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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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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 ftRA LW
372 205RA 76,260 square feetRA
12TA bh
12 116 372TA 21,576 square feetTA
Page 125
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
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
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
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..
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
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
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
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
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
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
DONE
Objectives
• Formulas from Geometry• Solving for a Variable• Other Formulas
#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
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
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
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