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Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 14.3 Unit Rate existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 14.3 Unit Rate

Exercises

SOLVE

The graph below shows the distance Isabella rides her bicycle over time.

Question 1.
Which coordinates (point) on the graph show the unit rate?
Answer:
The coordinate (1,10) shows the unit rate on the graph.

Question 2.
What is the unit rate shown in the graph?
Answer:
The unit rate shown in the graph is 10.
The value for y at which the x coordinate is equal to 1 is called the unit rate.

Question 3.
If the graph also included a line to show how far Isabella walked over time at a rate of 3 miles per hour, would that line be above or below the one on the graph?
Answer:
If the graph also included a line to show how far Isabella walked over time at a rate of 3 miles per hour then the line be below the one on the graph.

Question 4.
Are the quantities in the table above proportional?
Answer:
Yes, the quantities in the above table are proportional.

Question 5.
What is the unit rate?
Answer:
The unit rate shown in the above table is 3.00.
The value for price at which the gallons of milk is equal to 1 is called the unit rate.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 14.2 Graphing Relationships existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 14.2 Graphing Relationships

Exercises

SOLVE

Vicki can read 2 pages in 3 minutes. Bruce can read 4 pages in 3 minutes.

Question 1.
Complete the chart below to show this relationship.

Answer:

Pages read by Vicki in 1 minute:
3 minutes = 2 pages
1 minute = y pages
3 x y = 2 x 1
3y = 2
Divide both sides of the equation by 3.
3y/3 = 2/3
y = 2/3
Pages read by Vicki in 6 minute:
3 minutes = 2 pages
6 minute = z pages
3 x z = 6 x 2
3z = 12
Divide both sides of the equation by 3.
3z/3 = 12/3
z = 4
Pages read by Bruce in 1 minute:
3 minutes = 4 pages
1 minute = y pages
3 x y = 4 x 1
Divide both sides of the equation by 3.
3y/3 = 4/3
y = 4/3
Pages read by Bruce in 6 minute:
3 minutes = 4 pages
6 minute = z pages
3 x z = 6 x 4
3z = 24
Divide both sides of the equation by 3.
3z/3 = 24/3
z = 8

Question 2.
Graph the relationship on the grid. Label each axis. Be sure to use different colors or styles of line so that you can tell the difference. Include a legend to describe which hue is for which person.

Answer:

The time is represented in x-axis and pages read by Vicki and Bruce are represented in y-axis as we can observe in the above image.

Question 3.
Which of the following correctly describes the relationship?
(a) Vicki reads twice as fast as Bruce.
(b) Bruce reads twice as fast as Vicki.
(c) Vicki’s reading rate is one-third of Bruce’s.
(d) Bruce reads half as fast as Vicki.
Answer:
Option B, Bruce reads twice as fast as Vicki correctly describes the relationship.

Question 4.
A store sells 5 pounds of potatoes for $1.00. How much would 1 pound cost? Construct a graph for potato price on the grid below. Label each axis.

Answer:

A store sells 5 pounds of potatoes for $1.00.
5 pounds = $1.00
1 pound = s $
5 x s= $1 x 1
5s = $1
Divide both sides of the equation by 5.
s = $1/5
The cost of 1 pound potatoes are $0.2.
In the above graph we can see price of potatoes are represented in y axis and pounds are represented in x axis.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 14.1 Plotting Ordered Pairs existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 14.1 Plotting Ordered Pairs

Exercises

PLOT ORDERED PAIRS

Give the coordinates for each point on the graph.

A ______________
Answer:
The coordinates (2,3) represents point A on the above graph.

B ______________
Answer:
The coordinates (4,-4) represents point B on the above graph.

C ______________
Answer:
The coordinates (-6,-6) represents point C on the above graph.

D ______________
Answer:
The coordinates (5,3) represents point D on the above graph.

E ______________
Answer:
The coordinates (3,5) represents point E on the above graph.

F ______________
Answer:
The coordinates (1,2) represents point F on the above graph.

G ______________
Answer:
The coordinates (-2,2) represents point G on the above graph.

H ______________
Answer:
The coordinates (-5,4) represents point H on the above graph.

I ______________
Answer:
The coordinates (9,7) represents point I on the above graph.

J _______________
Answer:
The coordinates (-6,8) represents point J on the above graph.

Plot the following ordered pairs on the graph:

A (3, 2)
B (7, -7)
C (-7, -7)
D (-7, 7)
E (7, 7)
F (9, 2)
G (6, 7)
H (-8, -6)
I (-2, 2)
J (-5, -5)
K (3, -4)
L(-2, -2)
Answer:

The given order pairs are plotted on the provided graph as we can observe in the above image.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 13.3 Rates existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 13.3 Rates

Exercises Solve

Question 1.
The two rectangles are similar (proportional). Given the information about rectangle EFGH, what is the length of side CD?

Answer:
The above two rectangles ABDC and EFGH are similar (proportional).
Consider length of the side CD = x
BD/CD = FG/ GH
30/x = 6/16
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
30 × 16 = 6 × x
480 = 6x
Divide both sides of the equation by 6.
480/6 = 6x/6
80 cm = x
Length of the side CD is equal to 80 cm.

Question 2.
George likes to sweeten his ice tea. When he drinks a 20-ounce ice tea, he puts in two teaspoons of sugar. If he is making a gallon of ice tea, how many teaspoons of sugar should he add?
Answer:
We know that 1 gallon is equal to 128 ounces.
20/2 = 128/x
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
20 × x = 128 × 2
20 x = 256
Divide both sides of the equation by 20.
20x/20 = 256/20
x = 12.8 teaspoons
If George makes a gallon of ice tea, he will add 12.8 teaspoons of sugar.

Question 3.
The two triangles are similar. Calculate the length of side s.

Answer:
The above two triangles are similar.
3/9 = 2/s
To find the s value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
3 × s = 2 × 9
3s = 18
Divide both sides of the equation by 3.
3s/3 = 18/3
s = 6
Length of the side s is equal to 6.

Question 4.
Chuck took a long hike where he started at point B and followed the path in the diagram shown. If Chuck wants to take a shorter hike where he starts at B and walks 4 miles east, turns and walks 3 miles north and then returns to point B, how long B would his last leg be? Solve this problem using proportions even though you can solve it using the Pythagorean Theorem.

Answer:
Consider chuck last leg as x.
6.25/5 = x/4
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
6.25 × 4 = x × 5
25 = 5x
Divide both sides of the equation by 5.
25/5 = 5x/5
x = 5

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 13.2 Proportions and Cross-Multiplying existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 13.2 Proportions and Cross-Multiplying

Exercises
Cross-Multiply
Round to the hundredths place.

Question 1.
\(\frac{x}{5}\) = \(\frac{30}{15}\)
Answer:
Given equation is \(\frac{x}{5}\) = \(\frac{30}{15}\)
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
15x = 30 × 5
15x = 150
Divide both sides of the equation by 15.
\(\frac{15x}{15}\) = \(\frac{150}{15}\)
x = 10

Question 2.
\(\frac{z}{3}\) = \(\frac{24}{6}\)
Answer:
Given equation is \(\frac{z}{3}\) = \(\frac{24}{6}\)
To find the z value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
6z = 24 × 3
6z = 72
Divide both sides of the equation by 6.
\(\frac{6z}{6}\) = \(\frac{72}{6}\)
z = 12

Question 3.
\(\frac{14}{z}\) = \(\frac{100}{50}\)
Answer:
Given equation is \(\frac{14}{z}\) = \(\frac{100}{50}\)
To find the z value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
14 × 50 = 100 × z
700 = 100z
Divide both sides of the equation by 100.
\(\frac{700}{100}\) = \(\frac{100z}{100}\)
7 = z

Question 4.
\(\frac{56}{4}\) = \(\frac{y}{20}\)
Answer:
Given equation is \(\frac{56}{4}\) = \(\frac{y}{20}\)
To find the y value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
56 × 20 = y × 4
1120 = 4y
Divide both sides of the equation by 4.
\(\frac{1120}{4}\) = \(\frac{4y}{4}\)
280 = y

Question 5.
\(\frac{45}{9}\) = \(\frac{w}{3}\)
Answer:
Given equation is\(\frac{45}{9}\) = \(\frac{w}{3}\)
To find the w value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
45 × 3 = w × 9
135 = 9w
Divide both sides of the equation by 9.
\(\frac{135}{9}\) = \(\frac{9w}{9}\)
15 = w

Question 6.
\(\frac{14}{x}\) = \(\frac{70}{10}\)
Answer:
Given equation is \(\frac{14}{x}\) = \(\frac{70}{10}\)
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
14 ×10 = 70 × x
140 = 70x
Divide both sides of the equation by 70.
\(\frac{140}{70}\) = \(\frac{70x}{70}\)
2 = x

Question 7.
\(\frac{33}{r}\) = \(\frac{11}{3}\)
Answer:
Given equation is \(\frac{33}{r}\) = \(\frac{11}{3}\)
To find the r value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
33 × 3= 11 × r
99 = 11r
Divide both sides of the equation by 11.
\(\frac{99}{11}\) = \(\frac{11r}{11}\)
9 = r

Question 8.
\(\frac{72}{9}\) = \(\frac{24}{z}\)
Answer:
Given equation is \(\frac{72}{9}\) = \(\frac{24}{z}\)
To find the z value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
72 × z = 24 × 9
72z = 216
Divide both sides of the equation by 72.
\(\frac{72z}{72}\) = \(\frac{216}{72}\)
z = 3

Question 9.
\(\frac{3}{2}\) = \(\frac{x}{5}\)
Answer:
Given equation is \(\frac{3}{2}\) = \(\frac{x}{5}\)
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
3 × 5 = x × 2
15 = 2x
Divide both sides of the equation by 2.
\(\frac{15}{2}\) = \(\frac{2x}{2}\)
7.5 = x

Question 10.
\(\frac{700}{50}\) = \(\frac{35}{w}\)
Answer:
Given equation is \(\frac{700}{50}\) = \(\frac{35}{w}\)
To find the w value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
700 × w = 35 × 50
700w = 1750
Divide both sides of the equation by 700.
\(\frac{700w}{700}\) = \(\frac{1750}{700}\)
w = 2.5

Question 11.
\(\frac{36}{4}\) = \(\frac{x}{6}\)
Answer:
Given equation is \(\frac{36}{4}\) = \(\frac{x}{6}\)
To find the x value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
36 × 6 = x × 4
216 = 4x
Divide both sides of the equation by 4.
\(\frac{216}{4}\) = \(\frac{4x}{4}\)
54 = x

Question 12.
\(\frac{84}{12}\) = \(\frac{q}{4}\)
Answer:
Given equation is \(\frac{84}{12}\) = \(\frac{q}{4}\)
To find the q value we have to perform cross multiplication. Multiply the numerator of the first fraction by the denominator of the second fraction and write product on left hand side of the equation. Next multiply the denominator of the first fraction by the numerator of the second fraction and write the product on the right side of the equation.
84 × 4 = q × 12
336 = 12q
Divide both sides of the equation by 12.
\(\frac{336}{12}\) = \(\frac{12q}{12}\)
28 = q

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 13.1 Ratios existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 13.1 Ratios

Exercises Compare

Question 1.
Jennifer is making a cake that requires 2 cups of flour and 1 cup of milk. What is the ratio of flour to milk?
Answer:
Jennifer is making a cake that requires 2 cups of flour and 1 cup of milk.
2 cups of flour/ 1 cup of milk
A cake that requires the ratio of flour to milk is 2:1 or 2/1.

Question 2.
At the movie theatre the manager wants to know which movie is selling the most tickets. He finds that movie A sold 150 tickets and movie B sold 100 tickets. What is the ratio of sales of movie A to movie B? Reduce your answer.
Answer:
At the movie theatre movie A sold 150 tickets and movie B sold 100 tickets.
Movie A/movie B = 150/100 = 3/2
The ratio of sales of movie A to movie B is 3:2 or 3/2.

Question 3.
Judy and Nancy have been swimming at the pool for one hour a day for the last month. Judy averages 25 laps every hour while Nancy averages 18. What is the ratio of Judy’s average to Nancy’s average?
Answer:
Judy averages 25 laps every hour.
Nancy averages 18 laps every hour.
25 laps/18 laps
The ratio of Judy’s average to Nancy’s average is 25:18 or 25/18.

Question 4.
Michael and Leon get ice cream cones, Michael orders chocolate chip and Leon orders yogurt chip. They each count the number of chocolate chips and yogurt chips that they eat until they finish. Michael had 32 and Leon 17. What is the ratio of yogurt chips to chocolate chips in the ice cream?
Answer:
Michael had 32 chocolate chips in the ice cream.
Leon had 17 yogurt chips in the ice cream.
The ratio of yogurt chips to chocolate chips in the ice cream is 17:32 or 17/32.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 12.5 Estimating Decimal Quotients existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 12.5 Estimating Decimal Quotients

Exercises Estimate

Use the estimating method from Example 1 to complete questions 1-8.

Question 1.
313.2 ÷ 3.322 =
Answer:
313.2 ÷ 3.322 = 94.280.

Explanation:
The quoitent of 313.2 ÷ 3.322 is 94.280

Question 2.
10.71 ÷ 2.31 =
Answer:
10.71 ÷ 2.31 = 4.636.

Explanation:
The quoitent of 10.71 ÷ 2.31 is 4.636.

Question 3.
33.3 ÷ 21.2 =
Answer:
33.3 ÷ 21.2 = 1.570.

Explanation:
The quoitent of 33.3 ÷ 21.2 is 1.570.

Question 4.
33.67 ÷ .837 =
Answer:
33.67 ÷ .837 = 40.227.

Explanation:
The quoitent of 33.67 ÷ .837 is 40.227.

Question 5.
631.23 ÷ 1.61 =
Answer:
631.23 ÷ 1.61 = 392.068.

Explanation:
The quoitent of 631.23 ÷ 1.61 is 392.068.

Question 6.
2.87 ÷ 130 =
Answer:
2.87 ÷ 130 = 0.022.

Explanation:
The quoitent of 2.87 ÷ 130 is 0.022.

Question 7.
7.3 ÷ 2.31 =
Answer:
7.3 ÷ 2.31 = 3.160.

Explanation:
The quoitent of 7.3 ÷ 2.31 is 3.160.

Question 8.
111.131 ÷ 5.23 =
Answer:
111.131 ÷ 5.23 = 21.248.

Explanation:
The quoitent of 111.131 ÷ 5.23 is 21.248.

Use the estimating method from Example 2 to complete questions 9-17.

Question 9.
13.7123 ÷ 4.38 =
Answer:
13.7123 ÷ 4.38 = 3.130.

Explanation:
The quoitent of 13.7123 ÷ 4.38 is 3.130.

Question 10.
37.23 ÷ 8.12 =
Answer:
37.23 ÷ 8.12 = 4.585.

Explanation:
The quoitent of 37.23 ÷ 8.12 is 4.585.

Question 11.
12.67 ÷ 1.86383 =
Answer:
12.67 ÷ 1.86383 = 6.797.

Explanation:
The quoitent of 12.67 ÷ 1.86383 is 6.797.

Question 12.
211.111 ÷ 8.2311 =
Answer:
211.111 ÷ 8.2311 = 25.647.

Explanation:
The quoitent of 211.111 ÷ 8.2311 is 25.647.

Question 13.
43.31 ÷ 2.30001 =
Answer:
43.31 ÷ 2.30001 = 18.830.

Explanation:
The quoitent of 43.31 ÷ 2.30001 is 18.830.

Question 14.
71.36 ÷ 3.33 =
Answer:
71.36 ÷ 3.33 = 21.429.

Explanation:
The quoitent of 71.36 ÷ 3.33 is 21.429.

Question 15.
66.2882 ÷ 10.101 =.
Answer:
66.2882 ÷ 10.101 = 6.562.

Explanation:
The quoitent of 66.2882 ÷ 10.101 is 6.562.

Question 16.
87.23 ÷ 9.1101 =
Answer:
87.23 ÷ 9.1101 = 9.575.

Explanation:
The quoitent of 87.23 ÷ 9.1101 is 9.575.

Question 17.
78.3 ÷ 4.20101 =
Answer:
78.3 ÷ 4.20101 = 18.638.

Explanation:
The quoitent of 78.3 ÷ 4.20101 is 18.638.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 12.1 Dividing Decimals by Whole Numbers existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 12.1 Dividing Decimals by Whole Numbers

Exercises Divide

Round to the nearest ten-thousandth.

Question 1.

Answer:
45.7÷5 = 9.14000.

Explanation:
The quotient of 45.7÷5 is 9.14000, so the nearest ten thousand will be 9.1400.

Question 2.

Answer:
29.34÷4 is 07.33500.

Explanation:
The quotient of 29.34÷4 is 07.33500, so the nearest ten thousand will be 07.3350.

Question 3.

Answer:
78.9÷11 is 07.17272.

Explanation:
The quotient of 78.9÷11 is 07.17272, so the nearest ten thousand will be 07.1727.

Question 4.

Answer:
41.8÷-7 is -5.97142.

Explanation:
The quotient of 41.8÷-7 is -5.97142, so the nearest ten thousand will be -5.9714.

Question 5.

Answer:
2.3÷2 is 1.15000.

Explanation:
The quotient of 2.3÷2 is 1.15000 so the nearest ten thousand will be 1.1500.

Question 6.

Answer:
456.2÷-3 = -152.06667.

Explanation:
The quotient of 456.2÷-3 is -152.06667, so the nearest ten thousand will be -152.0667.

Question 7.

Answer:
33.3÷7 is 04.75714.

Explanation:
The quotient of 33.3÷7 is 04.75714, so the nearest ten thousand will be 04.7571.

Question 8.

Answer:
89.34÷6 = 14.89000.

Explanation:
The quotient of 89.34÷6 is 14.89000, so the nearest ten thousand will be 14.8900.

Question 9.

Answer:
745.2÷14 = 053.22857.

Explanation:
The quotient of 745.2÷14 is 053.22857, so the nearest ten thousand will be 053.2286.

Question 10.

Answer:
41.9÷7 = 053.22857.

Explanation:
The quotient of 41.9÷7 is 053.22857, so the nearest ten thousand will be 053.2286.

Question 11.

Answer:
22.56÷18 =z 01.25333.

Explanation:
The quotient of 22.56÷18 is 01.25333, so the nearest ten thousand will be 01.2533.

Question 12.

Answer:
71.45÷-8 = -08.93125.

Explanation:
The quotient of 71.45÷-8 is -08.93125, so the nearest ten thousand will be -08.9313.

Question 13.

Answer:
963.5÷4 = 240.87500.

Explanation:
The quotient of 963.5÷4 is 240.87500, so the nearest ten thousand will be 240.8750.

Question 14.

Answer:
65.1÷9 = 07.23333.

Explanation:
The quotient of 65.1÷9 is 07.23333, so the nearest ten thousand will be 07.2333.

Question 15.

Answer:
964.5÷-50 = -19.29000.

Explanation:
The quotient of 964.5÷-50 is -19.29000, so the nearest ten thousand will be -019.2900.

Question 16.

Answer:
715.1÷75 = 009.53466.

Explanation:
The quotient of 715.1÷75 is 009.53466, so the nearest ten thousand will be 009.5347.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 11.3 Estimating Decimal Products existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 11.3 Estimating Decimal Products

Exercises Estimate

Question 1.
505.2 × -.322
Answer:
505.2 × -.322 = -162.6744.

Explanation:
The product of the decimals 505.2 × -.322 is -162.6744.

Question 2.
10.71 × 2.31
Answer:
10.71 × 2.31 = 24.7401.

Explanation:
The product of the decimals 10.71 × 2.31 is 24.7401.

Question 3.
55.5 × 21.2
Answer:
55.5 × 21.2 = 1176.60.

Explanation:
The product of the decimals 55.5 × 21.2 is 1176.60.

Question 4.
35.67 × .437
Answer:
35.67 × .437 = 15.58779.

Explanation:
The product of the decimals 35.67 × .437 is 15.58779.

Question 5.
631.23 × -1.61
Answer:
631.23 × -1.61 = -1016.2803.

Explanation:
The product of the decimals 631.23 × -1.61 is -1016.2803.

Question 6.
2.87 × 950
Answer:
2.87 × 950 = 2726.50.

Explanation:
The product of the decimals 2.87 × 950 is 2726.50.

Question 7.
7.3 × .51
Answer:
7.3 × .51 = 37.23.

Explanation:
The product of the decimals 7.3 × .51 is 37.23.

Question 8.
111.159 × .23
Answer:
111.159 × .23 = 25.56657.

Explanation:
The product of the decimals 111.159 × .23 is 25.56657.

Question 9.
81.453 × -1.8
Answer:
81.453 × -1.8 = -146.6154.

Explanation:
The product of the decimals 81.453 × -1.8 is -146.6154.

Question 10.
245.459 × -.37
Answer:
245.459 × -.37 = -90.8193.

Explanation:
The product of the decimals 245.459 × -.37 is -90.8193.

Question 11.
93.7123 × .54
Answer:
93.7123 × .54 = 50.604642.

Explanation:
The product of the decimals 93.7123 × .54 is 50.604642.

Question 12.
37.25 × 4.12
Answer:
37.25 × 4.12 = 153.4700.

Explanation:
The product of the decimals 37.25 × 4.12 is 153.4700.

Question 13.
12.67 × .96543
Answer:
12.67 × .96543 = 12.2319981.

Explanation:
The product of the decimals 12.67 × .96543 is 12.2319981.

Question 14.
1.111 × -4.2519
Answer:
1.111 × -4.2519 = -4.7238609.

Explanation:
The product of the decimals 1.111 × -4.2519 is -4.7238609.

Question 15.
5.51 × .50001
Answer:
5.51 × .50001 = 2.7550551.

Explanation:
The product of the decimals 5.51 × .50001 is 2.7550551.

Excel in your academics by accessing McGraw Hill Math Grade 7 Answer Key PDF Lesson 11.2 Multiplying Money existing for free of cost.

McGraw-Hill Math Grade 7 Answer Key Lesson 11.2 Multiplying Money

Exercises Multiply

Question 1.

Answer:
$5.65×3.2 = $18.080.

Explanation:
The product of $5.65×3.2 is $18.080.

Question 2.

Answer:
8.4×$96.25 = $808.500.

Explanation:
The product of 8.4×$96.25 is $808.500.

Question 3.

Answer:
34.2×$2.25 is $76.950.

Explanation:
The product of 34.2×$2.25 is $76.950.

Question 4.

Answer:
$3.01×5.6 = $16.856.

Explanation:
The product of $3.01×5.6 is $16.856.

Question 5.

Answer:
6.3×$2.45 = $15.435.

Explanation:
The product of 6.3×$2.45 is $15.435.

Question 6.

Answer:
$5.57×0.15 = $0.8355.

Explanation:
The product of $5.57×0.15 is $0.8355.

Question 7.

Answer:
64.3×$7.88 = $506.684.

Explanation:
The product of 64.3×$7.88 is $506.684.

Question 8.

Answer:
$12.33×10.35 = $127.6155.

Explanation:
The product of $12.33×10.35 is $127.6155.

Question 9.

Answer:
$13.50×85.3 = $1151.550.

Explanation:
The product of $13.50×85.3 is $1151.550.

Question 10.

Answer:
$5.01×96.85 = $485.2185.

Explanation:
The product of $5.01×96.85 is $485.2185.

Question 11.

Answer:
$45.55×3.25 = $148.0375.

Explanation:
The product of $45.55×3.25 is $148.0375.

Question 12.

Answer:
$78.10×2.1 = $164.010.

Explanation:
The product of $78.10×2.1 is $164.010.

Question 13.

Answer:
$89.99×0.011 = $0.98989.

Explanation:
The product of $89.99×0.011 is $0.98989.

Question 14.

Answer:
$56.37×2.7 = $152.199.

Explanation:
The product of $56.37×2.7 is $152.199.

Question 15.

Answer:
$85.14×6.62 = $563.6268.

Explanation:
The product of $85.14×6.62 is $563.6268.

Question 16.

Answer:
2.3×$9.63 = $22.149.

Explanation:
The product of 2.3×$9.63 is $22.149.