problem solving division lesson 2.9 answer key

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GO MATH! 2.9 Problem Solving Division

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An ostrich at the zoo weighs 3 times as much as Mark’s pet dog. Together, the ostrich and the dog weigh 448 pounds. How much does the ostrich weigh?

A student drew this diagram to solve a word problem. Which statement matches what she drew?

A man weighs 12 times as much as a cat.

A cat weighs 12 times as much as a man.

A man weighs 13 times as much as a cat.

A cat weighs 13 times as much as a man

Captain James offers a deep-sea fishing tour. He charges $2,940 for a 14-hour trip. How much does each hour of the tour cost?

An athlete ran 1,200 laps around a track over a two-month period. He ran 4 times the number of laps in the second month than he ran in the first month. How many laps did the athlete run the first month?

The diagram below compares the weights of a boy and a fish. Tell what the diagram means.

The fish weighs 11 times as much as the boy.

The boy weighs 11 times as much as the fish.

Yani drew this diagram to solve a word problem. Which statement matches what she drew?

A hamster weighs 7 times as much as a duck.

A duck weighs 7 times as much as a hamster.

A hamster weighs 6 times as much as a duck.

A duck weighs 6 times as much as a hamster.

An oak tree in Jake’s yard is 13 times taller than a model tree that he is making for a performance set. Together, the oak tree and model tree are 112 feet tall. How tall is the model tree?

Raj drives his car 5 times farther than Gary drives his car. Together they drive 624 miles. How many miles does Raj drive?

Multi-Step The total number of books in the community’s mobile library is 1,155. There are 20 times more nonfiction books than fiction books. How many fiction books and how many nonfiction books are in the mobile library?

110 fiction and 1,045 nonfiction

55 fiction and 1,100 nonfiction

55 fiction and 1,045 nonfiction

1,100 fiction and 55 nonfiction

Multi-Step Charlie’s weight is 5 times as great as the weight of his little sister, Tali. If the total weight of the two children is 132 pounds, how much does each child weigh?

Tali: 22 pounds; Charlie: 110 pounds

Tali: 110 pounds; Charlie: 22 pounds

Tali: 44 pounds; Charlie: 88 pounds

Tali: 88 pounds; Charlie: 44 pounds

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Multiply by Tens - Lesson 3.1

Estimate Products - Lesson 3.2

Area Models and Partial Products - Lesson 3.3

Multiply Using Partial Products - Lesson 3.4

Multiply with Regrouping - Lesson 3.5

Choose a Multiplication Method - Lesson 3.6  

Problem Solving - Multiply 2-Digit Numbers - Lesson 3.7

Multiplication Comparing Using Algebra - Lesson 2.1

Comparison Problems - Section 2.2

Multiply Tens, Hundreds & Thousands - Lesson 2.3

Estimate Products - Lesson 2.4

Multiply with the Distributive Property - Lesson 2.5

Multiply Using Expanded Form - Lesson 2.6

Multiply Using Partial Products - Lesson 2.7

Multiply Using Mental Math - Lesson 2.8

Problem Solving With Multistep Multiplication - Lesson 2.9

Multiply 2-Digit Numbers With Regrouping - Lesson 2.10

Multiply 3 and 4-Digit Numbers With Regrouping - Lesson 2.11

Solve Multistep Problems Using Equations - Lesson 2.12

Lines, Rays, and Angles - Lesson 10.1

Classify Triangles - Lesson 10.2

Parallel and Perpendicular Lines - Lesson 10.3

Classify Quadrilaterals - Lesson 10.4

Line symmetry - lesson 10.5, find and draw lines of symmetry - lesson 10.6, problem solving: shape patterns - lesson 10.7   , relate tenths and decimals - lesson 9.1.

Relate Hundredths and Decimals - Lesson 9.2  

Equivalent Fractions and Decimals - Lesson 9.3

Relate Fractions, Decimals, and Money - Lesson 9.4

Problem Solving With Money - Lesson 9.5

Add Fractional Parts of 10 and 100 - Lesson 9.6

Compare Decimals - Lesson 9.7

Add & Subtract Parts of a Whole - Section 7.1

Write Fractions as Sums - Section 7.2

Adding Fractions Using Models - Section 7.3

Subtract Fractions Using Models - Section 7.4

Add and Subtract Fractions - Section 7.5

Rename Fractions and Mixed Numbers - Section 7.6

Add & Subtract Mixed Numbers - Section 7.7

Subtraction with Renaming - Section 7.8

Fractions and Properties of Addition - Section 7.9

Problem Solving Using Multistep Fractions - Section 7.10

Chapter 7 Test Review on Add and Subtract Fractions

Equivalent Fractions - Section 6.1

Gener ate Equivalent Fractions - Section 6.2

Fractions in Simplest Form - Section 6.3

Common Denominators - Section 6.4

Problem Solving With Equivalent Fractions - Section 6.5  

Compare Fractions - Section 6.6

More Comparing Fractions - Section 6.7

Compare and Order Fractions - Section 6.8

Chapter 6 Review on Fraction Equivalence and Comparison

Modeling Factors of Numbers - Section 5.1

Factors and Divisibility - Section 5.2  

Problem Solving With Common Factors - Section 5.3

Factors and Multiples - Section 5.4

Prime and Composite Numbers - Section 5.5

Number Patterns - Lesson 5.6

Review For Test on Chapter 5; Factors, Multiples, and Patterns

Estimate Quotients Using Multiples - Lesson 4.1

Remainders - Lesson 4.2

Interpret the Remainder - Lesson 4.3

Divide Tens, Hundreds, and Thousands - Lesson 4.4

Estimate Quotients Using Compatible Numbers - Lesson 4.5

Division and the Distributive Property - Lesson 4.6

Divide Using Repeated Subtraction - Lesson 4.7

Divide 3 Digits by 1 Using Partial Quotients - Section 4.8

Divide 3 Digits by 1 With Regrouping - Section 4.9

Divide 3 Digits by 1 Using Place Value - Section 4.10

Divide By 1 Digit Numbers Using Place Value - Section 4.11

Multi-Step Problem Solving with Whole Numbers-Section 4.12

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Perimeter - Lesson 13.1

Area - Lesson 13.2

Area of Combined Rectangles - Lesson 13.3

Find Unknown Measures - Lesson 13.4

Problem Solving - Find the Area - Lesson 13.5

Angles and Fractional Parts of a Circle - Lesson 11.1

Degrees - Lesson 11.2

Measure and Draw Angles - Lesson 11.3

Joint and Separate Angles - Lesson 11.4

Problem Solving: Unknown Angle Measures - Lesson 11.5

Measurement Benchmarks - Lesson 12.1

Customary Units of Length - Lesson 12.2

Customary Units of Weight - Lesson 12.3

Customary Units of Liquid Volume - Lesson 12.4

Line Plots - Lesson 12.5

Show What You Know (Pre-Chapter 1)

Vocabularly Builder (Pre-Chapter 1)

Model Place Value Relationship - Lesson 1.1

Read and Write Numbers - Lesson 1.2

  Compare and Order Numbers - Lesson 1. 3  

Rounding Numbers - Lesson 1.4

Renaming Numbers - Lesson 1.5

Adding Whole Numbers - Lesson 1.6

Subtracting Whole Numbers - Lesson 1.7

Problem Solving - Lesson 1.8

Multiples of Unit Fractions - Lesson 8.1

Multiples of Fractions - Lesson 8.2

Multiply Fractions by Whole Numbers - Lesson 8.3  

Multiply Fractions and Mixed Numbers by Whole Numbers 8 4

Comparison Problem Solving with Fractions - Lesson 8.5

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9.2: Solve Equations using the Division and Multiplication Properties of Equality

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Learning Objectives

By the end of this section, you will be able to:

• Solve equations using the Division and Multiplication Properties of Equality
• Solve equations that require simplification
• Translate to an equation and solve
• Translate and solve applications

Before you get started, take this readiness quiz.

• Simplify: \(−7(\frac{1}{-7})\). If you missed this problem, review Exercise 1.6.13 .
• Evaluate \(9x+2\) when \(x=−3\). If you missed this problem, review Exercise 1.5.34 .

Solve Equations Using the Division and Multiplication Properties of Equality

You may have noticed that all of the equations we have solved so far have been of the form \(x+a=b\) or \(x−a=b\). We were able to isolate the variable by adding or subtracting the constant term on the side of the equation with the variable. Now we will see how to solve equations that have a variable multiplied by a constant and so will require division to isolate the variable.

Let’s look at our puzzle again with the envelopes and counters in Figure \(\PageIndex{1}\).

In the illustration there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

How do we determine the number? We have to separate the counters on the right side into two groups of the same size to correspond with the two envelopes on the left side. The 6 counters divided into 2 equal groups gives 3 counters in each group (since \(6\div 2=3\)).

What equation models the situation shown in Figure \(\PageIndex{2}\)? There are two envelopes, and each contains xx counters. Together, the two envelopes must contain a total of 6 counters.

We found that each envelope contains 3 counters. Does this check? We know \(2\cdot 3=6\), so it works! Three counters in each of two envelopes does equal six!

This example leads to the Division Property of Equality .

THE DIVISION PROPERTY OF EQUALITY

For any numbers a , b , and c , and \(c\neq 0\),

\[\begin{array} {llll} {\text { If }} &{a} &{=} &{b} \\ {\text {then}} & {\frac { a } { c }} &{=} &{\frac { b } { c }} \end{array}\]

When you divide both sides of an equation by any non-zero number, you still have equality.

Doing the Manipulative Mathematics activity “ Division Property of Equality ” will help you develop a better understanding of how to solve equations by using the Division Property of Equality.

The goal in solving an equation is to ‘undo’ the operation on the variable. In the next example, the variable is multiplied by 5, so we will divide both sides by 5 to ‘undo’ the multiplication.

Example \(\PageIndex{1}\)

Solve: \(5x=−27\).

Try It \(\PageIndex{2}\)

Solve: \(3y=−41\).

\(y = -\frac{41}{3}\)

Try It \(\PageIndex{3}\)

Solve: \(4z=−55\).

\(y = -\frac{55}{4}\)

Consider the equation \(\frac{x}{4} = 3\). We want to know what number divided by 4 gives 3. So to “undo” the division, we will need to multiply by 4. The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.

THE MULTIPLICATION PROPERTY OF EQUALITY

For any numbers a , b , and c ,

\[\begin{array} {llll} {\text {If}} &{a} & {=} &{b} \\ {\text {then}} &{a c} &{=} &{b c} \end{array}\]

If you multiply both sides of an equation by the same number, you still have equality.

Example \(\PageIndex{4}\)

Solve: \(\frac{y}{-7} = -14\)

Here y is divided by −7. We must multiply by −7 to isolate y.

Try It \(\PageIndex{5}\)

Solve: \(\frac{a}{-7} = -42\)

\(a = 294\)

Try It \(\PageIndex{6}\)

Solve: \(\frac{b}{-6} = -24\)

\(b = 144\)

Example \(\PageIndex{7}\)

Solve: \(-n = 9\)

Try It \(\PageIndex{8}\)

Solve: \(−k=8\).

Try It \(\PageIndex{9}\)

Solve: \(−g=3\).

Example \(\PageIndex{10}\)

Solve: \(\frac{3}{4}x = 12\)

Since the product of a number and its reciprocal is 1, our strategy will be to isolate x by multiplying by the reciprocal of \(\frac{3}{4}\).

Try It \(\PageIndex{11}\)

Solve: \(\frac{2}{5}n=14\).

Try It \(\PageIndex{12}\)

Solve: \(\frac{5}{6}y=15\).

In the next example, all the variable terms are on the right side of the equation. As always, our goal in solving the equation is to isolate the variable.

Example \(\PageIndex{13}\)

Solve: \(\frac{8}{15} = -\frac{4}{5}x\)

Try It \(\PageIndex{14}\)

Solve: \(\frac{9}{25} = -\frac{4}{5}z\)

\(z = - \frac{9}{5}\)

Try It \(\PageIndex{15}\)

\(\frac{5}{6} = -\frac{8}{3}r\)

\(r = -\frac{5}{16}\)

Solve Equations That Require Simplification

Many equations start out more complicated than the ones we have been working with.

With these more complicated equations the first step is to simplify both sides of the equation as much as possible. This usually involves combining like terms or using the distributive property.

Example \(\PageIndex{16}\)

Solve: \(14−23=12y−4y−5y\).

Begin by simplifying each side of the equation.

Try It \(\PageIndex{17}\)

Solve: \(18−27=15c−9c−3c\).

\(c=−3\)

Try It \(\PageIndex{18}\)

Solve: \(18−22=12x−x−4x\).

\(x = -\frac{4}{7}\)

Example \(\PageIndex{19}\)

Solve: \(−4(a−3)−7=25\).

Here we will simplify each side of the equation by using the distributive property first.

Try It \(\PageIndex{20}\)

Solve: \(−4(q−2)−8=24\).

\(q=−6\)

Try It \(\PageIndex{21}\)

Solve: \(−6(r−2)−12=30\).

\(r=−5\)

Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.

PROPERTIES OF EQUALITY

When you add, subtract, multiply, or divide the same quantity from both sides of an equation, you still have equality.

\[\begin{array} {ll} {\textbf {Subtraction Property of Equality}} &{\textbf{Addition Property of Equality}} \\ {\text{For any real numbers a, b, and c,}} &{\text{For any real numbers a, b, and c,}} \\ {\text{if }a = b,} &{\text{if }a = b,} \\ {\text{then }a - c = b - c} &{\text{then }a + c = b + c} \\ {\textbf {Divition Property of Equality}} &{\textbf{Multiplication Property of Equality}} \\ {\text{For any real numbers a, b, and c,}} &{\text{For any real numbers a, b, and c,}} \\ {\text{if }a = b,} &{\text{if }a = b,} \\ {\text{then }a - c = b - c} &{\text{then }a + c = b + c} \end{array}\]

Translate to an Equation and Solve

In the next few examples, we will translate sentences into equations and then solve the equations. You might want to review the translation table in the previous chapter.

Example \(\PageIndex{22}\)

Translate and solve: The number 143 is the product of −11 and y .

Try It \(\PageIndex{23}\)

Translate and solve: The number 132 is the product of −12 and y .

132=−12y;y=−11

Try It \(\PageIndex{24}\)

Translate and solve: The number 117 is the product of −13 and z .

117=−13z;z=−9

Example \(\PageIndex{25}\)

Translate and solve: n divided by 8 is −32.

Try It \(\PageIndex{26}\)

Translate and solve: nn divided by 7 is equal to −21.

\(\frac{n}{7}=−21; n=−147\)

Try It \(\PageIndex{27}\)

Translate and solve: n divided by 8 is equal to −56.

\(\frac{n}{8}=−56;n=−448\)

Example \(\PageIndex{28}\)

Translate and solve: The quotient of yy and −4 is 68.

Begin by translating the sentence into an equation.

Try It \(\PageIndex{29}\)

Translate and solve: The quotient of q and −8 is 72.

\(\frac{q}{-8}=72;q=−576\)

Try It \(\PageIndex{30}\)

Translate and solve: The quotient of p and −9 is 81.

\(\frac{p}{-9}=81;p=−729\)

Example \(\PageIndex{31}\)

Translate and solve: Three-fourths of p is 18.

Begin by translating the sentence into an equation. Remember, “of” translates into multiplication.

Try It \(\PageIndex{32}\)

Translate and solve: Two-fifths of f is 16.

\(\frac{2}{5}f=16; f=40\)

Try It \(\PageIndex{33}\)

Translate and solve: Three-fourths of f is 21.

\(\frac{3}{4}f=21; f=28\)

Example \(\PageIndex{34}\)

Translate and solve: The sum of three-eighths and x is one-half.

Try It \(\PageIndex{35}\)

Translate and solve: The sum of five-eighths and x is one-fourth.

\(\frac{5}{8} + x = \frac{1}{4}; x = -\frac{3}{8}\)

Try It \(\PageIndex{36}\)

Translate and solve: The sum of three-fourths and x is five-sixths.

\(\frac{3}{4} + x = \frac{5}{6}; x = \frac{1}{12}\)

Translate and Solve Applications

To solve applications using the Division and Multiplication Properties of Equality, we will follow the same steps we used in the last section. We will restate the problem in just one sentence, assign a variable, and then translate the sentence into an equation to solve.

Example \(\PageIndex{37}\)

Denae bought 6 pounds of grapes for $10.74. What was the cost of one pound of grapes?

\[\begin{array} {ll} {\text{What are you asked to find?}} &{\text{The cost of 1 pound of grapes}} \\\\ {\text{Assign a variable.}} &{\text{Let c = the cost of one pound.}} \\\\ {\text{Write a sentence that gives the}} &{\text{The cost of 6 pounds is }$10.74} \\ {\text{information to find it.}} &{} \\\\ {\text{Translate into an equation.}} &{6c = 10.74} \\ {\text{Solve.}} &{\frac{6c}{c} = \frac{10.74}{6}} \\ {} &{c = 1.79} \\\\ {} &{\text{The grapes cost }$ 1.79 \text{ per pound.}} \\ \\ {\text{Check: If one pound costs }$1.79, do} &{} \\ {\text{6 pounds cost }$ 10.74?} &{} \\\\ {6(1.79) \stackrel{?}{=} 10.74} &{} \\ {10.74 = 10.74\checkmark} &{} \end{array}\]

Try It \(\PageIndex{38}\)

Translate and solve:

Arianna bought a 24-pack of water bottles for $9.36. What was the cost of one water bottle?

Try It \(\PageIndex{39}\)

At JB’s Bowling Alley, 6 people can play on one lane for $34.98. What is the cost for each person?

Example \(\PageIndex{40}\)

Andreas bought a used car for $12,000. Because the car was 4-years old, its price was \(\frac{3}{4}\) of the original price, when the car was new. What was the original price of the car?

\[\begin{array} {ll} {\text{What are you asked to find?}} &{\text{The original price of the car}} \\\\ {\text{Assign a variable.}} &{\text{Let p = the original price.}} \\\\ {\text{Write a sentence that gives the}} &{$12000\text{ is }\frac{3}{4} \text{ of the original price.}} \\ {\text{information to find it.}} &{} \\\\ {\text{Translate into an equation.}} &{12000 = \frac{3}{4}p} \\ {} &{\frac{3}{4}(12000) = \frac{4}{3}\cdot \frac{3}{4}p}\\ {}&{16000 = p} \\{\text{Solve.}} &{} \\\\ {} &{\text{The original cost of the car was }$ 16000.} \\ \\ {\text{Check: Is }\frac{3}{4} \text{ of }$16000 \text{ equal to }$12000} &{} \\\\ {\frac{3}{4}\cdot 16000 \stackrel{?}{=} 12000} &{} \\ {12000 = 12000\checkmark} &{} \end{array}\]

Try It \(\PageIndex{41}\)

The annual property tax on the Mehta’s house is $1,800, calculated as \(\frac{15}{1000}\) of the assessed value of the house. What is the assessed value of the Mehta’s house?

Try It \(\PageIndex{42}\)

Stella planted 14 flats of flowers in \(\frac{2}{3}\) of her garden. How many flats of flowers would she need to fill the whole garden?

Key Concepts

• The Division Property of Equality —For any numbers a , b , and c , and \(c\neq 0\), if \(a=b\), then \(\frac{a}{c} = \frac{b}{c}\). When you divide both sides of an equation by any non-zero number, you still have equality.
• The Multiplication Property of Equality —For any numbers a , b , and c , if \(a=b\), then \(ac = bc\). If you multiply both sides of an equation by the same number, you still have equality.

Be Prepared

180 , 096 180 , 096

1 , 024 1 , 024

x − 8 x − 8

4 4 and 215 215

2 , 3 , 5 , 6 , 10 2 , 3 , 5 , 6 , 10

• ⓐ 18 plus 11; the sum of eighteen and eleven
• ⓑ 27 times 9; the product of twenty-seven and nine
• ⓒ 84 divided by 7; the quotient of eighty-four and seven
• ⓓ p minus q ; the difference of p and q
• ⓐ 47 minus 19; the difference of forty-seven and nineteen
• ⓑ 72 divided by 9; the quotient of seventy-two and nine
• ⓒ m plus n ; the sum of m and n
• ⓓ 13 times 7; the product of thirteen and seven
• ⓐ fourteen is less than or equal to twenty-seven
• ⓑ nineteen minus two is not equal to eight
• ⓒ twelve is greater than four divided by two
• ⓓ x minus seven is less than one
• ⓐ nineteen is greater than or equal to fifteen
• ⓑ seven is equal to twelve minus five
• ⓒ fifteen divided by three is less than eight
• ⓓ y minus three is greater than six
• ⓑ expression
• ⓐ expression
• ⓐ 4 · 4 · 4 · 4 · 4 · 4 · 4 · 4
• ⓑ a · a · a · a · a · a · a
• ⓐ 8 · 8 · 8 · 8 · 8 · 8 · 8 · 8
• ⓑ b · b · b · b · b · b

The terms are 4 x , 3 b , and 2. The coefficients are 4, 3, and 2.

The terms are 9 a , 13 a 2 , and a 3 , The coefficients are 9, 13, and 1.

9 and 15; 2 x 3 and 8 x 3 ; y 2 and 11 y 2

4 x 3 and 6 x 3 ; 8 x 2 and 3 x 2 ; 19 and 24

4 x 2 + 14 x

12 y 2 + 15 y

• ⓑ 11 a − 14
• ⓐ 4( p + q )
• ⓑ 2( x − 8)

x + 1 = 7; x = 6

x + 3 = 4; x = 1

21 ⋅ 3 = 63

2( x − 5) = 30

2( y − 4) = 16

x + 7 = 37; x = 30

y + 11 = 28; y = 17

z − 17 = 37; z = 54

x − 19 = 45; x = 64

Divisible by 2, 3, 5, and 10

Divisible by 2 and 3, not 5 or 10.

Divisible by 2, 3, not 5 or 10.

Divisible by 3 and 5.

1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

1, 2, 4, 5, 8, 10, 16, 20, 40, 80

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5, or 2 4 ⋅ 5

2 ⋅ 2 ⋅ 3 ⋅ 5, or 2 2 ⋅ 3 ⋅ 5

2 ⋅ 3 ⋅ 3 ⋅ 7, or 2 ⋅ 3 2 ⋅ 7

2 ⋅ 3 ⋅ 7 ⋅ 7, or 2 ⋅ 3 ⋅ 7 2

Section 2.1 Exercises

16 minus 9, the difference of sixteen and nine

5 times 6, the product of five and six

28 divided by 4, the quotient of twenty-eight and four

x plus 8, the sum of x and eight

2 times 7, the product of two and seven

fourteen is less than twenty-one

thirty-six is greater than or equal to nineteen

3 times n equals 24, the product of three and n equals twenty-four

y minus 1 is greater than 6, the difference of y and one is greater than six

2 is less than or equal to 18 divided by 6; 2 is less than or equal to the quotient of eighteen and six

a is not equal to 7 times 4, a is not equal to the product of seven and four

5 · 5 · 5 5 · 5 · 5

2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2

Section 2.2 Exercises

15 x 2 , 6 x , 2

10 y 3 , y , 2

x 3 and 8 x 3 ; 14 and 5

16 ab and 4 ab ; 16 b 2 and 9 b 2

17 x 2 + 20 x + 16

5 ( x + y )

He will pay $750. His insurance company will pay $1350.

Section 2.3 Exercises

x + 2 = 5; x = 3

x + 3 = 6; x = 3

23 − 19 = 4

2( n − 10) = 52

3 y + 10 = 100

p + 5 = 21; p = 16

r + 18 = 73; r = 55

d − 30 = 52; d = 82

u − 12 = 89; u = 101

c − 325 = 799; c = 1124

Section 2.4 Exercises

2, 4, 6, 8, 10 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48

6, 12, 18, 24, 30, 36, 42, 48

8, 16, 24, 32, 40, 48

10, 20, 30, 40

Divisible by 2, 3, 4, 6

Divisible by 3, 5

Divisible by 2, 3, 4, 5, 6, 10

Divisible by 2, 4

Divisible by 2, 5, 10

1, 2, 3, 4, 6, 9, 12, 18, 36

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72,144

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 49, 84, 98, 147, 196, 294, 588

Section 2.5 Exercises

2 ⋅ 2 ⋅ 3 ⋅ 11

3 ⋅ 3 ⋅ 7 ⋅ 11

3 ⋅ 3 ⋅ 5 ⋅ 5 ⋅ 11

2 ⋅ 2 ⋅ 2 ⋅ 7

2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3

2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 5

2 ⋅ 3 ⋅ 5 ⋅ 5

3 ⋅ 5 ⋅ 5 ⋅ 7

2 ⋅ 2 ⋅ 3 ⋅ 3

2 ⋅ 5 ⋅ 5 ⋅ 7

Review Exercises

3 times 8, the product of three and eight.

24 divided by 6, the quotient of twenty-four and six.

50 is greater than or equal to 47

The sum of n and 4 is equal to 13

8 ⋅ 8 ⋅ 8 ⋅ 8

y ⋅ y ⋅ y ⋅ y ⋅ y

12 n 2 ,3 n , 1

3 and 4; 3 x and x

10 y 2 + 2 y + 3

x + 3 = 5; x = 2

7 + 33 = 40

2( n − 3) = 76

x + 8 = 35; x = 27

q − 18 = 57; q = 75

3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48

2, 3, 5, 6, 10

1, 2, 3, 5, 6, 10, 15, 30

1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180

2 ⋅ 2 ⋅ 3 ⋅ 7

Answers will vary

Practice Test

15 minus x , the difference of fifteen and x .

• ⓑ 3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3 = 243

y − 15 = 32; y = 47

2 3 ⋅ 3 3 ⋅ 5

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• Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
• Publisher/website: OpenStax
• Book title: Prealgebra 2e
• Publication date: Mar 11, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/prealgebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/prealgebra-2e/pages/chapter-2

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Eureka Math Grade 2 Module 2 Lesson 9 Answer Key

Engage ny eureka math 2nd grade module 2 lesson 9 answer key, eureka math grade 2 module 2 lesson 9 problem set answer key.

Explanation: Completed the chart by first estimating the measurement around a classmate’s body part and then finding the actual measurement with a meter strip as shown above.

a. Which was longer, your estimate or the actual measurement around your classmate’s head? _Actual measurement___ Answer: The actual measurement around my classmate’s head is longer than my estimate,

Explanation: As seen the chart the actual measurement around my classmate’s head is longer than my estimate because actual measurement around my classmate’s head is 54 cm and my estimate is 45 cm only, So by 9 cm the actual measurement around my classmate’s head is longer than my estimate.

Explanation: Upon Using a string to measure all three paths Path 1 is 15 cm, Path 2 is 14 cm and Path 3 is 13 cm respectively.

a. Which path is the longest? __Path 1_____ Answer: Path 1 is the longest,

Explanation: As Path 1 is 15 cm, Path 2 is 14 cm and Path 3 is 13 cm respectively, So Path 1 is the longest.

b. Which path in the shortest? _Path 3_____ Answer: Path 3 is the shortest,

Explanation: As Path 1 is 15 cm, Path 2 is 14 cm and Path 3 is 13 cm respectively, So Path 3 is the shortest.

Explanation: Estimated the length of the path as 18 centimeters long.

b. The actual length of the path is __22___ cm. Answer: The actual length of the path is 22 cm,

Explanation: Used my piece of string to measure the length of the path. Then, measured the string with my meter strip, So the actual length of the path is 22 cm.

Eureka Math Grade 2 Module 2 Lesson 9 Exit Ticket Answer Key

Explanation: Using my string to measure the two paths, The length in centimeters is Path M is 10 cm long and Path N is 6 cm long.

Question 2. Mandy measured the paths and said both paths are the same length. Is Mandy correct? Yes or No? ___No____ Explain why or why not. _____________________________________________________________ _____________________________________________________________ Answer: No, Mandy is incorrect,

Explanation: Actually Path M is 10 cm long and Path N is 6 cm long, but Mandy measured the paths and said both paths are the same length, So Mandy is incorrect because both path’s are not the same therefore Mandy is incorrect.

Eureka Math Grade 2 Module 2 Lesson 9 Homework Answer Key

Explanation: Drawn a tape diagram comparing the measurements of the basketball and the bottom of the glue bottle as basketball is 41 – 8 = 33 cm more than bottom of a glue bottle.

Explanation: Measured the path, Path A is 6 cm long.

b. Path B is ____7_____ cm long. Answer: Path B is 7 cm long,

Explanation: Measured the path, Path B is 7 cm long.

c. Together, Paths A and B measure ___13___ cm. Answer: Together, Paths A and B measure 13 cm,

Explanation: As measured Path A is 6 cm and Path B is 7 cm, So together Path A and Path B measure 6 + 7 = 13 cm.

d. Path A is ____1___ cm (shorter/longer) than Path B. Answer: Path A is 1 cm shorter than Path B,

Explanation: As Path A is 6 cm and Path B is 7 cm, So Path A is 7 – 6 = 1 cm shorter than Path B.

Question 3. Shawn and Steven had a contest to see who could jump farther. Shawn jumped 75 centimeters. Steven jumped 9 more centimeters than Shawn. a. How far did Steven jump?” ____84_____ centimeters Answer: Steven jump 84 centimeters,

Explanation: Given Shawn and Steven had a contest to see who could jump farther. Shawn jumped 75 centimeters. Steven jumped 9 more centimeters than Shawn means Steven jumped 75 + 9 = 84 centimeters.

b. Who won the jumping contest? ___Steven_____ Answer: Steven won the jumping contest,

Explanation: As Shawn jumped 75 centimeters and Steven jumped 84 centimeters so on comparing Steven jumped more by 84 – 75 = 9 centimeters, therefore Steven won the jumping contest.

Explanation: Drawn a tape diagram to compare the lengths that Shawn and Steven jump, Steven jump more by 84 – 75 = 9 centimeters as shown above.

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