Solutions Manual

21. cos 0 - cos 0 tan 0 = 0(60) cos 0(1 - tan 0) = 0

COS 0 = 0o = 90°,270°

I-tanO=O

tan 0 = 1o = 45°,225°

Since tan 90° and tan 270° are undefined, theexpression does not make sense when 0 = 90° oro = 270°.

o = 45°,225°

22. sin 40 + 1 = 0(52)

sin 48 = -1

40 = 270°,630°,990°,1350°

o = 67.5°,157.5°,247.5°,337.5°

23. -90° < 0 < 90°~~

Arctan [tan (-120°)] = Arctan (tan 60°) = 60°

24. csc2 (- 37r) + tan347r _ sin2 (_ 37r)~~ 2 2

= (csc ~r+ (tan 0)3 - (sin ~r= (1)2 + (0)3 - (1)2 = 0

25. (f - g)(5100) = sec2 (510°) - tan2 (510°)(24,48)

= sec2 150° - tan2 150°

= (-sec 30°)2 - (-tan 30°)2

(-~r-(-~r43

~ = 13

26. 21n (x - 1) = In (3x - 5)(59)

In (x - 1)2 = In (3x - 5)

(x - 1)2 = 3x - 5

x2 - 2x + 1 = 3x - 5

x2 - 5x + 6 = 0

(x - 2)(x - 3) = 0

x = 2,3

Advanced Mathematics, Second Edition

Problem Set 69

27. 3 log2 5 - log2 x = log- x(59) . 3

log2 5 = log2 x + log2 x

log2 125 = Iog2 x2

125 = x2

x = 5-JS

28. e3 In5 _ 1OIog3+2 log 2 _ In e-3(59)

eln53 _ 1OIog3+log22 - (-3)

eln 125 _ 1OIog(3)(4) + 3

125 - 12 + 3 = 116

29.(53,56)

5 in.

H = 6 sin 50° = 4.5963 in.

Area = !(5)(4.5963) = 11.491 in.22

11.491 in.2 x 1 ft2

~ = 0.080 ft2144 in.

30. (a) antilog, 3 = 53 = 125(67)

(b) antilog j d = 34 = 81

Problem Set 69

1. (a) {ReTe = Dc(38) (b) RBTB = DB

Te = 2 + TB, Te + TB = 6

Te + TB = 6

(2 + TB) + TB = 6

2TB = 4

TB = 2To = 2 + TB = 2 + (2) = 4

(a) ReTe = Dc

Re(4) = 160

Re = 40mph

(b) RBTB = DB

RB(2) = 160

RB = 80mph

217

Problem Set 69 Solutions Manual

.J;;;i2. p = --

(18) X2

Directrix: y = k - p

y = 0 - (-1~)3

y = 10Axis of symmetry: x = h = 0

y _ k = _1 (x _ h)24p

1 2Y - 0 = ( 3) (x - 0)

4--10

5 2Y = --x6

=2.J;;;4y2 .J;;;y2

x2 = 32 -r = 32p

4

Therefore, p is multiplied by 32.

3. 4(90 - A) + 40 = 2(180 - A)(1)

360 - 4A + 40 = 360 - 2A

2A = 40

A = 20°

4. Distance = x(28)

Time = kNew time = k - 20

5Parabola: y = - - x26

N • distanceew rate =

new timex mi

k - 20 min

D.. 3irectrix. y = -10Axis of symmetry: x = 0

y

5.(18)

s + 1.2s + (1.2)[(1.2)s] = 29,120

3.64s = 29,120

s = $8000--4-.,.........---+---- X

(6~j 1-:~I(-4)(2) - (5)(6) = -8 - 30 = -38

(6~j I ; x ~ 11 = 4

x(x - 1) - (3)(2) = 4

x2 - x - 10 = 0

9. Vertex: (h, k) = (-3,2)(68)

Focus: (h, k + p) = (-3, -1)

k + P = -1(2) + p = -1

P = -3Directrix: y = k - p = (2) - (-3) = 5x =

-(-1) ± ~(_1)2 - 4(1)(-10)=

2(1) Axis of symmetry: x = h = -3

1 ± -J4i2

8. Vertex: (h, k) = (0,0)(68)

1 2Y - k = -(x - h)4p

Y - 2 _1_ [x - (-3)]24(-3)

y = - ~ (x + 3)2 + 212

Focus: (h, k + p) = (0 -~)'10

k+p=310

310

310

Directrix: y = 5

Axis of symmetry: x = -3

Parabola: y = -~(x + 3)2 + 212

O+p=

p =

218 Advanced Mathematics, Second Edition

Solutions Manual

10 M - -2 + 3 - 4 + 5 - 1 + 0 - 0 17. ean- -.(61) 6

. -1 + 0Median = = - 0.5

2

Mode: All the numbers appear only once. Thereforethe mode does not exist.

Variance

~ [(-2 - ~r+ (3 - ~r+ (-4 - ~r+ (5 - ~r+ (-1 - ~r+ (0 - ~r]

= 9.14

Standard deviation = .)9.14 = 3.02

11. Mean = 0, a = 1(61)

In a normal distribution, 68% of the data would liebetween -1 and 1, or within a of the mean. Since thedata is evenly distributed, 34% lie between -1 and O.Likewise, 95% of the data would lie between -2 and2, or within 20' of the mean. So, 47.5% lie betweeno and 2. Therefore, the percentage between -1 and 2,is 34% + 47.5% = 81.5%.

12. y = logs x(65) 3"

y

321

-1I I f I I • x

13. Function = cos x(66)

Centerline = 5

Amplitude = 4

Period = n

Phase angle =3n8

Coefficient = 2n = 2n

y = 5 + 4cos2(X + 3:)Advanced Mathematics, Second Edition

Problem Set 69

14. Function = sin ()(66)

Centerline = -3

Amplitude = 10

Period = 120°

Phase angle = 20°

360°Coefficient = -- = 3

120°y = -3 + 10 sin 3«() - 20°)

15. (6 cis 215°)(2 cis 205°) = 12 cis 420°

(64) = 12 cis 60° = 12(cos 60° + i sin 60°)

=12(±+i~)=6+6F3i

16. (a) 6 - 2i(64)

y

Ie::: ,~ I I • X

2

R = ~62 + (_2)2 =

26

() = 18.43°

2M = 6.32

tan ()

The polar angle is 360° - 18.43° = 341.57°.

6.32 cis 341.57°

(b) 5 cis (_ 1~n ) = 5 cis ( _ 5: )

5( n ., n)= - cos "4 + I Sill "4

= 5[(- '7) + {'7)] =5-5 5-5.

--+-12 2

17.(62)

(a) {ax + by = c(b) px + qy = d

-pea) -pax - pby = -pc

a(b) pax + qay = ady(aq - pb) = ad - pc

ad - pcy =

aq - pb

219

Problem Set 69

18. {y ~ (x - 3)2 + 1(56) y < X

(parabola)

(line)

Solutions Manual

The area is beneath the line and on or above theparabola.

y y=x/

.;

21.(60)

(tan e - .J3)(tan e + .J3) = 0

tane-.J3=O tane+.J3=O

tan e = .J3 tan e = -.J3

54321

e ,. 4,.3'3

. e = 2,. 5,.3 '3

~IIII.; 1 2 • x)I 3 4

19. x - y + 3 = 0(58)

y = x + 3Equation of the perpendicular line:

y = -x + b(8) = -(-3) + b

b = 5

y = -x + 5

Point of intersection:

-x + 5 = x + 3

2x = 2x = 1

y = x + 3 = (1) + 3 = 4

(1,4) and (-3,8)

D = ~[1 - (_3)]2 + (4 - 8)2 f32 4.fi

e ,. 2,. 4,. 5,.3'3'3'3

20.(56)

22. sin x + 2 sin2 x = 0(60)

sinx(1 + 2 sin x) = 0

sin x = 0x = 0,,.

+ 2 sin x = 0

2sinx =-1

H = 60 sin 70° = 56.3816 em

Asegment = Asector - Atriangle

= (1l00)(,.)(60)2 _ (!)(60)(56.3816)360° 2

= 3455.76 - 1691.45 = 1764.31 cm2

220

sinx =12

7,. 11,.6'6x =

x = 0," 7,. 11,., ,-6 6

23.(60)

2 1tanx--=O3

(tan x - ~ )(tan x + ~) = 0

1tan x - .J3

1tan x + .J3o o

tan x1

.J3

,. 7,.6'6 x

tan x1

-.J3

5,. 11,.6'6x =

,. 5,. 7,. 11,.x------

- 6' 6 ' 6' 6

24. 0° ~ e ~ 180°(32)

Arccos (cos 210°) = Arccos (- ~) = 150°

25.(48)

cot2 (-510°) - csc3 (-510°) - tan2 (-510°)

= cot2 (210°) - csc ' (210°) - tan2 (210°)

= (cot 30°)2 - (- csc 30°)3 - (tan 30°)2

= (.J3)2 - (_2)3 - (~ r1 2 32

= 3 - (-8) - - = 10- = -3 3 3

Advanced Mathematics, Second Edition

Solutions Manual Problem Set 70

26. [x - (-J2 + i)][ x - (-J2 - i)] = 0(46)

(a) 20 = ml000 + b-(b) -30 = -m3000 - b

-10 = -m2000

m = 0.005

(x - -J2 - i)(x - -J2 + i) = 0

x2 - -J2x + ix - -J2x + 2 - -J2 i. r;;2. ·2 0- IX + -YL.I - I =

x2 - 2-fix + 3 = 0

(a) 20 = ml000 + b20 = (0.005)1000 + b

20 = 5 + bb = 15

S = 0.005A + 15

= 0.005(6000) + 15 = 45 squirrels

27.(56)

2. RJTJ + RMTM = jobs(25)

(5 jObS) ."4 ~ (6 hr) + RM(6 hr) = 9 Jobs

30 jobs + RM(6 hr) = 9 jobs4

IIII

:H0.5m I

III

55° _CI- - - --0.18 m

H = 0.5 sin 55° = 0.4096 m

112Area = -BH = -(0.18)(0.4096) = 0.037 m2 2

228. - log, 8 - 3 log, X = log, 4(59) 3

6 3 .RM( hr) = - Jobs

2

RM = .!. jobs4 hr

Mike would take 4 hr to complete one job.

2- 3log, 83 - log, x = log, 4

log, 4 - log, x3 = log, 4

log, 4 - log, 4 = log, x3

4 3log, - = log, x4

x3 = 1

x = 1

3. (a) {(B - W)Tu = Du(36) (b) (B + W)TD = DD

B = lOW, Tu = TD - 2

(b) (B + W)TD = DD

(lOW + W)TD = 264

llWTD = 264

WTD = 24

29. (a) antilog72.78 = 72.78 = 223.55(67)

(b) antilogj , 2.78 = 132.78 = 1249.58

30. (f ° g)(x) = In (log x)(34)

f(g(x)) = In (log x)f(x) = loxg(x) = logx

(a) (B - W)Tu = Du

(lOW - W)(TD - 2) = 180

9W(TD - 2) = 180

9WTD - 18W = 180

9(24) - 18W = 180

18W = 36

W = 2mphProblem Set 70

l.S=mA+b(62)

B = lOWB = 10(2)

B = 20mph(a) {20 = mlOOO + b(b) 30 = m3000 + b

Advanced Mathematics, Second Edition 221