Problem Set 68 Solutions Manual

24.(59)

29. x2 - 4x + 6 = 0(46)

In x + In x = In (4x - 3)

Inx2 = In (4x - 3)

x2 = 4x - 3

x2 - 4x + 3 = 0

(x - l)(x - 3) = 0

x = 1,3

x =4 ± ~(_4)2 - 4(1)(6)

2(1)

4 ± ..J16 - 24 4 ± 2.fii2 ± Pux = = =

2 2

(x - 2 - -!ii)(x - 2 + -!ii)

25. 3 log-, x = log- S! - log73(59) ;~j (80:)( l~~m )C ~O~m )( 2.~~n~m)

x C~~~.)(5~8~ft)

= (80)(1000)(100) mi = 49.71 mi(2.54)(5280)(12) hr hr

log7 x3 = log , C31)

x3 = 27

x = 3

26.(59)

2log2 3 _ 32log3 2 + 5log5 4+log5 6 _ 4 log 105

= 3 - 3log3 22 + 5log5 4(6) _ log 1020

= 3 - 4 + 24 - 20 = 3 1. C = mN + b(62)

Problem Set 68

{y ~ (x - 3)2 + 2

27. 2(56) Y > x

(a) {250 = m20 + b(b) 325 = m30 + b

3(a) 750 = m60 + 3b-2(b) -650 = -m60 - 2b

100 = b

(a) 250 = m20 + b250 = m20 + (100)

m20 = 150

m = 7.5

C = 7.5N + 100 = 7.5(50) + 100 = $475

(parabola)(parabola)

The region must be on or above the parabolaY = (x - 3)2 + 2 and above the parabolaY = x2

.

y

~\ 5\ 4\ 3\

~2 \ 2Y = x- \\ 1

~-4~'~+-+-~~-x

y = (x - 3)2 + 2

2. ~ = 126(55) 5!4!

(3,2)

-2-1 2 3 4

3. ~ = 60(55) 3!2!

28.(56)

vIIII

HII 40cmIII~ ~~o _~~ __ ~~

4. r =(53) OJ

15 mihr 15 mi-min_---2.!!!....,- =

400 rad 400 hrmin

r =

28cm

(15 mi-min)(~)(5280 ft)(12 in.)

400 hr 60 min 1 mi 1 ft

(15)(5280)(12) in. = 39.60 in.(60)(400)

r =H = 40 sin 45° = 28.2843 em

Area = !BH = !(28)(28.2843) = 395.98 cm22 2

r =

214 Advanced Mathematics, Second Edition

Solutions Manual

5. Average rate = total distance

(44) total time

m + s m + s= - ---

u + u + 2 2u + 2

Time = 100m + s2u + 2

100(2u + 2) hrm + s

6. Rate = w3

a2

stereos(44) g dollars

Rate x price = stereos

(W3a2 stereosJ(750 d 11) 750w

3a

2-- --- 0 ars = stereos

g dollars g

7. N, N + 1, N + 2, N + 3(7)

(N + 1)(N + 3) = 2(N)(N + 2) + 3

N2 + 4N + 3 = 2N2 + 4N + 3

N2 = 0

N=O

0,1,2,3

8. Directrix: y = k - p = -3(68)

Focus: (h, k + p) = -(0, 3)

k + P = 3k - p = -3

2k = 0

k = 0

Vertex: (h, k) = (0,0)

k + P = 3

(0) + P = 3

p = 3

1y - k = -(x - h)24p

1y - (0) = -[x - (0)]24(3)

1y = _x2

12

1Parabola: y = _x2

12Vertex: (0, 0)

Advanced Mathematics, Second Edition

Problem Set 68

9. Vertex: (h, k) = (0, 1)(68)

Focus: (h, k + p) = (0,3)

k + P = 3(1) + p = 3

p = 2Directrix: y = k - p = (1) - (2) = -1

Axis of symmetry: x = h = 0

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

1y - (1) = - [x - (0)]2

4(2)

1 2Y - 1 = -x81

y = _x2 + 18

1Parabola: y = _x2 + 1

8Directrix: y = -1

Axis of symmetry: x = 0

y

1 2 3 4 5

432

(0, 1)I I I I I I I I I I I •. x

-5-4-3-2-1

10. y = _1_x2(68) 100

14p

1 1-=-

100 4p

p = 25

a

The receiver should be placed 25 ft above the vertex.

11. (a) Approximately 68% of the scores lie between 74(61) and 82, which is within one standard deviation of

the mean.

(b) The scores are symmetrically distributed aroundthe mean, so 34% of the scores are between 78and 82.

215

Problem Set 68

12. STEM(61) 2

3

4

LEAF53,94,11,75,92,52

78,12,87,10,66,96,51

22,08,00

331 E.'3 01 = 2:::3. co::< = . ._1

... ..,. 1(1 t'1ed = 331 coIJ):: = t, ...:·• .._1

p·,in 21 1 0:;: = 3131 co= .._1

p·,a::( = 4·-'·-'LL

~I I I I I

211 283.5 331.5 391.5 422Min 01 Med 03 Max

13. y = log , x(65) 2

y

321

-1--1--t-4-+-+-+-~ x

14. Function = cos B(66)

Centerline = 3Amplitude = 3

Period = 2

Phase angle = 0

ffi . 2nCoe icient = - = n

2y = 3 + 3 cos nB

15. Function = - sin x(66)

Centerline = 1

Amplitude = 3

Period = 100°

Phase angle = 0°

C ffici 360° 18oe icient = -- =100° 5

3. 18y = 1 - sm-x

5

16. (3 cis 80°)(5 cis 310°) = 15 cis 390°(64) = 15 cis 30° = 15(cos 30° + i sin 30°)

= 15(-J3 + i!) = 15.,j3 + 15 i2 2 2 2

216

Solutions Manual

17. x2 + i + lOx - 2y = 1(63)

(x2 + lOx + ) + (y2 - 2y + ) = 1

(x2 + lOx + 25) + (y2 - 2y + 1) = 27

(x + 5)2 + (y - 1)2 = (.fi7)2

Center = (-5,1); radius = .fi7

y

-11-HH---4---4-+-+-+-+-+-+-f--+---X

18. (a) {ax - dy = s(62) (b) bx + wy = k

w(a) wax - wdy = wsd(b) bdx + wdy = kd

x(wa + bd) = ws + kd

ws + kdx =

wa + bd

{3X - 5y 10

19. 2(66) 4x + 2y = 2

a = 3, b = 4, d = 5, w = 2, s = 10, k = 22ws + kd

x =wa + bd(2)(10) + (22)(5) 20 + 110 130 = 5= = =(2)(3) + (4)(5) 6 + 20 26

20. (tan x)(2 sin x + 1) = 0(60)

tan x = 0 2 sin x + 1 = 0

x = O,n 2 sin x = -1

sinx = 12

'lr: l lx6'6x =

7n 11nx = 0,n'6'6

Advanced Mathematics, Second Edition

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 40 = -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 (- 3n) + tan34n _ sin2 (_ 3n)~~ 2 2

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

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

= sec2 150° - tan2 150°

= (-sec 30°)2 - (-tan 30°)2

(-~r-(-~r4 _ ! = 1"3 3

26. 2 In (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.(59)

3 Iog2 5 - Iog2 x = Iog2 x. 3

Iog2 5 = Iog2 x + Iog2 x

Iog2 125 = Iog2 x2

125 = x2

x = 5-JS

28. e3 In5 _ 1OIog3+2 log2 _ In e-3

(59)eln 53 _ 1OIog3+log 22 _ (-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

. 2 1 ft2 211.491 Ill. x 2 = 0.080 ft

144 in.

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

(b) antilog j d = 34 = 81

Problem Set 69

1. (a) {RGTG = DG(38) (b) RBTB = DB

TG = 2 + TB, TG + TB = 6

TG + TB = 6

(2 + TB) + TB = 6

2TB = 4

TB = 2

TG = 2 + TB = 2 + (2) = 4

(a) RGTG = DG

RG(4) = 160

RG = 40mph

(b) RBTB = DB

RB(2) = 160

RB = 80mph

217