Solution manual for college algebra 3rd edition by young

To be an active member in a fraternity or sorority, you must also be a student at the university... Forgot to apply the power to “2”.. In standard form.. Cutting a square with dimensions

CHAPTER 0 Section 0.1 Solutions -

1 rational (integer/integer) 2 rational (integer/integer)

3 irrational (doesn’t repeat) 4 π =3.14159 irrational (doesn’t repeat)

5 rational (repeats) 6 rational (repeats)

7 5 =2.2360 irrational (doesn’t repeat) 8 17 =4.1231 irrational (doesn’t repeat)

9 a Rounding: 1 is less than 5, so 7 stays 7.347 b Truncating: 7.347 1 7.347

10 a Rounding: 9 is greater than 5, so 4 rounds up to 5 9.255 b Truncating: 9.254 9 9.254

11 a Rounding: 9 is greater than 5, so 4 rounds up to 5 2.995 b Truncating:2.994 9 2.994

12 a Rounding: 1 is less than 5, so 5 stays 6.995 b Truncating:6.995 1 6.995

13 a Rounding: 4 is less than 5, so 4 stays: 0.234

6 11

5 2 3 7

5 6 7

11 7 4

+ ⋅ −+ −

− =

18

N NN

20 18 22

terms inside the parenthesis

80 Forgot to distribute “- 1” through the

2nd term in (1 – y)

81. False Not all student-athletes are honors students

82. True To be an active member in a fraternity or sorority, you must also be a student at the university

17 8 3

1 2

1 2

5⋅ ⋅32= ⋅ =5 2 10

19 12 1

9 3

32 2

2.08 10 ft

15.761.32 10 ft

73 Should be adding exponents here: y5 74 Forgot to apply the power to “2”

5 15 Degree 0 6 In standard form Degree 0

7. y− =2 y1− Degree 1 2 8 In standard form Degree 1

69 Cutting a square with dimensions x from the four corners of the material create sides

with lengths 15 2x− and 8 2x− , and height x The resulting volume is

Thus, the perimeter of the track is P=(2πx+4x+10 feet)

b The areas of the semi-circular pieces are each 1( )2 2 2

2 πx =π2 x feet The area of the rectangular piece is 2

(2 )(2x x+5) feet Thus, the area of the track is ( 2 2 ) 2

4 10 feet

A= πx + x + x

72 a The volume of the hemisphere is 1(4 3) 2 3 3

2 3πr = 3πr units The volume of the cylinder is πr2(2 )r =2πr3 units3

So, the volume of the silo is 8 3 3

3πr units

b The surface area of the cylinder is 2πr(2 )r =4πr2 units2 The surface area of the hemispherical top is 1( 2) 2 2

2 4πr =2πr units The surface area of the bottom of the silo is πr2 units2

So, the surface area of the silo is 7πr2 units2

73

2

( )(3 ) 3 3(10 ) 100 100

75. Forgot to distribute the negative sign through the entire second polynomial

x− y x− y = x − xy+ y 81 Add the degrees, so m + n

82 Take the larger of the two degrees, so

( 1)( 3) ( 1)( 1)( 3)

=+

−

=+

y y

y y

−

151

x x

1

t t

t t t

+

Note: 1

t t t

+ =+ Note: y≠ − 92

+

26 2

42( 2)

x x

++ Note: x≠ −2

x x

154

x x

+

=+

Note: x≠ −4

29 (3x+1) (2x−1) (x+5) (2x−1)

3 15

x x

+

=+

3 13

x x

+

=+

++ Note: x≠ −1, 2

32 3 4

2

x x

x x

−

=

−Note: 0, 5x≠ ±

35 2(x− 1) ( 1)

3

x x x

+

⋅(x+ 1) (x− 1)

2 3 Note:x 1,0,1

12

= Note: 0, 1x≠ ±

t t

− + ( 2) 3( 32)

Note: 2,3

t t t

t

−

= + +

−

−

53( 8)( 8)

y y y

−

=++

( 3)14( 3)

a a a

y y y

−

=

−Note: 2, 3y≠ −

44 (t+3) (t−2) (t+2) (t−2)

Note: 0, 2t≠ ±

45 3x (x−5)

2 x(x+5) (x−5)

(2 3)( 5)3

⋅

2

( 5)(2 3)( 5)

+

46

5 14

t t

⋅ (4t+3) (4t−3) (4t+3)

5 14(4 3)

t t

2Note: 0, , 9

x x

x+

=Note: x≠ ±2

52 5(x+6)10

8(x 6)⋅

− 520 (x+6)

15(x 6)

=

−Note: x≠ ±6

55 − (p− 2 ) (p− 1 ) (p+ 1 )

(p+ 1 )

⋅

( )

2 p− 2 ( ) ( )

1

2 1 1

4( 4)

x x

−

⋅+ 3 (4−x)

13(x 4)

=+

Note: x≠ ±4

57 (6 −n) (6 +n)(n− 3) (n+ 3)

(n+ 3)

⋅

(n+ 6)

6 3

n n

−

=

− Note:n≠ − − 6, 3, 3

58 (7−y) (7+y) (y−5) (y+5)

y

y y

Note: 0, 220

w w

( )

2 2

−

=( )

2 5 1

w

2 1 5

1

w w w

+

= +

(x− 4) (x+ 5)

⋅(x− 9) (x+ 7)( 3)( 4)

Note: 7, 5, 2, 4, 9 ( 2)( 9)

=

+

66 (x+ 3) (x− 9) (x+ 7) (2x− 1)

69 ( ) ( )

2 2

(5 41 8) ( 9)( 2) (5 1)( 8) ( 9)( 2)

Note: 2 2

x x

x x

+ +

(5 6)( 2)

18 30

Note: , 2(5 6)( 2)

x x

1 3

3 1 (3 1)

Note: 0,

x x x

9 5

x

x x

2 2

1 1

1

1 ( 1)( 1) 1 ( 1)1

Note: 0, 1

x x x

( 7)( 7)

6Note: 1

1 1

pq f

++

x x

+

96 Cannot cancel the 1’s Must factor

first x+ 1(x+ 1)( )

1 1

+ +

103 The graph is as follows There is a

hole at x= −7 It agrees with Exercise 23

104 The graph is as follows There is a

hole at x=2 It agrees with Exercise 21

105 a

1 1

1 1

3 1

2 1

x y

y

x y =

8

216

2 8

2

8

216

77 False If a=3,b= then 4, 2 2

a +b =

25= while 5, a+ b = 3+ ≠ 2 5

78 False −4 is not a real number

79 Multiply the exponents to get a mnk 80 Multiply the exponents to get a

81

2

7 4(1)(12) 12(1) 2

2

2 2

+ +

( )

2 2

61 ( )

2 2

3, 1, 21

1

t

t t

(x+ 2) ( )x− 1

⋅

(x+ 3) (x− 2)

( )( )( )2

, 3,1,2 3

x x

x x x x

+

⋅ + +

( )( ) ( )

2 2 , 3, 2, 1, 0 3

1 4(5 15) 3

5 15

x

x x

+

2 2

3x 1

x

2, 0,

x

x x

3 13 6 2

16

44

x x

86 Simplifying the radical using the calculator gives the approximation 3.605551275 Since there is no discernable pattern, it seems that the number is irrational

87 1.945 10× −6 88 1.5625 10× 3

161

( 4)( 4)( 4) 4

++

+

b

c The graphs agree as long as x≠0

(x−4) (x+4)

⋅

(2x+3)

3 2

(2 3)( 4)

Note: 4, ,15( 15)

x x

5

1 3

N o te : , 3

x

x x

+

= − +

( )( )

2 2