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Copyright © 2013 Pearson Education, Inc. Section 2.4 Formulas

Section 2.4

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Section 2.4. Formulas. 205 ft. 372 ft. 116 ft. Example. Page 125. A residential lot is shown. Find the area of this lot. Solution The area of the rectangle: The area of the triangle: Total area = 76,260 + 21,576 = 97,836 square feet. Example. - PowerPoint PPT Presentation

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Page 1: Section 2.4

Copyright © 2013 Pearson Education, Inc.

Section 2.4

Formulas

Page 2: Section 2.4

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 ftRA LW

372 205RA 76,260 square feetRA

12TA bh

12 116 372TA 21,576 square feetTA

Page 125

Page 3: Section 2.4

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+2l32062 ww

3208 w

40w

Page 4: Section 2.4

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 1805 5x

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: Section 2.4

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: Section 2.4

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: Section 2.4

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 mpn

p q m

Page 129

Page 8: Section 2.4

Solving a Formula for a VariableEXAMPLE

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

mxx

by

mx

by

Page 9: Section 2.4

Solving a Formula for a Variable

543

yxSolve for x

54443

yyyx

543

yx

)54(33

3 yx

1512 yx

)(21 baA

Solve for length b

baA 2

)(2122 baA

abaaA 2

baA 2aAb 2

Page 10: Section 2.4

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 dGPAa b c d f

Slide 10

Page 130

Page 11: Section 2.4

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 8 418 22 8 4 0

GPA

15852

3.04

Page 130

Page 12: Section 2.4

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

5 329

C F

25 329

3C

5 99

C

Page 130

Page 13: Section 2.4

DONE

Page 14: Section 2.4

Objectives

• Formulas from Geometry• Solving for a Variable• Other Formulas

Page 15: Section 2.4

#20

#24

hha

hhbA

2

Solving a Formula for a Variableproblem 20 on page 144

)(21 bahA

Solve for a

Mult by 2 to remove 1/2

)(2 bahA

hbhaA 2

hbhbhahbA 2

hahbA 2

bhA

hhbAa

2or 2

Page 16: Section 2.4

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

wlw

wA

lwA

pmDT

Solve for m

pmDDDT

pmDT

ppm

pDT

wAl

pDTm

Page 17: Section 2.4

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,

DMG

6

2378.

27.6 miles per gallon.

Page 124

Page 18: Section 2.4

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

852

V 1125

32V

110.45 cubic inchesV

Page 128