lesson study guide 1 - mr. chan's classroom · 23 algebra 1 chapter 1 resource book evaluate...

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10Algebra 1Chapter 1 Resource Book

Evaluate algebraic expressions and use exponents.

VocabularyA variable is a letter used to represent one or more numbers.

An algebraic expression, or variable expression, consists of numbers, variables, and operations.

To evaluate an expression, substitute a number for the variable, perform the operation(s), and simplify the result if necessary.

A power is an expression that represents repeated multiplication of the same factor.

A power can be written in a form using two numbers, a base and an exponent. The exponent represents the number of times the base is used as a factor.

GOAL

Study GuideFor use with pages 227

LESSON

1.1

Evaluate algebraic expressions

Evaluate the expression when x 5 5.

a. 7x

b. 12 1 x

Solution

a. 7x 5 7(5) Substitute 5 for x.

5 35 Multiply.

b. 12 1 x 5 12 1 5 Substitute 5 for x.

5 17 Add.

Exercises for Example 1

Evaluate the expression for the given value of the variable.

1. 15 2 a when a 5 3 2. 3b when b 5 7

3. 11 1 c when c 5 10 4. 28}d

when d 5 4

5. 1}2

n when n 5 18 6. 0.4f when f 5 8

EXAMPLE 1

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11Algebra 1

Chapter 1 Resource Book

Evaluate an expression The cost of filling a car’s gas tank can be represented by the expression xy where x is the price per gallon of gasoline and y is the number of gallons purchased. You purchase 10 gallons of gasoline when the price per gallon is $2.35. Find the total cost.

Solution

Total Cost 5 xy Write expression.

5 2.35(10) Substitute 2.35 for x and 10 for y.

5 23.50 Multiply.

The total cost is $23.50.

Exercises for Example 2 7. You purchase 5 gallons of gasoline when the price of gasoline is $2.26 per

gallon. Find the total cost.

8. You purchase 8 gallons of gasoline when the price of gasoline is $2.20 per gallon. Find the total cost.

EXAMPLE 2

Read and write powers

Write the power in words and as a product.

a. 83

b. m6

Solution

a. eight to the third power, or eight cubed; 8 p 8 p 8b. m to the sixth power; m p m p m p m p m p m


Write the power in words and as a product.

9. 48 10. 11}3 24 11. x2

EXAMPLE 3

Study Guide continuedFor use with pages 227

LESSON

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Use the order of operations to evaluate expressions.

VocabularyThe order of operations was established to evaluate an expression involving more than one operation.

Order of Operations

STEP 1 Evaluate expressions inside grouping symbols.

STEP 2 Evaluate powers.

STEP 3 Multiply and divide from left to right.

STEP 4 Add and subtract from left to right.

GOAL


LESSON

1.2

Evaluate expressions

Evaluate the expression 42 p 5 2 62.

Solution

42 p 5 2 62 5 16 p 5 2 36 Evaluate powers.

5 80 2 36 Multiply.

5 44 Subtract.


Evaluate the expression.

1. 20 2 32 1 7 2. 5 p 23 4 6 3. 4 p 6 2 21 4 3

EXAMPLE 1

Evaluate expressions with grouping symbols


a. 47 2 2(9 1 12) b. 6[23 1 (13 2 8)]

Solution

a. 47 2 2(9 1 12) 5 47 2 2(21) Add within parentheses.

5 47 2 42 Multiply.

5 5 Subtract.

b. 6[23 1 (13 2 8)] 5 6[8 1 (13 2 8)] Evaluate power.

5 6[8 1 5] Subtract within the parentheses.

5 6[13] Add within the parentheses.

5 78 Multiply.

EXAMPLE 2

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23Algebra 1


Evaluate an algebraic expression

Evaluate the expression4y 1 8}2 1 y

when y 5 3.

Solution4y 1 8}2 1 y

5 4(3) 1 8}

2 1 3Substitute 3 for y.

5 12 1 8}2 1 3

Multiply.

5 20}5 Add.

5 4 Divide.


Evaluate the expression when w 5 9.

10. 17 2 3w 11. w2 2 13

12. 5w

}w 1 6

13. 7(13 2 w)

14. 2w2 2 15 15. 5w 2 1}3

w

EXAMPLE 3



4. 3(14 2 5) 5. 6(9 2 14)

6. (7 1 5) 2 (8 1 4) 7. (33 2 6) 4 3

8. 42(2 1 8) 9. 9[15 4 (2 1 3)]


LESSON

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19Algebra 1


Add positive and negative numbers.

VocabularyThe number 0 is the additive identity.

The opposite of a is its additive inverse.

Rules of Addition

Same signs To add two numbers with the same sign, add their absolute values. The sum has the same sign as the numbers added.

Different signs To add two numbers with different signs, subtract the lesser absolute value from the greater absolute value. The sum has the same sign as the number with the greater absolute value.

Properties of Addition

Commutative Property The order in which you add two numbers does not change the sum.

Associative Property The way you group three numbers in a sum does not change the sum.

Identity Property The sum of a number and 0 is the number.

Inverse Property The sum of a number and its opposite is 0.

GOAL

Add real numbersa. 17 1 (224) 5 ⏐224⏐ 2 ⏐17⏐ Rule of different signs

5 24 2 17 Take absolute values.

5 27 Subtract. Use the sign of the number with the greater absolute value.

b. 21.3 1 (25.8) 5 ⏐21.3⏐ 1 ⏐25.8⏐ Rule of same sign

5 1.3 1 5.8 Take absolute values.

5 27.1 Add and use the sign of the numbers added.


Find the sum.

1. 23.6 1 7.1 2. 5.3 1 (29) 3. 20.2 1 (20.6)

4. 231}2 1 122

1}5 2 5. 11 1 (215) 1 8 6. 25 1 (28) 1 6

EXAMPLE 1

Study GuideFor use with pages 73 –79

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Identify the properties of addition

Identify the property being illustrated.

Statement Property Illustrated

a. 215 1 0 5 215 Identity property of addition

b. 12 1 (217) 5 217 1 12 Commutative property of addition



7. 9 1 (2x) 5 2x 1 9

8. 5.1 1 (25.1) 5 0

9. [3 1 (22)] 1 1 5 3 1[(22) 1 1]

EXAMPLE 2

Study Guide continuedFor use with pages 73 –79

LESSON

2.2

Solve a multi-step problem Gas Prices The table shows the changes in gas prices for two companies. Which company had the greater total change in gas prices for the three weeks?

Week Price change for company A Price change for company B

1 $.05 2$.06

2 2$.08 $.13

3 $.11 2$.04

Solution

STEP 1 Calculate the total change in gas prices for each company.

Company A: Company B:Total change 5 0.05 1 (20.08) 1 0.11 Total change 5 20.06 1 0.13 1 (20.04)

5 20.08 1 (0.05 1 0.11) 5 0.13 1 [2(0.06 1 0.04)]

5 20.08 1 0.16 5 0.13 1 (20.1)

5 0.08 5 0.03

STEP 2 Compare the total change in gas prices: 0.08 > 0.03.

Company A had the greater total change in gas prices.

Exercise for Example 310. In Example 3, suppose that the changes in gas prices for week 4 are 2$.06 for

company A and $.07 for company B and the changes in gas prices for week 5 are $.04 for company A and 2$.03 for company B. Which company has the greater total change in gas prices for the five weeks?

EXAMPLE 3

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Multiply real numbers.

VocabularyThe number 1 is called the multiplicative identity.

The Sign of a Product

p The product of two real numbers with the same sign is positive.

p The product of two real numbers with different signs is negative.

Properties of Multiplication

Commutative Property The order in which two numbers are multiplied does not change the product.

Associative Property The way you group three numbers when multiplying does not change the product.

Identity Property The product of a number and 1 is that number.

Property of Zero The product of a number and 0 is 0.

Property of 21 The product of a number and 21 is the opposite of the number.

GOAL

Multiply real numbers

Find the product.

a. (27)(23) b. 4(22)(26)

Solution

a. (27)(23) 5 21 Same signs; product is positive.

b. 4(22)(26) 5 (28)(26) Multiply 4 and 22; product is negative.

5 48 Same signs; product is positive.


Find the product.

1. 25(4) 2. 9(28)

3. 212(26) 4. (24)121}2 2

5. 4(28)(5) 6. 28(22)(28)

EXAMPLE 1

Study GuideFor use with pages 87–93

LESSON

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43Algebra 1


Identify the properties of multiplication


Solution

Statement Property Illustrated

a. x p 5 5 5 p x Commutative property of multiplication

b. 26 p (21) 5 6 Multiplicative property of 21

c. (23 p x) p 2 5 23 p (x p 2) Associative property of multiplication



7. 25 p 1 5 25

8. [5 p (22)] p (23) 5 5 p [(22) p (23)]

9. 0 p 11 5 0

10. 23 p (212) 5 212 p (23)

EXAMPLE 2

Use properties of multiplication

Find the product (23x) p (22). Justify your steps.

Solution

(23x) p (22) 5 (22) p (23x) Commutative property of multiplication

5 [22 p (23)]x Associative Property of multiplication

5 6x Product of 22 and 23 is 6.


Find the product. Justify your steps.

11. 29(2x)

12. w(23)(12)

13. (28)(5)(2z)

EXAMPLE 3

Study Guide continuedFor use with pages 87–93

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43Algebra 1


Translate verbal sentences into equations or inequalities.

VocabularyAn open sentence is a mathematical statement that contains two expressions and a symbol that compares them.

An equation is an open sentence that contains the symbol 5.

An inequality is an open sentence that contains one of the symbols <, ≤, >, or ≥.

When you substitute a number for the variable in an open sentence, the resulting statement is either true or false. If the statement is true, the number is a solution of the equation, or a solution of the inequality.

GOAL

Write equations and inequalities

Write an equation or an inequality.

a. 8 times the quantity of 11 plus a number x is 112.

b. The product of 7 and a number y is no more than 31.

c. A number z is more than 8 and at most 15.

Solution

Verbal phrase Equation or inequality

a. 8 times the quantity of 11 plus a number x is 112. 8(11 1 x) 5 112

b. The product of 7 and a number y is no more than 31. 7y ≤ 31

c. A number z is more than 8 and at most 15. 8 < z ≤ 15


Write an equation or an inequality.

1. The difference of 73 and a number x is 17.

2. The product of 8 and the quantity of a number y plus 6 is less than 21.

3. The quotient of a number w and 5 is at most 4.

4. The sum of a number z and 2 is greater than 15 and less than 23.

EXAMPLE 1


LESSON

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Check possible solutions

Check whether 5 is a solution of the equation or inequality.

Equation/inequality Substitute Conclusion

a. 3x 2 7 5 12 3(5) 2 7 0 12 8 Þ 12 ✗

5 is not a solution.

b. 9 1 2x ≤ 23 9 1 2(5) ≤? 23 19 ≤ 23 ✓

5 is a solution.


Check whether the given number is a solution of the equation or inequality.

5. 13 1 a 5 17; 4 6. 7b 2 3 5 10; 2 7. 4c < 15; 3

8. 21 2 3d ≥ 11; 2 9. 4g 1 6 ≤ 14; 3 10. 7 < m 1 8 < 15; 6

EXAMPLE 2

Solve a multi-step problem A soccer team is selling pizzas for $6 each. Each pizza costs $4 to make. The team has 10 players and wants to raise $900 for equipment and uniforms. How many pizzas does the team need to sell? How many pizzas will each player sell if every player sells the same number of pizzas?

Solution

STEP 1 Write a verbal model. Let p be the number of pizzas sold. Write an equation.

1Price ofpizza 2 Cost to make

each pizza 2 3 1Number ofpizzas sold 2 5 Profi t

(6 2 4) 3 p 5 900

STEP 2 Use mental math to solve the equation (6 2 4)p 5 900, or 2p 5 900. Think: 2 times what number is 900? Because 2(450) 5 900, the solution is 450.

The team needs to sell 450 pizzas.

STEP 3 Find the number of pizzas each player sells: 450 pizzas}10 players

5 45 pizzas per player

Each player will sell 45 pizzas.

Exercise for Example 3 11. Your family is driving 188 miles to visit a relative. Your father drives 63 miles

then stops for a break. How many more miles are left in the trip? Your father drives 50 miles per hour. How long will the remainder of the trip take? Write a verbal model for the situation, then solve.

EXAMPLE 3


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Solve multi-step equations.GOAL

Solve an equation by combining like terms

Solve 17x 2 11x 1 8 5 20.

Solution

17x 2 11x 1 8 5 20 Write original equation.

6x 1 8 5 20 Combine like terms.

6x 1 8 2 8 5 20 2 8 Subtract 8 from each side.

6x 5 12 Simplify.

6x}6 5

12}6 Divide each side by 6.

x 5 2 Simplify.


Solve the equation. Check your solution.

1. 9x 2 13x 1 7 5 31

2. 13 2 5x 1 8x 5 22

3. 15x 2 9 2 8x 5 12

4. 18 2 2x 2 4x 5 224

EXAMPLE 1


LESSON

3.3

Solve an equation using the distributive property

Solve 4x 1 3(2x 2 1) 5 17.

Solution

METHOD 1 Show All Steps METHOD 2 Do Some Steps Mentally

4x 1 3(2x 2 1) 5 17 4x 1 3(2x 2 1) 5 17

4x 1 6x 2 3 5 17 4x 1 6x 2 3 5 17

10x 2 3 5 17 10x 2 3 5 17

10x 2 3 1 3 5 17 1 3 10x 5 20

10x 5 20 x 5 2

10x}10

5 20}10

x 5 2

EXAMPLE 2

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33Algebra 1




5. 3(x 2 4) 1 4x 5 16

6. 9x 2 6(3x 2 3) 5 9

7. 22x 1 7(3x 21) 5 31

8. 5(2x 1 8) 2 6x 5 16

Multiply by a reciprocal to solve an equation

Solve 3}4

(5x 2 4) 5 12.

Solution

3}4 (5x 2 4) 5 12 Write original equation.

4}3 p 3}

4 (5x 2 4) 5

4}3 p 12 Multiply each side by

4}3, the reciprocal of

3}4 .

5x 2 4 5 16 Simplify.

5x 5 20 Subtract 4 from each side.

x 5 4 Simplify.



9. 1}2 (x 2 11) 5 9

10. 23}2 (2y 1 6) 5 15

11. 215 5 5}7 (4z 2 1)

12. 36 5 23}4 (5m 1 12)

EXAMPLE 3


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Solve equations with variables on both sides.

VocabularyAn equation that is true for all values of the variable is an identity.

GOAL

Solve an equation with variables on both sides

Solve 13 2 6x 5 3x 2 14.

Solution

13 2 6x 5 3x 2 14 Write original equation.

13 2 6x 1 6x 5 3x 2 14 1 6x Add 6x to each side.

13 5 9x 2 14 Simplify.

27 5 9x Add 14 to each side.

3 5 x Divide each side by 9.

The solution is 3. Check by substituting 3 for x in the original equation.

CHECK 13 2 6x 5 3x 2 14 Write original equation.

13 2 6(3) 5 3(3) 2 14 Substitute 3 for x.

25 5 3(3) 2 14 Simplify left side.

25 5 25 ✓ Simplify right side. Solution checks.



1. 9a 5 7a 2 8 2. 17 2 8b 5 3b 2 5 3. 25c 1 6 5 9 2 4c

EXAMPLE 1


LESSON

3.4

Solve an equation with grouping symbols

Solve 4x 2 7 5 1}3

(9x 2 15).

Solution

4x 2 7 5 1}3 (9x 2 15) Write original equation.

4x 2 7 5 3x 2 5 Distributive property

x 2 7 5 25 Subtract 3x from each side.

x 5 2 Add 7 to each side.

The solution is 2.

EXAMPLE 2

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45Algebra 1


Identify the number of solutions of an equation

Solve the equation, if possible.

a. 4(3x 2 2) 5 2(6x 1 1) b. 4(4x 2 5) 5 2(8x 2 10)

Solution

a. 4(3x 2 2) 5 2(6x 1 1) Write original equation.


12x 5 12x 1 10 Add 8 to each side.

The equation 12x 5 12x 1 10 is not true because the number 12x cannot be equal to 10 more than itself. So, the equation has no solution. This can be demonstrated by continuing to solve the equation.

12x 2 12x 5 12x 1 10 212x Subtract 12x from each side.

0 5 10 Simplify.

The statement 0 5 10 is not true, so the equation has no solution.

b. 4(4x 2 5) 5 2(8x 2 10) Write original equation.


Notice that the statement 16x 2 20 5 16x 2 20 is true for all values of x. So, the equation is an identity.


Solve the equation, if possible.

7. 11x 1 7 5 10x 2 8

8. 5(3x 2 2) 5 3(5x 2 1)

9. 1}2 (6x 1 18) 5 3(x 1 3)

EXAMPLE 3



4. 2m 2 7 5 3(m 1 8)

5. 1}5 (15n 1 5) 5 8n 2 9

6. 7p 2 3 5 3}4 (8p 2 12)


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Solve multi-step inequalities.GOAL

Solve a two-step inequality

Solve 24x 1 3 > 15. Graph your solution.

Solution

24x 1 3 > 15 Write original inequality.

24x > 12 Subtract 3 from each side.

x < 23 Divide each side by 24.Reverse inequality symbol.

The solutions are all real numbers less than 23.0 1 2212223242526Check by substituting a number less than 23 in

the original inequality.

CHECK 24x 1 3 > 15 Write original inequality.

24(25) 1 3 >? 15 Substitute 25 for x.

23 > 15 ✓ Solution checks.


Solve the inequality. Graph your solution.

1. 7x 1 8 > 22 2. 27 ≥ 22x 1 9 3. 2.3x 2 6.9 < 7.13

EXAMPLE 1


LESSON

6.3

Solve a multi-step inequality

Solve the inequality.

a. 21}3 (x 1 12) < 5 b. 9x 1 2 < 5x 2 18

Solution

a. 21}3 (x 1 12) < 5 Write original inequality.

2x}3 2 4 < 5 Distributive property

2x}3 < 9 Add 4 to each side.

x > 227 Multiply each side by 23. Reverse the inequality symbol.

The solutions are all real numbers greater than 227.

b. 9x 1 2 < 5x 2 18 Write original inequality.

9x < 5x 2 20 Subtract 2 from each side.

4x < 220 Subtract 5x from each side.

x < 25 Divide each side by 4.

The solutions are all real numbers less than 25.

EXAMPLE 2

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39Algebra 1


Identify the number of solutions of an inequality

Solve the inequality, if possible.

a. 5(3x 2 2) < 15x 1 7 b. 9 2 28x > 4(5 2 7x)

Solution

a. 5(3x 2 2) < 15x 1 7 Write original inequality.

15x 2 10 < 15x 1 7 Distributive property

210 < 7 Subtract 15x from each side.

All real numbers are solutions because 210 < 7 is true.

b. 9 2 28x > 4(5 2 7x) Write original inequality.

9 2 28x > 20 2 28x Distributive property

9 > 20 Add 28x to each side.

There are no solutions because 9 > 20 is false.


Solve the inequality, if possible.

7. 2m 2 7m 2 4 > 1 2 5m

8. 3n 2 13 < 3(n 2 2)

9. 11p 2 3p 1 6 ≥ 4(2p 2 1)

EXAMPLE 3


Solve the inequality.

4. 3(2x 2 7) > 15

5. 10 2 3x ≤ 5x 2 14

6. 1}2 (8x 1 6) <

1}3 (9x 2 15)


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Solve and graph compound inequalities.

VocabularyA compound inequality consists of two separate inequalities joined by and or or.

GOAL

Write and graph a compound inequality

Translate the verbal phrases into an inequality. Then graph the inequality.

a. All real numbers that are less than or equal to 7 or greater than or equal to 10.

Inequality: x ≤ 7 or x ≥ 10

3 4 5 7 8 9 10 116

b. All real numbers that are greater than 21 and less than or equal to 1.

Inequality: 21 < x < 1

0 1 2 3 421222324


Translate the verbal phrases into an inequality. Then graph the inequality.

1. All real numbers that are less than 23 or greater than 0.

2. All real numbers that are less than 9 and greater than or equal to 7.

3. All real numbers that are greater than or equal to 14 or less than or equal to 10.

EXAMPLE 1


LESSON

6.4

Solve a compound inequality with and

Solve 7 ≤ x 2 4 ≤ 12. Graph your solution.

Solution

7 ≤ x 2 4 ≤ 12 Write original inequality.

7 1 4 ≤ x 2 4 1 4 ≤ 12 1 4 Add 4 to each expression.

11 ≤ x ≤ 16 Simplify.

The solutions are all real numbers greater than or equal to 11 and less than or equal to 16.

9 10 11 13 14 15 16 1712

EXAMPLE 2

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53Algebra 1


Solve a compound inequality with or

Solve 3x 1 4 < 16 or 5x 2 12 > 13. Graph your solution.

Solution

Solve the two inequalities separately.

3x 1 4 < 16 or 5x 2 12 > 13 Write original inequality.

3x 1 4 2 4 < 16 2 4 or 5x 2 12 1 12 > 13 1 12 Use addition or subtraction property of inequality.

3x < 12 or 5x > 25 Simplify.

3x}3 <

12}3 or

5x}5 >

25}5 Use division property of

inequality.

x < 4 or x > 5 Simplify.

The solutions are all real numbers less than 4 or greater than 5.

0 1 2 3 4 5 7621

Exercises for Examples 2 and 3

Solve the inequality. Graph your solution.

4. 9 < 2x 1 3 < 15

5. 30 ≥ 27x 2 12 > 16

6. 28 ≤ 4(2x 2 3) ≤ 68

7. 3x 2 7< 11 or 2x 1 4 < 15

8. 1}2 (x 1 18) > 6 or 7x 1 5 < 251

9. 3x 1 8 > 7x 2 12 or 9(x 2 2) > 8x 2 9

EXAMPLE 3


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85Algebra 1


Write equations in function form and rewrite formulas.

VocabularyAn equation in x and y is written in function form when the dependent variable y is isolated on one side of the equation.

A literal equation is an equation that contains two or more variables.

GOAL

Rewrite an equation in function form

Write 9x 2 4y 5 8 in function form.

Solution

To write an equation in function form, solve the equation for y.

9x 2 4y 5 8 Write original equation.

24y 5 8 2 9x Subtract 9x from each side.

y 5 22 1 9}4x Divide each side by 24.

The equation y 5 22 1 9}4 x is written in function form.

EXAMPLE 1


LESSON

3.8

Solve a literal equation

The formula for the volume of a rectangular prism is V 5 lwh. Solve the formula for l.

Solution

V 5 lwh Write original equation.

V}wh

5 lwh}wh

Assume w Þ 0 and h Þ 0. Divide each side by wh.

V}wh

5 l Simplify.

The rewritten equation is V}wh

5 l.

Exercises for Examples 1 and 2

Write the equation in function form.

1. 7x 1 y 5 12 2. 3y 2 9x 5 21 3. 5y 2 2x 5 15

Solve the literal equation.

4. I 5 Prt for P 5. A 5 1}2 (b1 1 b2)h for b2

EXAMPLE 2

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Solve and use a geometric formula

The area A of a triangle is given by the formula A 5 1}2

bh where b is the base and h is the height.

12 in.

b

a. Solve the formula for the base b.

b. Use the rewritten formula to find the base of the triangle shown, which has an area of 106.8 square inches.

Solution

a. Solve the formula for b.

A 5 1}2bh Write original formula.

2A 5 bh Multiply each side by 2.

2A}h

5 b Divide each side by h.

b. Substitute 106.8 for A and 12 for h in the rewritten formula.

b 5 2A}h

Write rewritten formula.

b 5 2(106.8)}

12Substitute 106.8 for A and 12 for h.

b 5 17.8 Simplify.

The base of the triangle is 17.8 inches.


The surface area S of a sphere is given by the formula S 5 4πr2 where ris the radius of the sphere.

6. Solve the formula for r.

7. Use the rewritten formula from Exercise 6 to find r when S 5 314 square meters. Use 3.14 for π.

EXAMPLE 3


LESSON

3.8L

ES

SO

N 3

.8

lesson study guide 1 - mr. chan's classroom · 23 algebra 1 chapter 1 resource book evaluate...

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