Problem Set 75 Solutions Manual

(b) 2ln e - 2 log 102 = 2(1) - 2(2) = -25. gC3 =

8!= 56 triangles

I (75) (8 - 3)!3!(c) 2 In e4 + 4ln e2 - log 105

= 2(4) + 4( ±) - 5 = 5 6. gC4 =8!

= 70(75) (8 - 4)!4!

30. x2 + 20 = 01-

3-;1(46)

x2 = -20 -4 -9 - 16 257. x = I~-;1= = --

(74) 9 - (-8) 17x = ±--/-20 = ±2-/5 i

(x - 2JSi)(x + 2JSi) I~-31-4 -12 - (-6) 6y = I~ = = --

-; 19 - (-8) 17

Problem Set 7S25 6x = - 17; y = --

171. e = 4.40 + 1.10(m - 3)(44) = 4.40 + 1.10m - 3.30

8.= $1.10 + $1.10m (73)

2.(44)

D DP - 20 p

Dp D(p - 20)(p - 20)p pep - 20)

20D dollarsp2 _ 20p student

Rafter - Rbefore =

=

km23. p =-

(18) ~

k(4m)2=-J9N

Therefore, p is multiplied by 16.3

4. Distance = (mileS )(gallOnSof gas)(25) gallon

DJ = (7 miles)(1 gallon) = 7 milesgallon

D2 = [; miles ](1 gallon)- gallon3

= 21 miles

D2 - DJ = 14 miles

240

() = .!. (3600) = 1802 10

A = 10 cos 18° = 9.511 emx = 10 sin 18° = 3.09 em

1Area~ = -(3.09)(9.511) = 14.6945 cm2

2

Area = (20)(14.6945) = 293.89em2

9.(73)

= 22.5°

A = = 4.828 emtan 22.5°

1Area~ = -(2)(4.828)2

Area = (16)(4.828)

Area = 77.25 em2

Radius = 4.83 em

= 4.828 cm2

Advanced Mathematics, Second Edition

Solutions Manual

10. SL = 2,5SS(5)

2SL = 5SS

SL = ~Ss 2

AL = (SL)2 = (SL)2 = (~)2 = 25As (Ssf Ss 2 4

11. A = 180° - 70° - 60°(72) A = 500

Problem Set 75

14.(68)

1y = __ x216

1 1_x2 = __ x24p 16

p = -4

Focus: (O,p) = (0,-4)

Directrix: y = -p = -(-4) = 4

Focus = (0, -4)Vertex = (0, 0)Directrix: y = 4

a

sin 50°8

sin 60°y

8 sin 50°sin 60°

a = 7.08

a = 1II±o:-!+I........t::III •• x

(0, -4)b 8

sin 70° sin 60°

b = 8 sin 70°sin 60°

b = 8.6815. y = 1 + log2 x(65)

12. 5x2 + 8y2 = 40(71)

y

x2 y2-+- = 18 5

4321Let x = 0

o 2- + L = 08 5

y = ±-J5

Let y = 0

x2 0- + - = 18 5

x=±-J8

-1 -t/G,o) 3 4 5 6 7-2

I I ~ I I I I I I I •• x

y

3+(0, -vs) 16. Function = cos x(66)

Centerline = 3Amplitude = 4Period = 540°

Phase angle = -90°

C ff 360° 2oe icient = -- =540° 3

2y = 3 + 4 cos - (x + 90°)

3

(\Is, 0)~ I I I I I ~ •• x

-3+(0, --vs)

\3 k \13. = 1

(69) k + 1 k

3(k) - (k + l)k = 1

3k - k2 - k = 1

k2 - 2k + 1 = 0

(k - 1)2 = 0

k = 1

17. Coefficient = 21r = !(66) 41t 2

y = 10 + 6 sin ~(x - ~)Advanced Mathematics, Second Edition 241

Problem Set 75 Solutions Manual

18. [5 cis (-120°)](2 cis 660°) = 10 cis 540°(64)

= 10 cis 180° = 10 (cos 180° + i sin 180°)

= 10(-1 + Oi) = -10 + Oi

22. sin2 8 - 4 sin 8 + 3 = 0(60)

(sin 8 - 3)(sin 8 - 1) = 0

sin 8 = 3 sin 8 = 1

8 = 90°no answer

19. (a) 5 + 6i(64)

2 14n 2 14n . 2 14n 2 14n23. cot - - csc - + Slfl - + cos -(48) 3 3 3 3

2 2n 2 2n . 2 2n 2 2n= cot - - csc - + sm - + cos -3 3 3 3

= (- cot ~ r - (csc ~ r+ (sin ~r+ (-cos ~r

y

R = ~52 + 62 = 7.81

tan9=~5

8 = 50.19°

7.81 cis 50.19°

143 1= ---+-+-=0

3 3 4 4

(b) -7 cis (-585°) = -7 cis (-225°)

= -7(-cos 45° + i sin 45°)3+6+9-2+0+0

24. Mean = = 2.67(61) 6

Median = 1.5

Mode = 0

Variance = .! [(3 - 2.67)2 + (6 - 2.67)26

+ (9 - 2.67)2 + (-2 - 2.67)2

+ (0 - 2.67)2 + (0 - 2.67)2]

= 14.56

Standard deviation = "'14.56 = 3.82

= -7(- ~ + ~ i)7..[i 7..[i.

= -- - --I2 2

20.(60)

(sin 8 - ~)( sin 8 + ~) = 0

-f3sin 8 = sin 8 =

28 = 60°, 120° 8 =

{y s x2 - 425.

(56) x2 + i :s; 9(parabola)(circle)

The region must be below or on the parabola andinside or on the circle.

y

21.(52)

2. 8sin "3 - 1 = 0

.81sm - =

3 2

8 = 30038 = 90° (0, -4)

242 Advanced Mathematics, Second Edition

Problem Set 76Solutions Manual

26. Equation of the perpendicular line, using the point-{58} slope formula:

1 3 1- - -(c) log 102 + In e2 - 6 In e3

1 3- -

= log 102 + In e 2 - In e2

1 3=-+--2=02 2

Y - YI = m(x - Xl)

1Y - 1 = -- (x - 1)

31 4

Y = --x + -3 3 29. (a) antilog, (-2) = 6-2 ={67}

136164

Point of intersection:(b) antilog- (-6) = r6 =1 4--x + -3 3

10-x =3

= 3x + 223

15

Y = 3x + 2 = 3 ( - ~) + 2 = ~

30. 2(x2 - 1x + 3) = 0{46} 2

x =% ± ~¥- 4(1)(3) = 1+ -J23i

24- 4

2(x - ~ - ~ i)( x - ~ + ~ i)x =

(-~, ~) and (1, 1)

D = .{l + ~r+ (1- ~r= ~ 36 + 4 = /40 = 2M

25 25 ~2s 5

Problem Set 76

1 c 1O! 45 .• 10 8 = - = comrmttees{75} 8!2!

10!2. lOC6 = - = 210 groups{75} 6!4!

27.{59}

1- In 8 + 2ln x = In (-3x + 2)3

1

In 83 + lnx2 = In (-3x + 2)

In 2x2 = In (-3x + 2)

2x2 + 3x - 2 = 0

3. 1.2P = 480{R} P = 400

1.15P = 1.15(400) = $46076(460) = $34,960

(2x - 1)(x + 2) = 0

1x = - -2

2'

12

(x :;:. -2)4. (a) {MN = RN + 1

{25} (b) 3(MN - 8) = 2(RN - 8) + 6

(b) 3(MN - 8) = 2(RN - 8) + 6

3MN - 24 = 2RN - 16 + 6

3MN - 2RN = 14

3(RN + 1) - 2RN = 14

RN = 11

(a) MN = RN + 1 = (11) + 1 = 12

MN + 17 = M = 29 yr

RN + 17 = R = 28 yr

x =

28. (a) h310gh 2-logh 2-logh 6{59}

= hlogh23_logh 2-logh 6

8 2= h

10gh(2)(6) = h

1ogh"3 23

1

(b) 5 log 103 + 2 log 102 - In e2

= log 1015 + log 10 - In e2

= 15 + 1 - 2 = 145. Rate = distance

{28} ti =lIne

y yards-----m - 15 minutes

Advanced Mathematics, Second Edition 243