copyright © 2013 pearson education, inc. section 2.4 formulas

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Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

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Page 1: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Copyright © 2013 Pearson Education, Inc.

Section 2.4

Formulas

Page 2: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

A residential lot is shown.Find the area of this lot.SolutionThe area of the rectangle:

The area of the triangle:

Total area = 76,260 + 21,576 = 97,836 square feet.

205 ft

372 ft

116 ft

RA LW372 205RA 76,260 square feetRA

12TA bh

12 116 372TA 21,576 square feetTA

Page 125

Page 3: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

32033 wwww

A rectangular swimming pool is three times as long as it is wide. If the perimeter of the pool is 320 feet, what are the pool’s dimensions.?

Divide 8

Length is 3 times 40 length= 120

3w

wP=2w+2l

32062 ww

3208 w

40w

Page 4: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

In a triangle, the smaller angles are equal in measure and are one-third of the largest angle. Find the measure of each angle.SolutionLet x represent the measure of each of the two smaller angles. Then the measure of the largest angle is 3x, and the sum of the measures of the three angles is given by

3 180x x x

5 180x

5 180

5 5

x

36x

The measure of the largest angle is 3x, thus 36 ∙ 3 = 108°.

The measure of the three angles are 36°, 36°, and 108°.

Page 124-5

Page 5: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Find the volume and the surface area of the box shown.

SolutionThe volume of the box isV = lwhV = 12 ∙ 6 ∙ 5 V = 360 cm3

The surface area of the box is

Example

12 cm

6 cm

5 cm

2 2 2S LW WH LH 2(12)(5) 2(5)(6) 2(12)(6)S 120 60 144S 324 square centimetersS

Page 127

Page 6: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

Solve each equation for the indicated variable.a. b.

Solutiona.

Multiply by LCD which is 5

Subtract y

3 for 5

y zx z

for np nm nq p

3 5

y zx

15 x y z 15 x y z

15 z x y

Page 129

5 5..

Page 7: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

Solve each equation for the indicated variable.a. b.

Solutionb.

Subtract nm

GCF is n

Divide by GCF

3 for 5

y zx z

for np nm nq p

for np nm nq p

np nq nm

( )np n q m

( )n q mp

n

p q m

Page 129

Page 8: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Solving a Formula for a Variable

EXAMPLE

SOLUTION

Solve the formula y = mx + b for m

y = mx + b Think of m saying, “I really want to be alone.”

y – b = mx + b – b Subtract b from both sides.

y – b = mx Perform the addition. b – b = 0.

Divide both sides by x to find m.x

mx

x

by

mx

by

Page 9: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Solving a Formula for a Variable

543

yx

Solve for x

54443

yyyx

543

yx

)54(33

3 yx

1512 yx

)(2

1baA

Solve for length b

baA 2

)(2

122 baA

abaaA 2

baA 2

aAb 2

Page 10: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Other Formulas

To calculate a student’s GPA, the number of credits earned with a grade of A, B, C, D, and F must be known. If a, b, c, d, and f represent these credit counts respectively, then

4 3 2.

a b c dGPA

a b c d f

Slide 10

Page 130

Page 11: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

A student has earned 18 credits of A, 22 credits of B, 8 credits of C and 4 credits of D. Calculate the student’s GPA to the nearest hundredth.SolutionLet a = 18, b = 22, c = 8, d = 4 and f = 0

The student’s GPA is 3.04.

4 18 3 22 2 8 4

18 22 8 4 0GPA

158

52 3.04

Page 130

Page 12: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

The formula is used to convert degrees Fahrenheit to degrees Celsius. Use this formula to convert 23°F to an equivalent Celsius temperature.

Solution

59 32C F

= −5°C

532

9C F

25

329

3C

59

9C

Page 130

Page 13: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

DONE

Page 14: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Objectives

• Formulas from Geometry

• Solving for a Variable

• Other Formulas

Page 15: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

#20

#24

h

ha

h

hbA

2

Solving a Formula for a Variableproblem 20 on page 144

)(2

1bahA

Solve for a

Mult by 2 to remove 1/2

)(2 bahA

hbhaA 2

hbhbhahbA 2

hahbA 2

bh

A

h

hbAa

2or

2

Page 16: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

1

3

Solving a Formula for a VariableCP 1, 3 on pages 136-137

wlA Divide by w

Solve for length l

Subtract D from both sides

Divide by p

w

lw

w

A

lw

A

pmDT

Solve for m

pmDDDT

pmDT

p

pm

p

DT

w

Al

p

DTm

Page 17: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

A tourist starts a trip with a full tank of gas and an odometer that reads 59,478 miles. At the end of the trip, it takes 8.6 gallons of gas to fill the tank, and the odometer reads 59,715 miles. Find the gas mileage for the car. SolutionThe distance traveled is 59,715 – 59, 478 = 237 miles and the number of gallons used is G = 8.6. Thus,

DM

G

6

237

8. 27.6 miles per gallon.

Page 124

Page 18: Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Example

A cylindrical soup can has a radius of 2 ½ inches and a height of 5 5/8 inches. Find the volume of the can.

Solution h

r

2V r h2

45

8

5

2V

1125

32V

110.45 cubic inchesV

Page 128