Course 2 1-9 Simplifying Algebraic Expressions 1-9 Simplifying Algebraic Expressions Course 2 Warm...

Course 2

1-9 Simplifying Algebraic Expressions1-9 Simplifying Algebraic Expressions

Course 2

Warm UpWarm UpProblem of the DayProblem of the DayLesson PresentationLesson Presentation

Course 2

1-9 Simplifying Algebraic Expressions

Warm UpEvaluate each expression for y = 3.

1. 3y + y2. 7y3. 10y – 4y4. 9y5. y + 5y + 6y6. 10y

122118273630

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Problem of the DayEmilia saved nickels, dimes, and quarters in a jar. She had as many quarters as dimes, but twice as many nickels as dimes. If the jar had 844 coins, how much money had she saved?$94.95

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Learn to simplify algebraic expressions.

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Vocabularytermcoefficient

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In the expression 7x + 9y + 15, 7x, 9y, and 15 are called terms. A term can be a number, a variable, or a product of numbers and variables. Terms in an expression are separated by + and –.

7x + 5 – 3y2 + y + x3

term term term termIn the term 7x, 7 is called the coefficient. A coefficient is a number that is multiplied by a variable in an algebraic expression. A variable by itself, like y, has a coefficient of 1. So y = 1y.

Coefficient Variableterm

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Like terms are terms with the same variable raised to the same power. The coefficients do not have to be the same. Constants, like 5, , and 3.2, are also like terms.

12

Like TermsUnlike Terms

3x and 2x

5x2 and 2xThe exponentsare different.

3.2 and nOnly one term

contains avariable

6a and 6bThe variablesare different

w and w7 5 and 1.8

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Identify like terms in the list.

Additional Example 1: Identifying Like Terms

3t 5w2 7t 9v 4w2 8v

Look for like variables with like powers.3t 5w2 7t 9v 4w2 8v

Like terms: 3t and 7t 5w2 and 4w2 9v and 8v

Use different shapes or colors to indicate sets of like terms.

Helpful Hint

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1-9 Simplifying Algebraic ExpressionsCheck It Out: Example 1


2x 4y3 8x 5z 5y3 8z

Look for like variables with like powers.

Like terms: 2x and 8x 4y3 and 5y3 5z and 8z

2x 4y3 8x 5z 5y3 8z

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x

Combining like terms is like grouping similar objects.

+ =x

x

x

x x

x x x

x x x x

x x x x x

4x + 5x = 9x

To combine like terms that have variables, add orsubtract the coefficients.

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Simplify. Justify your steps using the Commutative, Associative, and Distributive Properties when necessary.

Additional Example 2: Simplifying Algebraic Expressions

A. 6t – 4t6t – 4t

2t

6t and 4t are like terms.

Subtract the coefficients.

B. 45x – 37y + 87

In this expression, there are no like termsto combine.

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Additional Example 2: Simplifying Algebraic Expressions

C. 3a2 + 5b + 11b2 – 4b + 2a2 – 6

3a2 + 5b + 11b2 – 4b + 2a2 – 6

5a2 + b + 11b2 – 6

Identify liketerms.

Add or subtractthe coefficients.

(3a2 + 2a2) + (5b – 4b) + 11b2 – 6 Group liketerms.


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A. 5y + 3y5y + 3y

8y

5y and 3y are like terms.

Add the coefficients.

B. 2(x2 – 13x) + 6

There are no like terms to combine.

2x – 26x + 62 Distributive Property.

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C. 4x2 + 4y + 3x2 – 4y + 2x2 + 5

9x2 + 5

Identify liketerms.

Add or subtractthe coefficients.

4x2 + 4y + 3x2 – 4y + 2x2 + 5

Group liketerms.

(4x2 + 3x2 + 2x2)+ (4y – 4y) + 5


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Write an expression for the perimeter of the triangle. Then simplify the expression.

Additional Example 3: Geometry Application

x

2x + 3 3x + 2

2x + 3 + 3x + 2 + x

(x + 3x + 2x) + (2 + 3)

6x + 5

Write an expression usingthe side lengths.Identify and group like terms.Add the coefficients.

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x

2x + 12x + 1

x + 2x + 1 + 2x + 1

5x + 2

Write an expression usingthe side lengths.Identify and group like terms.Add the coefficients.

Write an expression for the perimeter of the triangle. Then simplify the expression.

(x + 2x + 2x) + (1 + 1)

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1-9 Simplifying Algebraic ExpressionsLesson Quiz: Part I


1. 3n2 5n 2n3 8n

2. a5 2a2 a3 3a 4a2


3. 4a + 3b + 2a

4. x2 + 2y + 8x2

2a2, 4a2

5n, 8n

6a + 3b

9x2 + 2y

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1-9 Simplifying Algebraic ExpressionsLesson Quiz: Part II

5. Write an expression for the perimeter of the given figure.

6x + 8y

2x + 3y

2x + 3y

x + yx + y

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1-9 Simplifying Algebraic Expressions1-10 Equations and Their Solutions

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Warm UpWarm UpProblem of the DayProblem of the DayLesson PresentationLesson Presentation

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Warm UpEvaluate each expression for x = 12.

1. x + 22. 3. x – 84. 10x – 45. 2x + 126. 5x + 7

1434

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1-10 Equations and Their Solutions

x4

1163667

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Problem of the DayAlicia buys buttons at a cost of 8 for $20. She resells them for $5 each. How many buttons does Alicia need to sell for a profit of $120?48 buttons

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Learn to determine whether a number is a solution of an equation.

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Vocabularyequationsolution

Insert Lesson Title Here

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Ella has 22 CDs. This is 9 more than her friend Kay has.

This situation can be written as an equation.An equation is a mathematical statement that two expressions are equal in value.An equation is like a balanced scale.

Right expressionLeft expression

Number of CDs Ella has

22

is equalto=

9 more thanKay has

j + 9

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1-9 Simplifying Algebraic ExpressionsJust as the weights on both sides of a balanced scale are exactly the same, the expressions on both sides of an equation represent exactly the same value.When an equation contains a variable, a value of the variable that makes the statement true is called a solution of the equation.

22 = j + 9 j = 13 is a solution because 22 = 13 + 9.22 = j + 9 j = 15 is not a solution because 22 15 + 9.

The symbol ≠ means “is not equal to.” Reading Math

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Course 2


Determine whether the given value of the variable is a solution of t + 9 = 17.

Additional Example 1A: Determining Whether a Number is a Solution of an Equation

26

26 + 9 = 17?

35 = 17?

26 is not a solution of t + 9 = 17.

Substitute 26 for t.

t + 9 = 17

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1-9 Simplifying Algebraic ExpressionsAdditional Example 1B: Determining Whether a

Number is a Solution of an EquationDetermine whether the given value of the variable is a solution of t + 9 = 17.

8

8 + 9 = 17?

17 = 17?

8 is a solution of t + 9 = 17.

Substitute 8 for t.

t + 9 = 17

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Determine whether each number is a solution of x – 5 = 12.

A. 22

22 – 5 = 12?

17 = 12?

22 is not a solution of x – 5 = 12.

Substitute 22 for x.

B. 8

8 – 5 = 12?

3 = 12?

8 is not a solution of x – 5 = 12.

Substitute 8 for x.x – 5 = 12

x – 5 = 12

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Mrs. Jenkins had $32 when she returned home from the supermarket. If she spent $17 at the supermarket, did she have $52 or $49 before she went shopping?

Additional Example 2: Writing an Equation to Determine Whether a Number is a Solution

$52m – 17 = 3252 - 17 = 32?

35 = 32? Substitute 52 for m.

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You can write an equation to find the amount of money Mrs. Jenkins had before she went shopping. If m represents the amount of money she had before she went shopping, then m - 17 = 32.

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1-9 Simplifying Algebraic ExpressionsAdditional Example 2 Continued

$49m – 17 = 3249 - 17 = 32?

32 = 32?Substitute 49 for m.

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You can write an equation to find the amount of money Mrs. Jenkins had before she went shopping. If m represents the amount of money she had before she went shopping, then m - 17 = 32.

Mrs. Jenkins had $49 before she went shopping.

Mrs. Jenkins had $32 when she returned home from the supermarket. If she spent $17 at the supermarket, did she have $52 or $49 before she went shopping?

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Mr. Rorke had $12 when he returned home from buying a hat. If he spent $47 at the hat store, did he have $61 or $59 before he bought the hat?

Check it Out: Additional Example 2

$61m – 47 = 1261 - 47 = 12?

14 = 12?Substitute 61 for h.

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You can write an equation to find the amount of money Mr. Rorke had before he purchased a hat. If m represents the amount of money he had before he purchased a hat, then m – 47 = 12.

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Mr. Rorke had $12 when he returned home from buying a hat. If he spent $47 at the hat store, did he have $59 or $61 before he bought the hat?

Check it Out: Additional Example 2 Continued

$59m – 47 = 1259 - 47 = 12?

12 = 12?Substitute 59 for h.

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You can write an equation to find the amount of money Mr. Rorke had before he purchased a hat. If m represents the amount of money he had before he purchased a hat, then m – 47 = 12.

Mr. Rorke had $59 before he purchased a hat.

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Which problem situation best matches the equation 5 + 2x = 13?

Additional Example 3: Deriving a Real-World Situation from an Equation

Situation A:Admission to the county fair costs $5 and rides cost $2 each. Mike spent a total of $13. How many rides did he go on?

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$2 per ride 2x

Mike spent $13 in all, so 5 + 2x = 13. Situation A matches the equation.

$5 for admission 5 +

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Additional Example 3 Continued

Situation B:

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Admission to the county fair costs $2 and rides cost $5 each. Mike spent a total of $13. How many rides did he go on?

$5 per ride 5xSince 5x is not a term in the given equation, Situation B does not match the equation.

The variable x represents the number of rides that Mike bought.

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Check It Out: Additional Example 3

Situation A:

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Admission to the baseball game costs $4 and souvenir hats cost $13 each. Trina spent a total of $25. How many souvenir hats did she buy?

$13 per souvenir hat 13xSince 13x is not a term in the given

equation, Situation A does not match the equation.

The variable x represents the number of souvenir hats Trina bought.

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Check It Out: Additional Example 3 Continued

Situation B:

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$4 per souvenir hat 4x

Trina spent $25 in all, so 13 + 4x = 25. Situation B matches the equation.

$13 for admission 13 +

Admission to the baseball game costs $13 and souvenir hats cost $4 each. Trina spent a total of $25. How many souvenir hats did she buy?

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1-9 Simplifying Algebraic ExpressionsLesson Quiz

Determine whether the given value of the variable is a solution of 5 + x = 47.

1. x = 42 2. x = 52

Determine whether the given value of the variable is a solution of 57 – y = 18.

3. y = 75 4. y = 39

5. Kwan has 14 marbles. This is 7 more than Drue has. Does Drue have 21 or 7 marbles?

noyes


no yes

7

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Course 2

1-9 Simplifying Algebraic Expressions2-6 Rates, Ratios, and Proportions

Holt Algebra 1

Lesson QuizLesson QuizLesson PresentationLesson PresentationWarm UpWarm Up

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1-9 Simplifying Algebraic ExpressionsWarm UpSolve each equation. Check your answer.1. 6x = 362. 3. 5m = 184. 5. 8y =18.4Multiply.6. 7.

648

3.6–63

2.3

7 10

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Write and use ratios, rates, and unit rates.Write and solve proportions.

Objectives

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ratio proportionrate cross productsscale scale drawingunit rate scale modelconversion factor

Vocabulary

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A ratio is a comparison of two quantities by division. The ratio of a to b can be written a:b or , where b ≠ 0. Ratios that name the same comparison are said to be equivalent.

A statement that two ratios are equivalent, such as , is called a proportion.

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1-9 Simplifying Algebraic ExpressionsExample 2: Finding Unit Rates

Raulf Laue of Germany flipped a pancake 416 times in 120 seconds to set the world record. Find the unit rate. Round your answer to the nearest hundredth.

The unit rate is about 3.47 flips/s.

Write a proportion to find an equivalent ratio with a second quantity of 1.

Divide on the left side to find x.

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1-9 Simplifying Algebraic ExpressionsCheck It Out! Example 2

Cory earns $52.50 in 7 hours. Find the unit rate.

The unit rate is $7.50.

Write a proportion to find an equivalent ratio with a second quantity of 1.

Divide on the left side to find x.

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A rate such as in which the two quantities are equal but use different units, is called a conversion factor. To convert a rate from one set of units to another, multiply by a conversion factor.

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1-9 Simplifying Algebraic ExpressionsExample 3A: Converting Rates

Serena ran a race at a rate of 10 kilometers per hour. What was her speed in kilometers per minute? Round your answer to the nearest hundredth.

The rate is about 0.17 kilometer per minute.

To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.

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Helpful HintIn example 3A , “1 hr” appears to divide out, leaving “kilometers per minute,” which are the units asked for. Use this strategy of “dividing out” units when converting rates.

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1-9 Simplifying Algebraic ExpressionsExample 3B: Converting Rates

A cheetah can run at a rate of 60 miles per hour in short bursts. What is this speed in feet per minute?Step 1 Convert the speed to feet per hour.

The speed is 316,800 feet per hour.

To convert the first quantity in a rate, multiply by a conversion factor with that unit in the second quantity.

Step 2 Convert the speed to feet per minute.

The speed is 5280 feet per minute.

To convert the first quantity in a rate, multiply by a conversion factor with that unit in the first quantity.

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1-9 Simplifying Algebraic ExpressionsExample 3B: Converting Rates

The speed is 5280 feet per minute.

Check that the answer is reasonable.• There are 60 min in 1 h, so 5280 ft/min is

60(5280) = 316,800 ft/h.• There are 5280 ft in 1 mi, so 316,800 ft/h

is This is the given rate in the problem.

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To convert the first quantity in a rate, multiply by a conversion factor with that unit in the second quantity.

Step 1 Convert the speed to feet per hour.

Check It Out! Example 3A cyclist travels 56 miles in 4 hours. What is the cyclist’s speed in feet per second? Round your answer to the nearest tenth, and show that your answer is reasonable.

Change to miles in 1 hour.

The speed is 73,920 feet per hour.

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Step 2 Convert the speed to feet per minute.To convert the second quantity in a

rate, multiply by a conversion factor with that unit in the first quantity.

The speed is 1232 feet per minute.Step 3 Convert the speed to feet per second.

The speed is approximately 20.5 feet per second.

To convert the second quantity in a rate, multiply by a conversion factor with that unit in the first quantity.

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Check that the answer is reasonable. The answer is about 20 feet per second. • There are 60 seconds in a minute so 60(20) = 1200 feet in a minute.• There are 60 minutes in an hour so 60(1200) = 72,000 feet in an hour. • Since there are 5,280 feet in a mile 72,000 ÷ 5,280 = about 14 miles in an hour.• The cyclist rode for 4 hours so 4(14) = about 56 miles which is the original distance traveled.

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In the proportion , the products a • d and b • c are called cross products. You can solve a proportion for a missing value by using the Cross Products property.

Cross Products Property

WORDS NUMBERS ALGEBRA

In a proportion, cross products are equal.

2 • 6 = 3 • 4

If and b ≠ 0

and d ≠ 0then ad = bc.

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1-9 Simplifying Algebraic ExpressionsExample 4: Solving Proportions

Solve each proportion.

3(m) = 5(9)3m = 45

m = 15

Use cross products.

Divide both sides by 3.

Use cross products.

6(7) = 2(y – 3)42 = 2y – 6+6 +648 = 2y

24 = y

A. B.

Add 6 to both sides.Divide both sides by 2.

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y = −20

–12 –124g = 23

g = 5.75

A. B.

Use cross products.


Use cross products.

Subtract 12 from both sides.


2y = –402(y) = –5(8) 4(g +3) = 5(7)

4g +12 = 35

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A scale is a ratio between two sets of measurements, such as 1 in:5 mi. A scale drawing or scale model uses a scale to represent an object as smaller or larger than the actual object. A map is an example of a scale drawing.

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1-9 Simplifying Algebraic ExpressionsExample 5A: Scale Drawings and Scale Models

A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft.A wall on the blueprint is 6.5 inches long. How long is the actual wall?blueprint 1 in. actual 3 ft.

x • 1 = 3(6.5)x = 19.5The actual length of the wall is 19.5 feet.

Write the scale as a fraction.

Let x be the actual length.

Use the cross products to solve.

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1-9 Simplifying Algebraic ExpressionsExample 5B: Scale Drawings and Scale Models

A contractor has a blueprint for a house drawn to the scale 1 in: 3 ft.One wall of the house will be 12 feet long when it is built. How long is the wall on the blueprint?blueprint 1 in. actual 3 ft.

The wall on the blueprint is 4 inches long.


Let x be the actual length.

Use the cross products to solve. 12 = 3x

4 = x

Since x is multiplied by 3, divide both sides by 3 to undo the multiplication.

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A scale model of a human heart is 16 ft. long. The scale is 32:1. How many inches long is the actual heart it represents?model 32 in. actual 1 in.

The actual heart is 6 inches long.


Use the cross products to solve. 32x = 192Since x is multiplied by 32, divide

both sides by 32 to undo the multiplication.

Let x be the actual length.Convert 16 ft to inches.

x = 6

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1-9 Simplifying Algebraic ExpressionsLesson Quiz: Part 1

1. In a school, the ratio of boys to girls is 4:3. There are 216 boys. How many girls are there?162

Find each unit rate. Round to the nearest hundredth if necessary.2. Nuts cost $10.75 for 3 pounds. $3.58/lb

3. Sue washes 25 cars in 5 hours. 5 cars/h4. A car travels 180 miles in 4 hours. What is the

car’s speed in feet per minute? 3960 ft/min

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1-9 Simplifying Algebraic ExpressionsLesson Quiz: Part 2


5.

6.

7. A scale model of a car is 9 inches long. The scale is 1:18. How many inches long is the car it represents?

6

16

162 in.

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1-9 Simplifying Algebraic ExpressionsSMUE DIRECTIONS

For Smue question 18,19,20,and21,• 18,write the response and tell how you got your

answer• 19,write the answer in expand if you has a answer

in numberic form• 20,21 compare and contrast the number

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Question 18Question 18

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