Solutions Manual Problem Set 34

(a) y + Z = 652

y + z = 130y = 130 - z

2. RET = concoctions(25)

concoctionsR =

ET

20(3)(5)

4 concoctions3 eater-hrR = =

Use substitution to solve.RET = concoctions

(b) z-Y=242

z - y = 48

z - (130 - z) = 48

2z = 178

z = 89

T= concoctionsRE

T=~=21hr(~}5) 10

y = 130 - zy = 130 - (89)

y = 41

3. RsT + RBT = jobs(25)

(~)T + (~)T = 10

29. ~ = 4(1) b

a = 4b

!2T= 1030

300 hrT = 17

A and B are equal. Therefore the answer is C.$120

4. B = -- = $7.50 per call(18) 16

Ix!~cA

30.(1J

3 3A = - B = - ($7.50) = $11.25 per call2 2

12A = 12($11.25) = $135

Since AB = BC, mLBAC = mLBCA = yO.

XO = mLBAC - mLDAC

= yO _ mLDAC

7

5. L 3 = 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21(34) i=l

XO < yO

Therefore the answer is B.2 3j 3(0) 3(1) 3(2)

6. L 1 - 2j = 1 - 2(0) + 1 - 2(1) + 1 - 2(2)(34) }=o1 3 9

= -+-+-1 -1 -3

= 1 - 3 - 3 = -5Problem Set 34

1. Distance = m(28)

Rate = z 7.(34)

From the graph of the line we pick two orderedpairs: (70, 850) and (90, 700).

m = 850 - 700 = -7.570 - 90

Mo = -7.5Zr + b700 = -7.5(90) + b

b = 1375

Mo = -7.5Zr + 1375

Time = mZ

New distance = m

New time = m _ 3 = m - 3zz z

m _New rate = m - 3z -

z

mz mi---

m - 3z hr

Advanced Mathematics, Second Edition 113

Problem Set 34Solutions Manual

8. From the graph of the line we pick two ordered(34) pairs: (102, 100) and (105, 120).

m = 120 - 100 = 6.67105 - 102

H = 6.67C + b100 = 6.67(102) + b

b = -580

H = 6.67C - 580

(b) y

a

(3~j I(x) = -fX tan ( Arcsin ~) = tan () = !g(x) = x + 1

(c) y

10.(33)

A property of a rectangle is that it has 4 right angles.The square also has 4 right angles. However, we arenot told anything about the sides or diagonals.Therefore the answer is A.

-+-r"T"T--- x

-3

111. Median = -(Bl + B2)(33) 2

3x + 5 = .!.[(2x - 1) + (6x - 9)]2

13x + 5 = -(8x - 10)

23x + 5 = 4x - 5

x = 10

-90° < () < 90°

1cos [Arctan (-3)] = cos () = .JlO = M10

12. 1 5 14. (a) x axis, yes.y = -x - -(31)

9x2 + (_y)2 = 9 is equivalent to 9x2 + y2 = 9.(32) 3 61 5 y axis, yes.--x = -y - -3 6

9(-x)2 + y2 = 9 is equivalent to 9x2 + y2 = 9.x = 3y +

5Origin, yes.-

29(-x)2 + (_y)2 = 9 is equivalent to

Interchange x and y.9x2 + y2 = 9.

y = 3x + 5 (b) x axis, no.-2

-y = x3 is not equivalent to y = x3.

y axis, no.13. (a) y y = (_x)3 is not equivalent to y = x3.(32)

Origin, yes.-y = (_x)3 is equivalent to y = x3.

x

15. Replace y with y + 3.(31)

-90° < () < 90° y + 3 = IxlArctan 1 = 45° y = Ixl - 3

114Advanced Mathematics, Second Edition

Solutions Manual Problem Set 34

16. g(x) = x2 - 2(31)

119. log2 - = a(26) 32

2a = J.32

2a = _125

a = -5

y

I \ I I I! I • x

20. log , P = -2(26) "5

17. -3/-135° - 2/-140°(30)

= -3/-135° + (-2/-140°)

y

P = (~r2=15-2 = 25

y

,I,I

/

21. (a) Not a function(21)

(b) Not a function

(c) Function, 1 to 1(d) Function, not 1 to 1/

/

/

DB

>tV'. I I ••• X .,v ". - I I •• X

A = 3 cas 45° = 2.1213

B = 3 sin 45° = 2.1213C = 2 cas 40° = 1.5321

D = 2 sin 40° = 1.2856

22. x + 11 ~ 0(21)

x ~ -11

x2 - x - 12 i' 0

(x - 4 )(x + 3) i' 0x i' 4,-3

{x E IR I x ~ -11, x i' 4, -3 }Resultant = (2.1213 + 1.5321) i

+ (2.1213 + 1.2856)J

= 3.6534 i + 3.4069 Jy 241

~-v3

cas 60° = !2

sin 30° = ~2

23.(24,29)

3.41sin 90° = 1

cas 90° = 03 cas 60° sin 30° - sin 90° cas 90°

V)- I I •• X

R = ~(3.6534)2 + (3.4069)2 = 5 = 3 (~)(~) - (1)(0) = ~tan () = 3.4069

3.6534() = 43° 24. (a) (fg)(x) = (3x - 5)(2x2) = 6x3 - 1Ox2

(24)(fg)(2) = 6(2)3 - 10(2)2 = 48 - 40 = 8

(b) (gf)(x) = (2x2)(3x - 5) = 6x3 - 1Ox2

(gf)(2) = 6(2)3 - 10(2)2 = 48 - 40 = 8

(c) (f ° g)(x) = 3(2x2) - 5 = 6x2 - 5

(f ° g)(2) = 6(2)2 - 5 = 24 - 5 = 19

Resultant = 5/43°, so

Equilibrant = 5 /223°

18. log, 125 = 3(26)

b3 = 125

1

b = 1253 = 5

Advanced Mathematics, Second Edition 115

Problem Set 35 Solutions Manual

25. First we label the equations.(19)

~h 04..[3

3-.J3

I = ~(3.J3)2 + (4.J3)2 = 5.J3

(a) {x2 + i = 10(b) 2x2 - i = 17

Next we use elimination to solve.

(a) x2 + y2 = 10

(b) 2x2 - y2 = 17

3x2 = 27

x2 = 9x = ±3

Asurface = Abase + 6Aside

= 6[~(6)(3.J3)] + 6[~(6)(5.J3)]

= 144-J3 cm2

(a) x2 + y2 = 10

y2 = 10 _ x2

y2 = 10 _ (±3)2

y2 = 1

y = ±1

(3,1), (3, -1), (-3, 1), (-3, -1)

29. x + 2y = ~ + 2y = 1 + 2y(1) x x x x

A and B are equal. Therefore the answer is C.

a e30.

b d(1)

ad be

o = be - ad126. f(x + h) - f(x) = -- - -

(21) x + h xa + e a (a + e)b - a(b + d)----b + d b b(b + d)

ab + be - ab - ad be - adb(b + d) b(b + d)

= 0x - (x + h)

x(x + h)-h

27.(15)

Therefore the answer is C.B

Problem Set 35

c 1. First we write three equations.(18)

(a) {4NW = 9NR + 10(b) NB = ~

NR 1(c) N B + N R + N W = 65

A D

STATEMENTS REASONS

1. BD bisects AC2. AD = CD

1. Given2. A bisector divides a segment

into two congruentsegments.3. Given4. Reflexive axiom5. SSS congruency postulate6. CPCTC

3. AB = CB4. BD = BD5. f'lABD = f'lCBD6. LA = LC

(b) NB 3NR 1

NB = 3NR

28.

~

(2)

v= ~(Abase)(h)

216 = ~ [6(~)c6)(3.J3)Jch)

h = 4.J3

116

(a) 4Nw = 9NR + 10

9 5Nw = -NR + -

4 2

Now we can use substitution to solve.

(c) NB + NR + Nw = 65

(3NR) + NR + (~NR + %) = 65

25 NR = 1254 2

NR = 10

Advanced Mathematics, Second Edition