) = -(3)(-2)
-()

Step 4.
Copy and complete the table using addition of the opposite and your results from Step 2.

Answer:

Math Journal Based on your observations in Step 1 to Step 4,

a) What do you observe about the sign of the product of two integers with the same sign?
Answer:
I observed the product of two integers with the same sign results ‘ + ‘ sign.

b) What do you observe about the sign of the product of a positive integer and a negative integer? of integers with different signs?
Answer:
I observed the product of a positive integer and a negative integer  with ‘ – ‘ sign.

Evaluate each product.

Question 1.
9(-8)
Answer:
9(-8) = – 72
Explanation:
Perform multiplication operation on above two numbers. Multiply 9 with -8 the product is -72. The product of a positive integer and a negative integer  is represented with ‘ – ‘ sign.

Question 2.
-7 • (-5)
Answer:
-7 • (-5) = -(7)(-5) = -(-35) = 35
Explanation:
Explanation:
Perform multiplication operation on above two numbers. Multiply -7 with -5 the product is 35. The product of two integers with the same sign results ‘ + ‘ sign.

Question 3.
3(-4)(6)
Answer:
3(-4)(6) = (-12)(6) = -72
Explanation:
Perform multiplication operation on above three numbers. First multiply 3 with -4 the product is -12. Next multiply -12 with 6 the product is -72. The product of a two positive integers and a negative integer  is represented with ‘ – ‘ sign.

Think Math
Will the product of three negative numbers be positive or negative? What about the product of four negative numbers? Explain your answers.
Answer:
The product of three negative numbers is negative.
The product of four negative numbers is positive.
Explanation:
Odd numbers of negative integers will have a negative product. Even numbers of negative integers will have a positive product.

Solve.

Question 4.
In a regional golf championship, Steven plays four rounds. The score for a round is recorded as positive (over par) or negative (under par). If Steven scores 6 points under par for all four rounds, what is his total score for his game?
• (-6) =
His score is points.
Answer:
4 • (-6) = -24
His total score is -24 points.
Explanation:
In a regional golf championship, Steven plays four rounds. It is represented with the number 4. Steven scores 6 points under par for all four rounds. Here under par is represented with ‘-‘ sign. 6 points is represented with -6. Multiply 4 with -6 the product is -24. His total score is -24 points.

Question 5.
The price of a stock falls $2 each day for 9 days. Find the total change in the price of the stock during this time.
Answer:
$2 • 9 = $18
The price of a stock falls by $18.
Explanation:
The price of a stock falls $2 each day for 9 days. Multiply $2 with 9 the product is $18. The total change in the price of the stock during this time falls by $18.

Evaluate each quotient.

Question 6.
-36 ÷ (-4)
Answer:
-36 ÷ (-4) = 9
The quotient is 9.
Explanation:
Perform division operation on above two numbers. Divide -36 by -4 the quotient is 9. Even numbers of negative integers will have a positive quotient.

Question 7.
-35 ÷ 5
Answer:
-35 ÷ 5 = -7
The quotient is -7.
Explanation:
Perform division operation on above two numbers. Divide -35 by 5 the quotient is -7. Odd numbers of negative integers will have a negative quotient.

Question 8.
45 ÷ (-3)
Answer:
45 ÷ (-3) = -15
The quotient is -15.
Explanation:
Perform division operation on above two numbers. Divide 45 by -3the quotient is -15 Odd numbers of negative integers will have a negative quotient.

Solve.

Question 9.
Find the change in elevation per minute of a hiker who descended 320 feet in 40 minutes.
Answer:
-320 ÷ 40 = -8 feet/min
Explanation:
A hiker who descended 320 feet in 40 minutes is represented with -320 feet in 40 minutes. The change in elevation per minute of a hiker who descended 320 feet in 40 minutes is -8 feet/min.

Math in Focus Course 2A Practice 2.3 Answer Key

Evaluate each product.

Question 1.
5 • (-7)
Answer:
5 • (-7) = -35
Explanation:
Perform multiplication operation on above two numbers. Multiply 5 with -7 the product is -35. The product of a positive integer and a negative integer  is represented with ‘ – ‘ sign.

Question 2.
12 • (-9)
Answer:
12 • (-9) = -108
Explanation:
Perform multiplication operation on above two numbers. Multiply 12 with -9 the product is -108. The product of a positive integer and a negative integer  is represented with ‘ – ‘ sign.

Question 3.
-6 • 8
Answer:
-6 • 8 = -48
Explanation:
Perform multiplication operation on above two numbers. Multiply -6 with 8 the product is -48. The product of a negative integer and a positive integer is represented with ‘ – ‘ sign.

Question 4.
-3 • 15
Answer:
-3 • 15 = -45
Explanation:
Perform multiplication operation on above two numbers. Multiply -3 with 15 the product is -45. The product of a negative integer and a positive integer is represented with ‘ – ‘ sign.

Question 5.
-4 • (-12)
Answer:
-4 • (-12) = -(4) • (-12) = -(-48) = 48
Explanation:
Perform multiplication operation on above two numbers. Multiply -4 with -12 the product is 48. The product of two integers with the same sign results ‘ + ‘ sign.

Question 6.
-8 • (-20)
Answer:
-8 • (-20) = -(8) • (-20) = -(-160) = 160
Explanation:
Perform multiplication operation on above two numbers. Multiply -8 with -20 the product is 160. The product of two integers with the same sign results ‘ + ‘ sign.

Question 7.
-14 . 0
Answer:
-14 • 0 = 0
Explanation:
Perform multiplication operation on above two numbers. Multiply -14 with 0 the product is 0.

Question 8.
0 • (-50)
Answer:
0 • (-50) = 0
Explanation:
Perform multiplication operation on above two numbers. Multiply 0 with -50 the product is 0.

Question 9.
-3 • 12 • 7
Answer:
-3 • 12 • 7= (-36)• 7 = -252
Explanation:
Perform multiplication operation on above three numbers. First multiply -3 with 12 the product is -36. Next multiply -36 with 7 the product is -252. The product of a two positive integers and a negative integer is represented with ‘ – ‘ sign.

Question 10.
8 • (-4) • 2
Answer:
8 • (-4) • 2= (-32)• 2 = -64
Explanation:
Perform multiplication operation on above three numbers. First multiply 8 with -4 the product is -32. Next multiply -32 with 2 the product is -64. The product of a two positive integers and a negative integer is represented with ‘ – ‘ sign.

Question 11.
20 • 5 • (-5)
Answer:
20 • 5 • (-5) = 100 • (-5)= -500
Explanation:
Perform multiplication operation on above three numbers. First multiply 20 with 5 the product is 100. Next multiply 100 with -5 the product is -500. The product of a two positive integers and a negative integer is represented with ‘ – ‘ sign.

Question 12.
-4 • 10 • (-6)
Answer:
-4 • 10 • (-6) = -40 • (-6) = 240
Explanation:
Perform multiplication operation on above three numbers. First multiply -4 with 10 the product is -40. Next multiply -40 with -6 the product is -240. The product of a two negative integers and a positive integer is represented with ‘ + ‘ sign.

Question 13.
-7 • (-2) • 10
Answer:
-7 • (-2) • 10 = 14 • 10 = 140
Explanation:
Perform multiplication operation on above three numbers. First multiply -7 with -2 the product is 14. Next multiply 14 with 10 the product is 140. The product of a two negative integers and a positive integer is represented with ‘ + ‘ sign.

Question 14.
9 • (-6) • (-4)
Answer:
9 • (-6) • (-4) = (-54) • (-4) = 216
Explanation:
Perform multiplication operation on above three numbers. First multiply 9 with -6 the product is -54. Next multiply -54 with -4 the product is 216. The product of a two negative integers and a positive integer is represented with ‘ + ‘ sign.

Question 15.
-2 • (-8) • (-7)
Answer:
-2 • (-8) • (-7) = 16 • (-7) = -112
Explanation:
Perform multiplication operation on above three numbers. First multiply -2 with -8 the product is -16. Next multiply 16 with -7 the product is -112. The product of a three negative integers is represented with ‘ – ‘ sign.

Question 16.
-5 • (-12) • (-3)
Answer:
-5 • (-12) • (-3) = 60 • (-3) = -180
Explanation:
Perform multiplication operation on above three numbers. First multiply -5 with -12 the product is 60. Next multiply 60 with -3 the product is -180. The product of a three negative integers is represented with ‘ – ‘ sign.

Question 17.
14 • 0 • (-15)
Answer:
14 • 0 • (-15) = 0 • (-15) = 0
Explanation:
Perform multiplication operation on above three numbers. First multiply 14 with 0 the product is 0. Next multiply 0 with -15 the product is 0.

Question 18.
-30 • (-2) • 0
Answer:
-30 • (-2) • 0 = 60 • 0 = 0
Explanation:
Perform multiplication operation on above three numbers. First multiply -30 with -2 the product is 60. Next multiply 60 with 0 the product is 0.

Question 19.
-6 • (-7) • 2 • 5
Answer:
-6 • (-7) • 2 • 5
= 42 • 2 • 5
= 84 • 5
= 420 
Explanation:
Perform multiplication operation on above three numbers.
First multiply -6 with -7 the product is 42.
Next multiply 42 with 2 the product is 84.
Multiply the obtained product 84 with 5 the product is 420.
The product of a two negative integers and two positive integer is represented with ‘ + ‘ sign.

Question 20.
-8 • (-2) • (-4) • 12
Answer:
-8 • (-2) • (-4) • 12
= 16 • (-4) • 12
= -64 • 12
= -768
Explanation:
Perform multiplication operation on above three numbers.
First multiply -8 with -2 the product is 16.
Next multiply 16 with 4 the product is -64.
Multiply the above product -64 with 12 the product is -768.
The product of a three negative integers and a positive integer is represented with ‘ – ‘ sign.

Question 21.
-9 • (-5) • (-4) • (-3)
Answer:
-9 • (-5) • (-4) • (-3)
= 45 • (-4) • (-3)
= -180 • (-3)
= 540
Explanation:
Perform multiplication operation on above three numbers.
First multiply -9 with -5 the product is 45.
Next multiply 45 with -4 the product is -180.
Multiply the above product -180 with -3 the product is 540.
Even numbers of negative integers will have a positive product.

Evaluate each quotient.

Question 22.
125 ÷ (-25)
Answer:
125 ÷ (-25) = -5
The quotient is -5.
Explanation:
Perform division operation on above two numbers. Divide 125 by -25 the quotient is -5. Odd numbers of negative integers will have negative quotient.

Question 23.
300 ÷ (-15)
Answer:
300 ÷ (-15) = -20
The quotient is -20.
Explanation:
Perform division operation on above two numbers. Divide 300 by -15 the quotient is -20. Odd numbers of negative integers will have negative quotient.

Question 24.
-100 ÷ 25
Answer:
-100 ÷ 25 = -4
Explanation:
Perform division operation on above two numbers. Divide -100 by 25 the quotient is -4. Odd numbers of negative integers will have negative quotient.

Question 25.
-32 ÷ 4
Answer:
-32 ÷ 4 = -8
Explanation:
Perform division operation on above two numbers. Divide -32 by 4 the quotient is -8. Odd numbers of negative integers will have negative quotient.

Question 26.
-480 ÷ (-12)
Answer:
-480 ÷ (-12) = 40
Explanation:
Perform division operation on above two numbers. Divide -480 by -12 the quotient is 40. Even numbers of negative integers will have positive quotient.

Question 27.
-144 ÷ (-24)
Answer:
-144 ÷ (-24) = 6
Explanation:
Perform division operation on above two numbers. Divide -144 by -24 the quotient is 6. Even numbers of negative integers will have positive quotient.

Question 28.
0 ÷ (-8)
Answer:
0 ÷ (-8) = 0
Explanation:
Perform division operation on above two numbers. Divide 0 by -8 the quotient is 0.

Question 29.
0 ÷ (-111)
Answer:
0 ÷ (-111) = 0
Explanation:
Perform division operation on above two numbers. Divide 0 by -111 the quotient is 0.

Solve. Show your work.

Question 30.
While returning to the glider port, Laura descended at minute for 3 minutes. Calculate her change in altitude.
Answer:

Question 31.
A scuba diver took 6 minutes to rise to the surface at a minute. How far was he below sea level?
Answer:

Question 32.
An elevator descends 1,500 feet in 60 seconds. Find the change in height per second.
Answer:
1,500 ÷ 60 = 25 feet/sec
Explanation:
An elevator descends 1,500 feet in 60 seconds. Here descends represented with negative sign. The change in height per second is -25 feet/sec

Question 33.
A scientist measures the change in height per second of a diving osprey as -198 feet per second. Find the change in position of the osprey after 2 seconds.
Answer:

The change in height per second of a diving osprey as -198 feet per second.
1 sec = -198 feet
2 sec = ?
-198 • 2 = -396 feet
The change in position of the osprey after 2 seconds is -396 feet.

Question 34.
Math Journal
Margaret wrote -25 ÷ (-100) = \(\frac{-25}{-100}\) = –\(\left(\frac{1}{4}\right)\) and
-2 • (-2) = -4. Discuss and correct her mistakes.
Answer:
-25 ÷ (-100) = 1/4
Perform division operation on above two numbers. Divide -25 by -100 the quotient is 1/4. Even numbers of negative integers will have positive quotient.
-2 • (-2) = 4
Perform multiplication operation on above two numbers. Multiply -2 with -2 the product is 4. The product of two integers with the same sign results ‘ + ‘ sign.

Question 35.
Math Journal Umberto has trouble solving -12 ÷ 3 • 2 ÷ (-4). Write an explanation to help him.
Answer:
To evaluate -12 ÷ 3 • 2 ÷ (-4)
Umberto can do it from left to right. Perform any two operations at one time, and use the rules for sign multiplication and division.
-12 ÷ 3 • 2 ÷ (-4)
= -4 • 2 ÷ (-4)
= -8 ÷ (-4)
= 2

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 2 Lesson 2.2 Subtracting Integers to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 2 Lesson 2.2 Answer Key Subtracting Integers

Math in Focus Grade 7 Chapter 2 Lesson 2.2 Guided Practice Answer Key

Hands-On Activity

Materials:

• counters

Subtract Integers

Work in pairs.

A zero pair can be added to any expression without changing its value. It is the same thing as adding 0 to a number.
Sometimes, you add zero pairs in order to subtract.

Step 1.
Use counters to model and complete the subtraction of a positive integer.

a) Evaluate 5 – (+2) and compare with 5 + (-2).

Answer:

b) Evaluate -5 – (+2) and compare with -5 + (-2).

Answer:

Step 2
Use counters to evaluate each expression.
a) 6 – 4 and 6 + (-4)
Answer:

b) -6 – 4 and -6 + (-4)
Answer:

Step 3
Use counters to model and complete the subtraction of a negative integer,

a) Evaluate 5 – (-2) and compare with 5 + 2.

Answer:

b) Find -5 – (-2) and compare with -5 + 2.

Answer:

Step 4
Use counters to evaluate each expression.

a) 7 – (-3) and 7 + 3
Answer:

b) -7 – (-3) and -7 + 3
Answer:

Math Journal Based on your results in step 1 to step 4, explain how you can subtract integers.

Copy and complete.

Question 1.
21 – 30
21 – 30 = 21 + Rewrite subtraction as adding the opposite.
Using absolute values,
|-30| – |21| = 30 – 21 Subtract the absolute values, because the addends have different signs.
= Simplify.
21 – 30 = 21 +
= Use a negative sign, because -30 has a greater absolute value.
=
Answer:
Rewrite subtraction as adding the opposite.
21 – 30 = 21 + (-30) 
Using absolute values,
|-30| – |21| = 30 – 21
Subtract the absolute values, because the addends have different signs.
= 9
21 – 30 = 21 + (-30)
= -9 Use a negative sign, because -30 has a greater absolute value.
= -9

Question 2.
A fishing boat drags its net 35 feet below the ocean’s surface. Then it lowers the net by an additional 12 feet. Find the new position relative to the surface of the fishing net.
-35 – = -35 + Rewrite subtraction as adding the opposite.
Using absolute values,
+ = Add the absolute values because the addends have the same sign.
-35 – = -35 + Use the common sign, a negative sign, for the sum.
= Simplify.
The fishing net’s new position is feet below the surface of the ocean.
Answer:
Rewrite subtraction as adding the opposite.
-35 – 12 = -35 + (-12) 
Using absolute values,
|-35| +  |-12| = 47
Add the absolute values because the addends have the same sign.
-35 – 12 = -35 + (-12)
Use the common sign, a negative sign, for the sum.
= -47
The fishing net’s new position is 47 feet below the surface of the ocean.

Solve.

Question 3.
A submarine was at 1,200 feet below sea level, It then moved to 1,683 feet below sea level. How many feet did the submarine descend?

Answer:
1,683 – 1,200 = 493 feet
The submarine descend to 493 feet below sea level.
Explanation:
A submarine was at 1,200 feet below sea level, It then moved to 1,683 feet below sea level. Subtract 1,200 feet from 1,683 feet the difference is 493 feet. The submarine descend to 493 feet below sea level.

Copy and complete.

Question 4.
17 – (-4)
17 – (-4) = 17 + Rewrite subtraction as adding the opposite.
= Add.
Answer:
17 – (-4)
Rewrite subtraction as adding the opposite.
= 17 + 4 
= 21
Explanation:
Perform subtraction operation one above numbers. Rewrite subtraction as adding the opposite. Add 17 with4 the sum is 21.

Evaluate each expression.

Question 5.
-25 – (-9)
Answer:
-25 – (-9)
Rewrite subtraction as adding the opposite.
= -25 + 9
= -16
Explanation:
Perform subtraction operation one above numbers. Rewrite subtraction as adding the opposite. Add -25 with 9 the sum is -16.

Question 6.
-19 – (-7) – (-6)
Answer:
-19 – (-7) – (-6)
= -19 + 7 + 6
= -19 + 13
= -6
Explanation:
Perform subtraction operation one above numbers. Rewrite subtraction as adding the opposite. Add 7 with 6 the sum is 13. Next add -19 with 13 the sum is -6.

Copy and complete.

Question 7.
Find the distance between 3 and -2.
Method 1
Use a number line to plot the points and count the units.

The distance between 3 and -2 is units.
Method 2
Use absolute value to find the distance between integers with opposite signs.
Distance between 3 and -2:
|3 – | = | | Rewrite subtraction as adding the opposite.
= units Add.
The distance between 3 and -2 is units.
Answer:
Use a number line to plot the points and count the units.

The distance between 3 and -2 is 5 units.
Method 2
Use absolute value to find the distance between integers with opposite signs.
Distance between 3 and -2:
Rewrite subtraction as adding the opposite.
|3 – (-2) | = | 3 + 2 |
= 5 units
The distance between 3 and -2 is 5 units.

Question 8.
A particular town has an elevation of 8 feet below sea level. Another town on top of a mountain has an elevation of 2,421 feet above sea level. What is the difference in the elevations of the two towns?

Elevation of town below sea level: ft
Elevation of town on top of mountain: ft
Difference between the two elevations:
| – | = | | Rewrite subtraction as adding the opposite.
= ft Add.
The difference in the elevations of the two towns is feet.
Answer:
Elevation of town below sea level: -8 feet
Elevation of town on top of mountain: 2,421 feet
Difference between the two elevations:
Rewrite subtraction as adding the opposite.
| 2,421 – (-8) | = | 2,421 + 8 |
= 2,429 feet 
The difference in the elevations of the two towns is 2,429 feet.

Math in Focus Course 2A Practice 2.2 Answer Key

Evaluate each expression.

Question 1.
7 – 18
Answer:
7 – 18 = -11
Explanation:
Perform subtraction operation on above two numbers. Subtract 18 from 7 the difference is -11.

Question 2.
20 – 30
Answer:
20 – 30 = -10
Explanation:
Perform subtraction operation on above two numbers. Subtract 30 from 20 the difference is -10.

Question 3.
53 – 109
Answer:
53 – 109 = -56
Explanation:
Perform subtraction operation on above two numbers. Subtract 109 from 53 the difference is -56.

Question 4.
45 – (-16)
Answer:
45 – (-16)
= 45 + 16
= 61
Explanation:
Perform subtraction operation on above two numbers. Rewrite subtraction as adding the opposite. Add 45 with 16 the sum is 61.

Question 5.
-7 – (-5)
Answer:
-7 – (-5)
= -7 + 5
= -2
Explanation:
Perform subtraction operation on above two numbers. Rewrite subtraction as adding the opposite. Add -7 with 5 the sum is -2.

Question 6.
-94 – (-68)
Answer:
-94 – (-68)
= -94 + 68
= -26
Explanation:
Perform subtraction operation on above two numbers. Rewrite subtraction as adding the opposite. Add -94 with 68 the sum is -26.

Question 7.
-6 – 8 – 12
Answer:
-6 – 8 – 12
= -14 – 12
= -26
Explanation:
Perform addition operation on above numbers. Add -6 with -8 the sum is -14. Add -14 with -12 the sum is -26.

Question 8.
-23 – 17 – 7
Answer:
-23 – 17 – 7
= -40 – 7
= -47
Explanation:
Perform addition operation on above numbers. Add -23 with -17 the sum is -40. Add -40 with -7 the sum is -47.

Question 9.
-8 – (-4) – 5
Answer:
-8 – (-4) – 5
= – 8 + 4 -5
= -13 + 4
= -9
Explanation:
Perform subtraction operation on above numbers. Rewrite subtraction as adding the opposite. Add -8 with -5 the sum is -13. Add -13 with 4 the sum is -9.

Question 10.
-5 – (-10) – 6
Answer:
-5 – (-10) – 6
= – 5 + 10 – 6
= 5 – 6
= -1
Explanation:
Perform subtraction operation on above numbers. Rewrite subtraction as adding the opposite. Add -5 with 10 the sum is 5. Add 5 with -6 the sum is -1.

Question 11.
-20 – (-16) – (-7)
Answer:
-20 – (-16) – (-7)
= -20 + 16 + 7
= -20 + 23
= 3
Explanation:
Perform subtraction operation on above numbers. Rewrite subtraction as adding the opposite. Add 16 with 7 the sum is 23. Add -20 with 23 the sum is 3.

Question 12.
-11 – (-8) – (-14)
Answer:
-11 – (-8) – (-14)
= -11 + 8 + 14
= -11 + 22
= 11
Explanation:
Perform subtraction operation on above numbers. Rewrite subtraction as adding the opposite. Add 8 with 14 the sum is 22. Add -11 with 22 the sum is 11.

Evaluate the distance between each pair of integers.

Question 13.
4 and 20
Answer:
Use a number line to plot the points and count the units.

The distance between 4 and 20 is 16 units.

Question 14.
16 and 52
Answer:
52 – 16 = 36
The distance between 16 and 52 is 36.

Question 15.
-15 and 36
Answer:
Use absolute value to find the distance between integers with opposite signs.
Distance between -15 and 36:
Rewrite subtraction as adding the opposite.
|-15 + 36 | = | 15 + 36 |
= 51 units
The distance between -15 and 36 is 51 units.

Question 16.
-7 and 41
Answer:
Use absolute value to find the distance between integers with opposite signs.
Distance between -7 and 41:
Rewrite subtraction as adding the opposite.
|-7 + 41 | = | 7 + 41|
= 48 units
The distance between -7 and 41 is 48 units.

Question 17.
-28 and -3
Answer:

The distance between -28 and -3 is 25 units.

Question 18.
-19 and -8
Answer:
Use a number line to plot the points and count the units.

The distance between -19 and -8 is 11 units.

Solve. Show your work.

Question 19.
Rick leaves to go skiing in Burlington, Vermont, when the temperature is -4°C. The temperature drops 10°C when a cold front moves in. What is the new temperature?
Answer:
-4°C – 10°C = -14°C
The new temperature is -14°C.
Explanation:
When the temperature is -4°C Rick leaves to go skiing in Burlington, Vermont. The temperature drops 10°C when a cold front moves in. Which is represented as -10°C. Add -4°C with -10°C the sum is -14°C. The new temperature is -14°C.

Question 20.
The water level of the Dead Sea dropped from 390 meters below sea level in 1930 to 423 meters below sea level in 2010. By how much did the water level drop from 1930 to 2010?
Answer:
423 – 390 = 33 meters
The water level drop 33 meters below sea level from 1930 to 2010.
Explanation:
The water level of the Dead Sea dropped from 390 meters below sea level in 1930 to 423 meters below sea level in 2010. Subtract 390 meters from 423 meters the difference is 33 meters.

Question 21.
Math Journal Florence has only $420 in her bank account. Describe how to find the amount in her account after she writes a check for $590.
Answer:
$420 – $590 = -$170
Explanation:
Florence has only $420 in her bank account. Then she writes a check of $590. Subtract $590 from $420 the difference is -$170. The amount in her account is -$170.

Question 22.
Math Journal Darren has trouble simplifying 15 – (-36). Write an explanation to help him.
Answer:
15 – (-36)
= 15 + 36
= 51
Explanation:
Perform subtraction operation on above numbers. Rewrite subtraction as adding the opposite. Add 15 with 36 the sum is 51.

Question 23.
The wind-chill temperature at 10 P.M. was -8°F. One hour later, the wind-chill temperature had fallen to -28°F. Write an expression to represent the change in temperature. Then find the change in temperature.
Answer:
(-28) – (-8) 
= -28 + 8
= -20°F
The change in temperature is -20°F.
Explanation:
The wind-chill temperature at 10 P.M. was -8°F. One hour later, the wind-chill temperature had fallen to -28°F. The expression to represent the change in temperature is -(28) – (-8). After solving the expression the change in temperature is -20°F.

Question 24.
The lowest point in North America is in Death Valley, California, which is 86 meters below sea level at its lowest point. The highest point is Denali, a mountain in Alaska, with an elevation of 6,198 meters above sea level. What is the difference in their elevations?

Answer:
6,198 m – (-86m)
= 6,198 m + 86 m
= 6,284 m
The difference in their elevations is 6,284 meters.
Explanation:
The lowest point in North America is in Death Valley, California, which is 86 meters below sea level at its lowest point. It is represented as -86 meters. The highest point is Denali, a mountain in Alaska, with an elevation of 6,198 meters above sea level. The difference in their elevations is 6,284 m.

Question 25.
Belinda has two freezers. Freezer A keeps frozen foods at a temperature of -20°F, while Freezer B keeps frozen foods at a temperature of -4°F. She transferred a package of frozen food from one freezer to the other.
a) What is the temperature difference between the two freezers?
Answer:
-4°F – (-20°F)
= -4°F + 20°F
= 16°F
The temperature difference between the two freezers are 16°F.

b) If the temperature of the package rises after the transfer, from which freezer was the package taken?
Answer:
The package was taken from freezer A. Freezer A keeps frozen foods at a temperature of -20°F, while Freezer B keeps frozen foods at a temperature of -4°F. The temperature -4°F is greater than -20°F. The temperature of the package rises after the transfer.

Question 26.
You and a friend are playing a video game. Your score so far is 340 points and your friend’s score is -220 points. What is the difference between your scores?
Answer:
340 – (-220)
= 340 + 220
= 560
The difference between our scores is 560 points.

Question 27.
Two record low monthly temperatures for Anchorage, Alaska, are -34°F in January and 31°F in August. Find the difference between these two temperatures.
Answer:
31°F – (-34°F)
= 31°F + 34°F
= 65°F
The difference between the two temperatures is 65°F.
Explanation:
Two record low monthly temperatures for Anchorage, Alaska, are -34°F in January and 31°F in August. Add 31°F with 34°F the sum is 65°F. The difference between the two temperatures is 65°F.

Question 28.
Town X is 120 feet above sea level, Town Y is 25 feet below sea level, and Town Z is 30 feet below sea level. How high is
a) Town X above Town Y?
Answer:
Town X is 120 feet above sea level. Above sea level is represented with +120 feet.
Town Y is 25 feet below sea level. Below sea level is represented with -25 feet.
-25 feet + 120 feet = 95 feet.
Town X is 95 feet higher than Town Y.

b) Town Y above Town Z? Town X
Answer:
Town Y is 25 feet below sea level. Below sea level is represented with -25 feet.
Town Z is 30 feet above sea level. Below sea level is represented with -30 feet.
-25 feet – (-30) feet = 5 feet.
Town Y is 5 feet higher than Town Z.

c) Town X above Town Z?
Answer:
Town X is 120 feet above sea level. Above sea level is represented with +120 feet.
Town Z is 30 feet below sea level. Below sea level is represented with -30 feet.
-30 feet + 120 feet = 90 feet.
Town X is 90 feet higher than Town Z.

Question 29.
Math Journal
a) Find |8 – 12| and |8| – |12|. Is |8 – 12| equal to |8| – |12|?
Answer:
|8 – 12|
= |-4|
= 4
|8| – |12|
= 8 – 12
= -4
No, |8 – 12| is not equal to |8| – |12|

b) Find |12 – 8| and |12| – |8|. Is |12 – 8| equal to |12| – |8|?
Answer:
|12 – 8|
= |4|
= 4
|12| – |8|
= 12 – 8
= 4
Yes, |12 – 8| equal to |12| – |8|

c) Joe thinks that to find the distance between two integers m and n, he can write |m| – |n| or |n| – |m|. Use your answer in a) and b) to explain why you agree or disagree.
Answer:
Disagree
Explanation:
The distance between any two integers is a positive number. So, we need to find |m – n| or |n – m|.
For example, if the numbers are 8 and 12, |8 – 12| = 4, and |12 – 8| = 4.
So, the distance between 8 and 12 is 4. But |8| – |12| = 8 – 12 = -4.
Since distances must always be positive, So Joe is incorrect.

Question 30.
Math Journal Use the data in the following table. Which two gases have boiling points that are closest in value? Explain.

Answer:
In the above table we can observe gases and temperatures. Two gases having boiling points that are closest in value are Oxygen and Nitrogen. The boiling point for oxygen is -297°F and the boiling point for Nitrogen is -321°F.

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 2 Lesson 2.1 Adding Integers to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 2 Lesson 2.1 Answer Key Adding Integers

Math in Focus Grade 7 Chapter 2 Lesson 2.1 Guided Practice Answer Key

Copy and complete.

Question 1.
Evaluate -3 + (-2).
Method 1
Use a number line to model the sum of two negative integers.

Answer:

Start at -3. Then add -2. Because you are adding a negative integer, -2, you make a jump of 2 to the left.

Method 2
Use absolute values to find the sum of two negative integers.
|-3| = Write the absolute value of each integer.
|-2| =
|-3| + |-2| = + Add the absolute values.
= Simplify.
-3 + (-2) = Use the common sign, a sign, for the sum.
Answer:
|-3| = 3
|-2| = 2
|-3| + |-2| = 3 + 2
= 5
-3 + (-2) = -5 Used the common sign, a negative sign, for the sum.

Question 2.
Evaluate -15 + (-7).
Method 1
Use a number line to model the sum of two negative integers.

Answer:

Start at -15 then add -7. Because we are adding a negative integer, -7, we make a jump of 7 to the left.
Method 2
Use absolute values to find the sum of two negative integers.
|-15| =
|-7| =
|-15| + |-7| = +
=
-15 + (-7) =
Answer:
|-15| = 15
|-7| = 7
|-15| + |-7| = 15 + 7
= 22
-15 + (-7) = -22

Evaluate each sum.

Question 3.
9 + (-9)
Answer:
9 – 9 = 0
Explanation:
Perform addition operation on above two numbers 9 and -9. Add 9 with -9 the sum is o.

Question 4.
-21 + 21
Answer:
-21 + 21 = 0
Explanation:
Perform addition operation on above two numbers -21 and 21. Add -21 with 21 the sum is o.

Hands-On Activity

Materials:

• counters

Find the sum of integers with different signs.

Work in pairs.

A zero pair has a value of zero.
(+1) + (-1) = (-1) + (+1)
= 0
Removing a zero pair does not change the value of an expression. For example:
(+1) + (-1) + 2 = 2 because
(+1) + (-1) + 2 = 0 + 2
= 2

Step 1.
Use counters to model and find the sum of two integers with different signs.
a) Evaluate 3 + (-2).

Answer:

A zero pair has a value of zero.
(+2) + (-2) = (-2) + (+2)
= 0
Removing a zero pair does not change the value of an expression.
(+2) + (-2) + 1 = 1 because
(+2) + (-2) + 1 = 0 + 1
= 1

b) Evaluate (-3) + 2

Answer:

A zero pair has a value of zero.
(+2) + (-2) = (-2) + (+2)
= 0
Removing a zero pair does not change the value of an expression.
(-2) + (+2) + (-1)  = -1 because
(-2) + (+2) + (-1) = 0 + (-1)
= -1

Step 2.
Use counters to model and find each sum.
a) 7 + (-2) and (-7) + 2
Answer:

b) (-8) + 5 and 8 + (-5)
Answer:

Math Journal
Explain how to add two integers with different signs. How are the absolute values of the addends related to the sum?

Evaluate each sum.

Question 5.
-10 + 3
Answer:
-10 + 3 = -7
Explanation:
Perform addition operation on above two numbers -10 and 3 the sum is -7.

Question 6.
-9 + 2
Answer:
-9 + 2 = -7
Explanation:
Perform addition operation on above two numbers -9 and 2 the sum is -7.

Question 7.
11 + (-23)
Answer:
11 + (-23) = -12
Explanation:
Perform addition operation on above two numbers 11 and -23 the sum is -12.

Evaluate each sum.

Question 8.
10 + (-3) + 6
Answer:
10 + (-3) + 6
= 16 + (-3)
= 13
Explanation:
Perform addition operation on above numbers. First add 10 with 6 the sum is 16. Next add the sum 16 with -3 the sum is 13.

Question 9.
-7 + (-23) + 15
Answer:
-7 + (-23) + 15
= -30 + 15
= -15
Explanation:
Perform addition operation on above numbers. First add -7 with -23 the sum is -30. Next add the sum -30 with 15 the sum is -15.

Solve.

Question 10.
A submarine ¡s at 400 feet below sea level. If it ascends 150 feet and then descends 320 feet, how far is ¡t above or below sea level?

You can think of the submarine ascending as an adding a positive integer, and descending as adding a negative integer. So, the verbal description can be translated as -400 + 150 + (-320).

Answer:
-400 + 150 + (-320) = -570
A submarine is 570 feet below sea level.
Explanation:
The submarine ascending as an adding a positive integer, and descending as adding a negative integer. A submarine ¡s at 400 feet below sea level. It ascends 150 feet and then descends 320 feet. Add -400 with 150 and -320 the sum is -570 feet.

Math in Focus Course 2A Practice 2.1 Answer Key

Evaluate each sum using a number line.

Question 1.
-3 + (-9)
Answer:

Explanation:
To evaluate the sum of two negative integers use a number line as we can observe in the above image. Start at -3. Then add -9. Because i am adding a negative integer, -9, I make a jump of 9 to the left. The sum of two negative integers is -12.

Question 2.
-8 + (-4)
Answer:

Explanation:
To evaluate the sum of two negative integers use a number line as we can observe in the above image. Start at -8. Then add -4. Because i am adding a negative integer, -4, I make a jump of 4 to the left. The sum of two negative integers is -12.

Question 3.
7 + (-7)
Answer:

Explanation:
To evaluate the sum of two integers use a number line as we can observe in the above image. Start at 7. Then add -7. Because i am adding a negative integer, -7, I make a jump of 7 to the left. The sum of two integers is 0.

Question 4.
-9 + 9
Answer:

Answer:
To evaluate the sum of two integers use a number line as we can observe in the above image. Start at -9. Then add 9. Because i am adding a positive integer, 9, I make a jump of 9 to the right. The sum of two integers is 0.

Question 5.
-10 + 6
Answer:

Explanation:
To evaluate the sum of two integers use a number line as we can observe in the above image. Start at -10. Then add 6. Because i am adding a positive integer, 6, I make a jump of 6 to the right. The sum of two integers is -4.

Question 6.
-17 + 9
Answer:

Explanation:
To evaluate the sum of two integers use a number line as we can observe in the above image. Start at -17. Then add 9. Because i am adding a positive integer, 9, I make a jump of 9 to the right. The sum of two integers is -8.

Evaluate each sum using the absolute values.

Question 7.
-23 + (-9)
Answer:
Use absolute values to find the sum of above two negative integers.
|-23| = 23
The absolute value of -23 is 23.
|-9| = 9
The absolute value of -9 is 9.
Add the absolute values.
|-23| + |-9| = 23 + 9
= 32
-23 + (-9) = -32
Use the common sign, a negative sign, for the sum.

Question 8.
-11 + (-34)
Answer:
Use absolute values to find the sum of above two negative integers.
|-11| = 11
The absolute value of -11 is 11.
|-34| = 34
The absolute value of -34 is 34.
Add the absolute values.
|-11| + |-34| = 11 + 34
= 45
-11 + (-34) = -45
Use the common sign, a negative sign, for the sum.

Question 9.
-15 + (-7)
Answer:
Use absolute values to find the sum of above two negative integers.
|-15| = 15
The absolute value of -15 is 15.
|-7| = 7
The absolute value of -7 is 7.
Add the absolute values.
|-15| + |-7| = 15 + 7
= 22
-15 + (-7) = -22
Use the common sign, a negative sign, for the sum.

Question 10.
12 + (-18)
Answer:
Use absolute values to find the sum of above two integers.
|12| = 12
The absolute value of 12 is 12.
|-18| = 18
The absolute value of -18 is 18.
Add the absolute values.
|12| + |-18| = 12 + 18
= 30
12 + (-18) = -6
Use the common sign, a negative sign, for the sum.

Question 11.
-40 + 26
Answer:
Use absolute values to find the sum of above two integers.
|-40| = 40
The absolute value of -40 is 40.
|26| = 26
The absolute value of 26 is 26.
Add the absolute values.
|-40| + |26| = 40 + 26
= 66
-40 + 26 = -14
Use the common sign, a negative sign, for the sum.

Question 12.
-75 + 19
Answer:
Use absolute values to find the sum of above two integers.
|-75| = 75
The absolute value of -75 is 75.
|19| = 19
The absolute value of 19 is 19.
Add the absolute values.
|-75| + |19| = 75 + 19
=94
-75 + 19 = -56
Use the common sign, a negative sign, for the sum.

Evaluate each sum.

Question 13.
-8 + 4 + 5
Answer:
-8 + 4 + 5
= -8 + 9
= 1
Explanation:
Perform addition operation on above numbers. First add 4 with 5 the sum is 9. Next add -8 with 9 the sum is 1.

Question 14.
5 + (-10) + (-6)
Answer:
5 + (-10) + (-6)
= -5 + (-6)
= -11
Explanation:
Perform addition operation on above numbers. First add 5 with -10 the sum is -5. Next add -5 with -6 the sum is -11.

Question 15.
-6 + (-8) + (-12)
Answer:
-6 + (-8) + (-12)
= -14 + (-12)
= -26
Explanation:
Perform addition operation on above numbers. First add -6 with -8 the sum is -14. Next add -14 with -12 the sum is -26.

Question 16.
-13 + (-17) + 7
Answer:
-13 + (-17) + 7
= -30 + 7
= -23
Explanation:
Perform addition operation on above numbers. First add -13 with -17 the sum is -30. Next add -30 with 7 the sum is -23.

Question 17.
-20 + 16 +(-7)
Answer:
-20 + 16 +(-7)
= -4 + (-7)
= -11
Explanation:
Perform addition operation on above numbers. First add -20 with 16 the sum is -4. Next add -4 with -7 the sum is -11.

Question 18.
-11 + (-8) + 14
Answer:
-11 + (-8) + 14
= -19 + 14
= -5
Explanation:
Perform addition operation on above numbers. First add -11 with -8 the sum is -19. Next add -19 with 14 the sum is -5.

Solve. Show your work.

Question 19.
The temperature is originally recorded as -4°F. What will the temperature be if the temperature rises 20°F?
Answer:
-4°F + 20°F = 16°F
Explanation:
The temperature is originally recorded as -4°F. The temperature rises to 20°F. Add -4°F with 20°F the sum is 16°F. The temperature is 16°F.

Question 20.
Mr. Lawson parked his car in a parking garage 33 feet below street level. He then got in an elevator and went up 88 feet to his office. How far above street level is his office?
Answer:

Answer:
88 – 33 = 55 feet
His office is 55 feet above street level.
Explanation:
Mr. Lawson parked his car in a parking garage 33 feet below street level. He then got in an elevator and went up 88 feet to his office. Subtract 33 feet below street level from 88 feet the difference is 55 feet. His office is 55 feet above street level.

Question 21.
A hiker starts hiking in Death Valley at an elevation of 143 feet below sea level. He climbs up 400 feet in elevation. What is his new elevation relative to sea level?
Answer:
A hiker starts hiking in Death Valley at an elevation of 143 feet below sea level. Below sea level is represented with negative number which is -143 feet.
He climbs up 400 feet in elevation. The word ‘climbs up’ represents positive number which is 400 feet.
– 143 feet + 400 feet = 257 feet
His new elevation is 257 feet above seal level.

Question 22.
Elizabeth was playing a board game with her friends. On her first turn, she moved 6 spaces forward. On her second turn, she moved another 5 spaces forward. On her third turn, she moved 4 spaces backward. How many spaces forward or backward from her starting point was she after her third turn?
Answer:
6 + 5 – 4
= 11 – 4
= 7
After her third turn Elizabeth moved 7 spaces forward from her starting point.
Explanation:
Elizabeth was playing a board game with her friends. On her first turn, she moved 6 spaces forward. Forward spaces are represented with positive number. On her second turn, she moved another 5 spaces forward. Forward spaces are represented with positive number. Add 6 with 5 the sum is 11. On her third turn, she moved 4 spaces backward. Backward spaces are represented with negative number. Subtract 4 from 11 the difference is 7. After her third turn Elizabeth moved 7 spaces forward from her starting point.

Question 23.
In the U.S. Open Golf Tournament, each qualifying golfer plays four rounds. The score for a round is recorded as positive (over par) or negative (under par). If a golfer scores -4, 6, 3, and -2 in the four rounds, what is the golfer’s total score for the tournament?
Answer:
The golfer scores -4, 6, 3, and -2 in four rounds.
-4 + 6 + 3 -2
= 9 – 6
= 3
The golfer’s total score for the tournament is 3.

Question 24.
Math Journal
In a game, all scores with even numbers are recorded as positive numbers. Odd numbers are recorded as negative numbers. Explain how to find David’s total score in this game if his individual scores during the game are 9, 12, 7, 18, and 19.
Answer:
The individual scores during the game are 9, 12, 7, 18, and 19.
Here the scores with even numbers are recorded as positive numbers.
The scores with Odd numbers are recorded as negative numbers.
The scores are -9, 12, -7, 18, -19
David’s total score is calculated by adding the given scores.
– 9 + 12  -7 + 18 – 19
= 30 – 9 – 7 – 19
= 30 – 35
= -5
David’s total score in this game is -5.

This handy Math in Focus Grade 2 Workbook Answer Key Chapter 14 Practice 2 Reading and Writing Time detailed solutions for the textbook questions.

Math in Focus Grade 2 Chapter 14 Practice 2 Answer Key Reading and Writing Time

Write the time in words.

Example

Question 1.

Answer:

three fifteen or 15 minutes after 3.

 

Question 2.

Answer:

eleven forty or 40 minutes after eleven.

 

Question 3.

Answer:

one twenty-five or 25 minutes after 1

Question 4.

Answer:

three thirty five or 35 minutes after 3

Question 5.

Answer:

eight fifty-five or 55 minutes after 8.

 

Write the time

Example

 

Question 6.

The time is _____________.
Answer:

The time is 1:45

 

Question 7.

The time is _____________.
Answer:

The time is 2:00

 

Question 8.

The time is _____________.
Answer:

The time is 6:20

 

Question 9.

The time is _____________.
Answer:

The time is 9:05

 

Question 10.

The time is _____________.
Answer:

The time is 2:55

 

Draw the minute hand to show the time.

Example

The time is 3:55.

Question 11.

The time is 6:30.
Answer:

 

 

 

 

 

Question 12.

The time is 10:15.
Answer:

 

 

 

 

 

Question 13.

The time is 8:00.
Answer:

 

 

 

 

 

Question 14.

The time is 12:40.
Answer:

 

 

 

 

 

Question 15.

The time is 9:05.
Answer:

 

 

 

 

 

Draw the hour hand to show the time.

Example

The time is 10:00.

Question 16.

The time is 11:30.
Answer:

 

 

 

 

 

Question 17.

The time is 7:15.
Answer:

 

 

 

 

 

 

Question 18.

The time is 4:20.
Answer:

 

 

 

 

 

Question 19.

The time is 2:50.
Answer:

 

 

 

 

Question 20.

The time is 3:40.
Answer:

 

 

 

 

 

Draw the hands to show the time.

Example

The time is 7:15.

Question 21.

The time is 4:30.
Answer:

 

 

 

 

 

Question 22.

The time is 1:20.
Answer:

 

 

 

 

 

Question 23.

The time is 9:25.
Answer:

 

 

 

 

 

Question 24.

The time is 7:00.
Answer:

 

 

 

 

 

Question 25.

The time is 9:50.
Answer:

 

 

 

 

 

Draw the hands to show the time. Then write the time in words.

Example

Question 26.

Answer:

 

 

 

 

 

five ten or 10 minutes after 5

 

Question 27.

Answer:

 

 

 

 

 

eleven forty or 40 minutes after 11

 

Question 28.

Answer:

 

 

 

 

 

six fifty five or 55 minutes after 6

 

Question 29.

Answer:

 

 

 

 

 

one twenty-five or 25 minutes after 1

 

Question 30.

Answer:

 

 

 

 

 

twelve-fifteen or 15 minutes after 12

This handy Math in Focus Grade 2 Workbook Answer Key Chapter 14 Practice 1 The Minute Hand detailed solutions for the textbook questions.

Math in Focus Grade 2 Chapter 14 Practice 1 Answer Key The Minute Hand

Question 1.
Fill in the boxes with the number of minutes.

Answer:

 

Fill in the blanks.

Example

Question 2.

The minute hand points to ___________ minutes.
Answer:

The minute hand points to 35 minutes.

 

Question 3.

The minute hand points to ___________ minutes.
Answer:

The minute hand points to 20 minutes.

 

Question 4.

The minute hand points to ___________ minutes.
Answer:

The minute hand points to 55 minutes.

 

Write the time.

Question 5.

Answer:

3 o’clock                                                      5 minutes after 3 o’clock

Question 6.

Answer:

7 o’clock                                                20 minutes after 7 o’clock

Question 7.

Answer:

10 o’clock                                                     45 minutes after 10 o’clock

 

Draw the minute hand to show the time.

Question 8.
15 minutes after 4 o’clock

Answer:

 

Question 9.
40 minutes after 6 o’clock

Answer:

 

Question 10.
50 minutes after 1 o’clock

Answer:

 

Question 11.
35 minutes after 10 o’clock

Answer:

This handy Math in Focus Grade 2 Workbook Answer Key Chapter 15 Multiplication Tables of 3 and 4 detailed solutions for the textbook questions.

Math in Focus Grade 2 Chapter 15 Answer Key Multiplication Tables of 3 and 4

Math Journal

These items are sold in a supermarket. Use the items to write a multiplication story.

Example
I want to buy 4 boxes of cereal.
I will have to give the cashier $16.

Story
__________________
Answer:

I want to buy cracked which costs $3

I want to buy cereal which costs $4.

 

Put On Your Thinking Cap!

Challenging Practice

Question 1.
Steve starts reading a book on page 7. He reads the book for 4 days. He reads 3 pages each day. Which page will Steve stop at on the 4th day?
(Hint: Use a diagram to help you solve.)
Answer:

Given,

Steve starts reading a book on page 7.

He reads the book for 4 days.

He reads 3 pages each day.

So, 4 x 3 = 12

12 + 7 = 19

Therefore, on the 4th day, steve will stop on the 19th page.

 

Question 2.
The music teacher is selecting children to sit in the front row at a concert. 100 children are given numbers 1 to 100. The teacher first picks the child with the number 3. He then skip-counts by tens to pick the other children. What are the numbers of the other children who are picked?
The numbers are ___________________
Answer:

Given,

The music teacher is selecting children to sit in the front row at a concert.

100 children are given numbers 1 to 100.

The teacher first picks the child with the number 3.

He then skip-counts by tens to pick the other children.

 

Put On Your Thinking Cap!

Problem Solving

Solve the riddle.

Question 1.
I am a two-digit number. I am more than 20 but less than 30. I can be found in both the multiplication tables of 3 and 4. What number am I?
Answer:

Given,

I am a two-digit number.

I am more than 20 but less than 30.

I can be found in both the multiplication tables of 3 and 4.

Let the number be a

20 < a > 30

the numbers that are both multiple of 3 and 4 and between 20 and 30 is are 24 only ( 3 x 8 = 4 x 6 )

Therefore, the number is 24.

 

Chapter Review/Test

Vocabulary

Fill in the blanks with words from the box.

skip-counting
dot paper
related multiplication facts

Question 1.

____________ is fun!
Answer:

 

Question 2.
are examples of _____________.
Answer:

Concepts and Skills

Skip count to find the missing numbers.

Question 3.
9 × 3 = ____________
Answer:

Question 4.
___________ × 3 = 24
Answer:

Question 5.
_________ × 4 = 16
Answer:

Question 6.
_________ × 4 = 36
Answer:

Find the missing numbers.

Question 7.
8 groups of 3 = __________ × 3
= ___________
Answer:

Question 8.
7 groups of 4 = ___________ × 4
= ___________
Answer:

Use dot paper to find the missing numbers.

Question 9.
6 × 4 = 5 groups of 4 + ___________ group of 4
= ___________ + 4
= ___________
Answer:

Problem Solving

Use skip-counting or dot paper to solve.

Question 10.
Caleb ties sets of 3 medals with a piece of ribbon. He ties 10 sets of medals. How many medals does Caleb have in all?
Answer:

Use related multiplication facts to solve.

Question 11.
Gail has 32 star-shaped key chains. She puts 4 key chains equally into some boxes. How many boxes are there?
Answer:

Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 1 Lesson 1.3 Introducing Irrational Numbers to finish your assignments.

Math in Focus Grade 7 Course 2 A Chapter 1 Lesson 1.3 Answer Key Introducing Irrational Numbers

Math in Focus Grade 7 Chapter 1 Lesson 1.3 Guided Practice Answer Key

Hands-On Activity

Materials:

• paper
• ruler
• scissors

Find the value of \(\sqrt{2}\) using a square.

Work in pairs.

Step 1.
Draw a square that has a length of 2 inches on a piece of paper. Then cut out the square.

Step 2.
Find the area of the square (square A).

Step 3.
Fold the four vertices of square A towards the center to form square B as shown below.

Step 4.
State how the areas of square A and square B are related. State the area of square B. How can you represent the length of a side of square B?

Step 5.
Using your answer in step 4, find the length of a side of square B with a calculator. Round your answer to 2 decimal places.

Math Journal
Place an edge of square B alongsi& a ruler to measure its length. Explain why the reading from the ruler is different from the answer in step 5.

Copy and complete.

Question 1.
Graph \(\sqrt{5}\) on the number line using rational approximations.
Which two whole numbers is \(\sqrt{5}\) between? Justify your reasoning. Using a calculator, \(\sqrt{5}\) = .
Graph an interval where \(\sqrt{5}\) is located.
The value of \(\sqrt{5}\) with two decimal places is
is closer to than to . So, \(\sqrt{5}\) is located closer to .

By using an approximate value of \(\sqrt{5}\), locate \(\sqrt{5}\) on the number line.

Answer:
The \(\sqrt{5}\) is between two whole numbers. The two whole numbers are 2 and 3.
By using calculator \(\sqrt{5}\) = 2.236067977….
The value of \(\sqrt{5}\) with two decimal places is 2.24.
The decimal 2.24 is closer to 2.2 than to 2.3.
So, \(\sqrt{5}\) is located closer to 2.2.

By using an approximate value of \(\sqrt{5}\),  located  \(\sqrt{5}\) on the number line as we can observe in the below image.

Copy and complete.

Question 2.
Graph –\(\sqrt{2}\) on the number line using rational approximations.
Which two integers is –\(\sqrt{2}\) between? Justify your reasoning. Using a calculator, –\(\sqrt{2}\) = .
Graph an interval where –\(\sqrt{2}\) is located.
The value of –\(\sqrt{2}\) with two decimal places is ? .
is closer to than to . So, –\(\sqrt{2}\) is located closer to ?

By using an approximate value of –\(\sqrt{2}\), locate –\(\sqrt{2}\) on the number line.

Answer:
The –\(\sqrt{2}\) is between two whole numbers. The two whole numbers are -1 and -2.
By using calculator –\(\sqrt{2}\) = -1.414213562…..
The value of –\(\sqrt{2}\) with two decimal places is -1.41.
The decimal -1.41 is closer to -1.4 than to -1.5.
So, –\(\sqrt{2}\) is located closer to -1.4.

By using an approximate value of –\(\sqrt{2}\),  located  –\(\sqrt{2}\) on the number line as we can observe in the below image.

Solve.

Question 3.
Graph –\(\sqrt{7}\) on the number line using rational approximations.
Answer:
The –\(\sqrt{7}\) is between two whole numbers. The two whole numbers are -2 and -3.
By using calculator –\(\sqrt{7}\) = -2.645751311…..
The value of –\(\sqrt{7}\) with two decimal places is -2.64.
The decimal -2.64 is closer to -2.6 than to -2.7.
So, –\(\sqrt{7}\) is located closer to -2.6.
By using an approximate value of –\(\sqrt{7}\),  located  –\(\sqrt{7}\) on the number line as we can observe in the below image.

Math in Focus Course 2A Practice 1.3 Answer Key

Locate each positive irrational number on the number line using rational approximations. First tell which two whole numbers the square root is between.

Question 1.
\(\sqrt{3}\)
Answer:
The \(\sqrt{3}\) is between two whole numbers. The two whole numbers are 1 and 2.
By using calculator \(\sqrt{3}\) = 1.732050807….
The value of \(\sqrt{3}\) with two decimal places is 1.73.
The decimal 1.73 is closer to 1.7 than to 1.8.
So, \(\sqrt{3}\) is located closer to 1.7.

By using an approximate value of \(\sqrt{3}\),  the positive irrational number \(\sqrt{3}\) is located on the number line as we can observe in the above image.

Question 2.
\(\sqrt{7}\)
Answer:
The \(\sqrt{7}\) is between two whole numbers. The two whole numbers are 2 and 3.
By using calculator \(\sqrt{7}\) = 2.645751311….
The value of \(\sqrt{7}\) with two decimal places is 2.64.
The decimal 2.64 is closer to 2.6 than to 2.7.
So, \(\sqrt{7}\) is located closer to 2.6.

By using an approximate value of \(\sqrt{7}\),  the positive irrational number \(\sqrt{7}\) is located on the number line as we can observe in the above image.

Question 3.
\(\sqrt{11}\)
Answer:
The \(\sqrt{11}\) is between two whole numbers. The two whole numbers are 3 and 4.
By using calculator \(\sqrt{11}\) = 3.316624790….
The value of \(\sqrt{11}\) with two decimal places is 3.31.
The decimal 3.31 is closer to 3.3 than to 3.4.
So, \(\sqrt{11}\) is located closer to 3.3.

By using an approximate value of \(\sqrt{11}\),  the positive irrational number \(\sqrt{11}\) is located on the number line as we can observe in the above image.

Question 4.
\(\sqrt{26}\)
Answer:
The \(\sqrt{26}\) is between two whole numbers. The two whole numbers are 5 and 6.
By using calculator \(\sqrt{26}\) = 5.099019513….
The value of \(\sqrt{26}\) with two decimal places is 5.09.
The decimal 5.09 is closer to 5.1 than to 5.0.
So, \(\sqrt{26}\) is located closer to 5.1.

By using an approximate value of \(\sqrt{26}\),  the positive irrational number \(\sqrt{26}\) is located on the number line as we can observe in the above image.

Question 5.
\(\sqrt{34}\)
Answer:
The \(\sqrt{34}\) is between two whole numbers. The two whole numbers are 5 and 6.
By using calculator \(\sqrt{34}\) = 5.830951894….
The value of \(\sqrt{34}\) with two decimal places is 5.83.
The decimal 5.83 is closer to 5.8 than to 5.9.
So, \(\sqrt{34}\) is located closer to 5.8.

By using an approximate value of \(\sqrt{34}\),  the positive irrational number \(\sqrt{34}\) is located on the number line as we can observe in the above image.

Question 6.
\(\sqrt{48}\)
Answer:
The \(\sqrt{48}\) is between two whole numbers. The two whole numbers are 6 and 7.
By using calculator \(\sqrt{48}\) = 6.928203230….
The value of \(\sqrt{48}\) with two decimal places is 6.92.
The decimal 6.92 is closer to 6.9 than to 7.0.
So, \(\sqrt{48}\) is located closer to 6.9.

By using an approximate value of \(\sqrt{48}\),  the positive irrational number \(\sqrt{48}\) is located on the number line as we can observe in the above image.

Locate each negative irrational number on the number line using rational approximations. First tell which two integers the square root is between.

Question 7.
–\(\sqrt{5}\)
Answer:
The –\(\sqrt{5}\) is between two integers. The two integers are -2 and -3.
By using calculator –\(\sqrt{5}\) = -2.236067977…..
The value of –\(\sqrt{5}\) with two decimal places is -2.23.
The decimal -2.23 is closer to -2.2 than to -2.3.
So, –\(\sqrt{5}\) is located closer to -2.2.

By using an approximate value of –\(\sqrt{5}\), the negative irrational number –\(\sqrt{5}\) is located on the number line as we can observe in the above image.

Question 8.
–\(\sqrt{6}\)
Answer:
The –\(\sqrt{6}\) is between two integers. The two integers are -2 and -3.
By using calculator –\(\sqrt{6}\) = -2.449489742…..
The value of –\(\sqrt{6}\) with two decimal places is -2.44.
The decimal -2.44 is closer to -2.4 than to -2.5.
So, –\(\sqrt{6}\) is located closer to -2.4.

By using an approximate value of –\(\sqrt{6}\), the negative irrational number –\(\sqrt{6}\) is located on the number line as we can observe in the above image.

Question 9.
–\(\sqrt{17}\)
Answer:
The –\(\sqrt{17}\) is between two integers. The two integers are -4 and -5.
By using calculator –\(\sqrt{17}\) = -4.123105625…..
The value of –\(\sqrt{17}\) with two decimal places is -4.12.
The decimal -4.12 is closer to -4.1 than to -4.2.
So, –\(\sqrt{17}\) is located closer to -4.1.

By using an approximate value of –\(\sqrt{17}\), the negative irrational number –\(\sqrt{17}\) is located on the number line as we can observe in the above image.

Question 10.
–\(\sqrt{26}\)
Answer:
The –\(\sqrt{26}\) is between two integers. The two integers are -5 and -6.
By using calculator –\(\sqrt{26}\) = -5.099019513….
The value of –\(\sqrt{26}\) with two decimal places is -5.09.
The decimal -5.09 is closer to -5.1 than to -5.0.
So, –\(\sqrt{26}\) is located closer to -5.1.

By using an approximate value of –\(\sqrt{26}\), the negative irrational number –\(\sqrt{26}\) is located on the number line as we can observe in the above image.

Question 11.
–\(\sqrt{53}\)
Answer:
The –\(\sqrt{53}\) is between two integers. The two integers are -7 and -8.
By using calculator –\(\sqrt{53}\) = -7.280109889….
The value of –\(\sqrt{53}\) with two decimal places is -7.28.
The decimal -7.28 is closer to -7.3 than to -7.2.
So, –\(\sqrt{53}\) is located closer to -7.3.

By using an approximate value of –\(\sqrt{53}\), the negative irrational number –\(\sqrt{53}\) is located on the number line as we can observe in the above image.

Question 12.
–\(\sqrt{80}\)
Answer:
The –\(\sqrt{80}\) is between two integers. The two integers are -8 and -9.
By using calculator –\(\sqrt{80}\) = -8.944271909….
The value of –\(\sqrt{80}\) with two decimal places is -8.94.
The decimal -8.94 is closer to -8.9 than to -9.0.
So, –\(\sqrt{80}\) is located closer to -8.9.

By using an approximate value of –\(\sqrt{80}\), the negative irrational number –\(\sqrt{80}\) is located on the number line as we can observe in the above image.

Use a calculator. Locate each irrational number to 3 decimal places on the number line using rational approximations.

Question 13.
\(\sqrt{47}\)
Answer:
By using calculator \(\sqrt{47}\) = 6.855654600….
The value of \(\sqrt{47}\) with three decimal places is 6.855.
The decimal 6.855 is closer to 6.86 than to 6.85.
So, \(\sqrt{47}\) is located closer to 6.86.

The given irrational number \(\sqrt{47}\) is located on the number line by using rational approximation as we can observe in the above image.

Question 14.
–\(\sqrt{15}\)
Answer:
By using calculator –\(\sqrt{15}\) = –3.872983346….
The value of –\(\sqrt{15}\) with three decimal places is –3.872.
The decimal -3.872 is closer to –3.87 than to –3.88.
So, –\(\sqrt{15}\) is located closer to -3.87.

The given irrational number –\(\sqrt{15}\) is located on the number line by using rational approximation as we can observe in the above image.

Question 15.

Answer:
By using calculator = 4.54683594….
The value of with three decimal places is 4.546.
The decimal 4.546 is in between 4.5 and 4.6.
So, is located on 4.55.

The given irrational number is located on the number line by using rational approximation as we can observe in the above image.

Locate each irrational number on the number line using rational approximations.

Question 16.
\(\sqrt{101}\)
Answer:
By using calculator \(\sqrt{101}\) = 10.049875….
The value of \(\sqrt{101}\) with two decimal places is 10.04.
The decimal 10.04 is closer to 10 than to 10.1.
So, \(\sqrt{101}\) is located closer to 10.

The given irrational number \(\sqrt{101}\) is located on the number line by using rational approximation as we can observe in the above image.

Question 17.
–\(\sqrt{132}\)
Answer:
By using calculator –\(\sqrt{132}\) = – 11.489125….
The value of –\(\sqrt{132}\) with two decimal places is –11.48.
The decimal -11.48 is closer to –11.5 than to –11.4.
So, –\(\sqrt{132}\) is located closer to –11.5.

The given irrational number –\(\sqrt{132}\) is located on the number line by using rational approximation as we can observe in the above image.

Question 18.
\(\sqrt{2,255}\)
Answer:
By using calculator \(\sqrt{2,255}\) = 47.4868….
The value of \(\sqrt{2,2551}\) with two decimal places is 47.48.
The decimal 47.48 is closer to 47.5 than to 47.4.
So, \(\sqrt{2,255}\) is located closer to 47.5.

The given irrational number \(\sqrt{2,255}\) is located on the number line by using rational approximation as we can observe in the above image.

Solve.

Question 19.
Locate the value of the constant, π, on the number line using rational numbers.
Answer:
We know that π = 3.14159265

In the above image we can observe the value of the constant, π, is located on the number line using rational numbers.

Question 20.
3.1416 and \(\frac{22}{7}\) are two rational approximate values of π.

a) Graph 3.1416, \(\frac{22}{7}\), and π on the number line.
b) Which of the two rational approximate values is closer to π?
Answer:
a)We know that 22/7 = 3.1428, π = 3.14159

In the above image we can observe 3.1416, 22/7 and π on the number line.
b) The two rational approximate values closer to π are  3.1416 and 22/7.

Question 21.
A triangle is cut from a square as shown in the diagram. The area of the square is 59 square inches. Approximate the height of the triangle to 3 decimal places.

Answer:

Question 22.
Math Journal When do you need to approximate an irrational number with a rational value? Explain and illustrate with an example.
Answer: