Post on 01-Sep-2019
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