Solution manual for finite mathematics 11th edition by sullivan

Use the points 0, 0 and –2, 1 to compute the slope of the line: Since the y-intercept 0, 0 is given, use the slope-intercept form of the equation of the line: 102122 This is the general

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Chapter 1 Linear Equations

7. The point-slope form of the equation of a line

with slope m containing the point (x1, y is 1)

y−y =m x−x

8. If the graph of a line slants downward from left

to right, its slope m is negative

11 The set of points of the form, (2, y), where y is a

real number, is a vertical line passing through 2

on the x-axis The equation of the line is x = 2

12 The set of points, (x, 3), where x is a real

number, is a horizontal line passing through 3 on

the y-axis The equation of the line is

−

2

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15 2x – y = 6

0

x y

We interpret the slope to mean that for every 2

unit change in x, y changes 1 unit That is, for every 2 units x increases, y increases by 1 unit

unit change in x, y changes –1 unit That is, for every 2 units x increases, y decreases by 1 unit

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unit change in x, y changes by –1 unit That is, for every 1 unit increase in x, y decreases by 1

unit change in x, y changes by 1 unit That is, for every 3 units x increases, y increases by 1 unit

A slope of 3 means that for every 1 unit change

in x, y will change 3 units

A slope of 1 means that for every 1 unit change

in x, y will change 1 unit

− means that for every 2 unit

increase in x, y will decrease –1 unit

3 means that for every 3 unit

increase in x, y will increase 2 units

A slope of zero indicates that regardless of how

x changes, y remains constant

0( 5) 4 9

A slope of zero indicates that regardless of how

x changes, y remains constant

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Since the y-intercept, (0, 0), is given, use the

slope-intercept form of the equation of the line:

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44 Use the points (0, 0) and (–2, 1) to compute the

slope of the line:

Since the y-intercept (0, 0) is given, use the

slope-intercept form of the equation of the line:

102122

This is the general form of the equation

45. Use the points (1, 1) and (–1, 3) to compute the

slope of the line:

1( 1) 1 2

Now use the point (1, 1) and the slope m = –1 to

write the point-slope form of the equation of the line:

46. Use the points (–1, 1) and (2, 2) to compute the

slope of the line:

47 Since the slope and a point are given,

use the point-slope form of the line:

48 Since the slope and a point are given,

use the point-slope form of the line:

This is the general form of the equation

49 Since the slope and a point are given, use the

point-slope form of the line:

50 Since the slope and a point are given, use the

point-slope form of the line:

51 Since we are given two points, (1, 3)

and (–1, 2), first find the slope

52 Since we are given two points, (–3, 4) and (2, 5),

first find the slope

( )43 52 15

m= − =

− −Then use the slope, one of the points, (–3, 4), and the point-slope form of the line:

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53 Since we are given the slope m = –2 and the

y-intercept (0, 3), we use the slope-intercept

form of the line:

54 Since we are given the slope m = –3 and the

y-intercept (0, –2), we use the slope-intercept

form of the line:

55 We are given the slope m =3 and the x-intercept

(–4, 0), so we use the point-slope form of the line:

56 We are given the slope m = –4 and the

x-intercept (2, 0) So we use the point-slope

form of the line:

This is the general form of the equation

57 We are given the slope 4

5

m= and the point

(0, 0), which is the y-intercept So, we use the

slope-intercept form of the line:

405

58 Since we are given the slope 7

3

m= and the

point (0, 0), which is the y-intercept, we use the

slope-intercept form of the line:

703

59 We are given two points, the x-intercept (2, 0)

and the y-intercept (0, –1), so we need to find

the slope and then use the slope-intercept form

of the line to get the equation

60. We are given two points, the x-intercept (–4, 0) and the y-intercept (0, 4), so we need to find the

slope and then use the slope-intercept form of the line to get the equation

4 0

0 ( 4)44

61. Since the slope is undefined, the line is vertical The equation of the vertical line containing the

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66 y = –3x + 4

slope: m = –3; y-intercept: (0, 4)

67 To obtain the slope and y-intercept, we

transform the equation into its slope-intercept

form by solving for y

1

12

1

23

123

223

70 To obtain the slope and y-intercept, we

332

11

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22

73 The slope is not defined; there is no y- intercept

So the graph is a vertical line

74 slope: m = 0; y-intercept: (0, –1)

75 slope: m = 0; y-intercept: (0, 5)

76 The slope is not defined; there is no y-intercept

So the graph is a vertical line

32

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(continued)

81 To graph an equation on a graphing utility, first

solve the equation for y

1.3 2.7 82.7 1.3 8

Ymin = –10; Ymax = 10

The x-intercept is (2.83, 0)

The y-intercept is (0, 2.56)

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86 To graph an equation on a graphing utility, first

23

151515

x y

ππ

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93. Using the intercepts (–2, 0) and (0, 2),

94. Using the intercepts (1, 0) and (0, 1),

95. Using the intercepts (3, 0) and (0, 1),

2

y= − x− General form:

97 a The equation is C = 0.54x, where x is the

number of miles the car is driven

98 a Each week it costs $224 to rent the truck and

an additional $0.52 per mile for each mile the truck is driven So the total cost for a weekly rental is given by the equation

C = 0.52x + 224, where x is the number of

miles the truck is driven

e The y-intercept $224 represents the fixed cost

of renting the truck

99 a The fixed cost of electricity for the month is

$8.23 In addition, the electricity costs

$0.10438 (10.438 cents) for every

kilowatt-hour (KWH) used If x represents the number

of KWH of electricity used in a month, the total monthly charge is represented by the equation

0.10438 8.23, 0 400

c The charge for using 100 KWH of electricity

is found by substituting 100 for x in part (a):

( )

0.10438 100 8.2310.438 8.23 18.668 $18.67

d The charge for using 300 KWH of electricity

is found by substituting 300 for x in part (a):

( )

0.10438 300 8.2331.314 8.23 39.544 $39.54

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e The slope of the line, m = 0.10438, indicates

that for every extra KWH used (up to 400 KWH), the electric bill increases by 10.438 cents

100 a The fixed monthly cost of electricity is $5.69

In addition, the electricity costs $0.08735 (8.735 cents) for every kilowatt-hour (KWH)

used If x represents the number of KWH of

electricity used in a month, the total monthly charge is represented by the equation 0.08735 5.69, 0 1000

e The slope of the line, m = 0.08735, indicates

that for every extra KWH used (up to 1000 KWH), the electric bill increases by 8.735 cents

101 a Since we are told the relationship is linear, we

will use the two points to get the slope of the line:

b To find the Celsius measure of 68 ºF,

substitute 68 for F in the equation and

simplify:

5

68 32 209

102 a. K = C° + 273

b Since we only have relations between Kelvin and Celsius, and Celsius and Fahrenheit, we first use the relationship between Celsius and Fahrenheit, and then substitute 5( )

2735

32 2739

5255.229

103 a If t = 0 represents December 21, then January

20 is represented by t = 30 Use the points

(0, 102.7) and (30, 104.1) to find the slope of the equation

We will use the slope and the point (0, 102.7)

to write the point-slope form of the equation

102.7 0.0467 00.0467 102.7

d. A = 0.0467(41) + 102.7 = 104.6147

The model predicts that there will be 104.6147 billion gallons of water in the reservoir on January 31, 2010

e. The reservoir will be full when A = 265.5

0.0467 102.7 265.50.0467 162.8

3486.1

t t t

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(continued)

We use the point (4000, 100,000) and the slope 100 to obtain the point-slope form of the equation of the line:

b To determine how much advertising is needed

to sell 200,000 boxes of cereal, we will let N

be 200,000 in the equation from part (a) and

solve for A

200, 000 100 300, 000

500, 000 1005000

A A A

=

So, the company would need to spend $5000

on advertising to sell 200,000 boxes of cereal

c. The firm will sell 100 extra boxes of cereal for

every additional dollar spent on advertising

=

106 We are given two points and are told that the

number of diseased mice is linearly related to the number of days since exposure In this

problem we will let n represent the number of diseased mice in the cage and let t represent the

number of days after the first exposure We first compute the slope of the line:

We now use the point (4, 8) and the fact that the

slope m = 3 to obtain the point-slope form of the

equation of the line

t t t

x x x

=

=Dan would need to have sales that generate

$9146.20 in profit to earn the median amount

108 a. Since the relationship between time and depletion is linear, we use the points to find the slope of the line

b. To determine the year when the fields run dry,

let A = 0 and solve for t

0 0.14 284.760.14 284.76

284.76

20340.14

t t

t

=

The fields are predicted to run dry in 2034

c. The slope represents the annual depletion rate These fields decrease by 140 million barrels per year

d. Divide the current reserves in the Jack Field

by the annual depletion rate (that is, the slope)

15 billion 15

107.14

140 million=0.14≈The Jack Field will last about 107 years

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109 a. Since the rate of increase is constant, we use

the points to find the slope of the line

110 a Since we assume the rate of increase is

constant, we use the points to find the slope of

111 a Since we assume the rate of increase is

constant, we use the points to find the slope of

c. The slope is the annual average increase in the percentage of people over 25 years of age who have a bachelor’s degree or higher For every

unit change in the year t, P increases by 0.5

112 a. Since we assume the relationship between time and number of degrees conferred is linear, we use the points to find the slope of the line

c. The slope is the average annual increase in the number of bachelor’s degrees awarded

113 a Since the cost of the houses is linear, we first use the points to find the slope of the line

b The projected average cost of a house in 2011

is found by letting t = 2011 in the equation

6543 2011 13, 233,167 $75,194

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114 Two points are given, (h1, w1) = (67, 139) and

(h2, w2) = (70, 151), and we are told they are linearly related So we will first compute the slope of the line:

116 a We first find the slope

117 a If the Smiths drive x miles in a year, and their

car averages 17 miles per gallon of gasoline, then they will use about

17

x

gallons of gasoline per year In 2010, the Smith’s annual fuel cost is given by

e. The difference in the annual costs is about

$2417 – $1630 = $787 The Smiths spend

$787 more at the 2010 price than at the 2009 price

118. a If the Jones’s drive x miles in a year, and their

car averages 39 miles per gallon of gasoline, then they will use about

39

x

gallons of gasoline per year In 2010, the Jones’s annual fuel cost is given by

2.739 0.0702 39

64.33

N N m

Then we use the slope and the point (0, 954.7)

to write the point-slope form of the line

( )

954.7 64.3 064.3 954.7

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b. The slope indicates that credit and debit cards

in force are increasing at an average rate of 64.3 million cards per year

d. To estimate the year that the number of credit

cards will first exceed 1.5 billion, let N = 1500

(since the equation is written in millions), and

solve for t

1500 64.3 954.7545.3 64.3

545.38.564.3

t t t

120. From the graph we can see that the line has a

positive slope and a y-intercept of the form (0, b) where b is a positive number Put each of

the equations into slope-intercept form and choose those with both positive slope and

121. From the graph we can see that the line has a

negative slope and a y-intercept of the form (0, b) where b is a positive number Put each of

the equations into slope-intercept form and choose those with negative slope and positive

122. Answers will vary

123. A vertical line cannot be written in

slope-intercept form since its slope is not defined

124 Not every line has two distinct intercepts A line

passing through the origin has the point

(0, 0) as both its x- and y-intercept Usually vertical lines have only an x-intercept and horizontal lines have only a y-intercept

125 Two lines that have equal slopes and equal

y-intercepts have equivalent equations and

identical graphs

126. If two lines have the same x-intercept and the same y-intercept, and the x-intercept is not

(0, 0), then the two lines have equal slopes

Lines that have equal slopes and equal

y-intercepts have equivalent equations and identical graphs

127. If two lines have the same slope, but different

x-intercepts, they cannot have the same y-intercept If Line 1 has x-intercept (a, 0) and Line 2 has x-intercept (c, 0), but both have the same slope m, write the equation of each line

using the point-slope form then change them to slope-intercept form and compare the

129 The line y = 0 has infinitely many x-intercepts,

so yes, a line can have two distinct x-intercepts

or a line can have infinitely many x-intercepts

130. Yes, a line can have no x-intercepts For example, the line y = 2 has no x-intercepts

No, a line cannot have neither an x-intercept nor

a y-intercept It must have one or the other or

both

131 monter (t.v.) to go up; to ascend, to mount; to

climb; to embark; to rise, to slope up, to be uphill; to grow up; to shoot; to increase (source:

Cassell’s French Dictionary)

We use m to represent slope The French verb

monter means to rise, to climb or to slope up

1.2 Pairs of Lines

1 parallel

2. intersect

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3. To determine whether the pair of lines is

parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts

: 10 10

The slopes of the two lines are the same, but the

y-intercepts are different, so the lines are

parallel

parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their

parallel

parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their

two lines are the same, the lines are coincident

8 To determine whether the pair of lines is parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts

two lines are the same, the lines are coincident

9. To determine whether the pair of lines is parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their slopes, and, if necessary, compare their y-intercepts

parallel

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10 To determine whether the pair of lines is

: 4 2 7

2 4 7

7 2

parallel

11. To determine whether the pair of lines is

12 To determine whether the pair of lines is

parallel, coincident, or intersecting, rewrite each equation in slope-intercept form, compare their

slopes, and, if necessary, compare their

slope: m = 0, y-intercept: (0, –2)

Since the slopes of the two lines are different, the lines intersect

14. L: x = 4 slope: not defined; no y-intercept M: x = –2

slope: not defined; no y-intercept

These are two vertical lines Since they have

different x-intercepts, they are parallel

15. To find the point of intersection of two lines, first put the lines in slope-intercept form

Since the point of intersection, (x0, y0), must be

on both L and M, we set the two equations equal

to each other and solve for x0 Then we substitute the value of x0 into the equation of one

of the lines to find y0

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16 To find the point of intersection of two lines,

first put the lines in slope-intercept form

0 0

3 31

x x

17 To find the point of intersection of two lines,

0 0

18 To find the point of intersection of two lines, first put the lines in slope-intercept form

0 0

The point of intersection is (1, 3)

0 0

4 41

x x

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0 0

M x y

= − +

0 0

The point of intersection is (2, 1)

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0 0

6 6 1

x x

25 L is the vertical line on which the x-value is

always 4 M is the horizontal line on which

y-value is always –2 The point of intersection is (4, –2)

26 L is the vertical line on which the x-value is

always 0 It is the equation of the y-axis.M is the horizontal line on which y-value is always 0 It

is the equation of the x-axis The point of

intersection is (0, 0), which is the origin

27 L is parallel to y = 2x, so the slope of L is m = 2

We are given the point (3, 3) on line L Use the

point-slope form of the line

28 L is parallel to y = –x, so the slope of L is

m = –1 We are given the point (1, 2) on line L

Use the point-slope form of the line

general form: 3x+ = −y 1

31 We want a line parallel to 2x− = −y 2 Find the slope of the line and use the given point, (0, 0)

to obtain the equation Since the y-intercept

(0, 0) is given, use the slope-intercept form of the equation of a line

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32 We want a line parallel to x−2y= −5 Find the

slope of the line and use the given point, (0, 0)

to obtain the equation Since the y-intercept

(0, 0) is given, use the slope-intercept form of the equation of a line

33 We want a line parallel to the line x = 3 This is

a vertical line so the slope is not defined A parallel line will also be vertical, and it must contain the point (4, 2) The parallel line will

have the equation x = 4

34 We want a line parallel to y = 3 This is a

horizontal line; so it has a slope m = 0 The parallel line will also have a slope m = 0 but will

contain the point (4, 2) Use the point-slope form of the equation of a line, or recognize the fact that horizontal lines have an equation of the

form y = b The equation of the line is y = 2

35 To find the equation of the line, we must first

find the slope of the line containing the points (–2, 9) and (3, –10):

( ) ( )

19x+5y= −63

36 To find the equation of the line parallel to the

line containing the points (–4, 5) and (2, –1), first find the slope of the line containing the two

Solve for y to get the slope-intercept form:

7

y= − −x

Rearrange terms to obtain the general form of

the equation: x + y = –7

37 We will let x = the number of caramels the box

of candy, and y = the number of creams in the

box of candy Since there are a total of 50 pieces

of candy in a box, we have x + y = 50, or

y = 50 – x Each caramel costs $0.10 to make,

and each cream costs $0.20 to make So, the cost

of making a box of candy is given by the equation:

0.1 0.20.1 0.2(50 )

38 Let x denote the number of pounds of cashews

in the mixture, and let y denote the number of

pounds of pecans in the mixture Use a table to organize the data

(continued on next page)

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(continued)

Pounds

Cost per Pound

Total Value

Pecans y = 60 – x $7.50 7.5(60 – x)

Mixture x + y + 40 = 100 $4.89 4.89(100)

The last column gives the information needed to

solve the problem since the sum of the values of the ingredients must equal the total value of the mixture

6.5 7.5(60 ) 80 4896.5 450 7.5 80 489

530 1.0 48941

60 41 19

x x y

We need 41 pounds of cashews and 19 pounds

of pecans to be mixed with the 40 pounds of peanuts to make 100 pounds of mixture worth

$4.89 per pound

39 Investment problems are simply mixture

problems involving money We will use a table

to organize the information Let x denote the

amount Mr Nicholson invests in AA bonds, and

y denote the amount he invests in S & L

Certificates

Investment

Amount Invested

Interest Rate

The last column gives the information we need

to set up the equation solve since the sum of the interest earned on the two investments must equal the total interest earned

0.1 0.05(150, 000 ) 10, 0000.1 7500 0.05 10, 0000.05 7500 10, 0000.05 2500

40 The only difference between this problem and

Problem 41 is that the total interest earned must equal $12,000 The equation which will give

Mr Nicholson’s distribution of funds is:

0.1 0.05(150, 000 ) 12, 0000.1 7500 0.05 12, 0000.05 7500 12, 0000.05 4, 500

To earn the extra $2000 in interest, Mr

Nicholson should increase his investment in the higher yielding AA Bonds to $90,000 and reduce his investment in the Savings and Loan Certificates to $60,000

41 Let x denote the amount of Kona coffee and y

denote the amount of Columbian coffee in the mix We will use the hint and assume that the total weight of the blend is 100 pounds

Coffee

Amount Mixed

Price per Pound

Total Value

Columbian y = 100 – x 6.75 6.75(100 – x)

Mixture x + y = 100 10.80 10.80(100)

The last column gives the information necessary

to write the equation, since the sum of the values

of each of the two individual coffees must equal the total value of the mixture

22.95 6.75 100 10.80 10022.95 675 6.75 1080

16.2 40525

100 25 75

x x y

Mix 25 pounds of Kona coffee with 75 pounds

of Columbian coffee to obtain a blend worth

$10.80 per pound

42 Let x denote the amount of corn meal and y

denote the amount of soybean meal in the mix

We will use a table to organize the information

Amount Mixed

Protein Content

Amount of Protein

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