Algebra 2
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Algebra 2
An Incremental Development
THIRD EDITION
SAXON
Algebra 2
Solutions Manual
Third Edition
Acknowledgments
Saxon Publishers thanks those staff members who helped this book reach its final form.
Editorial @ Brian E. Rice
Rodney Clint Keele
Andrew C. Kershen
Sean G. Douglas
Chris Davey
Candice L. Holcombe
Production Eric S. Atkins
Brenda M. Bell
Jack W. Day
Nick Key
Tonea D. Morrow
Nancy J. Rimassa
Lois A. Rossman
Debra Sullivan
Eric A. Sullivan
HO
HHH
HH
$6666
Darlene Terry
© 2003 Saxon Publishers, Inc.
All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in
any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written
permission of the publisher.
Printed in the United States of America
ISBN: 1-56577-143-5
P2545 6-7-8 862: 13 12711410509 03-07-06
Preface
This manual contains solutions to every problem in the third edition of John Saxon’s Algebra 2
textbook. The solutions are designed to be representative of students’ work, but please keep in
mind that many problems will have more than one correct solution. We have attempted to stay
as close as possible to the methods and procedures outlined in the textbook. Early solutions of
problems of a particular type contain every step. Later solutions omit simpler steps. The final
answers are set in boldface for ease of grading.
Digitized by the Internet Archive
in 2022 with funding trom
Kahle/Austin Foundation
https://archive.org/details/algebra20000mcco
Problem Set A
PROBLEM SET A 12. -3? - 3 - (3)
1. y + 65 = 180 : ei er ioe is og)
= 180 - 65 = 115
y 13. 362 = ole CD) eee)
2. x + 40 = 90 = -3(1) - [-5(-2) + 3(-6)]
x = 90 — 40 = 50 = -3 - [10 - 18] = -3 —-[-8] = 5
3. x + 89 = 180 1h el ee
x = 180 — 89 ===4 8 216 = 30
pee hD|
Since vertical angles are equal, ce aed ra tale Ey ao AC Se IE
= 2,—69-- 8 = 4
y == $9
p=91 16) 1-30). 3). 27 1-6 = 3 |eeea
4. z+ 100 = 180 = -|-9|-4=-9-4=-13
z = 180 — 100 = 80 175) ee ie ee
Since vertical angles are equal, = -16
a = 80
ee a) 18... =3[=1-= 2(-1. ~ 1)|[=3(-2),— 1]
4y = 100 Sen3 ie 2-2)(6 — 1] = —3|-10 4 46-1
eee = -3[3][5] = -45
5. Angle + Supplement = 180° 19. =3[-3(-4/5,)) =3 —4)]
Angle + 40° = 180° = =S3l3(-3)— (-2)|
Angle = 180° — 40° = -3[15 + 7] = —3[22] = -66
Angle = 140°
20. -2[(-3 + 1) — (-2 — 2)(-1 + 3)]
6. Angle + Complement = 90° -2[(-2) — (-4)(2)] = -2[-2 + 8]
Angle + 40° = 90° -2[6] = -12
Angle = 90° — 40°
Angle = 50° 21. 2|-2-- et? |(El2r= —218ns 81C2)
—2[0](—2) = 0
Fo 88. CD ee 2 = 0
DD Ee BS (Ok
8 3--2))=3-2=5 23g 9 1446
#2 = -13
90 <9 = 3-2 oy 9S * 7) 2a al o|Seta) = = 41517}
20 ee 36-4}
a 5(2) = ~{-[35]} 4.35
=F = €1) 2010)
eo Wie Ua 24, -5 — |agi2
= GLB
4]
si = [27| = 9 = 3
6. 1265 4 Ge 3)
Il 35 =: Pa gaara24
= -[-2(-3) - (-5)]
= -(6 + 5] = -11 25, 3(-2 + SP-C 3) = 4-2|
11. 2 + C8)
2 4260)2!= 307) = AC)=e
ap 10 | S9 44ye
Algebra 2, Third Edition
Problem Set B
—5.—\(—2) + 8 — 465) 1
26. 6. Vpyramid = 3 YPrism
6 4(-3)
_-5+2+8-20 -15 5
= $lABase x height]
_ 6 + 12 ras 6
PAL En[I-3 = 4,= 5| - 2 Cn] = | (0x4) cm? x 10 om |
32
= (-2)[|-12| —- 8+ 1] = (-2)[12 - 8 + 1]
= —2[5] = -10 - (120) cm? = 40cm?
Soto
Oe (=5)
28.
7(|-3
+ 4I) 7. ABase = jana) + 5x2)? |m?
= Sige peeree
= (12 + 22) m* ~ 18.28 m*
7(1) u
Veylinder = ABase X height
29. 4(-2)[-7 - 3)(5 - 2)2] = -8[-4)G)@)] = (12 + 22)m? x 8m
= -8[-24] = 192 = (96 + 162) m* ~ 146.24 m?
30. Ae (4) = 5G = 1) 4 342)
2
=4 +44 — 5(2) + 3(4)\-8) 8. Vsphere = 3 Cylinder
=4+4- 10 —- 96 = -98 pi) ;
= 3 ABase x height]
PROBLEM SET B = S[n06? cm? x 12cm]
Ashaded = Agquare — Acircle
= [(8)(8) — 2(4)*] m? = $(362)(12) cm? ~ 904.32 cm?
= (64 — 162) m2 ~ 13.76 m2
S.A. = 4nr? = 4n(6)? cm? = 452.16 em?
Aghaded = ATotal — ANot Shaded
= |5120c9) |m2
Area of 72° sector fees (Acircle)
360
72 anh2
E ec ? +(3\6) 59x) |m2 atu
= — |7(10)*] em
= 60m? — 39m? = 21 m2 72
= 36 (100) cm? = 62.8 cm?
Ashaded = ACircle — ATriangle
= Go. an 5(168)| cm? 10. Perimeter = [Fxm2) +16 + 4|yd
= (642 — 64) cm? = 136.96 cm? + 520129] yd
Perimeter = |52xm2) + 14|m
= (47 + 20) yd =~ 32.56 yd
= (27 + 14)m =~ 20.28m z+ 70 = 180
z = 180 - 70
5 40 2 = 110
Area of 40° sector = 560 “Acircie)
Since vertical angles are equal,
_= tO re
360 UO) ]m 2x = 70
LS S5
40 2 2
= —(25
360° 7) m = §.7 2m
y = 110
Algebra 2, Third Edition
Practice Set 1
12. 5A + 40 = 90
5A = 90 — 40 28. a5] = 2y6 = 43 = 16191)
5 — |(4)(-3)|
5A = 50
ete dae oP eae)
Ae ae 10 c: 5 — 19)
5 — 5 +2+6-0 8
13. 2B + 140 = 180 5 wi tel
2B = 180 - 140
2B = 40 29. ao = (G3 22) 2)
=4(2s = 3)(F2)
jens sy
2 eee
—4(1)(-2)
14. Angle + Complement = 90°
= uv = 0
Angle + 10° = 90° 8
Angle = 90° — 10° = 80°
30. £9(3 454 6) 212"2), ee 5
a5. Angle + Supplement = 180° = 25) — 4-2) 6
Angle + 60° = 180°
=10+8+6-5=19
Angle = 180° — 60° = 120°
ooF 32 (3)? 2 PRACTICE SET 1
= A Be de) 418
a. mZA = mZC because they are angles opposite
2-8
| 24) 4) = H4-284
44 = 4 equal sides.
mZC = 35r
23) es = 8 9 = 15
mZB = 180° — mZA — mZC
-4 — (-3)3 - 27 + |-4| mZB = 180° —"35° —.345°
=-4+27-4+4=23 mZB = 110°
2374 6) 9-22-13 i x = 180 - 145
c= 35
oe eee ope 4s 3) 8
= FAl-7) 2 5 43 "= 28 5+ 3 =.26 y = 180 - x - 40
y = 180 - 35 - 40
~2[-1 - (-5)] - [-6(-2) + 3] y = 105
= 5-1 2°) — 412. 4-3)
= —2[4] = [15] = -8 - 15 = -23 c. A = 180 - 130
oo ee) — | Blac 2A 8 P= Daw —-16 ne 50
Caeh
723-4
8 2421) = 4 Cal
= 32. — 4 = =]
2B = 180 -C
25. -|-2 - 3 - 4] - |-2| = -[-9| - |2I (180 -50)
= -~9 —— = -11 — &
B = 65
26. staan ac Ral Seil 2)Ie)
6 — 4(-3
) d.
SOS
Nn
27. (-2)[|-3 + 4 - 5] - 22 - -1)] an
NnIa
n| — Nn
= (-2)[|-4| — 8 + 1] = (-2)[4 - 8 + 1]
= (-2)(-3) = 6 Oussoo
|
t=
©):
Algebra 2, Third Edition
Problem Set 1
1
PROBLEM SETI Awhite Triangles = |505065) +95 GS)
1. Since angles opposite equal sides are equal angles,
1
ae 4ax3)| cm 2
Reeds
2
y + 45 + 45 = 180 = 25cm
y = 180 - 90 = 90 Agshaded = ATotal — Awhite Triangles
= (24 + 25m) cm? — 25 cm?
2. Since angles opposite equal sides are equal angles, = (250 — 1) em2 ~ 77.5 cm”
aro:
1
y + 55 + 55 = 180 7. Perimeter = [Soncms) + 48)ft
y = 180 - 110 = 70 = (4m + 48) ft = 60.56 ft
S202 + 70 = 180 8 Veylinder = ABpase X height
ORM il) = [n(4)* ft? x 8 ft]
Gi=s55 = 1287 ft? ~ 401.92 fe
Since lines are parallel, B = 110 and A = 70. Vv. 2
Sphere — B Cylinder
402% SFi==3 2 4 3
5 = 3 (1282) ft? = 267.95 ft
SF = =
2
cate 9. p + 30 = 180
3.x SF =x p = 180 — 30 = 150
a($) = 2 dy.4 30h 000
5 2y = 90 - 30
= ry y= 30
Since vertical angles are equal angles, x = 30.
pe al Se
5. Atop Shaded = [5a(4) “s 58x14), aul 10. p + 60 = 180
= (87 es 16) cm2 p= 180 — 60 = 120
Since vertical angles are equal angles,
Apsotiom
ottom Shaded
Shaded = o>[a(4)?] enn”
360 10x = c 60
= 6
60 ; “
= 36000 cm 4y = 120
= 30
Aghaded = ATop Shaded + ABottom Shaded
Suse ie) cme 60° Gn) cue 11. Angle + Complement = 90°
® Angle + 17° = 90°
Angle = 90° — 17° = 73°
7 [x a on = 16cm? = 17.49 cm?
1 '
125 Vee 3 ABase x height)
6. Atotal = [58%6) i Sm) i
1 = 5x0 + 16) m2 x 7m
+ —n(4)* + 5#3) |em? 3,2
Vl
: 1/9
= (24 + 252) cm? ra AS + 30 (7)m?> ~ 102.97 m?
Algebra 2, Third Edition
Problem Set 1
13. r= ;side of square -|-4| - (-3) + 7 - 6(4 - |7 - 11)
7 - (3)(-2)|
Asquare er
ape) _ -4+34+7-
6(4 - |-4/)
Sar . 7—
|-6|
_ =44:3+7-6(4
- 4)
Asquare =a (4r)? = 16r? cd T =6
14. -[-2(-3 - 2) - (2 - 3)] = -[-2-5) - -5)] eee St Beek)
1
= -(10 + 5] = -[15] = -15
peed ete D) 8. eG
15. -2[-2 - 3(-2 - 2)][-2--4) - 3] 1 1
= -2[-2 - 3-4)][-2--4) - 3] 25. 3 = 2)(75 3 2) ee 3)"
= -—2[-2 + 12][8 - 3] = -2[10][5] = -100
= =(65)(F75 = |95)|)' Seer PG) a *5)kaS
= 5-12) - 9 = -60 - 9 = -69
16. ae [2 + 3-2)] =. 2"|
= 4 — §[-2 — 6] — (28) = 4*Rejegj2 8 26. (-3)[|-2 = 7 = 2| — C3 =D)
=-4+ 64-8 = 52 = =3[|-11)°9'+ 2] ==3(1f = 97+ 2]
= -—3[4] = -12
17. -3 — 2 - 4 - |-2 - 3(2)|
=-3-8-16-|2-6| =4 — (-3)-+ 7 = .6(2)
27.
= -3 - 8 — 16 — |-8| TESS) =Z)
= -3- 8 - 16-8
= -35 We 4 tt 1 le a 0 ee 0)
7+6 13 13
18. {ia ary Ser bl
= -{-2(4 + 2)[-3-1)]} = -{-2161(3)} 3 5 2? — ei3 eB)
= -{-36} = 36 = 3-5 — 4 - 16(-1)(-3- |-7| - 3)
= 3-5 - 4 - 16(-1)(-3- 7 - 3)
19. Mica Peat en, 4-3) eid 3-5-4 — 16(-1)-13)
mee Fer Rie (21) <3 = 3-5-4 - 208 = -214
seer 814 27. 5 87
29. 28 s(S3\(E2)2la Ee) Sem 292)
(4) = 4|-2)en?? 4 tl = 4 = 484. (-3\(4)s-e 7a)
= 2j6 — 40) — § +7 |=215| = 48 — 12 "7 4:12 =215
= =16= 8 =8 + 15 ==17
30. 6(-3)[415 - 4)(6 — 2)3] = -18[-1(4)3)]
21. 6 — {[3? - 8 + (-2)][-@ - 6-3)? + 2]2} = -18[-12] = 216
= 6 — {[9 - 8 — 2)[--2)9) + 2]2}
= 6 — {[9 - 8 — 2][18 + 2]2}
= 6 — {{-1][20]2} = 6 — {-40}
= 6+ 40 = 46
22. -[+-2)] - |-4 - 3/27 - 4
= -(2] - |-7]4 -4 =-2- 74) - 4
= -2 - 28 -4 = -34
23. (Jp 7-3 — 2) + 2Y]
= —3[-5(-5) + 4] = -3[25 + 4]
= —3[29] = -87
Algebra 2, Third Edition
Problem Set 2
Since lines are parallel,
PRACTICE SET 2
2C
= 4A
OF seee a1 1
ea 1
ig a ea 2C
= 80
C = 40
1 1
b 24)" = = — = —-— 5
=te4 ye —(16) 16 A + 60 = 180
A = 180 — 60 = 120
Since vertical angles are equal angles,
Viva VX VY XS VEX 2B = 60
fae ee B = 30
x'y* x
3C = 120
d. A = ar? = 49cm C = 40
r2 = 49 cm?
RS 1 ein 6. Agguare = s* = 16cm?
C = 2ar = 22(7 cm) = 147cm GS AIC cm” = 4cm
1 1
r= 7 ) = —(4cm)
—(s) a ) = lcm
PROBLEM SET 2
Acirele = Hr? = 3.14(1 cm)* = 3.14 em?
1. 4x SF = 6
oe aS2 ‘ Veylinder = Agase X height
4 2502 cm? = [2(5)*] cm* x height
3x SF=x 2502 cm? = 25mcm? x height
ax 2 =X
2502 cm?
height 7 = 10 cm
2 257 cm
9
Vote
‘ 8. 1 |
Voone = 3 X Apase X height
2. Since angles opposite equal sides are equal angles,
c= 238. - 4]
Sey
502) + (2)(2)
y + 38 + 38 = 180 ip sme |m? x 4m
y = 180 — 38 — 38 = 104
= As a 2 \a)m? =~ 8.76 m?
3. Vcylinder = Apase X height 3
60 ) 1 2
= Ex
|—(7(4ON eGo
+ —(4)(7 )|m x 8 m é xx2 (xy ee xx7 x oy —2 ote
xx G28 x2 x 510 ay
60 3 3
= |—~(16z) + 14((8)m- = 178.99m <3 xoy 12
360
i 2 p?(m aya m2 p°m ope
4. Since vertical angles are equal angles, 10.
m he p i(m = pe): mp ie. an ou
4A = 80
—2 2
A = 20 a ees mp3
m'1p
2B + 80 = 180
2B = 100 i
(x*y) xy ais xo yxy = Thay xy!
B = 50 ; x
We 2
(y aE
As Se yA
x*y x?y 6
Algebra 2, Third Edition
Practice Set 3
12. (a*b°)? ab? = a*b° ab? ? = ab ‘Igy. xy aye = Brey rye = xT?
a*b~ (ab)? — a2 b-2g2p-8 > g48 Gees. tye ia
= ab® ~ 1
xy
eer
13, ea ne ae xm zene al” ae
m5y3
= 2
24. _— = _23 = -§8
* De
14. Lod = cd = cad?
25) 372 (a Ce |
ee ead oeNee | ie |
= Hg wseO ys (adler cates
26, 22{ (23 Seri |[=2 3a
(m2n7>)-2 m(n°)? m+n! mn® a = : +
15, ee ar ae = -2{[-3 + 4][-2 + 6]} = -2{[III4]} = -8
373,10 : 29, 21-3 [C5--92}3) = 2]|
ae es = 2{-1[(-7)-3) - 2]} = 2{-1[21 - 2]}
= 2{-1[19]} = 2{-19} *="-38
yr ey, py xt
Dn a a ae ed 28, -3[4"
= -3[1
- 72+7 —- 4]3) - 2°] = -3[1 - 7-1) - 4]
yo
= -3[4] = -12
= as 29. -|-2 - 3| - (5) - 3° = -|-5| + 5 - 27
=-5 + 5 - 27 = -27
(fans co b~®c6e-3 b-®¢3
ad (epey ey pep ® 30. [232 = 2] 2° 3) Seiko .— 2] Sy es
3 sp +|4i1 |) bbs Semele Hi30e 9
pe
PRACTICE SET 3
1 (abc) c*b a abc 3c7b 2 ies ge
SE th ade yetiee, Br pc? a. “ab? = b= (3) = C3) = 29) + 3 = 21
_ p-3,-3
ape b. xy — (xy + y) = (2)(-3) - [-()(-3) + C3)]
=-6- [6-3] = 9
5 eee PRD 2 ci ke
sp pare
Oe) Tek SCO peg he 3 3
(Roby Qa x Sx a
Gs ee =- =I
5) 3 m am TT aX
20 - ee Aa
( 0,2)-3 34 = _symgt ~RE
594-83 i ik
er DDS
ee 5x a’m a’m 5 3x
Pi aom a~m Xx x am
z
cs Sel NR
gaiprey
ae (x3 yz73)? xy? i xP y276 xy? 4 x7 >y2z~6
(ns)? xy x yz xy x? yz®
x3 7-2
ti
Algebra 2, Third Edition
Problem Set 3
PROBLEM SET 3 (a — b) - ab) = GS - 3) - GS)C3)
= —§ - 15 = -23
1. 4x SF=6
Wa een!
4
—a(a — ax)(x — a)
—(-2)[-2 - -2A4HI]I4 - 2]
5x SF =x 2(-2 + 8)(4 + 2) = 2(6)(6) = 72
5) &X S =x
2 a’ — y(a - y’)y
15 (2)? — -3)3[-2 - C3713)
x= —
2 4 + 27(-2 -— 9\(-3) = 4 + 27-1 DC3)
Since lines are parallel, 4 + 891 = 895
O72 =)81
= 9
10. _p? — p(a — p*) = -23% - [4 - B37]
= =9 + 34-9) = -9 +355) =gem
3y + 9z = 180
=
3y + 81 = 180
3y
= 99 11. a? — (a — yy? = (2)? - 33-2 - 3?)3?
y= 30 = 4 — 27(-2 — 9)9 = 4 — 27--11)9 = 2677
1 : 12. @ axa) = 17 = 12 1
Veone = 3 ABase x height)
= 1 —1@:— 2)-= t— 10) = 1—0:-93
1
482m° = gla x 9m]
13.
1442 m? = ar? x 9m
3 e
1442 m?
pea es Sam —=— Pee a
9nm
Circumference = 27r = 27(4m) = 25.12m i 7A)J H =
Since angles opposite equal sides are equal angles,
A = 40. 14.
B + 40 + 40 180
B 180 — 40 — 40 = 100
i :
Voone = 3 ABase x height)
Beas
= 5 per y+ (518) cm* 2 x 10 cm
15.
1} 60 3 3
= —|=| —(25
ent mu) + 20 i1(10 ) cm = 110. 28 cm
7 9 > 2
Perimeter = 2rg + 2rg + 276 + 2rs5 Pe 3x*y?m4+ —-2x2ym3
16. 2xy°m — —=— + ——=—
2(8cm) + 2(8cm) + 2(6cm) m>x mx ~
2
+ 2(5 cm) 2
= 2xy’m — 3xy°m + 2ym* = 2ym* — xy“m
9 om >
l6cm + l6cm + 12cm + 10cm
ANE 3) 5 20°
(xy?) xy ae 6) or 259
54 cm 17.
(x?)-4(y2)9(p3)? r x8) 0,-6
pee xy =? — tes = (9)3) xy 2
L70 x8)-8,6
= —2 —2(Q9) — 6 = —2 — 18 — 6 = —26 oy
Pp
Algebra 2, Third Edition
Problem Set 4
jg, . Ge
7
ey eee
pee en Ae
a
xty* pps? PRACTICE SET 4
(p°xmy?)? xp? pp ®°x? m2 yO xp? 2a~4b9(2ab%c a?
2 4-4
ah opt re :
= m?x4y4 45,102 a -
7 # st sa Wabic® 2a-*b* a
Py xx? (x re
2)-2 ( mpx*) a 4 3 xx 2x44 p-4x-8 om cx 2 i cb =4
(2°? (2) (x2mp2y3 x 9x9, 6m 3p = 4b? b25E
2 xm *p~4 = x mnt a>x? a°c
A en .
Pp
b (2 ~ st] = -(-4 + 2]
a pix *k5(p%4)” : p22 5 p42 2 9 4
hots COS St A68
yd px 7-325 24-2
= Pax Kk 2,-10,3 i
En a px 1
oe! Dock fe Bk Aes
me)?
won GD
et
yx 14 6
ae
piers
“4
-+_= Laer seee
x” AP 13s
1%, (x2y2p)-3
2 p? ere = 66 Opn? aaa
9
F2( (p ay). x 2x pty10
oa gy ey? = x by16,2
i ch aad PROBLEM SET 4
4 (x72y2)5(x%y2)-4 : x710510,0),8 1. Acicte = Or 2 .
(x~4yy?)?2(p~4) x8 y2y4p9 Radius = rom: Area = 2(rcm)? = ar? cm?
+10, 18
ia Se ee x tyl2 Radius = 2rcm: Area = 2(2rcm)? = 4ar? cm?
x By
2. Since lines are parallel, 2x = 78
os Fe et,4 13> x = 39
a-2 3? 9
6z + 2x = 180
25. -(-23 a == iit Bi (3)? 6z + 78 = 180
(—3) 67-= 102
— 8 _ 9 = a | Z= 17
% te = 2 Ne - 5 2x SF =3
= —3[-2 - 2 + 3](-5) = -3[-1]C5) = -15 SF = ;
WM, -X-3 + 7) = |(2 —- 3){ = -2-3 + 1) - [-5|
= -2(-2) -5=4-5=-1 y= ix SF
2g. |-2| - 3? - 3 -2=2-9-
(27) -2 Seine lesa
mtn 04 OT oh tts a eae Ea
29, -2{[-3 — 2K-DIe2 - 3)}
= -2{[-5(-2)](-5)} = -2{[10](-5)} = 100
30. -—[(-3)(-2) - (-3)(-2 + 4)| = -[6 + 3(2)]
= -(6 + 6] = [12] = -12
Algebra 2, Third Edition
Problem Set 4
3. x + y + 100 = 180 8. 3608 =,3), ae rs
x+y = 80 =6x'— 9 24 Bx + Se 2
Since angles opposite equal sides are equal angles, —6x = 3a 3x + 3
x=
y = 40; -13 — 3 = 3x + &
x +) P-="180 —16
= 9x
40 + P = 180
C=
16
P = 140 9
Q+R+
P= 180
9, 20x 23)=- a8 = Sxu—3144)
Q+R
+ 140 = 180
-4x+6-8-3=x+4
Q+ R= 40
-4x -S5=-x+4
Since angles opposite equal sides are equal angles,
-4-S5=-x+4
Or="K = 20;
—9 = 3x
=
4. Agquare SS = (6 cm)? = 36 cm?
x= 2 = -3
6 cm hee 3
Radius of circle = Z =
Ashaded = Agquare — Aati circles 10 rae ae = 32
3 2 5
= 36 cm? — 9n(1)* cm?
13 17/ 1
= (36 — 92) cm” ~ 7.74 cm?
3 5 D
5. Acircle = Atriangle 13 34 >
3 10 10
nr? = aoe
2 13 39
—x = —
3 10
m(1)* cm* = =G cm)(H)
y= ee Ue
a a em = 2.09 cm Lome 130 10
11. — + “ — ie
6. 15(4 — 5b) = 16(4 — 6b) + 10 5 7 8
60 — 75b = 64 — 96b + 10
Lye 35
ee 2
96b — 75b = 64 + 10 — 60
8 Th
Anse 114!
ee 245 kBy Se
ogbari 16
eee ee 56 56
21 3
oy 229
56
jee oe ge Be
6 3 fs 229 =) — _ 1145
10 2, 5 56. 3. F168
3 3 6
10
caine =
—4 5
we
Hane eee
x93 x(y?)® x
3 6 6
10 tars1 _ yxy” 3xy?y?
Seid years”
a
3 6
jane. $3 3 1
X= --°-— = — = fs eax od
6 10 60 20 x72 ae
10
Algebra 2, Third Edition
Practice Set 5
24. “ba — b\(b — a)
13, 2p es a g
ay? pe
ie C9) Sewer]
| ae eT
4-25
3ay sith:ayn
“a aol
—2(-1 + 2)(-2 + 1) = -2(1)C--1) = 2
_= WP ‘ = a
33)
pay Pp 25. ab(a” — b)a — b
me (3x7) -ayty3 u 81x 41,045 3 81x 45> = (-3)-D[-3 - (1-3) - C1)
= 39 + 1)-3) + 1 = 3(10)-3) + 1 = -89
(9y)* yy*x3 Oy *yy2x-9 Qyx~3
= gx-ly4
26. Sone
( eg ee |
=e) rll
1 ee
“ Oye) yx? 2 y x4 yx? " y x®
= bh wl oll ape Saee
(x*)9 yx? Ay grt a2 Ay=3x*
a xty? 27. 22/3 =o ee
4 = -2{2[C7) = 4] = 2} = -2{2[-11] 2)
ee) (220 a 48
- 2(x 2)? x24 7 ee r 2x5y-2
© 9442475 (2)3 x9 xx2x 5x6 x4
28. = 392 (22 = 3 = D3 Em eee)
= 7 = -1 — (-6)(-3) - (6) — 6
=-1-
18 +6-6=-19
(x yInp oy ey ee
92 16
(x ay 4x0 xy? Ax*y* 29. 216) ene
===ot718)
et are
= ae) ee
peed
x xey
4
30.
18 a ay - (2x? xy" x - foxy?
5 eiep Be Oh me xoty 2 x esi
= 12x°y4
19 2X7 XX _ 3x7y4 TAK Pyne PRACTICE SET 5
° xy! yy x ty4
3(-4 + 5N) = 7N — 212
= 2x?y? — 3x7y" + io a = 6x*y? Bee eae aw
8N = -200
3x | y ax N= 05
20. — — 7x*x + 2y*y
y
E Bz Py yy If -were Victorian, 7 were not.
7 y y x Es
=)
e x 1624 = NV
41. 2ay 2 x Sa Jae
a bya?
xy
x ay ~ ay 5% 203 = NY.
7a 2 ay : 4 2X ay" , 2xy
Favors, 24 1015 pieces = NV
x x a x a
22. a(a — ab) = (-2)*[-2 - (-2))] = 4(2 + 6)
= 4(4) = 16
23. x yx(xy — x2) = (3)°-DC3)-3C-1) - C3)*]
= 33-9) = 3(-6)*= -18
Algebra 2, Third Edition 11
Problem Set 5
PROBLEM SET 5 8. K + 110 180
K 180 — 110 = 70
1. QN - 9/4 = 10N - 8
8N — 36 = 10N — 8 Since lines are parallel, Q = 70, P = 110, and
-36 +8 = 10N — 8N D=x = 70.
ee RoR 2N Since C is a straight angle, C = 180.
N= -14
Circumference I N 9
2. 3Np + ll = py ae 167 in. = 2ar
1+ 13 = 4Np - 3Np 8 in. WW =
24 ducks = Np Acircle = 2r? = 18 in.)?
3. (SN —- 8)4 = 6N — 116
= 647 in2 ~ 200.96 in.”
20N — 32 = 6N — 116 Veylinder = Apase X height
20N — 6N = —116 + 32 (647) in.2 x 5in.
14N = -84 3207 in.? ~ 1004.8 in.?
N= -6
10. 3x = 3) stl 2°) 12 = Ss 2
6h + 94 = 2 = (0, m0
—6x + 8 = 10x -— 10
i xe Ca= 1
8 10 + 8 = 10x + 6x
8 18
= 16x
(C= 1D = Z = 96 clowns
18 9
x=— =
Sh Exc Ofa= is 16 8
2
7 x 140,000 = Nr 11. (2 sy C= 2) SG ee)
8 + 4x -3 +2 = -2x - 6
40,000 horde members = Np
Axc +7 = 2x — 6
2x + 4x =-6-7
6. Apor = —bH
2 6x = -13
27 in = 5(6in.)(H) x=
13
—-—
6
9in.
= A
AQ = PQ - PA = 9in. — 6in. = 3in. ee ap De = ae
ey) 4 8
I
Agas = 50H 1
= 5(2)3) in.” = 3 in.” i) 1 9
-—-x = -—--—- —
2 8 4
7. Since angles opposite equal sides are equal angles, —
7 — Ce
1 ee
18
2 8 8
y = 523
-19
x+y
+ 52 = 180 -x = —
8
x + S2e+552 = 180
x = 180 - 52 - 52 = 76
ee eee
8 I 56 28
12 Algebra 2, Third Edition
Problem Set 5
13. 5(6 — 8) + A(R ~ 12) = ax 21.= p?m*(p*)(2p)> __ p?m5p323p
3-4x
+ 6x-9 = 4x
Ox <0 sax
~6 - 6 = 4x
-12 = 2x
x = -6
22.
ra & arn as
14. -3 — 3° > Ses 5) SF 3) SOY eax ys Pee steam yoo Ixay
= -—x°(x.—8)
sea 2aax2 5x?
—3 -1—- 18¥ + 45 +2x+3= —x +3 23.
Ey he x nee 2"
a
+ 44 = —x
—l6x +3
= 3427 + 2a*x ~ 5a2x3 = 2ax - 8a2x3
—-l6x+x= 3 - 44
—15x= —41
24. mx — m(m — mx?)
41
a= = -2(-1) - (-2)[-2 - -2)-1)"]
15
Wl 2+ 2-2 +2) = 2+20) =2 + 022
15. 23 De Re) = 6x = 2
297%
25. a b(a -— b) = (-5] = “{-3 = 3)
—8 + 4x +8 -3x =2+4 2) Wane %
x= 6
4
16. 3lx 232 =.32)) + 2[x.= 3GL.2)]
=X 3) 1 3 4 5 7
= — + S| =. SS HS SS
4 16 16 16 16
=3y + 6.4 18 + 2% — Ox 12° = Foe— 35
—7x + 36 = 7x — 35
26. a — ba(a* — b)
—/x — 7x = —35 = 36
=A
71
x=—
14 as + -5+5)- 3-23)
Do eleAt wa WS BA
7 2 ee s ey ” 2abc2a7! — 6abac
“ ¢2 b b , c7b c2b Di- 2? = 1°16 B16. M6
me 2be*, 6a"be Seon”
c7b c2b c ale (2032) = 2i23)] =)=2r 89) 16)
2) 3 2 3 2
=22(=11) -2[6] = -22.- 12 = 10
18. _ ax bax Pa A _ ax cts 4 Sant ax
b a’ ba b
1
oe Aine 5 3a7x? 28. Ae a ee enn, ye wn = ist
= ——— + =—-xy> + -2? 4 4
ba? b b
_ (xm=2)9x%m —- 4.x%m xm 4 29. alae ene) |
xx2m9(2x)~? xx2m x2 x Sj} =e ie) Sell — 4 + 40!
:
= —1[35] = -35
“ 4c2de3(2cd~*)* _ 4c2%dc327c*d*
; c8c3 (cd)? Cocae 30. 32 5) — (3 pee?) |
345
= : ss = c4d
= 3[4-4— 2] = -3[-2] = 6
Cad,
Algebra 2, Third Edition 13
Problem Set 6
PRACTICE SET 6 6. Integers: N, N+ 1, N + 2, N + 3
4(N + N + 3) = 6(N + 2) + 24
a. If 0.017 were red dwarfs, then 1 — 0.017 = 0.983
were not. 8N + 12 = 6N + 36
(0.983)(29,000) = NRD 2N
= 24
28,507 stars = NRD Nose2
The desired integers are 12, 13, 14, and 15.
Consecutive even integers: N, N + 2, N + 4
Surface area = 4ar2 = 46cm
3[N + (N + 4)] = 12(N+ 2) - 84
3[2N+ 4] = 12N + 24 - 84
6N + 12 = 12N - 60
72 = 6N
12=N
Since angles opposite equal sides are equal angles,
The desired integers are 12, 14, and 16.
A = By= 34:
K +A = 180
PROBLEM SET 6 K = 180 — 34 = 146
WD xX of = is Since angles opposite equal sides are equal angles,
0.016T = 480 Moa viT.
T = 30,000 teachers
3.x SF
3
If 0.653 were prophetic, then 0.347 were not
prophetic. sl ees
3 3 9
WD xX of = is
0.347(3000) = NP >Zz x SF =x
1041 statements = NP
15 11
— Xe
Z 9
-3N-—-7=-2N-4
—3=WN
ore
1? XS OFF = Since lines are parallel,
3B = 130
ai x BW = 175
2 pe 190
2 3
BW = 17/5 x = = 70 barge workers
2A +: 3B = 180
2A = 180 — 130
Odd integers: N, N + 2, N+ 4
ZA
T= OU
6(N + N + 4) = 8(N + 2) + 28 A =-25
12N + 24 = 8N + 44
10. A = 2(180 — A)
4N
= 20
A = 360 — 2A
N=5
3A
= 360
The desired integers are 5, 7, and 9. A = 120°
14 Algebra 2, Third Edition
Problem Set 6
11. 0.005x + 0.6 = 2.05 a®bc® (a~!p-!)? pa Hz ene
5x + 600 = 2050 : sre abaabe — a> 2c
5x =.1450 = a~>b->c7!
x= 20)
NN 2x) 4G) r x ox ey? 2 x8
13. 325 of it a 7 2xx9 x! xxy? 8(2)x*ty? 16x4y?
5 4 3 1
LL = 22 se 3 i? 16x12 y?
5 3 4
0, -1 —1,.2
Lyseen:Seles 21 Sayles
Simei | 15 y Coe
73 = —2xy at Sxy — She = 3xy a. Ere
17
—xr = —
5 12 pee
RB 5 365 », 2x xy 2xxx — 3xy
pee! _ ION y ity aay
lee 17 =. 204
= 3x3y? + 2x3)? — 3x3)? = -3x3y2 _ xy?
Bese 2 oy — 82)
= 590" x) 2x 23. xy xery Sy BE 2)E4) =9-2)* Gay 4)
air + 6— 3 243 4°6 = 10 = Sree 2x = 8 + 16°+°4=28
fy $05 = 10-— 7x oe
=a) Clok-slencan
24. a“b — a(a — b)
i= 5 =)
14, 3 2) = 2G DO 5)] 2 “3 DAE Rt
et re Oe ae 14] -1+(3)-3)- 12 a
—I7t-al Gx & 30 Ph a 8 8
Ata
yitvol
25 |
¥
= i ome
a GLAG)-(
en ‘ i F anes as
oakae:
2
72
4) Vs joa ls
Lie 1
5
2
15004 + 3) 2 = 3) -= 2x. — 4G & eh = 3? ae A ee oy
4x+12+x+3=2x-4+4%-9 26. —3°[237 — 2-2 = 3)][-2°] = 41/-90% a0y-1)
5x + 15 = 6x — 13 = =If1jf-1] =]
2s
27. =31G3)" + Gabe 9 0 | 1k 6
= =|
LY | oD, Dp —Ixyp 2xyp
16. 2{= | 2) ] 3 “
Pp xy x Y, <7 apry pxty Dees eo Aan oo eD|
= -3p~? + 2x? =-9+9- 16-4 = -20
v7, 2k HP _n) = Kee, Derk
07,0 0
m9, 32-2 .
9=2 416-1
PA\Is px p 2
— cePy 2Pe tly op ep Cee eed
2.,0\-2 ,,,.-2 4,,0,,,-2 2 1 |
ay 6xxxy”(y™“)
= a Ae
Ax-y"yoe ao
4x-y Die) aoe See gue os
ua a2 4 28 Sb Se
oe 2 22 “1” wee 72
15
Algebra 2, Third Edition
Problem Set 7
PRACTICE SET 7 PROBLEM SET 7
100% 30% 70%
1) eed, =e aes
(pa em No
100 100
100 130
WN 226 1ses
20
Since one part of 130 is 26 for 20%, the other part
iy
must be 104 for 80%.
Before After
100% 20% 80%
0 Ee WHI 93 _——————— 7), Na
100
100 30s y at 63
30 100 30
WN = 310
- ia
100% 2000% Before 96
"OF FF
After
of 390 7800 9, ERset rs 1400
100 100
we = 1400 - 2 ~ goo%
Before After 350
WP x 390 = 7800 100% 400%
100 cso I {f = —— eee p,
oO
390
x 100
Sx 390 = 7800 x 390
Se
of 350 c 1400
WP = 2000%
Se = Wil OS ely
Before After
Sy
= 333
oe ie 20
3. 100
— x of =]
= 18 See 100
a WN == 460
Ses A
j3} = SxGuh)
= wil
wn = 460 - 1% ~ 2300
B= 34 20
Al=—B Since one part of 2300 is 460 for 20%, the other part
A = 34 must be 1840 for 80%.
100% 20% 80%
Se d\n
of 2300 1840
Before AS 460
After
16 Algebra 2, Third Edition
Problem Set 7
WP 10. Since lines are parallel, aie =
4 oo ae —— WP 20 = 680
(3x + 20) + (Sx) = 180
we = 680 - 1% — 3400% 8x + 20 = 180
20
Scl=- 160
100% 3400% 6 = PA)
Relabel the angles a, b, c, and d as having measures
x°, (3x)°, (6x)°, and (2x)° respectively.
X +, 3x + 6x + 2x = 180
12x" = 180
Before After
are ils)
P 1900
5. —— Xof=is == — xX WN = 380 mZa = 15°; °mZb = /45°;
100 f 100
mZc 90°; mZd = 30°
Wn = 390 -22% = 2
1900
12. The sum of x°, y°, z°, and the three 60° angles that
100% 1900% touch the vertex must equal 360°, a full circle.
x+y+z+
60 + 60 + 60 = 360
x+y+z=180
Satay mites ° =e S0>
13; 3p(=2°= 53) F pee 2 = (pear) = De
Before After
6p + 99 + p-4=-2p-4-1
6. Odd integers: N, N + 2,N + 4 lop — 4 = -2p - 5
7(N + N + 4) = 1O(-N — 2) - 120 Spit
14N + 28 = -10N — 140
24N = -168
qh)18
N=-7
14. 0.005x — 0.07 = 0.02x + 0.0032
The desired integers are -7, —5, and —3.
50x — 700 = 200x + 32
7. Evenintegers: N, N + 2, N + 4 3 2e= LOX
6(N + N + 4) = 14(N + 2) - 8
ep150 See
25
ate
12N + 24 = 14N + 20
A EAN
2 ay 15. aa A araay ee
5 2 20
The desired integers are 2, 4, and 6.
| |
8. -2N + 5 —N
be N
9. (4x + 15) + (Sx + 10) + @ + 5) = 180
10x + 30 = 180
1Oe—s 50
yisd
200
fy a7
100
yes 1h
17
Algebra 2, Third Edition
Practice Set 8
16. Lhe ON Gyomy a Oy! 25. x = xy)
SeS(ort 2) 3! (-2) — (-2)[(-2)(-4) - --4)]
-8 + 2(8 + 4) = -8 + 24 = 16
Byes Dy esosg ts Bie il ae Al eye = OPS a
2x+4=-5x-9 | | | |
26. S | is) |
g Paes =g es
iT] We) | NO
Nl NlR Nl
Ne,
<j
i-e— 15 Saas
| I
13 We) ahs JR | Jr at = ll ie) + wo ae
x= =—
eee, Nl
aN|e
See BIR
7
ll bd
2 2 e Bis |- 2xypx! z 6x ypx
17.
ee NYoD, ype Sey py yy
aE 6x2yp7!
27. 29{ [20m 3G?) — [23 — 2
otlet + 6 [-2-5)-2]}
—2{[5] = [-20] } = —2{25} = -50
Any 2kx? _ 3xy
18.
k y k 28. OO 0 oy Ae?
252 OD) 5)
Si BV ETI
oh ger por lye? =-1-2-4+4+8+4+4+4-2=7
ky kk
29. 3002 a Dee G2) 2 sates)
19. Cpe ae ees = (1)(-5)3)C2) - 4-2) = 30 + 8 = 38
nya eae yey
y © Bex
30. {2° eer) 15°F = G2) S832 —
=
20 Pe 2-10
2x“y~ (222) = a
x Sy? “4
al) "2 = ad
20. a eeGe a ae
PRACTICE SET 8
3x + 4y = 12
eles, Txyp! 4 2XXXp 4y = -3x + 12
21.
P ee y! 3
ooieees eee
- 4
7 3x3y Tx? y 2x7y 2 2x3y
The y-intercept is at (0, 3); the slope is —3.
Pp P Pp Pp
22.
px Pp
= —Ayp? 4+ 3p* — Dep"
= -6xp* + 3p’
Hee lol)
= ll 1 eee1 ee 1
=n Ea ay
23.
A s(-2 s =| - {-2] aihae
ZeEGy 56 2h 46 12
24.
sete CARAT -f
4-944) 2
32 16
1
18 Algebra 2, Third Edition
Problem Set 8
PROBLEM SET 8 —
P x = —
340
ae a
100 Of = 15 100 x 56 = WN
WD X of =
0.36 x K = 828 = al - 56 = 190.4
100
K = sens = 2300 knights 100% 340%
0.36
Even integers: N, N + 2,N+4,N+6
10(N + N + 6) = 9N +2+N + 6) + 24
20N + 60 = 18N + 96
aN
= 36 po P[m
Before After
N= 18
The desired integers are 18, 20, 22, and 24. A = 2(90 — A) + 3
Ag——1o0s a An aS
3N — 7 = -2N - 72 SAg——L85
SN
= -65 A = 61°
N = -13
3x + 4y = 8
POLO, Sis 4y = -3x + 8
7 3
— x WN = 420 = —-—x +2
16 4 4
The y-intercept is 2; the slope is -2.
WN =F420 x ° = 960 warriors
Odd integers: N, N + 2, N + 4
5(N + N + 4) = 2(-N — 2) + 108
10N + 20 = -2N + 104
12N
= 84
N=7
The desired integers are 7, 9, and 11.
Se Soares 20 x WN = 86
100 100
10. (2x + 2) + (4x + 4) + (3x + 3) = 180
WN = 86 - ee 430
20 Ox + 9 = 180
One|
Since one part of 430 is 86 for 20%, the other part
must be 344 for 80%. aD
100% 20% 80% 11. AGiole = mr = O70 m2
OA, a ea ass
A}
a 257m sak
5
ll BD = AC =5m
12. Since lines are parallel,
ti fe
Before 86 (5x — 19) + (3x — 1) = 180
she = PAN) Ss alte)
After
Sx — 9200
sf = Hs)
Algebra 2, Third Edition 19
Problem Set 8
13. (x + 5) + 155= 180 5x72 (y?x?)? x 20x 2 yx?
x = 180 — 155 -— 5 = 20 Qtr ny) xyx? xy
-11,-6
(y +10) + (x + 5) + 40 = 180 = Oe a ryt
y + 10 + 20 + 5 + 40 = 180 xy
y = 180 - 75 = 105 3
22. 3x-2y + 5x2y! - =F
xxy
14. 0.003x + 0.02x — 0.03 = 0.177
= 3x*y + 5x2y! ~ 3x*y = 5x2y7!
3x + 20% — 30 = 177
23x
= e207 23. daryFy, ty?
2y _ Tay?
see xy x iy
= Ixy? + 2xy? -— Txy = 4xy3 - Ty
2,
15. pales pala
3 =)
2
5) 24. Bre pee! ) A (-5\($]
Sistales) ae 2)4
i
re
_ SS
3i7/ 8
1 (1 1 1 4 1
3 5 5) = —— + —| — = —— + — = —— lrYS
8 32 32 a2
7
—-x
= —
29
3 5
x=>—:--
= —
OS. 87
5 Tf 35
25.
16. -—3(-2 + 2x)
= 49 — 2? - (-23 -3 -(2- 2x) - 4
ee
ey es eae ay ee eae |
6 — 6x = 2x - 6
Sxl 2
26.
x=
iz ==
3B
aie -4-($-3)=- (= - 3)
6 9 295) 6 Gg 18 18
ie 2 4) ol eae 1 (-=) 3 vi 4 2
SERS aad EO 16 = —— — | ——] = -— YH OSC
6 18 (8: 1S 18-5 —D
Ln td2S ee
Seo
ee enoSEES
45 SSS Sa
Vee Nes 6s Ion: 16 27. -3[(-2° - 4) - 2)3)]
_ bl 2 _ 302 _ 151 = [Ger =<2)5 = 2a]
ipa ie (56 =5 3-5 a0) ales a)
23211] — (9)
18. (2 “y =] = 33-9 =24
p\ y p°
tue =3 Dees
YDS 0) 4
a
2 2
yxy rey = Ay yay 3 = 9 gatey 5 33-3)
py pp
2-3 41 Led al
S000 ee9 + 9 +3+3 =
19. ae [-ax+ 3) - eee!
p p p
= 3x - 5p*xy
29. 2)00| 2? ee ae eae ee
£4
5, ADO DI Sieyay aye! 30.
327! x2 Gx?) 16x7y3 =-l-1-l+#ft= 4
20 Algebra 2, Third Edition
Probiem Set 9
PRACTICE SET 9 100% 20% 80%
—————————— aS OF
a. If 60% were neophytes, 40% were not.
40 Acs 2%)
100 of T/ $78,000
100, 40°. = 299 . 100
40 100 40
Before RM
A = 500 people
After
1 80
b. — = S17 1500 — x =eisg =) ——_x TT = $78,000
100 00 100 :
100 | 245 ry = 171,500 . 1 100
245 ~=100 245
LY = 70,000 widgets
Integers: VN, N + 1, N + 2
-5(N
+ N + 1) = 2(N + 1) - 43
PROBLEM SET 9 -10N — 5 = 2N - 41
1 100% 40% 60% 36 = 12N
SSS _
3S
The desired integers are 3, 4, and 5.
5 If 84% staggered to their feet, then 16% did not
get up.
a
Before After
= XO a=suiS
fe Ae “45 a ae eABI 160
100 100 16. sp = 40,000
100
100
AB = 160 - — = 400 100 :
40 SP = 40,000 x 1% = 250,000 soldiers
400 — 160 = 240 wood nymphs
If he could not see z of the Trojans, then he could
DA. 100% 145% see ; of the Trojans.
———————_———= SSS
F XS0s <=s
of LY 140,795
1 x 7 = 1400
8
T = 1400 x . = 11,200 Trojans
Before After
le 145
aa =i5 is
100 x of = =e ——
100 xX LY = 140'795
100
EYe=s409/ 95a 145 = 97,100 bushels
m = 180 — 80 — 50 = 50
p + 50 + 60 = 180
p = 180 — 50 - 60 = 70
Algebra 2, Third Edition 21
Problem Set 9
8 y + 140
= 180 — y= 40 12, 3y+6=-%x
(7x + 6) + (6x + 4) + y = 180 ll See
—
13x + 10 + 40 = 180
3
13x-=).130
x= 10
9. (5x + 10) + (7x + 50) = 180
12x + 60 = 180
yO)
x = 10
5x + 10 = 5(10) + 10 = 50 + 10 = 60
Since vertical angles are equal angles,
yo 1 ='60
vo =_59
z + 60 = 180
13. 0.02 — 0.003x + x = 5.005
z = 120
20 — 3x + 1000x = 5005
10. ACicle = mr = On m2 997x = 4985
9n m2 Rte
r= = 3m
1
1 2
—3=x + 10 = ar
Circumference = 2”r = 27(3 m) = 67m 14.
Le y= 2x + 3.= 0
16 38° 671
Yea 2 3
2 ee 26
: : mz 16, _ 380 _ 639
The y-intercept is —3; the slope is 7 5 90 90
16 259
a Out
ae ->(-2)= 1295
_ 259
90 16 1440 =288
TS, BCS yr 100491) | Se ae
—2[-x + 7(0)] = -3x + 6
2x = —3x + 6
ax = 6
oee
5
16. =2" (2x. — 3)-<ta=
2x — 3°
—2x, + 3.— 4 = 29h]
x= 1 = 28h
Ot=a4y
Q= x
oye [2 ‘ =| n xy p* x9y2y D
Uz:
po. Yea BE JS ep ee ee
=p'-p'=0
22
Algebra 2, Third Edition
Problem Set 10
18.
ak (24t x)_ 2akk4 ~—3ak-2k PRACTICE SET 10
a a+ 5* =
= 2h - 3a!
a’ + 25 = 81
19.
(2x*ya)
ya? — x~® yaya? xy a’ = 56
x*y(ay)? y 8x?ya*y*y — 8x7a a = \56
. $4 2
8x8y2
Dryer
(x2z73)3 8x2?
ege | anneal
8y>z
2 2 4 —2
21. cS a ey — = 3x - 2xy + 4x
y (x*)
= 7x — 2xy
2xy _ S5xxx 3xp!
22. p (xy y 1
Soe Say + ee
ne Biy
0 (IC)
P Pp
2 ‘
2 4 2
PROBLEM SET 10
If 20% were in a festive mood, then 80% were not
in a festive mood.
=a eT = 1400
25. ala — bab — b) = -2-2 — 3)[-2G) - 3] P = 1400 x = 1750 people
= —2(-5)(-6 — 3) = -2(-5)(-9) = -90
a(x — ax”) = (-2)[-4 - (-2)(-4)7] EF x-of =tis
= 4(-4 + 32) = 4(28) = 112
3— x S$ = 26,000
27. -2(-3 — 2°) — 2-2? - 2) = -2-4) - 1-6)
=8+6=14 S = 26,000 x < = 8000 soldiers
3s + 24-2) - GB -2) - 2]
= ~3[(-3)(-2) - (1) - 2] = -3[6 + 1 - 2] F xX. of = 3s
= —3[5] = -15
| YM = 590
10
29. _29(-2 = 3°) — 2) - |-31
= -]1(-3) + 8 - 3 M = 590 x = = 5900
-34+8-3=8
Menafter = Mentotai — Menxitted
30. aoe pe 8 PSR HS -9E7 Menafter = 5900 — 590 = 5310 men
a Pe
23
Algebra 2, Third Edition
Problem Set 10
4. Odd integers: N, N + 2, N + 4 10. Asquare = 4 * Ashaded
= eae 2
4N = 3(N+2+N+4)-8 = 4x 9m‘ = 36m
4N = 6N + 10 Agee sa se = ON:
-2N = 10 s = \36m* =6m
N =-5
11. 2y = 3x +2
The desired integers are —5, -3, and -1. 5
y= =xt+1
2
5. — x of =is 4
100 The y-intercept is 1; the slope is Zi
260 OA = 10,400
100
OA = 10,400 - a = 4000 minas
iP.
6. — x of = is
100 f
14
—xA
100 Total = 4200
100
Atotal = 4200 - VE a 30,000
Avjidden = ATotal — Aseen
Ayjidden = 30,000 — 4200 = 25,800 Argives 12. y=-3
7. Circumference = 27r = 6”%cm
67
r= —em = 3:em
20
Area = ar? = m3 cm)? = 97cm
8. Since angles opposite equal sides are equal angles,
x = 60.
x + y + 60 = 180
y = 180 - 60 - 60 = 60
y + ZBDC = 180
ZBDC = 180 —- 60 = 120
13.
z+ 20 + 120 = 180
z 180 — 20 — 120 = 40
9. (17x + 20) + (20x - 25) = 180
Bi xe son = 180
Syibe SIGS
eS
Since vertical angles are equal angles,
P = 17x + 20
P= 176)74"20%= 105
Since lines are parallel,
QO = PRi=al 5.
24
Algebra 2, Third Edition
Problem Set 10
14. 2,-3
a yen te See
22. 3x + se ay eae
Kay
z x72 yz?
_ 9x7y
= 5x*y axe), ey Ixy
Z Zz Z
z
(A +x + 3) + (B+x+ 2) +(C +x +1)
= 30cm
Since: A»= "x +.3; B-=x +*2:and C = x + 1:
Q(x + 3) + 2(x + 2) + 2x + :‘1) = 30cm
2x +6 + 2x +4 4+ 2x + 2 = 30cm
15. 25.
wo A8)-CF)
——F¥ fe
20 4
a els
ye
tg 26.
16. = + oy = U2y + 2)
—-52 + 30y = 2y + 4
28y = 56
y=2
4x(2 — 3°) + (-2)(x - 5) |=
iA
27.
17. Pe Ww= + ‘S)—
= -3x - 2
Paercnic)
8x — 4x — 2x + 10
2x + 10 = —3x - 2
Sx = 12
12
x= -—
5
12xyy7 8xyx
18. als 3y7 _ 2x) mmx m-ym
me? mex mn
28.
= 12 - 8x’m
2a? : ss _ 4x°yp _ 6x°ppyxy
=-§+8+4-2=2
19.
P yi Pp py -|-2| - |-29| -3? - 33°
= 4 — 6xy"p = 15
27 +
=—-2-1-9
20.
33, 2Qn)s
= 3r’y!
Ox ty a ey 30.
Gl-tae: ree
(ae) re Pp
¢ ae 7 op ep - <p hs:
21. x 2 ox9x 3p 3 “5 5-2
5ai aE =
hae
25
Algebra 2, Third Edition
Problem Set 11
PRACTICE SET 11 4. If 30% sat around, 70% worked.
ak y mz* 3akz 3y ee x of = is
a —m + — - = = ee i 100
BP bz bz 3bz 3bz 3bz 70
3 mz 4 + = bs 3y 100 x P = 1400
—
3bz
P = 1400 x - = 2000
Zz 2mn ab*z ab*k = 2mn
Di ee. ae D ape t ae Psat = 2000 - 1400 = 600 people
_ ab?z - ab?k + 2mn
ab? Sr 700 x of = 1S
340
c. = 2(40)
y == 100
— x RP = 6800
yas 180 100
(4x + 20) + (5x + 80) = 360 — 80 kage =
Ae as COREY;so 2000
9x + 100 = 280 Total = 6800 + 2000 = 8800 plums
9x = 180
20 6. Odd integers: N, N+ 2,N+4,N+6
C=
—-4(N + N + 6) = 10(-N—- 4) + 10
—8N — 24 = -10N — 30
PROBLEM SET 11 aN =EG
1. If 13% believed, 87% did not believe. N= 3
P The desired integers are —3, -1, 1, and 3.
— X of =is
100 7. Since angles opposite equal sides are equal angles,
87
ESET by Total a 220 Z = a2
100
100 y+z+ 32 = 180
Lota) = 5220 x eyain 6000 y = 180 — 32 - 32 = 116
x+y = 180
LBelieved = L Total — Lot believed <1 7 = 180
Lgelieved = 6000 — 5220 = 780 people g y
x = 64
2. Even integers: N, N + 2, N + 4, N + 6 Since the measure of an arc of a circle equals the
(IN angie. 6) see “At oe measure of the central angle, p = 64.
atN sald = N= 24 8. Ashaded = Asquare — A3 Triangles
i N=4
al = (4\(4) -2ya) - 22298) —peLana)
(
The desired integers are 4, 6, 8, and 10. = 16 le 3.54
= 10 units?
Name i Re
100 x of = is 9. Acircle = Atriangle
2 2 1
2)
100
x OP = $5599 Te ee
OP = $5599 x —
100 = $2545
mam)? = +(¢cm)(H)
>.
220 55
re 20 cm a 2
Increase = $5599 — $2545 = $3054 malnaasrreirmage
26
Algebra 2, Third Edition
Problem Set 11
10. 2(3x + 2) + 2(4x — 2) + 23x - 2) = 36m 16. (a) 4x + 3y -6=0
6x + 4+ 8x -4+4+ 6x —-4 = 36m 3y = -—4x + 6
20x - 4 = 36m 4
Saleem Liphkigd
20x = 40m ee
of Sa Dan (b)
x =-3
3x + 2 = 3(2m) + 2 = 8m
11.
kx? + abcx -— m
—__——
ax
12. Pre ae.De
ak 4k 4ak
re 4ack
4ak
3a2
4ak
4p - 4ack + 3a?
4ak
Be = je = 34
3
13.
heh7}Nn geen tap
139 Bet914.
3 4 4
7
ye 1
4c? px
3 4
4c?m*x — 12cp* - Sp
4c? px
-{-3)=4
* *a Fig 28
14. c= a fp 18. 0.03x —x + 2 = -0.91
112 = 42 + x? 3x — 100x + 200 = -91
121 = 16 + x? -97x = -291
105 = x? iS
105 =x
19. 3x(2°— 3°) & 7 2.208 a it ennd
15.
3x —-1=-4x +2
iva 3
#3
ie
-2
*(30°y - 25)i" 3a +2
a*y* 2a 2s
20.
y a y ya
a3 = 2a-4
oe - 2) P 6x “ypk " 4x ypyk
21.
Dea+ k \yp p kyp kp
D? = 25 + 64 = 6 ~ dy”
D? = 89 ee
2p*a eas Ee g ates
OA, (a
iowa
22.
D = V89 (2 pa~™) “ap pa
Algebra 2, Third Edition 27
Problem Set 12
Deed eee
23. Integers: N,N + 1, N + 2, N + 3
Oxsy\SrXyae yon 5. y + 3) = 3-N - 2) - 40
AN +N+1+N
xp? 3y7! 1 te Sa 3 3x, 2% 6N + 8 = -3N - 46
24. $a (3 me SE eee 3 ON= -54
if ip PPPYy yP YP YP
N = -6
The desired integers are —6, -5, —4, and —3.
25. — = -3ka + 3ka - Ska F X of = is
44 x 5 = 5100
4
-)- Cee
26. 4
S = 5100 x — = 1200
17
F x of = is
0-0 IEC a8
2—-xN=1
27. Nee
11 11
5(-N)
+ 25 = 8N + 90
-13N
= 65
28. N=-5
= -72
If the train completed 30%, then 70% remained.
29. — xX of = is
30. 20) Seo = 3/2 © 5) ae x TL = 6300
100
Sie Soc Teas}
100
TL = 6300 x 70 = 9000 miles
PRACTICE SET 12 z + (180 — 140) + (180 — 70) = 180
Every point on this line is 5 units below the x-axis, z+ 40 + 110 = 180
so the equation is y = —5. are)
The y-intercept is —1. The slope is negative with a (4x + 25) + (7x — 20) = 360 - 40
rise over the run of } for any triangle drawn. The Pieter — 2520
equation is y = -3x - 1.
Iilige=s3ih5
315
x= —
PROBLEM SET 12 11
If 40% were monochromatic, 60% were variegated. 1
y = —(40)
5) = 20
a X of = is
60
Asector = 36007") = 36m cm?
Oy = 2400
100 2 _ §(362) cm?
HG,
V = 2400 x oe = 4000
60 r
V¥216 cm = 66 cm
Monochromatic = 4000 — 2400 = 1600 vases Diameter = 2r = 2(6V/6) cm = 12/6 cm
28 Algebra 2, Third Edition
Problem Set 12
10. Relabel angles A, B, and C as having measures of
3x, 2x, and x respectively.
3x + 2x + x = 180
6x
= 180
a4 30
A = 90; B = 60; C = 30
2 3 2
1. m+>+—
ae ic
= mbcx MO ee
c Baad, bex bcx bex
n mbex? + bx? + c? I-—I—I—1—
Oanr
WP
bex?
17. (a) The y-intercept is +2. The slope is positive and
12, — - — - — = —— - —— - —— the rise over the run for any triangle drawn is 3.
1
=—x +2
ope
a*be (b) Every point is 2 units below the x-axis.
Y=
1 ee ee 2 te
Saat eae b 18. gi + hy = u
4 p. 8
144. C=aa+p? P)
=e —
1
=| —_
66
—
137 = 5% + k? 2 8 8
169 = 25 + k? 5
S36 Coa —_—
65
2 8
144 = k?
=k x ah 8oe 2 I1S0S E13:
Sry 40 4
15. 19. 0.001 + 0.02x — 0.1 = 0.002x
1 + 20x — 100
= 2x
Lexa 99
= il =" 5.5
2
20. -3(-2 ~ 2°) Gh) 2 B32 + 4)
6+ 3x +2+4+4+
6x = -2x - 8
Sie cb Ih, = oe — 13}
l1lx = -20
_ _20
11
a> x( xxx? a> xxxx~? 3a>x
sees tg lk ae ae
yy ene, Vay ey. ny,
= a xy? — 3a 3xy3
D = 3713 = —2axy?
2. aa ; pe |
16. (a) 3y+x-9=0 0-2 2 x2
3y = x +9 y
0-2 ,2 0-2 ,2
1 = oe Ss EE | ae
= ——-x +3
“Gey 8 3 yx
Algebra 2, Third Edition 29
Problem Set 13
Oy Ree oy OG b.
jG ee ae ee
ew Ces Seay?
ey any) 3 3
ORR 2 = 2
he ss —
eee ee = 64a2b> 2 2
OeT \s b
2.0 a 4 1
:ors 4 H+ 2 = 3
Cl erp ere = ee Boe eee
Ds, a aa x x 3 >
=i H* =%§
= 2ax
H=NV5
5a 3
26. abc - DP We Area = 5H = 5(4y(V5) = 2/5 m?
= abc — Sabc — 3abc = —Tabe
perry €i JICiL)re€1 I2)) PROBLEM SET 13
2 2 2\4 1. If 60% had blue sails, 40% did not have blue sails.
i 3) (2) 11
See ce 0 ee P
4 8 A\ 8 32 — x of = is
100
2 2
TA te Le1 oe)1 ee 1
ee 1 40
pba =
ur ey) | 5) ( 5) (3)| i00, ee
So Le eee be
= (3 *) (5) 16 B = 300 x 100
7 = 750
29. -ab(a - a’b — b) Blue sails = 750 — 300 = 450 boats
= -2(-3)[2 - (2)°(-3) - --3)]
20 hex =
= 62 +1204 3) =6U7).= 102 ea
3
— x C = 93,750
Bre a3 al) — (2s) 16
1 2
ea Gare C = 93,750 x : = 500,000 citizens
=1(—1 Whe) —a6e 0 220 3
Sa Ee eet a van WE 3. Even integers: N, N + 2,N +4, N + 6
SAO 9 a,Oi 3 (S223 3(N + N + 6) = 5(N + 4) + 14
6N + 18 = 5N + 34
N = 16
PRACTICE SET 13
The desired integers are 16, 18, 20, and 22.
a. 2(y + 7) + 3y = 4
2y + 14+ 3y =4 4. 3(N + 14) = 2(-N) + 67
Sy = -10 3N + 42 = -2N + 67
te 5N = 25
enya? en eS N=5
The solution is (5, —2).
5. Ashaded = ARectangle oe Atriangle
(1)(4) + 53)
4 + 6 = 10 units”
30
Algebra 2, Third Edition
Probiem Set 13
6. (10x + 10) + (6x + 10) 180 9. (a)23x
— y = 22
16x + 20 180 (a’) "y= Bx = 22
16xo= 160
(b) 2x + 3y = -11
10
Substitute (a’) into (b) and get:
(b) 2x + 3(3x = 22) = =11
llx -— 66 = -11
UNS fo
i eS
(a) y = 3(5) — 222 27
I < (5, -7)
10. (a) x + y = 20
@’) yy = 44120
Since lines are parallel, (b) 5x + 10y = 200
P = 10x + 10
Substitute (a’) into (b) and get:
P = 10(10) + 10 (b) 5x + 10(-x + 20) = 200
P-=-110 —5x + 200 = 200
—5x = 0
7. 5% = (AB)* + 4 io)
25 = (AB)? + 16
(a’) y = +0) + 20 = 20
9 = (AB)?
3 = AB (0, 20)
AC — AB = BC 11. (a) e+ y = 20
i2.-3 = BC (a) y = —x + 20
9 = BC (b) 25x + 10y = 395
(DCY = 9 + 4 Substitute (a’) into (b) and get:
(DC)? = 81 + 16 (b) 25x + 10(-x + 20) = 395
DC : 15x + 200 = 395
15x
= 195
8 (ayx=a y+! see 1
(b) 3x + 2y = 8 (a’) y = -(13) + 20 = 7
Substitute (a) into (b) and get: (13,7)
(b) 3(y + 1) + 2y = 8
YY Ac8 on. Ane 2
By +3 + 2y = 8 Aepgee "eee elt
a a a
Sy= 5
ya a k 4a? 4k? Fk
Se he eS ee ee
k 4 4k 4k 4k
(a) w= y+ 1
4a®
+ 5k?
x= (1) +1 =2
4k
(2,1)
2 am? p? amp m
[LORE ge Oreee 5G a
ap ap ap ap
e am?p* + amp + m
ap”
Algebra 2, Third Edition 31
Problem Set 13
15. 19. 0.03 + 0.03x = 0.003
2 + 30K 3
A0ne= 27
Pe27 = 209
Sa
20. 40
3? (oy = 2) eendx
—-| —-9 + 2x +. 2-=-/x
21.
16. (a) y = -3 22.
—3y
= -2x + 9
ma mp m
-3_,-2n-1
= —2a*p* + 4a~m™“p
xa*(x°a me o Aa a
23. Gx)? x4
ET VR)
mM” Pp
Pxx (Xx ) = — 3 = 9m*p
=
“x
i
(3p)? xpx
3x2 a ri 2x
25. xa — 5 —
= = *%a — 3xa + 2xa = 0
be 3a? m* é 5 pa
26. m! pa m
= amp"! ~ 3am2p"! + 5pam~~
17. (a) Every point is 5 units to the right of the y-axis.
= -2am*p + Spam?
eas
(b) The y-intercept is -2. The slope is negative and 27.
the rise over the run for any triangle drawn is . Ww N
I | | | | | | nN | |
N|H
aeae, Nile
Nee, ee
(=. Nl
eae
(WES
x= eee)
3 = -ptaletal- alas
8 ae 2 Zz NI
coffe 4% 2N2
gyno! & qe =S---
bi 4 2h
+t m—e—elUcTt
SOS
3 9
8 8 8 8
i] 65 39
a’ — ax(x — ax) = 47 — 4(-3)[-3 - (4)-3)]
aks — —_—_—_—_— ————
o 9 28.
265,23 52 16-4 12-38
x=eqre-cee
9 7 63 16 + 12(9) = 124
32 Algebra 2, Third Edition
Problem Set 14
—3(2N + 7) = 3(-N) + 9
=8+8-4-1
-6N
—- 21 = -3N
+9
-3N = 30
N = -10
30, 3° = 20-2) 2 )22 49.55] ou 4-6 = <1
Odd integers: N, N + 2,N +4
PRACTICE SET 14 4(N + 2 +N + 4) = 10N + 2
8N + 24 = 10N + 2
-2N
= -22
IW 1
The desired integers are 11, 13, and 15.
(a) 3x — ay = 21
(b) 2x —y = 12
(6 )cyr=207— 12
Substitute (b’) into (a) and get:
(a) 3x — 3(2x — 12) = 21
3x — 6x + 36 = 21
-3x = -15
Using (-2, 4), (4) = +—-2) + b bea)
4=2+b
(b’) y = 265) - 12 = 2
e=
The equation is y = -x + 2.
(5, -2)
(a) 4x -— y = 22
2
b y =—-—x+b
ae (a’) y = 4x — 22
(6)
6) =
2
ac
——(-3) + b
(b) 2x + 3y = 4
Substitute (a’) into (b) and get:
62.27)
(b) 2x + 3(4x — 22) = 4
4=b
2x + 12x - 66 = 4
ae 2 (ae = 70)
The equation is y = 34 + 4,
52 = 1S)
(a’) y = 4(5) - 22 = -2
PROBLEM SET 14
(5, -2)
Ie
1. — x of = is (a) x+y = 28
100 f
(a’) y = -x + 28
ib x R-= 6578
100 (b) Sx + 10y = 230
Rea= 65 15 vied = 2530 rats Substitute (a’) into (b) and get:
260
(b) 5x + 10(—x + 28) = 230
P 5x — 10x + 280 = 230
2 — x of = is
100 f —5x = -50
x = 10
Ba x 4900 = S$
100 (a’) y = -(10) + 28 = 18
_ 350 x 4900 = 17,150 sheep (10, 18)
100
33
Algebra 2, Third Edition
Problem Set 14
14. (a)
x = -4
(a)x + y = 22
(b) 3y + 2x = 6
(a’) x = -y + 22
3y = -2x
+6
(b) 100x + 25y = 2050 2
y = —-—x+2
3
Substitute (a’) into (b) and get:
(b) 100(-y + 22) + 25y = 2050
-100y + 2200 + 25y = 2050
-75y = -150
Vee
GQ) xe=-(C) +-22 ="20
(20, 2)
reap. One
Z 7 2
y cy cy cy cy
Fr cy’x + cyx? = 3x
Ss cy” 15. (a) The y-intercept is +4. The slope is positive and
the rise over the run for any triangle drawn is Z.
2
10. —+4 eae re = —-xr+4
x x as
(b) Every point on this line is 4 units below the x-axis.
Cc 4x Cc cx? y
11. 44+—-oy = —+-—- yield
x 25 x x
_ 4x +e - cx’y 16. Graph the line to find the slope.
x
12. oa
1sa
81 = H2 +4
17 = H2
V71=
B
Area = === = EE So cm”
13.
y=-lx
+ b
4 = -l(-3)
+b
l=b
Since m = -1 and b = 1, y = -x + 1.
3
17. y =) 7" 2D
3
6 = -—(-4)
as ) + b
3=5b
Since
m = ——ee
3
and b = 3 ae ee Forte
3
34 Algebra 2, Third Edition
Problem Set 15
18. A + 134 = 180
3xy 7x2x “ 8m_!x7!
A = 46 26.
m y xy
B + 46 + 90 = 180 SS peeky 1X. _ 82Y ape Lixyeny 7x
B= 44 m ym m m ym
C+ 44-= 180
pera ia alas he)
2 2
P406 bore ae |) Se | =
C = 136
a i ( 5) 2 Bye
1 1 1
1 =S--->-+—_—
19, A=
8B =—- 30 =915 9 2 (2
2
A=B=15
C + 150 = 180
28. atx — a(xa — a)
Gr=250
= (-1)(-3) - CD[-3CD - CD]
Since the measure of an arc of a circle is the same as —3+413+1)=3+4=1
the measure of the central angle, D = C = 30.
1
29. =59(2 = 3°) — 273 = 3 id eA
20. * ae ee (-2)?
=) 20 = -5
9 foe le
—-x = -——
20
a
20
30. a Lge pe IR) Sig ec:
Spee 5 eeSha
PRACTICE SET 15
21. 0.005x — 0.05 = 0.5
(a) 4x + 3y = 17
5x — 50 = 500
5x
= 550
(b) -3x + 4y = 6
x = 110 ola) tax A Oy 25]
4(b) -12x + l6oy = 24
22. aU g te ears (20-5) Py ak— vet
fe
=tiG-4 1204-4-9-— 8x + 20 y= 3
—10 = —10x + 22
4x + 3(3) = 17
—32 = —10x “We = tes
16 58
ie The solution is the ordered pair (2, 3).
ee 9 yx eee | xy’x 3xx Check: —3x + 4y = 6
23. —3(2) + 4(3) = 6
y —Xx y yx 9yy
—-6 + 12 = 6
=
x
— + —
x
Graa0
y 3
2x2 xy yxy? x Paxy?a?
24. PROBLEM SET 15
(xy)? xyxy Ax yx are
oe a y Ai aa STO ab
4x? y3 4x
es
20. Pp= 1400
25. mp'{m ile =4m |Seen Was
ip 4mp*m
ii 100
v Hl 1400 x = 7000 people
Algebra 2, Third Edition 5S
Problem Set 15
7. (a) 3x + y = 16
WP :
— xX of = is
100 (a’) y = -3x + 16
WP x 1000 = 200 (b) 2x - 3y = -4
100
Substitute (a’) into (b) and get:
WP = 200 x ao = 20 percent (b) 2x — 3(-3x + 16) = -4
2x + 9x — 48 -4
5(-N) — 7 = 2N - 35 11x 44
-5N — 7 = 2N —- 35 ie 4
-7N = -28 (a’) y = -3(4) + 16 = 4
N=4
(4, 4)
Integers: N, N + 1, N + 2
(a) x + 3y = -9
-4(N + N + 2) = 7-N —- 1) + 12
(a’) x = -3y - 9
-8N —- 8 = -7IN +5
(box — Dye = 23
Nea 3
N = -13 Substitute (a’) into (b) and get:
The desired integers are -13, -12, and -11.
(0) 3G3y = 7) = 2y = 23
(sy 145 = 25 23
(a) 3x + y = -17y
(b) 3x - 2y = 2 ay
2(a).6x.+R2y =.22
Cpe
(Dye a mac Ve 2 (3, -4)
9x =
a 3
2
= 3a” y? + 3xy? —- a~mx
(a) (3) + y = ili
3
Sat yee 1d
10.
y =3
11.
(a) 3x + 4y = 20 CXEG ce ar ac*x
(b) —4x + 3y = 15 x
4(a) 12x + l6y = 80 12. a” + be
3(b) -12x + 9y = 45 Aa 82
25y = 125 16 + 64
80
(a) 3x + 4(5) = 20 = Z| Sg
3x + 20 = 20
Shy=
(0, 5)
36 Algebra 2, Third Edition
Problem Set 15
16. Graph the line to find the slope.
Ron
\
i—|—I—
aoa
14. (a)
y = -3
(b) 3x - 4y = 8
—4y
= -3x + 8
3
y <2
re
17.
15. (a) Every point on this line is $ units above the N I || f No — + >
X-axis.
(b) The y-intercept is —1. The slope is negative and
the rise over the run for any triangle drawn is —4.
2
=-—x-1l
. 3
Since) m = —— and b = =,
Algebra 2, Third Edition 37
Practice Set 16
18. A + 134 = 180
A = 46 Oxc=1t-
8x =
B + 46 + 90 = 180
B= 44 ere:
8
Since vertical angles are equal angles,
C= B= 44, xy 2x%y » x*y? A lycs
25. xy (xo)? gyn y
k + 44 + 90 = 180
k = 46
8x yy eae 4 5m? x?
Since vertical angles are equal angles, 26. m2 x2 m2 x2 ye
y=k= 46. = 8m~y2 + 3m7y~ 5m~y" = 6m7y
19. A + B+ 140 = 180 ‘S
A+ B= 40 27. | Te= ll | LoS)
ae —
| | | Ww—
=
laN|e
Se Wile
eae N|e
pe
Since angles opposite equal sides are equal angles,
A = B = 20.
C + 140 = 180
C = 40
28. a(a°x — ax) = (-2)°[C2)°-3) = 2)C3)]
= 1(-3 — 6) = -9
Since the measure of an arc of a circle is the same as
the measure of the central angle, D = C = 40. 29. =" -3*)—=|-2 ae GC 2)]
= -1(1 - 9) - [-3] = -1(-8) + 3 = 8 + 3
sa, al
20. = Ul
2
6x + 4 = 9x - 17
3
3x
= -21 30. 3° — coy ee ~ SIS 27 2 = ies
rea7
P = 3(7) + 2 = 23 PRACTICE SET 16
{=e a =| 4x? + 9x + 34 + 8.
21.
p \ yeh pr” x — S)Ae =* 3,5 7hR tS
Ax? ~ 12x?
é. 3x —1_2
a 7 9x tee)
me ee Oxy3p?
Ox2 + 7x
Py Pp 9x? — 27x
34x + 3
22. aed A = jk 34x — 102
k 10 105
iW fe 8
Check:
5 10 +10
4x*(x — 3) _ Dez = 3) | 4G = 3)
17 81
—x = —
x-3 x -3 x -—3
5 10 105
x -—3
» 4x3 = 2x2 49x?
23. 0.07x — 0.02 = 0.4 cers
1x — 2,-=
40
x -—3
Ix
= 42
= 4x? = 3x? 4 ye 8
5
ee se)
38 Algebra 2, Third Edition
Problem Set 16
PROBLEM SET 16 6. (a) 7x - 2y = 13
1 Roeof =a . (b) 4x + Ty = 40
a1 x AN = 900 7(a) 49x - 14y = 91
? 2(b) 8x + 14y = 80
S/, = 1
AN = 900 x gh = 420 - 71
15 5et—
2. WD x of = is (b) 4(3) + 7y = 40
0.016 x 420,000 = D Ty = 28
6720 microbes = D Ua
(3, 4)
3. If 14% were uninvited, 86% had been invited.
UPR CS yo) oe Si ces
——en MOS = IF
100 (a Ly = 3x44 2
wo
86
x GA = 903 One >)
.
100
+p Substitute (a’) into (b) and get:
GA = 903 x ria 1050 guests (b) 2x0 53h + 2 7
Ines Moye. = A 7
ri ae rt oer ie 17
100
al
40
aE a ieee
(a joe = 311i 2 Sl
100 (1, -1)
C = 8600 x — = 21,500
40
8. (a) x -3y =4
Np = 21,500 — Ny = 21,500 — 8600
= 12,900 {a')\ x = By +4
(b) 4x - Jy = 16
5. (a) 4x - 3y = -l
a Substitute (a’) into (b) and get:
SAP ie ed (b) 4(3y #4) - Ty = 16
Ka) 4x.— Sy = =I 12) 4016 = 7y = 16
—2(b) -4x — 10y = -38 sy = 0
-13y y = -39 ae
yas
(a’) x = 30) +4 = 4
(b) 2x + 5(3) = 19
ox = 4 (4, 0)
x=2 9, (2x + 3)(2x2 — 4x + 3)
(2, 3) = dy? — 8x2 + 6x + 6x” — 12x + 9
= 4x> — 2x2 - 6x + 9
39
Algebra 2, Third Edition
Problem Set 16
15.
104
4x2 + 10x + 344 x-3
10 es a: a
Ax? — 12x?
10x? + 4x
10x? — 30x
34x + 2
34x — 102
104
Check:
Ax*(x — 3) e 10 x(x —- 3) . 34(x — 3)
x —-3 x —3 x —-3
104 Ax? — 2x2 + 4x +2
x -3 x -— 3
x74+ 4x +16 + — 4
11. pet, omer ee Ons
16. (a) x= Dye
x3 — 4x? a‘ 2
4x? + Ox
(b) x - 3y = 6
Ax? — 16x
16x — 8 3y
= -x+6
16x — 64
56 eee
Check:
x(x 4) A 4x(x — 4)
x-4 x-4
Loe Bt, 56
x-4 y= 4
x? — 4x2 + 4x2 — 16x + 16x — 8
x= 4
wy? $8
—4
a 4x? a 4x* +a
12. 2h ee = 5 17. (a) Every point on this line is 4 units below the x-axis.
Dig se Ds 2x
hae.
(b) The y-intercept is +3. The slope is negative and
13. Li i mee pp lbeg Ae 8s the rise over the run for any triangle drawn is —>.
Cx 4c*x Ac?x Ac? x 4c*x
1
_ 16c + 4c3x - 3 = —-—x
+ 3
eS
4c7x
14. Do eH”+ 12
4=H* +1
3 = H
eS el
Apeneeeee 2x V3 _ 3 in.”
40 Algebra 2, Third Edition
Problein Set 16
18. Graph the line to find the slope. 20. Areas = ars? = mcm?
roi lcm
ry = 2rg = 2cm
Area, = mr,” = m(2cm)* = 4aem?
21. Since angles opposite equal sides are equal angles,
A = 30.
B + 30 + 30 = 180
B = 120
C + 120 = 180
Seni oe C = 60
—7 7
2 Since the measure of an arc of a circle is the same as
y= “7% +b the measure of the central angle, D = C = 60.
-4=
eea) +-b 22.
1
3—x - —
1 = 3—
3
2 5 20
Preget
4
jet)
—
o3tterS
ee
2, 20 20
pete Aes, gett
iv ah ZT 5G
-=7 = VeTore
“2027
te Weo%
2-70
: 32
BINGE Bia peel DSS = 23. 0.05m — 0.05 = 0.5
32 Smo) =)50)
y=-—rx - —
7 7 TiS
m= 11
19. y = —x + dD
2, BAe yee S) = ws 2)
1 = —(-4) + b Ax +8 +18 =— +2
3 5x = —24
1=—— +b
5 ae 2
3 ee 12 = b
5 5 ace xy 18x7z 3xyz?
17 25. oom
=} Z y 7%
‘i A 2x yx2z x7? yxyz? 2 yz
E Fa ial 2
17 a
zy ONE 3x
Since m = z and b = a.
3 v7 ae DP 2 ae eye x4
dir aban ory Sexy) Genre ae yg
7 3x? xxy . Dae _ 1xxyx
wig Cabo eG) Ae
41
Algebra 2, Third Edition
Problem Set 17
28. epi (Dee) 20x + Sy + 40 + 2x + Sy = 180
ifi s 25] E (-2(3 = 22x + 10y = 140
40 + 2x + Sy + 110 = 180
2x +.5y = 30
+ 4 2 Mare AA 4 (a) 22x + 10y = 140
—2(b) —-4x -— 10y Ht | a o
29. 22) te28) o-DP
See (-3 7) | 18x = 80
ese 4 7-1-1064
Sy 1 ifs) =o 1s= or 22
30. ee esis 30°. |-30RtoI O5E3)
3
See 3 = 14 2 3 = 6
= M oS)=) l |
PRACTICE SET 17
—5(a) -5Nn = 5No = -300
(b) 5Ny + 10NQ = 310
5No = 10
No = 2
Ny + (2) = 60 PROBLEM SET 17
Nn = 58
If 2doubted, then 2 were nondoubters.
Feoxcof seins
Ry = 50
Rw = 70 25-600
5
Ty = 5 — Tw
= TOO0n x : = 1000 students
Rylu st RyTw = 380
(505 - Ty) + (70)Ty = 380
FOxropr= tis
250 = 50ly + 10Tw = 380
20Ty = 130 ORT= 340
13
pe ae13
oe 25 iP DINO) Sx ° = 390 teachers
jee io™see 13
AS ue 3 x 400 = §
3
I Bs
S = 1440 skeptics
Integers: N, N+ 1,N +2
71(N + 1 +N + 2) = 10N + 109
14N + 21 = 10N + 109
4N = 88
N= 22
The desired integers are 22, 23, and 24.
42
Algebra 2, Third Edition
Problem Set 17
(a) Ny + Np = 150 Substitute (b) and (c) into (a) and get:
(b) SNy + 10Np = 450 (a’) 50Ty + 80Ty = 250
-5(a) -SNy — 5Np = -750 Substitute Ty, = 5 — Ty into (a’) and get:
l(b) SNy + 10Np = 450
(a”) 50(5 — Ty) + 80Ty = 250
5Np = -300
250 — 50Ty + 80Ty = 250
30Ty = 0
(a) Ny + (-60) = 150
Nw = 210 Ty = 0
Ty = 5 - Tw
(a) Np + Np = 50
Ty = 5
(b) Np + 10Np = 140
(b’) Np = -10Np + 140 10. (a) ReTe = RywTw
Substitute (b’) into (a) and get: (b) Re = 200
(a) (-10Np + 140) + Np = 50
(c) Rw = 250
-ONp = -90
Np "10 (d) 9 - Tp = Ty
(a) Np + (10) = 50 Substitute (b) and (c) into (a) and get:
Np = 40 (a’) 2007; = 250Ty
Qxr 4 3)0x"
+ 254 2) Substitute (d) into (a’) and get:
= 4x° + 4x2 + 4x + 6x7 + 6x + 6 (a”) 2007; = 250(9 — Tz)
= 4x3 + 10x? + 10x + 6
2007, = 2250 - 250T;
$x71—
axe Bh— i 450Tp = 2250
Keb LI3x° + Ox? + Ox —.2 Tp = 5
3x?
+ 3x?
~3x?
+ Ox yao = Te
35283%
Ty = 4
3x —2
Spe cus
=)
11. (a) RyTy = ReTe
Check: (b) Ry = 8
oe Oe 3x(x + 1) (c) Rp = 2
x+1 x+1
f Aree vy
(hy earre 7
x+1 x+1 Substitute (b) and (c) into (a) and get:
_ 3x3 43x? — 3x? - 3x + 3x -2 (a’) 8Ty = 2Tr
x+1
Substitute (d) into (a’) and get:
357-1
x+1 (a”) 8Ty = 2(5 — Ty)
(a) RyTu + RyTw = 250
Ty = 10 ST
10Ty= 10
(b) Ry = 50
Ty =1
(c) Rw = 80
Gary 4 Tyi='5 Ti Ss Ty,
Type Te Tp= 4
Algebra 2, Third Edition 43
Problem Set 17
17. (a) The y-intercept is —-2. The slope is negative and
12. dpe oe ee 3x _ dax + 3x the rise over the run for any triangle drawn is —2.
a a a a
y=-2x -2
13) eee
»
(b) Every point on this line is 4 units to the right of
y np the y-axis.
7 -2np*x " cnp? xy ig Ixy? x = 4
np’y np*y — mpy 18. Graph the line to find the slope.
a —2np*x - cnp?xy + Tx2y?
npy
4. Paw + 6
49 = H* + 36
13 = H
V13 =H
ene I x Jal = 12m x J13m
a p}
= 6/13 m2
15.
5 5
Slopeope = = 5
5
ui = =—x
a +d
5
a
2=-—(4) +b
12=b
Since m = -2 and b = 12, y = ~Sx + 12.
2
19. y= oe +b
2
j= ea), + b
64 2.535= 29
16. The OP eae
Since nm= == audits ye bed oe
7 i 7 7
20. (2x + 10y) + (-2x + 6y) + 60 = 180
l6y = 120
. 45
<r
Since vertical angles are equal angles,
-2x + 6y = 50
—2x + 2] = 50
2
—2x = 5
eee
2
44 Algebra 2, Third Edition
Problem Set 18
(3)
z=
2x + 10y
PRACTICE SET 18
z Totals= 37+ 11
= 14
z=-5
+ 75 = 70 "Dye MG
14-350
21. Volume = zr*l = 2452 cm?
14M = 1050
Pe 2452cm?
M = 75 malefactors
(5cm)x
49
1 = —cm NS = 4000 — 800 = 3200
5
23200 NS
ay eee cee ee 4000 5000
5 4 20
4000NS = 16,000,000
11 *! 65
—rx = — + e's NS = 4000 kg
fe) 20 20
ee J. 3 yh 18
20 11 220 11
23. 0.003 — 0.03 + 0.3x = 3.3
3 — 30 + 300x = 3300
300x = 3327
$= 100
= 11.49
WA, 32x 5) + Ge 25) = (912)
x= 43
Oty |ty --ax4y2
25. 5 —= =
x-y m m
ns xy! 3x? 1 3x7m? PROBLEM SET 18
x3 ym x ym~* y?m y
2 zie
Me 97-3 ,0 (x?) 7 x2 a xty?
21 420
ig 3xyy Gey 8 2 + 420 = 21A
4.4 ; 840
Zils aS - a i, = xy? ~ 3xy? + Ty
40 Arabians
yy sae xy
= -2xy? + Ty
2500
8. sy x2 y)= 2(-5] - f 2-(-
]
3)
] 3000
2500 - 6000
=-] —- ZO) = -3
5000 kg
1 ;
29. -(-3 — 2) + 4(-2) + syst oa
1 7
54,000
0. 7) = = 4] = Sit20F 7]
54,000 x > = 200,000 natives
= -9 — 6 — 3(5) = -30
Algebra 2, Third Edition 45
Problem Set 18
9. (a) RrTp = RsTs
P
4. —— XxX Op = tS
100 f (b) Ts = 6
0.004 . TP = 40 (c) Tr = 5
100
(d) Rr - 16 = RS
TP = 40 x ee = 1,000,000 Substitute (b) and (c) into (a) and get:
0.004
(a’) 5Rpr = 6Rs
5. (a) Np + Ng = 200 Substitute (d) into (a’) and get:
(b) 10Np + 25Ng = 2750 (a”) SR = 6(Rp - 16)
-10(a) -10Np — 10Ng = -2000 5Rp = 6Rp - 96
1(b) 10Np + 25Ng = 2750 Rp = 96
I5Ng = 750 (d) (96) — 16 = Rs = 80
No = 50
(a) Np + (50) = 200 10. (a) RuTu = RrTr
Np = 150 (b) Ry = 8
(eRe
6. (a) Np + Np = 30
(anes Suri
(b) Np + 10Np = 291
Substitute (b) and (c) into (a) and get:
Substitute Np = 30 — Np into (b) and get:
(a’) 8Ty = 2Tp
(b’) (30 — Np) + 10Np = 291
Substitute
(d) into (a’) and get:
9Np = 261
(a”) 8Ty = 265 - Ty)
Np = 29
8Ty = 10 = 2Ty
(a) Np + (29) = 30
107, = 10
Np = 1
Ty = 1
7. (Qe. + 4G yt = 10) (d) Tr = 5 —- Ty
= 6x? — 4x? — 20x + 12x? - 8x - 40 Tr=4
= 6x> + 8x? — 28x - 40
11. (a) RgTg + RgTz = 100
5x? + 10x + 20 + —¥> Z) () ke =4
Gos ses Ox + OA
5x? — 10x? (c) Rg = 10
10x? + Ox (d) Tp = Tg + 3
10x? — 20x
20x - 1 Substitute (b) and (c) into (a) and get:
20x — 40 (a’) 47g + 10Tz = 100
39
Substitute (d) into (a’) and get:
Check:
(a”) 4Tg + 10(Tg + 3) = 100
5x2(x — 2) _ l0xla = 2), 20% - »
4Tg + 10Tg + 30 = 100
x-2 x-2 x= 2
14T, = 70
39 Sx3 — 10x? + 10x?
Tgt= §
x-2 x -2
se Belepet
20.105 140.9439" 5x° = 1 (d) TR Tgcti3
x —2 x-2 Tz = 8
46 Algebra 2, Third Edition
Problem Set 18
22222 Ae (a) Every point on this line is 3 units below the x-axis.
ee ee
xyz XYZ xyz Me tic)
F 7x2 y2z2 1
(b) The y-intercept is 0. The slope is negative and
xyz the rise over the run for any triangle drawn is —3.
ae: 3 y = -3x
13 eg RE es Ve
Dy xy xy xy
18. Graph the line to find the slope.
5 -3x7y? — cxy? + 7c
xy?
14. 6* = H* +5
36 = H* + 25
11 = H?
V11 cm = d
¥ee bx HH f' 10 cm x V11l.cm
2 2
= 5/11 cm?
15.
y=—x+b)D
5 = -(-3)
+b
= 2
Since m = —1 and b = 2, y = -x + 2.
5
19; = —x+ D
seal
—2= —(4) +b
a age
3
20"
68
16. 3 3
26 A;
3
26 5 26
Since
i m = and b= Fi
-—, = 3 3
-x-—.
20. 3x
SF =9
SF =3
a°=' 46h
ea 43) = 12
as SF
bpe=*5(6)) "15
47
Algebra 2, Third Edition
Problem Set 19
PRACTICE SET 19
21.
2 a. (a) N + D = 80
6x + 4 = 5x + 10
(b) 0.05N + 0.10D 6.50
=O
-5(a) -5N — 5D = —400
P = 3() + 2 = 20
100(b) 5N + 10D = 650
2
22. Area =. 129(gppyeale itn ea ses 1? 5D = 250
360 D = 50 dimes
N + (50) = 80
23. ee - ae: = i 2
6 a, ie N = 30 nickels
ian
6 24”
a es24 (a) R49 D' ="35
25% _ 45=45
25 (b) 12R + 4D = 300
A te -4(a) -4R — 4D = -140
(b) 12R + 4D = 300
8R = 160
24. —2[(@ — 2) -— 4x - 3] = -—C4 — 2x) R = 20
—2[-3x - 5] = 4 + 2x
(20) + D = 35 :
6x + 10 = 4 + 2%
D = 15 daffodils
Ave 6)
5 3SF
=5
x= -—
2
Shane3
ayy 2xyz7! a y*
Pan%
oe p Ee aS 5S >
BOvyi ee Gy 227xe 3
zp 2x pyz> z2
25
x= —
ao (a) (xy Ne e de wey
3
26. 5
ax? (y~*)? ay~* = 4|—
a x>y? y ]
a>
_ 20
Cae
ax ~ Ch elo 7 xxy
27. ve x y?
Da? 5x? Tp PROBLEM SET 19
7 2y = eee! (a) Ny + Np = 60
x-men Ht)- (4-2
y Mv
(b) 5Ny + 10Np = 500
28.
Substitute Np = 60 — Ny into (b) and get:
(b’) 5Ny + 10(60 — Ny) = 500
5Ny + 600 — 10Ny = 500
29. ~2|-2 = '5| + C3)|-2C2) = 3\s4 7 -5Ny = —100
= -2(7) + €3)d) + 7-= —-14 = 394 7-=)-10 Nw = 20 nickels
|
————_ +
3
——
(a) (20) + Np = 60
30. = § - 27 = -19
Np = 40 dimes
48 Algebra 2, Third Edition
Problem Set 19
2. (a) Nc ote Nu =" 20
tet ee pe 26
(b) 7INc + Ny = 86 x = 2)43x2
4 Ox? +7 0x — 2
x-2
Substitute Ny = 26 — Nc into (b) and get: —3x3 + 6x?
(b’) 1Nc at (26 - Nc) = 86 -~6x2 + Ox
—6x" + 12x
6Nc = 60 aioe ban?
Nc = 10 codfish -12x + 24
~26
2 49 294
Check:
49NS = 294(33) an? (xs 2) prox 2yt BLDC — 2)
x-2 ~ Xg? OF & et?
NS = 198 tons
26. 3x? + 6x? — 6x? + 12x
4. If 19% was used, 81% remained. x-2 x-2
r
Tae is
-12x + 24-26
3x3 -2
x-2 | ee 2
ome 1134
100 (a) RyTy ate RsTs = ite)
NA = 1134 x uy (b) Ry = 70
81
NA = 1400
(c) Rs = 20
Beginning amount = 1400 liters (d) Ty = Ts
Amount used = 1400 — 1134 Substitute (b) and (c) into (a) and get:
Amount used = 266 liters (a’) 707, + 20T; = 180
5. (-7N — 7)2 = -5N + 4 Substitute (d) into (a’) and get:
-14N — 14 = -5N +4 (a”) 70Ty + 20(Ty) = 180
-9N = 18 90Ty = 180
N = -2 Ty = 2
6. (a) 5x + 25y = -160 (djare = 2
(b) —3x + 2y = -23
10. (a) RrTp = RsTs
3(a) 15x + 75y = —480
(b) Ty = 6
5(b) —15x + 10y = -115
85y = —595 (c) Tr = 5
y=-] ‘(d) Rp — 10 = Rg
(b) -3x + 2(-7) = -23
Substitute (b) and (c) into (a) and get:
—3x = -9
(a’) 5Rp = 6Rs
Gt
Substitute (d) into (a’) and get:
(3, -7)
(a”) 5Rr = 6(Rp - 10)
7. Gr = 2a — 281-24) SR =" 6Rp=-60
le 4x
ar go Fy) De Ae
Rp = 60
+ 4x
-— 8
= x> — 2x4 — 4x3 + 8x7 + 4x - 8 (d) Ry = (60) — 10 = 50
Algebra 2, Third Edition 49
Problem Set 19
11. = wn
(a) RyTy = RrTr
(b) Ry = 8
(c) Rp =
(d) Tr 5 - Ty
Substitute (b) and (c) into (a) and get:
(a’) 8Ty = 2TrR
Substitute (d) into (a’) and get:
(a’”’) 8 Tu 2(5 — Ty)
Siyi= 10 OT),
10Ty = 10
Dea 5
D2
81
+ 25:
Ty = J D* = 106
(d) Tr 5-(1)=4 D = 106
ole —_—
i)
S
sal
= a +
gol
<r
ws
Exlle
po)a aSSie Res
|\
23]
seo
ay
Re Se
o R basispas
Ww ~
H] on
—3y = -5x +9
er) s| Sie A]
ne
bo
| i) | |
re nN eS N
QIN
Ss]
oR ee
alN
<<) I I
a IlnN
Py Pi
Pesceg JERSau2248
42702
nN
8 e ya
IlN as
RANA
eo
ERR
+ ia Qe) Cl 18h
=
ht a
RRR
Po
a
>lS
Aes
baie
<7
OSE peer28
SESS o
x
eo KO en LIN See
eS
nN
sae
NOUS
ane
NN
a aeas ac
4SoO=
17. (a) Every point on this line is 2 units above the x-axis.
(b) The y-intercept is 0. The slope is positive and
the rise over the run for any triangle drawn is 2.
y= 2%
18. Graph the line to find the slope.
50 Algebra 2, Third Edition
Problem Set 19
Slope = eee J Since angles opposite equal sides are equal angles,
+5 5 Ga="D,
9 C + D + (130) = 180
+ ihe ld Cs D/S50
Ga D =75
see ect
om eae 50 50
Ar ea SS 360
——_ a ar 2 = ——. 360 (3 3 cm) 2
—) === +79 ;
= = 3.93 cm
Beek: 25
Dat (18 YR apie eee
> 4 3 12
28
3 13
pve = ewokhe
4 12 12
ee y 5 :
a (2.6)13 eee
156 39
19, y =e 23; 0.07 0 p= 7
y 1x = 10%= -770
-5§ = =(3) cut 8 7x = -700
x = -100
Maen. 24, [02.2 6-2) 282)
CA eweP tery = =2[@ 2°2)2 = @x = 3)3i
7 ee ee {4 412-2)
= -2[2x < 4 - 6x & 9]
1: Palle -14 = -2[-4x + 5]
re 14 er 10
—4 = &x
Bie oe ant ha ee te
7 7 7 1
-—- =x
Z
20265c 4 15 =.6%
-—x = -20 25 erhe x Baye 3xy? Oxy?
* Seo oobi SgOF yr Ss
a= 20 P Pay xp y x
3
3 x
seal
5 =— —
ba uenie Py x
3
BD) 3 4
ge ped ee
9 x(y°y?) xy *
~ =y 7 2xtne Biya | -4x°y
9 ; y
eyVimex aes a Rea
Oe
s = iy = 2x2a = 3x7a rm 4 x2a ib 3x7a
9
y by y Ay
k=. 6% —-5_2 620) = 3 = 115
28. m — m’y(m -— y)
21. Since the measure of an arc of a circle is the same as 1 12/1 1 \
the measure ofthe central angle, A = 50. = “3 a (-3} (=\-3 = A
B + 50 = 180 Pe" s(-3)=-3 + Gwe 35
B = 130 : 3 «(54 Dalaiv SB = at08 108
Algebra 2, Third Edition 51
Problem Set 20
pone less ae joao ers (1a2)9)] 3.. (a) Ng = Nw + 3
=7i[3 - 4-3(2)] =-ll 3-446] (b) 7Ng + 2Nw = 111
= -l[-l] =1 Substitute Nz = Ny + 3 into (b) and get:
(b’) 1(Nw + 3) ae 2Nw =e len
30. zs Ee ee 1 16 ONy = 90
"2@y2 * =a Nw = 10 worthless ones
(a) Ng = (10) + 3
PRACTICE SET 20 Ng = 13 expensive ones
a. 4/40 -— 3140
P :
iy 6 Oe ean es Pa en aa, A Ae Kak mts
= 4/2/2J2V5 - 3V2V2V5V7 ya
= 8/10 - 635 700 * © = 200
b. 3V2(3V2 - V8) = 9V2V2 - 3/28 Oh
= 18 — 3¥2¥2/2 V2 = 18 - 12 = 6 +i
Nwypg = 600 — 240 = 360 tons
G 3s = 5
By Pes: 5. Even integers: N, N + 2, N + 4, N + 6
1 5 -4(N+N+
6) =-(N+2+N+4+4)+6
Jaeige e tex -8N — 24 = -2N
For the parallel line aON 5.28
\ N=-4
area t4 The desired integers are —4, -2, 0, and 2.
G) = 30)+b 6 eee
ae (b) 2y + 2x = 40
ye Substitute y = 2x + 8 into (b) and get:
(b’) 2(2x + 8) + 2x = 40
ysart2 6x + 16 = 40
Gx>= 24
x=4
PROBLEM SET 20 (a) y - 2(4) = 8
yas160 Se
640 (4, 16)
vere
160NC = 640(140)
NC = 560 kg
3 . i 4
2. If jo were virtuosos, 75 were not virtuosos.
F Xx of = 1s
Top a 38
10
Peas ox : = 40 performers
Ny = 40 — 28 = 12 virtuosos
52
Algebra 2, Third Edition
Problem Set 20
a 3K
Sry es pe 13. —-a-
cel Ja 2xe3 ee”2 Same et 2 ie x? 2a‘
-2x3 + 2x? 2a° 2a>x? 3x3
Ee) i
—2x?+ 2x
—3x,+2
—3x +3
«I
14. 42 = H2 + 32
Check: 16 = H2 +9
-2x*(x-
1) 2x(x- 1) 3(x -1) 7 = H
x-1 x-1 x -1 J7 =H
yp a ee SE oe ox ie 2x pram DRM. SAXNTR _ 57 gp
x-1 x -1
—3x+3-1 -2x3-x+2
x-1 A x-1l
15.
(a) RgTg + RgTg = 100
(b) Rg = 4
(c) Rg = 10
Ce ee ee
Substitute (b) and (c) into (a) and get:
(a’) 4Tg + 10Tz = 100 16.
Substitute (d) into (a’) and get:
(a”) 4Tg + 10(Tg + 3) = 100
14Tg = 70
Tg = 5
(d) Tz = (5) + 3
T, = 8
3./862-44/12 +25-/300
AY CV ee Gan pie poe ee
PO Liot © ON PREPPEDROR ER ERE
BA ger ao— S03 072 ~ SOUS 17. (a) Every point on this line is 1 unit above the x-axis.
as
10. 43(2V3 - V6) = 8V3V3 - 4V3V3V2
(b) The y-intercept is 0. The slope is negative and
= 24 - 12/2 the rise over the run for any triangle drawn is —1.
11. 5/5(2V5 - 3V10).= 10V5V5 - 15¥5V5V2 y=-x
Sod 4 thas Leip
m2 5 m m2 5x mx
12. ime oea Ma
b cag’ ¢ ax ax ax ax
Algebra 2, Third Edition
53
Problem Set 20
18. Write the equation of the given line in slope- he gh els
intercept form. 21.
12 6 24
3y —-x =3 widaeTie lel
3y =x +3 6 24 24
he, 6 150 25
Ue 39 gt dbl
1
* Scale, |.1 D4vibiedl
Since parallel lines have the same slope, 22. —0.04x -— x — 0.2x = 6.2
—4x — 100x — 20x = 620
1
Lala =-x+b
as x = 5
Beer 2S, —2[ C32: =x) +.2@ = 3)) = =2e
3 —2[6 + 3x + 2x - 6]
ca Deleae —10x ool Bal
3 3 x ll S
whoa; =x°y (2_ =) ba sii 3xy?
3 24. oa y y
y ys
: 1
Since m = — and b = -—= y = =x -— —.
3 3 3
25. (909)
a yyy? Ag NM ee
(ay) (2am) og neyo 2x?
19 Rae
12 26.
39 r (a°)a Se =~ 3%
SPO) + b
12
1
2) D
6 27. =
| = & =
NO | & —_
rn
oe
6 6
>
1 (-2)2)(-
ch, eis
1 my (all
ollieUlett
Z
C—O
1 1 ( 1
6 4, 32,0161 5 16 4 32 16
1 1 128 1 129
Since m = = and) Sey es Ae Sie 519% 251201, 6512
6 12
28. -29{ (7 shui) he
cee 2-2 = (-5)]}
20. A + 140 = 180
a at=a ey = ola} = -1{-19} = 19
> Il 40
B + 40 + 90 = 180
29. -2°|-3 - 7| + |(-5| — 2(-5) + 2
=-10+1+10+2 = 3
B = 50
Since vertical angles are equal angles, sowie tien Ane
30. = -27 + 54 = 27
C = B = 50. ~(-3)3 ~(-3)3
P+ 50. + 90 = 180
P = 40
Since vertical angles are equal angles,
y=
P = 40;
54 Algebra 2, Third Edition
Problem Set 21
PRACTICE SET 21 Si (a) 5Ny + 10Np = 575
N 4 (b) Ny + Np = 70
Y pss l(a) 5Ny + 10Np = 575
(b) VN,+ D.= 108) =. __D.= 108.— N -10(b) -10Ny —- 10Np = -700
Substitute D = 108 — N into (a) and get: —5Ny = -125
(a’) 41108 - N) = 5N Ny = 25
432 — 4N = 5N
Integers: N, N + 1, N + 2
432 = 9N &:
48= N —S(N + N + 2) = 4-N — 1) + 24
-10N - 10 = -4N + 20
(b) D = 108 —- (48)
-6N = 30
D = 60
N=-5
A+ B= 136 The desired integers are —5, —4, and —3.
A-B= 50
2A = 186 7 (a)_8y = 3x S222
A = 93 (b) 2y + 4x = 34
(93) + B = 136 Lye SY 9" =f
B = 43 —4(b) -8y — 16x = -136
-19x = -114
PROBLEM SET 21 x.=.6
(b) 2y + 4(6) = 34
(oes — 5N
= 3D 2y = 110
wae
Ve 2)
ON
+ Da. -— Ds52N
Substitute D = 96 — N into (a) and get: (6, 5)
(a’) 5N = 3(96 — N) 8. (4x2
— 2x + 2)(2x
— 3)
8N = 288 = 8x2 — 4x2 + 4x — 12x? + 6x - 6
N = 36 = 8x° — 16x” + 10x - 6
(b) (36) + D = 96
(a) RxeTx = RyTy
(b) Ry = 6
(a) L+ nA Il = (c) Ry = 3
(b) L—-S= 66 (d) Ty - Tr = 8
ZL, = 266
Substitute (b) and (c) into (a) and get:
L=
(a )6ly measly
(a) (133) + S = 200 Substitute Ty = Tx + 8 into (a’) and get:
= 67 (a”) 6Tx = 3(Tx + 8)
6Tx = 3Tx + 24
1500 NA
= —— 2400NA 3600(1500) 3Tr = 24
2400 3600
NA = 2250 kg pe i 8
(d) Ty - (8) = 8
If 20% combined, 80% did not combine.
Ty = 16
Seen = 740 3.(200 = 5/18 #°74/50;
100 10.
100 = 3/2J2V2V5V5 - 5V¥2V3V3
Ni= 740 x =—— = 925
80 + T2155
Nc = 925 - 740 = 185 kg =f304) DomelSV2r 3512 ue SOND,
Algebra 2, Third Edition )
Problem Set 21
11. 23 - 2V2(6V6 —3V2) 17. (a) Every point on this line is 2 units below the
x-axis.
243 +2/2(6V273 =3V/2)
4./2/3(6V2V3 — 3V2) vec
DAD} 21a(3,— don A De (b) The y-intercept is 0. The slope is negative and
= 144 - 24,3 the rise over the run for any triangle drawn
is —2.
12. Title Mie copies cael +
y = -2x
P Pp
m2 ie oe m2 > 3ax _ ax 18. Graph the line to find the slope.
13. a x2 ax x a x2 a x2 a x2
m2 — 3ax — a*mx
ax?
(0.0003 x 10%)(6000)
14.
(0.006 x 10!5)(2000 x 10°)
— (exdeene x 107) 18-10’
~ (6 x 10!2)(2 x 108) 12 x 10
1.5 x 10-8
15.
Since m = 0 and b = -5, y = -5.
19. Since parallel lines have the same slope,
WS SSre slp
16. (Q) xe =r4 2= -—(2)+)bD
(b) 4x — 3y = 12
4 Tart,
Y = —x%
ae —4
q
20 =p
——
0
Shien S082 and bie 2 wyee ee
oy 4 i
20. x +4=6
x=2
+ y= 10
y=10-(2)=8
56 Algebra 2, Third Edition
Problem Set 22
21. A + 130
= 180 aie i-oeho
29. ie |-44 3] = tt aed)
A = 50 = -1[0] = 0
B + 50 + 90 ll 180
B= 40 1 1
30. oe) ae a eee
Since vertical angles are equal angles,
C=B= 40.
D + 40 + 90 180 PRACTICE SET 22
D = 50
De
Since vertical angles are equal angles,
y= D= 50. —Dp
22 —- -—=:
Re = 14
Rp = 21
ReTgz = RpTp
Tp = Tr - 3
(14)Tz = (21)(Tg - 3)
147, = 217, - 63
7Tr = 63
23. 0.03(« — 4) = 0.02(x + 6) Tr = 9 hours
3(x — 4) = 2(x + 6) D = RgTr = (14)9) = 126 miles
34 — 12 = 2x 12
X-= 24 A eal?)
Ie ab
245164 — 143) = 6] OxE="95
= -2[(y - 43 - Qy - 5)] 35
x= —
—(15 - 1] = -2[3y - 12 — 2y + 5] 6
[14] = -2[y - 7] :
ee)
Ady 229-414 9 6
—28 = -2y 6y
= 45
14 =y
JIS
aero
be SIP (2 - 0-2 9} 2 2
le om
PROBLEM SET 22
26.
rit
=
ae —4
ORISAte
Uy
x | De |
a —2
a7 = Sieh = 53h ¥ 3-1 = 3xa — 5xa + 7Txa Do
XG
= 5xa
De = Dc so Regle = RcTc
Rr = 15; Rc = 30; Tc = Te - 3
28. x(x -— ax)x _= ss(od -+-3(-5
3 15Tr; = 30(Tg - 3)
15Tr = 30T; - 90
-15Tr; = -90
Tr = 6
Dc = Dg = ReTr = (15)(6) = 90 miles
Algebra 2, Third Edition 57
Problem Set 22
| Da |
x — 1)x? + Ox? — 4x + 2
ya OF
Dy
x? — 4x
Dy = Dw so RaT, = RyTw nevi iy
Ry = 9; T, = 4 Ty
=2 —3x +2
—=3x7+ 3
9(4) = Ry(2)
=
18 kph = Ry
Check:
Gee ee ONG 7D A(x =I, eer D3 =) _ 1
D 5
x-1 Cal x-1 x-1l
(b) Not D = 9600. —— =D = 9601=N
Substitute D = 960 — N into (a) and get: Par +e? = x= 30931
(a’) 5N = 7(960 — N) x—-1
12N = 6720 vr -—4x42
N = 560 x-1
(b) (560) + D = 960
D = 400 - Db — i= 1890 — x=18
10
(a) Tr = Ts + 500
15 33
(b) Tr + Ts = 6900 ay
10 es
ly ie y = —)
Substitute (a) into (b) and get:
(b’) (Ts + 500) + Ts = 6900 10. 2427 - 3475 = 2V3V3V3 — 3V3V5-~5
2T; = 6400 6V3 - 15/3 = -9/3
Ts = $3200
(a) Tr = (3200) + 500 = $3700 11. 3/2(2V2 - V6) - 4/3 +2
3. 2( 24/2 — iD S3.)nn A Set 2
(a) 5Py + 13Py = 109
(b) Pz + Pw = 9
1224/3 (24/2. =(/2-V8)e4 2
l(a) S5Py + 13Py = 109
24V2J2V3 - 12V2/2V3V3 + 2
-5(b) -5Py - SPw = -45 48/3
— 72 +2 = 48/3
- 70
8Pyw = 64
12. 2/3(5V3 - 2V6) = 2V3(5V3 - 223)
Py = 8 pecks
10V3V3 — 4/2/33 = 30 - 12/2
570 NI
0. 5000 —_ 600NI = 5000(570)
LX Leet
13. Se ie Oren |
NI = 4750 grams x Xx Xx x
(a) 5x + y = 24
5x2 By
(b) 7x — 2y = 20
14. pp-—=
Py py Py Py
Substitute y = 24 — 5x into (b) and get: ey 5x7 p + py — 3x
(b idx = 2@4 = 5x)4= 20
Py
7x — 48 + 10x
= 20
L7¥= 68
15.
(0.0035 x 10-4)(200 x 10°)
36 = A! (700 x 10°)(0.00005)
(a) 5(4) + y = 24 (35 % 10) 10° | =7xct0!
yond (7 x 10’)(5 x10) 35 x 102
(4, 4) = 2 x 10?
58 Algebra 2, Third Edition
Problem Set 22
(a)
x = -5
(b) 2x --y=4
-y=-2x
+4
y=2x-4
am
maeRoLGEE,
h d& wb
acne |
17. (a) Every point on this line is 4 units to the left of
the y-axis.
co —4 20. = =x +b
ates
(b) The y-intercept is +3. The slope is negative and Zz
the rise over the run for any triangle drawn = 50)
—-5 —(3) +b
: 3
1S —.
3 ay:
= —y +3 5
oes at
es saga
5 ‘.
18. Graph the line to find the slope.
ONE,
2
Since We rand hee eee
5 5 5
A + 140 = 180
A = 40
B + 40 + 90 = 180
B = 50
Since vertical angles are equal angles,
C=. B.= 50.
k + 50 + 90 180
k 40
M + 90 + 40 180
M 50
59
Algebra 2, Third Edition
Problem Set 23
2) 1 1 PRACTICE SET 23
me dey
oe sie 4
34 41 9
—x -—x = —
10 10 4
——
7 JX —
9
10 4
9 10 90
BF, One eh oe
4 7 28
0.02(p — 2) = 0.03(2p — 6)
2(p — 2) = 3(2p — 6)
2p - 4 = 6p - 18 (-1, -1)
14 = 4p Check: (-1) = 2(-1) + 1
3.55= Pp
=(e= =) +1
a ee aes|
24. G3 0C 1°) 6°] = -4/@ = 3)2]
-(9 — 1] = -4[2x - 6]
—-8 = -8x + 24
PROBLEM SET 23
See Ss 3
| Du |
k=
D;
pare | py ee er eeeh
25.
ae ‘.| ips =Pye Dy = Dy; so RyTy = RT;
= 1 3x4y7p2 Ry = 600; R; = 800; Ty = T; + 4
600(T; + 4) = 8007,
26. Keates eg Any 6007; + 2400 = 8007,
(2x7)? x“
- 4x1%-1,-1 2400 = 2007,
12 minutes = 7,
_ 3x’y Pe a 2 5xy- snag
27.
XX y xe xy 2.
= —6y
won ALB
Ds
28. Dp = Ds so RpTp = RTs
pes) yee)
Tr= 10; Ty = 12; Rp = Ry + 10
mle Eide
5
20 Wee 10), (
a aoe (-Zie pei
-20.\ 10 200 (Rs + 10)(10) = Rs(12)
10Rs + 100 = 12Ry
29. (5 - 7 - 2) - |-2- 7) - 2%) +2
S Siesta 100 = 2Ry
= =f[-14) 4.214 + 2 = 16
50 = Ry
Dr = Ds = (50)(12) = 600 kilometers
—3 2
30.
= —_-« = «277 + «216 =
a2
43
60
Algebra 2, Third Edition
Problem Set 23
3. @ S =i — jw=ip 9. 4/3 -3V12 - 2/3 = 413 - 3V2/2V3 - 2¥3
ek = 24/22/3133 = 14,3
(ob) N + D = 2304 —>D= 230—N
10. 3/75 - 4/48
12N = 11(230 — N) = 3V3V5V5 - 4/22/2243
12N = 2530 — 11N
23N = 2530 = 15V3 - 16V3 = -V3
N = 110 11. 2V5(5¥5 - 3V15) = 24/5 (Gy'5 = 31/35)
D = 230- 110= 120 = 10V5V5
— 6V¥3V¥5V5 = 50 - 30/3
110
Fraction = Ta vp
12. 3xy2m + 4 ¢ ooo
x Xx Xx
Seng
Zee
ea
(b) Ny — Ns = 50 x.
2Ny = 350 5x7 oe 5x*p =4p*m
;
Ny = 175 viands
13. —-4+4+ ——
= 4 --—
+ =
pm pm pom pom pom
(b) (175) - Ns = 50 _ 5x*p - 4p*m+c
Ns = 125 sandwiches p>m
5. (a) SNy + 25Ng = 500 14, £0:00003)( 0.006 x 10)
(b) Ny - No = 40 (1800 x 10!*)(100,000)
I(a) 5Ny + 25Ng = 500 _ Bx 10)(6 x10") _ 18 x 10°
25(b) 25Ny - 25Ng = 1000 (UR 100(1 c107) a 918 ee
30Ny = 1500 Flex 05”
Nn = 50 nickels 15. (a) 2y —~2x = 8
(b) (50) - Ng = 40
No = 10 quarters
6. If 20% was copper sulfate, 80% was not copper
sulfate.
= x of = is
hile A000 =e
100
NC = 320 tons
Ton (ay Se ove 70
(b) 30 — 2y =.10
8x = 80
aL
(b) 3(10) — 2y = 10 (a) 2y-2x = 8
Pere 2(b) 2y + 2x = -4
a 4y = 4
(10, 10) y=l
8. Gr Wer — x 4) (b) (1) + x = -2
B63 = 12 = 4° 4+ 2x? 28x xX=
= 6x° — 3x4 - 16x° + 2x7 + & (-3, 1)
61
Algebra 2, Third Edition
Problem Set 23
16. (a)
y - 2x 1 18.
y=2x+1
(Oe 2
DP = 4 + 2
D2
16+4
D?
= 20
Sop IS
PT=) o LS“o & oa & > Oo
=
Re}
a
Yee. ios} ro)
on
oo} — AN a3 =! N by
|i
|
A tt
Q
of G2
|
mloo
MIN aS TT | xe + >
se I = a
|
m|
SS r
tt i
|
we.om +
Ke
| nw
a 8
a
12
8
17. Graph the line to find the slope.
e©nrrane~
nntHno
Al
aS 13* = (BC)? + 5?
169 = (BC)? + 25
144 = (BC)
+] 12m = BC
Slope
a —IN
0=b
1 1
Since m =7 and
b = Oty:
y = =X.
2
62
Algebra 2, Third Edition
Problem Set 24
22. ce + i = oi PRACTICE SET 24
4 3
2x +3 oe
13
—x =
13*
— =)
abe
— 4 3. ood
= 6 6
1? gecioe youd ner oe
13
—,
19
—_——_
3 4
4 6 6x +9 — 16 =3
6x
= 10
x= ——
19 a
4 76
ae iat a
38
5
i SS
3
23. 0.3(2p - 4) = O.1(p + 3)
Aa LL tes
3(2p — 4) = \(p + 3) 3 5
6p - 12 =p +3 {sins LE Lyoeyigy fhe ages 8
2) 5
Sp
= 15
20x + 21 — 6x = 30
Diss3
14x=9
9
24. -[(-4 - 6)(-4) - 2] = -2% - 4) x= —
14
-(40 —- 2] = + +4
-38 = -x+ 4
ee.
gj 5)
42 Ne
= ©)
9
O22 0.2 XK = =
25. xy -2| (=>
x y - 305 | a es
)-4-4 >
x Xx 5a
ae.
= 1 - 3x! A © 02
2y
= 20
y (—3xy)? x7 y? ' ury “xy . xe:
y= 10
} (x2 wx" 9x6 yx? 9y
PROBLEM SET 24
3m 2x7! 5x2 m? 2 3m is 2m i
ah = 5 mm! " xem - oe x | De |
_ 6m Dg
x
Dr = Ds so Rplp = RTs
Tr = 6; Ty = 8; Rp = 60
1 1
28. ©) Ae (x = y) = 3 — (-+ = 2]
60(6) = Rg(8)
45 mph = Rs;
D,
20 eo 1 ns eR a es |
a Do
= -1[3 - 8 - 1] = -I-6] = 6 D,; = Dy so RT; = Roy
R, = 4; Ro = SieTy = Tp, + 1
ey 0 _90
y= 5+ 8 = 8 A(T, + 1) = ST)
30... —sien =
eRe a AS 4T, + 4 = ST)
4=T>
D, = Dy = 4(5) = 20 miles
Algebra 2, Third Edition 63
Problem Set 24
n & 3x7 + @x-12-—,
OS D
ee
7
ae 8 <1 42)
3
te2) ere S
Ox TrOx™:
(b) N + D = 120 OX ben OX
Substitute D = 120 - N into (a) and get: —ee
6x?2 2 —ne
12x rie
(a’) 7N = 5(120 — N) 12x + 24
12N = 600 -27
N = 50 Check:
(b) (50) + D = 120 _3x%(ex = 2) | Ox(-x = 2) | 12(-x = 2)
Di= 10 -x -2 -x -—2 -x -2
27
The fraction was a eSBs
~ 3x* + 6x7 = 6x7 — 125) telde 4°24 = 27
4. (a) Oc + Oy = 173 a BAPELe5
(b) Oc - Qn= 11 3x°—3
20¢ = 184 ey
Qc = 92 quarts
(b) (92) - Qy := 11 9, 43 ~5N2 OVID = 4439 542 622
Qy = 81 quarts = 120V2V2V2V3V3 = 7202
5. (a) 700Mp + 900Mp = 41,000 10. 4/63 -— 328 = 4/3V37 - 3/2V2V7
(b) Mp + Mp = 50 = 12V7 - 6V7 = 6V7
I(a) 700MR + 900Mp = 41,000 M1. 3V¥2(5V2 - 6V12) = 3V¥2(5V2 - 6V2V273)
~900(b)Oe -900Mp nes
- 900Mp = —45,000
OOM ps)
-200Mp = —4,000
= 15¥2/2 - 18/2/2213 = 30 - 366
Mp = 20 measures 12. 2/2(5V¥10 - 3V2) = 2/2(5V25 - 3V2)
‘ 48,300 _ NP = 10/2/25 = 6y 242 = 20/5 - 12
"49,000 4200 ih,
49,000NP = 4200(48,300) 13. 4m?yp + 2. - sells a se
NP = 4140 kilograms my my my
43
£ 4m" y*p + 6
7. (a) Ix + 9y = 119 m?y
(b)
b 2x + y
Use = 23 Sa
y == 23 os- 2x pi ere ers Oh g
Substitute y = 23 — 2x into (a) and get: 2p pc 2p*c 2 prc 2 pc
(a’) 7x + 9(23 — 2x) = 119 _ kpc + 2prc2 — 8
Ix + 207 — 18x = 119 2p7c
-llx = -88
eats 5, S20007
e100)
(0.00004 )(7,000,000)
b) 2(8 = 23
se a _ (7x 10°77)(4 x 10°) _ 28 x 10718
y=7 (4 x 10°°)(7 x 10°) 28 x 10!
(8, 7) =] x 10719
64
Algebra 2, Third Edition
Problem Set 24
16. (a) 3x + 2p =r ie Since angles opposite equal sides are equal angles,
2y = =3xr +. 12 C=D= 35.
3
= —-—x
+6 70 70
360 r = ——(m)(4
Area = ——ar* 3600 cm) cm)” = 9. cm
9.77 2
7 2
(b) 5x - 4y = 8
—4y
= -5x + 8
19. G21 )29=H + 2?
21 =H? +4
5
= —x -2 17 =
i ~
V17 = H
te. 4 in. _— in, _ 217 in.
20. Graph the line to find the slope.
2(a) 6x + 4y = 24
(b) 5x -4y = 8
fis = 32
_ 21
(a) 3)
32 +2 = 12
i i y = 2x44b
QV =
132
— -
96
— 5a=) 2(-2) + b
cauieis | dean
on
po Ly 18 Since m 2 and«eb-=-9.-y
=x + 9,
'Bais elhawy dias |
Gra)
Monit 21. =—-x+b
aes
2
Ly; (a) The y-intercept is +2. The slope is positive and —3
= 9)
—(5) +b
the rise over the run for any triangle drawn is 2.
Bite eee
=—-xr+2 9
ae 6
(b) Every point on this line is 4 units below the
igure
2
X-axis.
viens Sinton = and) = =e ye ee
"] 9 7
18. Since the measure of a central angle is the same as
the measure of an arc of a circle, 22. Bees Fe 4k + 12 = 27
3 4.
A = 70.
4k
= 15
B+
70 = 180
B = 110
k= 4
Algebra 2, Third Edition 65
Problem Set 25
PRACTICE SET 25
23.
8m2xy> + 6yx*m> - 2x ym?
= 2mxy(4y4 + 3mx - 1)
4m> + m2
m2
= 4m? +1
4SF
=7
hi 8 -— 4x Gee!
24. —+ 4
2 7
21x +°-16='8x%
= 42 a
ie ge EO ae
13x86 st is
35
Kee 5 + x ==4
25. 0.07 — 0.003x + 0.2 = 1.02 ie cea e
4
70 — 3x + 200 = 1020
15
—3x = 750 XS
4
x = -250
ll x
IES
aNSes
265 -2|x — (ea | Sf E31 =
=, 42
—2[x + 3 — 3] + [-3x + 6] = a er
2x - 3x +6=
La
-4x = ey
SS
aM = 24
Moe
7
21. ar
Pp
(ee
x
es |-4 533y
P
—2 PLD
=) Bee} a
Ney di
Bg
=1+ 5x%y
AN ==]
23
4x°y?x
ee
=
(-29)3
32x7y?
Se eee0
3-2
—O2y y yy" Roce4
29 3 p> x? xd 5 ppy”! 3 5 pox x"
xy x! xy PROBLEM SET 25
7 3px Ps 5p°x - 5p’x! 38 eed)
y y Y
14,440 36,100
14,440D = 38(36,100)
‘5 -2p*x Sp*x!
D = 95 days
Ba y
| Dg |
30. xa(ax - a) =
Dr
Dp = Dr SO RpTp = RyTr
Tp = 40; Tr = 30; Rp = Rp + 6
66 Algebra 2, Third Edition
Problem Set 25
40Rz = (Rp + 6)30 Check:
40Rz = 30Rz + 180 2h
x“(x = 5) op 5x(x -
- 7 5) £ 25(x a— 5) 5 123
10Rz = 180 oe x-5 tum 'S yo 08)
Rp = 18 kph ye = Sak + 5x7 2 254-4 25K — 2
Dy = Dg = 18(40) = 720 kilometers = TreeS
a co an?
3. (a) Ng = 5NG
ier
(b) Nz = 15NG — 100
Sty 2xy + 10xy* = xy(Sxy — 2 + 10y)
Substitute (a) into (b) and get: .
(b’) (SNG) = 15Ng - 100 9, x?y>m> + 12x3ym* - 3x2 y7m2
-10Ng = -100 = x*ym?(y2m?3 + 12xm? - 3y)
No = 10 girls
G . 10. 16m*p*y - 8y4mp? + 4m tptyr
(a) Ng = 5(10) = 50 boys
= 4mp7y(4mp - 2py? + my)
D
4. (@ ot = > — = 2, = 5D, i
x3y%23 " xyz" ~ 3x3 yz = xyz(xyz" a fea 3x)
* 12: px + pte - p*x = p?x(p?x? + px - 1)
(b) D; oe Ds = O8() Da =99 8 (ee D;
Substitute (b) into (a) and get: 13. 2V3 - 3V6 - 512
(a’) 2D, = 5(980 = D;) 2V3 - 323 - 5V2V2V3
7D, = 4900 30V2 V2 V2 V¥3V3V3 = 1806
D, = 700 minor disasters 14. 618 + 54/8 ='34/50.
ie (a) No + Nu =e)
6V2V3V3 + 5V2V2V2 - 3255
18y2 + 10/2 — 15/27 1342
(b) 25Ng + S5ONyq = 7500
-50(a) -50Ng — 50Nq = —10,000 15. V5 (GV15 — 245) = 2456345 — 25)
I(b) 25Ng + 50Nyq =~ 7,500 = 6V¥3V5V5 - 4V5V5 = 30V3 - 20
-25No = 2,500
a ab a _abt+a
= Se
No = 100 quarters 16. b b b b
(a) (100) + Ny = 200
ax? ax? cm’ p 2mp
Ny = 100 half-dollars ith oc
mp m ee m* pae m* p ee
m*BS
p
6. If 70% was sodium chloride, 30% was other _ ax? ~ cm*p + 2mp
chemicals. m? p
fits X of = is (38,000 x 10°)(300 x 107+)
100 18.
0.00019 x 107
saad x TW =, 660
100
(8 x 10°)(3 107°) ~ 14x 104
19 x 10710 ~ 19 x 19710
TW = 660 x = 2200 grams I 6 x 10!4
x74 Sx + 25 + 1%,
7. x — Spe + Oxe + (Ox—"S 2
ie a
5x7 + Ox
Sica W2Sx
25x - 2
25x
— 125
123
Algebra 2, Third Edition 67
Problem Set 25
19. (a) 3x — 2y = 10 J Xa S heal
—2y = -3x + 10 Sk aa
3
ad pfs ot ray
7
ep
a
280 280
22. : Arc leng th = 360
— - 2arr = 56 (
— -: 2aA(12cm )
= 58.61cm
33, Tel ees Datta
x-9
4 9) 10
se = 7/
4x -—8 2x —4
2
4. 5
+ amen
Ds
Substitute (b) into (a) and get: ea ie :
(a’) 3x — 2(-5 = 10 aed
Ee hee S19 95 gydled ceTe
6 3 5
3x = 9 5n + 15 — 10 = 12n —- 12
ee ~In = -17
17
(3-3)2 ma
20. Since parallel lines have the same slope, 26 : ,) - >=
4 >
mill. 9s 10x + 6 — 3 = 10
et 273 fOr?
1 ae
5 = —(3) +b Ty
6
9 8x2 + 12% 4x°(2x* +3)
{=p 27, ———_—— = = 2x?
2 4x3 4x3 oe
2,,-2 2 220
Sint = shald bs Sys ae ee. 28. ames [gay
6 2 6 2 zm \oye(Queeh 2
Ix*y*2
21. (a) 6x - 2y = 100 = SC xty? = ay - arty?
(b) 3x + 2y = 80 gs
Ox = 180 29, —3°((2 S40) ee Oe
x= 20 = -l[-11 - 5] = -1[-16] = 16
2
(b) 3(20) + 2y == 80 30. km(m2k — k) = 3(-4) (-z] (=)eek
Dye) 3\ 4 4) \3 3
Bea (1 6) 1 (5 ) 5
y = 10 FO Ape Rien! Gale cae eee
12\48 48 2 Ree k6 192
68 Algebra 2, Third Edition
Problem Set 26
PRACTICE SET 26 3. (a) SNe + 10N7 = 7000
a. x7 — 72 6 = x? - & =F Se Te + 1) (b) Np + Nr = 1200
l(a) 5Np + 10N; = 7,000
Deer eter
16 = -1)e bo
2 Oe © : Han Oteanslb) -10(b) -10Np — 10Np = —12,000
SGT — BYE a) —5Np = —5,000
c. 24x? + 2x9 — 3x4 = -3x2(x2 — 7x — 8) Np = 1000 fives
= -3x7(x - 8)(x + 1) (b) (1000) + Nr = 1200
Ae ae we Nr = 200 tens
A* = 16 + 25 4. If 14% were belligerent, 86% were eristic.
ee an Koop Suis
A= /41 100
ae a x Np = 4300
pea 100
5C = 32 Nr = 4300 x Pee 5000
32
C= — Nz = 5000 — 4300 = 700
BS [41 es 5. 1820 _ NCa
ae es 1960 2240
5B + Sai = 8VAI epee 20240
5B = 3V41 a=
: rams
cae 3a 6. If 10% 0 were nonabsorbent, 90% 0 were absorbent.
> as Diy Sus
100 ,
PROBLEM SET 26 ot a
Ty =be 7290
1. Ds : C = 7290 x a = $100 chemicals
Po ba we me x-1-—,
Ds
= Dg so RsTs
= RoTo Te x + I)x" + Ox) + Ox? + Ox - 2
T; = 20; To = 8; Ro = Rs + 60 Mi nae
sf ER g : -x3 + Ox?
20Rs5 = (Rs + 60)8 _x3 C4 x2
20Rs = 8Rs + 480 x> + Ox
12R, = 480 pee.
je
Rs = 40
—x - 1
Ds = Dg = (40)(20) = 800 miles =|
Check:
2. (a) Ng = 5Npr + 50 PPL AD) RE
(b) Nr = Ng - 210 . x+1 x+1 xh
Substitute (a) into (b) and get: eo aan 2 —
+ x +
Neel SNe 50)
= 210 :
(b’) oO ) Hed oak tito oe 1 |
—4Np = —160 = ee
Nr = 40 reds viata
(a) Ng = 5(40) + 50 = 250 blues pe er ge
69
Algebra 2, Third Edition
Problem Set 26
35x/ym = Tx>m?y* + 14y1x4m?
= Tx4y2m(5x3y> — xm + 2y>m)
6x2ym> - 2x7ym + 4xym
= 2xym(3xm4 - x + 2)
10. 4x2y4p° — 2xp>y7 + 8x4 p5y>
= 2xy‘p*(2xp - y? + 4x°y)
11. x7 +x-6
= (x + 3) - 2)
12. x” = 6x + 8S = dq — 2)
Substitute y = 2x — 3 into (a) and get:
13. —2ab + abx + abx* = ab(x + 2)(x - 1) (a’) 2(2% — 3) = x = 6
4x -6-x=6
Cnt— xy
6x“y t re 1) an
xy(6x —
14. 3x
= 12
xy xy
x= 4
15. 3/2 - 2V6 - 3V6 = 3V2 - 2V2V3 - 3V2V3 tb) y = 2G)c==3
= 18/2/2/233 y=
= 1082
(4, 5)
16. —3/12 + 5/27 — 8/25 21.
= -32 2/3 + 5V3V3V3 - 8V5V5
= -6/3 + 15/3 — 40 = 9/3
- 40
17. 3./2(5V¥3 — 2V2) = 15V¥2V3 - 6V2V2
= 15/6 - 12
18. an x2
et Cc
Pn
wer
cx? cx? CX
a a*cm = cx! = x4
cx?
19.
(3000 x 107!*)(0.00008)
(0.0002 x 10°)(200,000)
PAG g Sex 10>) -24 x 10788
(2 x 10!)(2 x 10°) 4 x 10° 22.
= 6 x 10°”
20. (a) 2y -x = 6
2y 20x FO
1
Sse
ee)
70 Algebra 2, Third Edition
Practice Set 27
Val 2= B+ Val 27.
Cea,’ ieah er
3 6
7V41
= 4B + 4/41
8 — 2x + 1 = 30
3/41= 4B
—2x = 21
3V41
—
ae
4 21
x= ——
By
Setec
4
felt 3 (Oar x 4 _
Sen) «0
Ls
i ge 28.
7
4
DR ee I AT eth Dy oe Dee 08
23. x + LO.=_90
-19x = 37
GU
y + 80 + 70 = 180
atl
~ 19
y= 30
x2(yz0) ) 8x2 y-? :
m + 30 = 180
29. 2-3 = sete Ge
m = 150 f-25 4) po yy
z+ 10 + 150 = 180
220 30. =
=t3)F*
:
9-3
(a — ba)
(a) 4x + y = 70
(b) 2x + 6y - 1 = 115
(b’) 2x + 6y = 116
ie)
=9-8(1)=9-8=1
D ee2
—6(a) —24x — 6y = —420
ib} 2x4 6y = 116
—22x = —304 PRACTICE SET 27
152 Ween He hes
x= — a.
11 mt+4 m>
(a) (=) + y= 70 _ m?*(m +3). m*(m+4)5 . (m+ 4)2
11
_ 710 608 _ 162
i‘ m2(m + 4) m(m + 4) m2(m + 4)
f 11 11 11
_ m+ 3m? + Sm? + 20m? + 2m + 8
cs m?(m + 4)
25.
bx
Atriangle - aS ae = 34m
H 2 _ 6m? + 23m? + 2m + 8
m?(m + 4)
» = 24(2) m*
9m
b. ‘Kei
> ages Pec
b = 12m = BC z+52+4 22+ 4)
AGircle = 2r? = m(12 m)? ~ 452.16 m?
eR Ties gy
(z + 4)(z + 1) z(z + 4)
Dax 6 — 3x
26. — + == 7 i" ae te) (at 1)
3 4
~ ez t4(ztl xz t+ 4(zt+ dD
Sri 18: 29 xe G4:
=i"00 z2 + 2z — |
x = -66 z(z + 4)(z + 1)
Algebra 2, Third Edition 71
Problem Set 27
PROBLEM SET 27 If 70% was silver iodide, 30% was not silver iodide.
Ng ;
— x of = is
100 !
reall x 2000 = NS
100
Dy = Dr so RyTy = ReTr
NS = 600 grams
Ry
= 4; Rr
= 20; Ty + Tyr = 18
4Ty = 20(18 — Ty) x? 42x +44 3,
4Ty = 360 — 20Ty x — 2)x3 + Ox? + Ox - 6
x3 — 2x?
24Ty = 360
2x”
+ Ox
Ty = 15 De AN
Dy = Dy = 15(4) = 60 miles 4x-— 6
4x-— 8
B. ah D;, D;
Check:
Dr 2
SACS ~
2), 2x(x =
2) , 4G =
A) 4 2
Dy = Dr SO RyTy = RrTr
x—-—2 x-2 x-—2 x-—2
Ry = 58 Rr = Di: Ty oF Tr =a) — we — 2x7 + 2x? - 4x4 4x -84+2
x-—2
or 6
51 ,,-=)200 — 207,
Bow)
25Ty = 200
Ty = 8 9m?x4p* + 3x2p°m4 - 6x4m>p*
Dy = Dr = 5(8) = 40 kilometers = 3m7x?p?(3x? + p4im? — 2x*m)
mx'y = mxy? a Amx?y a mxy(x? i y? = 4)
3. (a) NG = 2Nz + 6
(b) Nc + Ng = 36 Poe 4a>x3p - a’x'p = a*x*p(1 - 4a - x)
Substitute (a) into (b) and get:
4ax + ax” — 5a = ax + 5)x - 1)
(b’) (2Ng + 6) + Nz = 36
3NB = 30
8x? — x? — 15x = x = 5\(x — 3)
Nz = 10 boys
24ax — Sax* — ax? = -ax(x + 8)(@ - 3)
(b) Ng + (10) = 36
Ng = 26 girls ~ax* + 4ax? + 5ax? = -ax%™(x - 5)(x + 1)
4. Multiples of 7: 7N, 7N + 7, 7N + 14
S6p — 1Spx + px* = p(x — 8)(x - 7)
-3(7N + 7N + 14) = 5(C-7N — 7) - 21
-42N — 42 = -35N — 56 4xat+4x — 4x(a + 1) =
a+1
4x 4x 7
~IN = -14
N=2 17, 3/2 - 23 -3V12= 3/2 —-2V3 - 3V2/2V3
The desired integers are 14, 21, and 28. 3V2 - 6422/33
= 3) 2-36
: 600 __Ba
“3000 ~=9000 18. -3/20 + 2/125 + 5/45
3000Ba = 600(9000) SE PAG ON GE.
i NEWIG
Ba = 1800 grams = -6/5 + 10/5 + 15/5 = 195
72
Algebra 2, Third Edition
Problem Set 27
19. 2V3GV2 - 3V3) = 6V2v3 - 6V3V3 24. Since parallel lines have the same slope,
= 6/6 - 18
20. 26 3 AG)
Gb 3)
x x + p X(Npp) . X(X +p)
20 2p Ok LL Ok Zp
x(x +p) x(x
+ p) bo II | + >
x +3 3
21. aoe A heen
x+6 x2
x7(x +3) 5x2(x + 6) ‘ Sur + 0}
x(x + 6) x(x + 6) x*(x + 6)
tate yor be ot eT 8
x2(x + 6)
6x? + 33x72 + 3x + 18
x7(x + 6)
25. Az = 7? + 4
(0.00056 x 10*)(7 x 10%)
22.
(0.00049 x 10!°)(0.00002 x 107°)
_ (56 x107!)(7 x 10°) _ 392 x 10?
~ (49 x 10!})(2 x 10719) 98 x 10!
= 4x 10!
23. (a): 3x 4a 2ye =e 2
2y.=.—3x_ +12
y=" =——x + 6 V6 > = B+ V65
(b) 8x — 2y = 10 19x65. 1 Bt INOS:
5.65
= 7B
5/65 _
26. 3 OM
5
(a) 3x + 2y = 12 Als
co]
(b) 8x — 2y = 10
6+ x A
l1x = 22 27.
5 3)
x =-2
18 + 3x + 5
(a) 3(2)
+ 2y = 12
ane
= My)
2y= 6
ere
(2, 3)
S
Algebra 2, Third Edition
Problem Set 28
Multiples of 3: 3N, 3N + 3, 3N + 6, 3N +9
oe ene By
28.
3 6 5(3N+ 3N + 9) = 13(3N+ 6) - 6
6x -44+24=1 30N + 45 = 39N+ 72
6x = —19 -9N
= 27
19 N =-3
=
6
The desired integers are —9, —6, —3, and 0.
29) =
29. Laas
b
aD an
a“y
ay Fj
(a) Nave Neve 1
LL, , RED (b) Ng + Ng = 15
am ab azb a2b a*b
Substitute (a) into (b) and get:
30. 5 3) Aa (b’) (Ng - 1) + Ng = 15
= -(-7) + 8(-2)(4) - (-2) 4Np =ul6
=7-64+2=-55 Nz = 4 boys
(a) Nc = 3(4) - 1 = 11 girls
PRACTICE SET 28 (a) SNy + 10Np = 3000
ids p (b) Ny + Np = 500
Po gts m
Ye, a3 dG Pp Va Go ae l(a) SNy + 10Np 3000
Dp We ct BS -10(b) -10Ny -— 10Np = -5000
-SNn = —2000
m cee: Ny = 400 nickels
IG ap yy We =. m
Zz x+y 6 (b) (400) + Np = 500
x+y Z
Np = 100 dimes
4 2 Av? = 22 If 10 had seen a Dane, 190 had never seen a Dane.
BD > Aly 6 3
190 _ __NS
200 150,000
PROBLEM SET 28 200NS = 190(150,000)
| Dr | NS = 142,500 peasants
Dp If 16% was arsenic, 84% was silicon.
P
Dr = Dp SO RrIpr = RpTp
a
100 of = is
Rr = 60; Rp = 3; Tr + Tp = 21
84
60(21 - Tp) = 3Tp 100
— X EM = 7350
1260 - 607p = 3T7p
1260 = 63Tp EM = 7350 x 2@
84
20 = Tp Entire mixture = 8750 kg
Dp = 3(20) = 60 miles Ma, = 8750 — 7350 = 1400 kg of arsenic
74 Algebra 2, Third Edition
Problem Set 28
x24 Sx + 254 x-5
_118
7. x- 5)2 Oe toe Deg 19.
ats wis 60 Re
x? — 5x2
5x2 + Ox 4a pane 4a(a) , O(a + x)
5x? — 25x 20.
atx a a(a+x) alatx)
Ltt y
25x -— 125
_ 4a? + 6(a + x)
118 i‘ a(a + x)
Check: 2s a
21. »
Be
x*(x 2-— 5) e 5x(x -—
a 5) A 25(x -
= 5) ee 118 ob Dea byEhtis
eateel.5
x-5 x-5 x-5 x=5 a 2x i. 3(x + 1)
_— w= Sx? + S27 — 25+ 95x 2125 4118 (x+1)(x+1) («+ + 1)(% +1)
: x-5 2x+3x+3 SXigk3
£ 7 (x+D(x+1
(x +1)?
eee
22. (a) 5x + 2y =6
2x2y — 8x4y4 = 2x2y(1 - 4x2y3) 2y = -5x + 6
o
4x7y3p3 2 16x2y>p ks xty3p4 iy = =—=—Xot
oe 9
= x*y"p(4p? - 16 - x’p°) b) y y=—me
1
(b)
-35xy + 2x2y + xy = xy(x + 7)(x - 5)
~8a — Tax + ax* = a(x - 8)(x + 1)
2m + 3xm + m?x? = mx + 2)(x + 1)
~a? — a*x? — 2xa* = a(x + 1l@ + 1)
ake + ihe x(4x ei) >
4x +1
E %
15. Ad2T = 34/48 49075
= 4433/3 — 3¥2V2V2V2V3 + 237575 Substitute (b) into (a) and get:
= 12V3 - 12V3 + 10V3 = 10/3 a) Sx + 2| —
1
6
ee (5)
16. 35(V15 - 2V5) = 3V5(V3V5 - 2V5) 6x 6
= 33V5V5 - 6V5V5
= 15/3 - 30
(0.00077 x 10~7)(40 x 10°)
(3)
AT;
(0.00011 x 10°)(140,000)
_ (77 x 10-*)(4 x 10’) _ 308 x 1074
Po fina % 10 eis4 & 20° 23. (a) Every point on this line is 4 units below the x-axis.
=2x 10° A el
(b) The y-intercept is +2. The slope is negative and
x m= yp
the rise over the run for any triangle drawn is —3.
m+ D y - x
18. Wigks Titel aay 1
= —-—x +2
itp y ee <3
Algebra 2, Third Edition
thes
Practice Set 29
24. A+ 150 = 180 28.
A = 30
B + 30 + 90 = 180
B = 60
C = B =-60
D + 60 + 90 = 180
D = 30
E + 30 + 90 = 180 D2 728% 424?
E = 60 D? = 64 + 16
F + 60 + 40 = 180 D? = 80
F = 80
D = 80
D = 4/5
25. 3x + 90 = 180
3x
= 90 29.
x = 30
2x = 2(30) = 60
30.
60
pep een (ais)
Ashaded = 360 (7 m)?) ~ 2.09 m2
Length of BC = eo(2n(2 m)) =~ 2.09 m
26. —t+
10x + 3x —9 = 15
ISyo=o24 PRACTICE SET 29
x=
24
—
13
[ea
De
ae Dy asl
66
Tc = 8; Tv = 7
27 at a ee Biaeoe
3 5 10 Dc ats Dy 66
ee ee RcTc + RyTy = 66
=) 30 30
Rc(8) + (2Rc)(7) = 66
Ro + 14Ro = 66
22Re = 66
Rc 3 mph
76 Algebra 2, Third Edition
Problem Set 29
PROBLEM SET 29 If 17% contained xenophobes, then 83% did not
contain xenophobes.
100 x of = Is
RyTw + RrTpr = 76; Rw = 4;
83
Rr = 15s Twit Tr = 8 — xX HP = 116,200
100
4Ty tc 15(8 — Ty) =)
HP-= 116200 x 100
49 Pe DST = 76 83
—l1Ty = —44 = 140,000 hiding places
Tw = 4hr
x? + 2x + 5
De Dp x + 1)x? + 3x27 4 7x +5
x + x2
63
2x? + Tx
RcTc + RpTp = 63: Ix?
+ Ix
5x +5
Rc = 3; Tc = 9; Tp = 6
5x +5
3(9) + 6Rp = 63 0
6Rp = 36 Check:
Rp = 6 mph
ot), 2x(x + 1) een)
x +1 x+1 x+1
| Dr Dy |
x3 + x2 4+ 2x27 4+ 2x 45x45
7900 x+1
RrTr + RyTy = 7900: x2 + 3x7 47x 45
Ry = Rp — 400; Te = 6; Ty= 5 x+1
6Re + (Re — 400)5 = 7900 16x33 — 8x2y222 = 8x2y2z2(2xz
- 1)
6Re + SR — 2000 = 7900
2x7yp* = 6x>yp? < 2x? yp*
11Rp = 9900
Rp = 900 kph = 2x*yp*(p* — 3xp - 1)
Ry = (900) — 400 12a2x + a*x” + 35a” = a(x + T(x + 5)
Ry = 500 kph
~2m?x — m* — mx? = —-m*(x + Ll@ + 1)
(a) Ne = 2Np + 15
(b) Nr + Np = 255 x7k + 3kx — 40k = k(x + 8)(x - 5)
Substitute (a) into (b) and get: x* + ax Ser l-4 a)
lt+a
(b’) (2Np + 15) + Np = 255 xe xe
3Np = 240
Np = 80 14. 2175 — 548 POND
= 2/35 5 — 5y2N
2 ony ot 242/243
(a) Nr = 2(80) + 15 = 175 roses
=i0y3 — 2073 4493 =) 6N3
130, sy 3250
156,000 A IS. 2V3(3V6 - 4/3) = 2V3(3V2V3 - 473)
130H = 3250(156,000) = 6V2/3V3 — 8/33
H = 3,900,000 places = 18/2 - 24
Algebra 2, Third Edition 77
Problem Set 29
Substitute y = x + 3 into (b) and get:
(0.00052 x 10~*)(5000 x 107)
16. (b) (x + 3) + 2x = 6
(0.0026 x 102!)(10,000 x 1074)
C210 G10") _ 2260 x10" 3x = 4
G6 x10) 10-8), 26 x 10" x= 1
= 1x 107 @)iy = (De)
m Xx eke
x Th x Sree
M7:
mtx x mtx
(1, 4)
XxX Ue ROS
24. Since parallel lines have the same slope,
a Mie ae Oe 5)
=i——x+ D
18. m+ x b ay 2 oe
b Mo x b
m+ x b 3 = ——(2)
+b
19.
a. 412° svi 48
5/12 12 60 10 by SAEED
4
20.
5 i475 103 14N3 eet
B95. 225 225 45 4
3 9
4x 6 Siince m = -=aoa
and b = —-,mn
21. +
x+4 x+2
a 4x(x + 2) 6(x + 4) te il ! * !
@0
we
|
CeGet AG 2) —* ch Ae 2) a|o
_ 4x2 + 8x 4+6x +24 — 4x2 + 14x + 24 25. Area = mr? = 25002 m2
(x + 4)(x + 2) (x + 4)(x + 2)
had |25002 m2
3m - Sm 4
22.
m= +3m+2 m+1i1 r= 50m
‘e 3m _ __ 5m(m + 2)
36
(m+2)(m+1) (m+ 2)(m+1) Arcrc length
leng Peale
— (50 m)]
| 27(50
_ 3m- 5m? -10m _— -5m* - Im
Arc length = 107m = 31.4m
(m + 2)(m + 1) (m + 2)(m + 1)
23. (a) y-x = 3 26. AX = 9 + 6
y=x+3 A = 81 + 36
(b) y + 2x = 6 A*
= 117
y = -2x + 6 A = 117 = 3v/13
y 9.SF
= 13
vee
9
13
3V13 > = B+ 3V13
Ey
4
3
|
||
wp
oar
cles 2
9 3
78 Algebra 2, Third Edition
Problem Set 30
a7, 2% 5,442
2 uh 3
63x — 30 = 14x + 28
Rly + RpTp = 68;
49x = 58
Rp
= 6; Tr
= 6; Tp
= 4
58
x= — 6(6) + 4Rpz = 68
49
4Rpz = 32
Rz = 8 mph
28. oi = ae. ae =
5 10 15
(a) 5Np = 2Np
1 3 33
(b) Np + Np =
ll 35
a —_ —
10 15 15
Substitute Np = 35 — Np into (a) and get:
ee oe 20 (a’) 5Np = 2(35 — Np)
1
SNp = 70 - 2Np
ye Oe = =
29. eas ars z a yee mays? 7Np = 70
(x9 ie 2 y y? a
Np = 10 daisies
2 Di (b) (10) Wps= 35
30. x
ary(x y=eT
(5) ree
(=) (5 a Np = 25 prunes
(a) Np + No = 15
(b) 1ONp + 25Ng = 225
27 = 25
-10(a) -10Np — 10Ng = -150
108 108 108
(b) 10Np + 25Ng = 225
ISNg = 75
PRACTICE SET 30 No = 5 quarters
a. This is a valid argument because the conclusion (a) Np + (5) = 15
follows directly from the major premise. Np = 10 dimes
b. This is not valid, because we have reversed the
120in. = 10ft
order of the major premise (scholar ——> poor) (a) Area = 1X w = 40 ft x 10ft = 400 ft?
in the conclusion (poor ~~ _ scholar).
(b) Perimeter = 40 ft + 40 ft + 10 ft + 10ft
= 100ft
PROBLEM SET 30
x? - x+1+—
ee ee x +fldx® + O0ORe Ox £2
280 3360 x? + x?
—x?
+ Ox
2805S = 40(3360)
x? - &
S = 480 grams x+2
x + 1
1
2 “ x-of = is
Check:
Le M = 430 x(x + 1K CaGe Dm) (x tober 1
100 x +1 x +1 x +1 x+1
= xo txt?—x%@-xtxt+14+1— x3 42
M = 430 x = = 500 grams
i" x +1 x +1
Algebra 2, Third Edition
Problem Set 30
8. 35a? + 2a2x — ax? = -a%(x - 7)(x + 5) - we £42
. * tye Syat G te Xk He
9. 30 + 3x2 — 21x = 3(x — 5)(x — 2) oe 2 +3)
10. =4ap — 2SAb EE Cane abay— 5) <5) (EE DAB Oe
4x+2x+6 6x + 6
11. 14a7b? + 9a2xb* + x7a°b? ~GtDa ss) @ +G 3)
= a*b*(x + 7)(x + 2)
22. (a) 3x + 2y = 8
ee
Pied pie pane
eee y= —-x+4
13. 5/18 - 10/50
+ 3/72 (b) 2x + 3y = 6
= 5/2/33 - 10V2V5V5 + 3V2V6V6 rm
a |
Bis? — 5042 + 1842. =.-172, /
14. 3V18(4V2 -— 2/3)
= 3¥3¥3V2(4/2 - 2V3)
= 12/33/22 - 63323
= 72 - 186
15. ~
0.00025 x 10
rt ie ane) eigen ha) mae be
ra 25 x 107? OBS 10°?
= 7x 10”
3(a) 9x + 6y = 24
fe y —2(b) —4x - 6y = -12
Pe y pkey Deak ae = 12
x+y y xt+y LZ
ee ee badge
a at+b @ y= -3(2) Sete, 0
eee Pe ie 5 fo oe oes
x Pp a+b p
Pp
ea)
5°5
18. ——.-
0 N6® =
B26 =a “a8
V6 23. Graph the line to find the slope.
(9 lage 22 92:
5V18 18 90 90 15
20. 4a a+2
a+4 J 2a
3 4a(2a) ¥ (a + 2)(a + 4)
2a(a + 4) 2a(a + 4)
_ 8a +a? +6a+8 9a*+6a+8
- 2a(a + 4) ala + 4)
80
Algebra 2, Third Edition
Problem Set 31
4a*x?y a, 3p xy 2xa 2a7 yx?
29. p an x2 a yop Pp
| M4 4a*x7y 3a*x?y i 2a*xy 3 2a7x*y
< Il ~ + ~
W]e
Pee p P P
oS Mn I
oe i) —_— + > eS 2a*xy a*x*y
P Pp
30.
Since gabatee eat, See ty LO
24. Lateral Surface Area = Perimeter x Height
= (22) + 2m |x 10m 16
= (34 + 2)(10)m? ~ 114.2 m?
PRACTICE SET 31
25. (A) 21 + 22 = 180°; True.
y = -4x + b
(AB is a straight angle.)
at(1,2) (2) =-4(1) +b
(B42 —=) 272 rues
2=-4+b
(Alternate interior angles are equal.)
6=b
(C) 22 + Z3 = 180°; False.
y=-4x + 6
(It is not shown that 26 = 27.)
i) 22 = 26; Tne,
PROBLEM SET 31
(Corresponding angles are equal.)
Multiples of 5:
26. —--—= ON, SN + 5) 5N + 10, SN -k=t5
3 5 2
20 Gi I)
6(5N) = 2(5N + 5 + 5N + 15) + 40
a — 09 30N = 20N + 80
63 10N = 80
x= -—
5 N=8
The desired integers are 40, 45, 50, and 55.
Ife Pade nla egret
3 5 iS
1
Si
1 aS
5 15 15
apie= 218 RpTp + RoTo = 960; Rg = 40;
15% wl
Ro = 710; Tg = To + 2
2x -—3 2, 40(Tg + 2) + 70Tp = 960
28. ——
2 5
110T> = 880
10%—.5=—4
Tg = 8hr
Ore S111
Since the Orange Blossom Special left at noon, it was
11
x= — 8 p.m. when it and the bus were 960 miles apart.
10
81
Algebra 2, Third Edition
Problem Set 31
# x+y
x+y me oe
10.
m EVenateey, m
Xo y m
R,T; = RrTpr; Ry = 9;
Rp
= 30; 71; + Tp = 12
11. Byes
617 = 3002T)) Py Ge 10
36T, = 360
12. STNG aN
T; = 10 Zi Mas 6
Distance = R,;T; = 6(10) = 60 miles
13. 54175-= 34 3002+ 2427
(a) Np + Nr = 70 Il 5a)
3a Sal = Bx)Qa ON 34 NS ee yoy
ll 253 — 30V3 + 6V3 = V3
(b) 4Np + 6N7 = 360
Substitute N; = 70 — Np into (b) and get: 14. 3./2(5V2 — 4V6) = 3V2(5V2 — 4V2 3)
(b’) 4Np + 6(70 — Np) = 360 = 30 - 24/3
—2Np = —60 x+4x* — x(1 + 4x)
15. i Poetbe
Np = 30 pansy plants x Xx
(a) Nr = 70 — (30)
x 34+x
Nr; = 40 tomato plants 16. + SS
XQ xt Ae 4
P x(x + 2) 3+ x
100 x of ="1S
(x + 2)(x%+ 2) (x + 2)(x
+ 2)
x7 +3x +3
LEN 1200 (x + 2)?
100
100
N = 1200 x a = 12,000 kilograms 55 2x x? 2xee,
Ly: 4 oor SS + ———
K-30 4 = 3x x(x — 3) x(x — 3)
BO ae x?
+ 2x ee
2) eee
1440 4320 ~ Xe 3) vee aeceed
1440P = 40(4320)
18. -x? + 5x? — 6x = -—x(x — 3)(x — 2)
P = 120 grams
19. 2ax> — 18ax? + 40ax = 2ax(x - 5)(x — 4)
Since the slopes of perpendicular lines are negative
reciprocals of each other, 20. —3pax + pax? + 2pa = pa(x — 2)(x —- 1)
y= -3x +b
21. -10mc + 3mxc + mx2c = me(x + 5)(x — 2)
—3 = -3(2)
+b
3.=.b SMe Dilek yu.ty
22.
Since m = -3 and b = 3, y = -3x + 3.
6 2
2x +73 = 3x°='6
(0.0006 x 10~*7)(2000 x 1074) x= 3
(0.004 x 107!3) x=-3
EG Oe 10=)}
: 4 x 10-1
x Se 23.
3x +2
Su) dil St
DOT ATi
Sokal
30x +020 — 12 = 5x4 10
a b
DOt Gee et QI9xX=2
a+b b at+b
b at+b
Pe 2%
25
82 Algebra 2, Third Edition
Problem Set 31
24. (a) 2x + 3y = 18 27, a° = AesTF
ys t6 a* = 625
Us=25
(b) -12x + 6y = -18
24 x SF = 30
Gian4
aS TES
4
Eee
125 = 100 + 4c
1 ey
st Ls
/
ASABE
28, ax’ 13) 4x 2 So) aes er = 5)
Substitute y = 2x — 3 into (a) and get: 2+6-4x% —-4 =—2e +5
(a°) 2x + 32x — 3p= 18 —2x= 1
ae a HI) 1
ea
Veet 2
8
27 94. -8MS
wy=712)-3 29. B -= 180 - 30 a= 150
8 8 4
27 15 F + 150 = 180
8° 4 F = 30
en 1 et P + 30 + 90 = 180
De eT hac eRwOr? +x = 2 P = 60
xt + x?
Bee A = 120
+ Ox
x?
x? + ox C + 60 + 90 = 180
=x-= 2
C = 30
=k
— 1
“4 D = C = 30
26. (a) 6x + 4y = 11 E + 30 + 90 = 180
(b) 2x - 3y = -5 E = 60
l(a) 6x + 4y = 11
—3(b) =6r ¥ Fy = 15 A = 120; B = 150; C = 30; D = 30;
13y = 26 E = 60; F = 30; P = 60
mee 2
(b) 2 3(2) = -5 30. ACircle = ahem = 97 m
ox = 1 :
1 opie
ea Si 42 = 2 Fn? |= 6m
TU
(3.2)
1 | = (2d)°p) = (12 m)°2 = 144m
Agsquare 4
83
Algebra 2, Third Edition
Problem Set 32
(a) S5Ny + 10Np = 900
PRACTICE SET 32
(b) Np - Ny = 30
Substitute Np = Ny + 30 into (a) and get:
(a’) 5Ny + 10(Ny + 30) = 900
15Ny = 600
Ny = 40 nickels
(b) Np - 40 = 30
Np = 70 dimes
2/14 3V14 _ 4V14 _ 21vi4
7 a 4 14
= xX of = is
ie 300
100
S = ox M = 300. “ = 500 kilograms
ey,
p=4-1 20 _ NM
24 1440
pa A= 2
24NM = 20(1440)
ip =,3
NM = 1200 grams
PROBLEM SET 32 WD x of =
O87
Se— weLOI
Se wale = 2200 students
0.87
RwTw + R,T; = 56; Rw = 4;
R; = 8; Ty + Tj = 10 [2 #2 = SAS pk 2 8
3 2 Bes 2 v2 3
4Ty + 8(10 - Ty) = 56
ATy = —24 _ 2V6 , 3V6 _ 5v6
6 6 6
Tw = 6
Dy = 4(6) = 24 miles Se oll «pt
22 5 WT TL =
Dy, = 8(4) = 32 miles
2/35 3N35 _ 10935" 215/35
2. (a) N, = 3Ns + 30 7 5 35 35
Nearer
Oe Geer, ONS
= Ny,
ib
Substitute 6N; = N;, into (a) and get: 33 2V3 V5 _ 5v5 V3
(a’) (6Ns) = 3Ns + 30
a p03SaaS AE AB
3Ns = 30
_ 2N15~ S51 VONTS = 25405
ao 3) 15 15
Ns = 10 small ones
_19V15
(a) N; = 3(10) + 30 = 60 large ones 15
84 Algebra 2, Third Edition
Problem Set 32 _
10.
ce 2% 2)
x+3 x74 5x46
Xo 2) 2x —2
(x + 3)Go42) Bae 2)
Acircle = 2(2 m)? = 12.56 m? Xeerioae
— 2% +2
(x + 3)\(x
+ 2)
i. (180
— A) = 3(90
— A) - 30 x7 +2
180 — A = 270 - 3A - 30 (x + 3)(x
+ 2)
2A
= 60
A = 30° 1,
m-5 m* —5m m(m — 5) m(m — 5)
12. Since the slopes of perpendicular lines are negative m*—2
reciprocals of each other, m(m
— 5)
1
y = iy aetiO 2x3 + 8x? — 6x = -2x(x - 3)(x - 1)
2=—+(2) +b ~14x3 + 5x4 + x° = x3(x + 7)(x - 2)
3
ger aes5 Tax + ax? — 8ax* = ax(x -— 7)\(x - 1)
3 3
Be —|l2py + pxy + 4xpy = py(x + 6)(x —- 2)
3
1 8 1 8 ade
JO Ka, 3A
Siince m =-——a anandb=-,yzo = re t+ -.
--x 3 ae
5 2,
Que
— 5x 15s=F40
(0.0035 x 10!>)(0.002 x 10!7)
13. Se= Vil
7000
x 10°3
Mies 10 107). 7x 10" beet
+ 2843
id 7 x 107° 7 x 10°°
=x 107° xX Se poe
25.
2 A Oe
x+4y y
Ds
y Bes yea Ay
14.
x yi ches =ye= 30)
y yey x = -30
xy + 4x? y? _t xy t+ 4xy) ..
15. 1+ 4xy
xy xy
16. 3/125 — 245 + 3/20
= 3V5V5V5 — 2V3V3V5 + 3V2V2V5
= i545 — 645 61> = 15/5
a
Wed 4/5(2V10 - 3V5) = 4V5(2V2V5 - 3V5)
= 40,2 - 60
Algebra 2, Third Edition
85
Practice Set 33
26. (a) -x + 2y = 4 29.
y=—x+2
(b) x+y =-2
a
a
N
&
&
igi
z
td
|_|
si
ies
|
m Boe a0 ee 20) = 92) Lilfaieaeg®
Substitute y = (—x — 2) into (a) and get: 30. 3L( 2 3 2)(-3") ] | |
(er Pee PORE 4 = -3[(-4) -3[0] - 2 = -2
- 4] - 2 =(-1)
—3x
=8
8
asus PRACTICE SET 33
() ete PONE Lora m Sa mz + 3 P
% SL Bak OS i, WOM eee Pag
S58 a cM age
(-2.3) P p ps
=. mz tS
27. (4x + 2)(x? - 2x + 4) SZ
= Aika
tel6x fF 2x? — 4x 48
= 4x"
3 - 8x*2 + 12x
4 + 2x~ + 8
Nea |eB 2 SME are Oe
b. atm m_.atm_m matm
a a
23; (a) a+ 2y = 5 y+s yt+s
(Doyo =) my as
+ ms
m(a +m
(b’) y = 3x -7 a yanksOs+ sid
Cea Cre 2)
Substitute y = 3x — 7 into (a) and get: yor s
(AVR 4°20%-— 7) = 5 as+ms+my yrs
Ix <= 19 I) ae as
ES
a yt+s
aes a
Pighae ea
_ (as + ms + my)y + s)
y=
By ot(2)-S-5
19
hes =a
49 8 am(a
+ m)
86 Algebra 2, Third Edition
Problem Set 33
PROBLEM SET 33 sap) x s (F) + 228)
| oe) b _ at bb b\a+b
a2 AUS : a cK a
iT *H st+x St+x
4H = 17(104) bs +ax+bx s+x
H = 442 hedonists on ab + b* een”:
a s+x
s+x a
) | _M_ as (s + x)(bs + ax + bx)
~
wolnr 3600
ab(a
+ b)
\o 7(3600)
8. x + 40 = 180
=8 Hou2800 grams
x = 140
2y + 140 = 180
2y
= 40
y = 20
— x L = 1620
z= 40
mS |= 1620 x o = 2025 grams Arc length = = - 2(7)(6 cm) ~ 4.19 cm
9. z = 360 - 150 - 70
4. (a) Nr 8Nz + 10
z = 140
(b) Nr = 11Npz - 5
ger se
Bs
Substitute (b) into (a) and get:
140
(a’) 11Ng — 5 = 8Nz + 10 = — = 70
oe
3Npz = 15
10. Since the slopes of perpendicular lines are negative
Nz = 5 blues reciprocals of each other,
(a) Nr = 8(5) + 10 = 50 reds 5
= ake?
—(—4) 4D
noo
24,
3
RyTy + RcTc = 540; Ry = 40; Shice tke. One
3 3
ee
3
een
3
Ro =-00;-Fyp
+ Te
Rene
re ech) eee
40Ty + 60(11 - Ty) = 540 2 Pa give oc 45
~20Ty = -120 _ 3v5V2 2v2 45 — 3yi0 2v10
Ty = 6 WD ae fg? se FO 5
Dy = 40(6) = 240 miles 10 10 10
m 3 (=) 4 3 mx +3 p 5 6 _= 4V5 _ 2V6
12. [2-2/8 48 2
6 PA os a Path
na y Pa we _ 4/5V6 2V6 V5 _ 4V30 2730
Pp Pp Pp yy 6 16. SURG eG 5
_
=
WLS
rr
2030 12V30 _ 8V30 _ 430
30 30 30 15
Algebra 2, Third Edition 87
Problem Set 33
13. Perimeter = 11.28cm + 6cm + 6cm 3x — 2 2xit3 By
25.
= 23.28 cm ia 3
9x —6 -—-4x -6 = 24
14. 3(90 — A) = (180 — A) + 50
i) at 588)
210, Ar =e230) A
-2A = -40 Lies
75
A = 20°
(a) 2x -y = -5
(0.0027 x 10!>)(500 x 107°) 26:
15. Y= 2x75
900 x 10!4
Pe 100") 413.5 x.10° (b)x
+ y= 1
: 9 x 10/6 cor 10° y=x+1
= 1.5 x 10°”? y,
4x x* + 4x x
¥ van x ‘ l—xy _ x? + dxy
16.
= a! x 1 - xy
x x 1 — xy
rN
17. 3/18 + 2V50 - /98 BOCNCLO
feEN ot
= 342133 + 2V2V5V5 — V2V7V7 Een
EREEES
= 9./2 + 10V2 - 7/2 = 12/2
4x + 4xy ~ 4x(1 + y) = Wry
18. Substitute y = —x + 1 into (a) and get:
4x 4x
(a) 28 — Gx F P=
a b cx + 4 Shee a!
19.
a+y eo x+y 4
n ax e b(t :+7 y) x? (cx + 4) x=-—
3
(xty) (x+y) (x+y)
axe b(& af yy) + x (ex + 4)
‘ x(x + y)
4 6x —2
20.
ct 4 eo x7 +2x-8
€. te~ 2) 6x -2 a Divide 2x* — x by x — 2:
Pate ee = 2) eae 2) 2x?
+ 4x7 + 8x + 15 +
fy 85 8 IO aa —2x - 6 2)
2x4 Onder Owe Oe 4 0
~ (8 + 4)=2) ~ &@+4@
- 2) 2x4— 4x3
4x3
+ Ox?
21. 5x? + 4x3 — xt = x2 — 5)(x + 1) 4x3
— 8x?
Se — x
22. 10k? — 7k*x + k?x? = kx — 5)(x - 2) 8x2— 16x
15x + 0O
23. apx* — 20ap — apx = ap(x — 5)\(x + 4) 15x — 30
30
AAD 2482
24.
3 ge Check:
4x + 8 - 6x + 6 = 60 2x*(e = 2), 4x(x = 2) | 8x(x = 2)
—2x
= 46 et x-2 x-2
x = =23
, Ae 2) 30 sa
ie. x-2 x-2
88
Algebra 2, Third Edition
Problem Set 34
28. -39- x — y - y'-29 - 3) - [2 -|
-3° — 2) - €3)? - 3-4) - [+2 - 2)
-1+2-1 --0
9-4)
+ 36 = 36
-1+2-1 RxTx = RyTy; Rx = 10;
Ry
= 33:Ty + Ty.=,13
29. Perimeter = 2(2 ft) + 2(4 ft) = 12 ft
10Tx = 3(13 — Tx)
Area = 1 X w = 2ft x 4ft = 8 ft? 13Tx = 39
Tr = 3
Dx = 10(3) = 30 miles
PRACTICE SET 34
lz
100 X Ofe= is
au Kon ae
100
Rz = 30; Rr = 40; Tz = Tr+ 2
RrTr + 20 = RzTz s = 42 x 1 ~ 60 tons
70
(40)T; + 20 = (30)(Ty + 2)
40Ty + 20 = 307; + 60 a a
100 300
107; = 40
100S = 50(300)
Tr =4
S = 150 grams
Time = 7a.m. + 4hours = 11 a.m.
Integers:
N, N+ 1, N+ 2
S(N + N + 2) = 8(N + 1) + 14
PROBLEM SET 34
10N + 10 = 8N + 22
De 2Nag= 12
=r) N= 6
Dg
The desired integers are 6, 7, and 8.
RpTpz + 60 = ReTg; Re = 40;
——» _ | gee
Rp
= 50; Tg = Tg + 3
50Tz + 60 = 40(Tp + 3)
OE ee Ee ee
+ ae b- a ab b- a?
10Tz = 60 a ov ab baa
Tz = Obr
1 b lox SF os
Since Benita left at 11 a.m. and traveled for 6 hours,
she got within 60 kilometers of Elliot by 5 p.m. x Eee ee ee
Xx Xx Hg x?
x,
Dp
Bieicreere= Ven, = 3 ABase x Height)
1 :
De
RrTr +14 = RcTc; Rp => kay i
= 3 (2004) + (4)(4) + (12)(4)
Rc = Lie Tr = Tc
1.5(Tc) + 14 = 11T¢ i. hm? |w? x 8m
-3.5T¢ = -14 l
= 5(72 + 87)(8) m? 3 = 258.99 m 3
Tc = 4hr
Algebra 2, Third Edition 89
Problem Set 34
10. (4x + 20) = 2x + 20) a bx Cx
19. et
4x + 20 = 2x + 40 x(x + y) a x(x
es + eey) y
Dio 20 ax? bx? cx(x + y)
x= 10 x(x + y) x(x + y) (x + y)
ax? + bx? + cx(x + y)
11. Since the slopes of perpendicular lines are negative
x3(x + y)
reciprocals of each other,
1 2x -—1
eared
=—x+b 20.
x —-3 x2 —6x +9
«(BE Se ay 2x -1
Ce oe
D ~ (x — 3)(x - 3). (x = 3)(x
- 3)
0=b SAT
12 Del
1 1 (x - 39
Since m = — and b=0, y = =x.
2 2 x? ~ 9x + 13
(x = 3)
12.
2 C 2/2J9 3V9 V2
9)IE amie We ue a ph ara Mie
9 Quy V9 V9 « V2 V2 21. 4x? in? 4 Dae Ox + 2)ie = TD)
_ 2v18 — 3V18 — 4V18 — 274/18 22. ax?p — 8pa — 2axp = ap(x - 4) + 2)
Ly th 9 2 18 18
J B23g18)
4 2392 23. yx? — 4xy + 4y = ya - 2)@ —- 2)
18 6
fe es
Tis (Aint BVPaalBmen’s
a2
24.
D) 2
13.
2 +2/3 - We os 8 a X= Sie 38K HE 4 6
iS
3/6
flies pee Marae 13
B] 2) a
2
14. Ashaded = Acircle, Do Acircle, IG. = AGircle, 6
BG 2x —4
m(22 cm)? — 2(16 cm)* — 2(6 cm)" 23; ie = 5
3 2
(4842 -— 256m -— 362) cm?
2%. = OG ele oO)
= 602.88 1-cm-square floor tiles
—4x
= 18
(0.000032 x 10*)(700 x 107!*) i
9
15.
16,000 Dy
20.2,% 1087 x 10 ~) 26. (a) x =—"3y
="6
1.6 x 104 1
—12 y =-x-2
a
= eet See = 1.4 x 10716
1.6 x 10 (6): 2syrs 12
fee 5% 4y* +x y*
16. ioe eae ay pee ay
pee ol ye 3y? -1
y y aya
ie 8/27 — 2V75 + 2147
= 8/3V¥3V3
— 2V3V5V5 + 2V3V7V7
= 24/3 - 10/3 + 14/3 = 28/3
yee 5x7 y* ty x y(1 FOV) E
18. ® 1-5
xy xy
90 Algebra 2, Third Edition
Problem Set 35
Substitute y = —2x + 2 into (a) and get: . — Extend
6471/2 ae 1
(a) xe 36ex + 2)oae6
(be SA
642/3 — — 3164 = Bugg s
12
i)
7
= 8/8 = nn |
(b) y= -o{|p.=)
7
i14i erell101 -125° 79 *= get
es 1 a
Fe Aa Wij ce owlWe 25
27. (4-2)?
+ 2x7F 3)
= xt + 2x3 + 3x2 4 x9 + 2x? + 3x PROBLEM SET 35
= x44 3x3 + Sx? + 3x
Du
| 1200 ace
(a) See
Dy = 2
D,
(b) x - 3y =4
RyTy + 1200 = R,T,; Ry = 3Ry;
(b’) x = 3y + 4
Ty = T;, = 30
Substitute x = 3y + 4 into (a) and get: 30Ry + 1200 = 3Ry(30)
(a’) 3(3y + 4) - 2y = 2 —60Ry = —1200
Ty = -10 Ry = 20 yards per minute
ra) R;, = 3(20) = 60 yards per minute
eae t
(b’) (-2) rita Boe
7 7 7
(3-4)
7 oe
RrTr + RwTw = 56; Rr = 4;
Ry
= 6; Try + Tw = 12
AT, + 6(12 - Tz) = 56
29. (17)? = H? + 3?
17—9 = H* -2T; = -16
8 = H2
2/2 =H Tw = 12 - (8) = 4
They trudged for 8 hours and walked briskly for
Area = bey seals = 6/2 cm?
4 hours.
eyes 22 2) ex = xy Dp
—Dg
30. Slee
= —-l[-1 — 4 + 8 — 2](-3) +-3) — 3-4)
= Ns). 3 Sl? = os 12
RpTp = RpTp; Tp = 12;
Ty = 4; Rp = Rp + 6
PRACTICE SET 35 12Rp = 4(Rp + 6)
(14 — 2) x 180° = 12 x 180° = 2160° 8Rp = 24
Rp = 3
The sum of the measures of the exterior angles of
any polygon is always 360°. Dp = Dg = 3(12) = 36 miles
91
Algebra 2, Third Edition
Problem Set 35
(a) 5Np + 8Ny = 82 (b) x + 10(6) = 96
x = 36
(b) Np = 2Ny + 2
Substitute (b) into (a) and get: 5x SF=9
(a’) 5(2Ny + 2) + 8Ny = 82 Ware5
10Ny + 10 + 8Ny = 82
18Ny = 72
TxX—=A
Ny = 4 yellows
(b) Ng = 2(4) + 2 = 10 blues 63 _ 4
ofan
100 13. al 42/2Be = “+e
3/- OT Ae 2S=
5 7 5 V5 7, V1
40
— x A = 40
100 _ 3V35 , 2V35 _ 21V35 | 1035
ae 5 7 35 35
_ 31V35
wr 35
2020 Vv
400
20V = 20(400)
14. 1 V5 9f829 = as
WZis _ Fe
EE
V = 400 grams _ 2v10 _ 9V10 _ 4V10 — 4510
7 405 mM 10 10
Lateral S.A. = Perimeter x Height
_ 41/10
a (Fem) + (52x29) + 20|m 10
x 10m
GF» rea,
= (4m + 20)(10) m? ~ 325.6 m? b a b ay a ae
15.
xy xy Bb xy
1671/2 a 1] = 1 b b xy
1642 m4
Seay x+O6(x+y) x+y
ia sels at ot te A Cee
ee 16.
4
25 52 SY
a 4
at
x+y
4
10. 93/2 = (gl/2)3 - in SAGO
y) _ Tey
4 7 4
~6472/3 ae 1 2 1 1
11. 642/3 ad ~ (6413)? = gore 17. 2x + 10 = 80
2x = 70
12. (a) 3x - 4y 84
=woS
(b) x + 10y = 96
3y + 40 = 130
l(a) 3x - 4y= 84
3y
= 90
—3(b) —3x — 30y = -—288
y = 30
y= 6 k = 360 — 130 — 80 = 150
92 Algebra 2, Third Edition
Problem Set 35
18. 2x + 332 = 360 23. 35a - ax” — 2xa = -a(x + 7)(x - 5)
2x = 28
eed 24, 8x2 — 2x3 — x4 = 2% + AVX - 2)
y + 40 + 14 = 180 25.) 2V3 - V12 - 3V2 - J6 + 4/2(3V2 - V6)
y = 126 = 2/3/2V2V3 — 3J2V2V3 + 12V2V2
— 4/2/23
19. x = 2(105) — 112 = 98 12 — 6/3 + 24 - 8/3 = 36 - 14/3
x +88 98+ 88
= 93
2 2 26, (7000 x 10'*)(0.0002 x 1071!)
88 + 62 1400 x 10710
CEALE eis _ (7x 10792 x 10715) 14 x 10?
A TOES
9 1d 10-8
20. Graph the line to find the slope. 1 x 101°
x-3 2x
27.
a eee
4x - 12 - 14x = 140
-10x= 152
x= -—
28. 0.002x = 0.02 + 0.04
2x = 20 + 40
x = 30
Sicpe eye ea ’
: 29. 2x - 2x0 = 37
YS te? 7.
—x =—+—
Sie13: Wig
9 4 4
Ce
3 gene Liman be Wea
des ceva te 9
= ee
any 4
aly
3 30.
4 1 4 1
eee aa) =. eG
21. : 4 2x -—2
x (xy) x(x
4 y)
- 4 Qa =x «a2y* - 2x +4
x2 (x +ry) x? (x + y) x(x ¥, y)
3% Dax,
22. ——_ - —————_
x—-2 9% + xe- 6
* SA(X Foy” 2x
= 2d (> 2)05+ 3)
*OXe427)
= —2\(7
+3)
Algebra 2, Third Edition 93
Problem Set 36
PRACTICE SET 36
x2 — 6x +9 ie AO
x2 + 5x — 24 yr 72
RT; + 80 = RpTp; T; = Tp = 4; Ry =)
- ~3)G—F)
(x - NG+By _x-9
(e+ Ble 3) (xe 7X -:2) x —-2 30(4) + 80 = 4Rpz
200 = 4Rzg
x2 + 2x - 8 We tae:
50 mph = Rp
x? — 4x* — 21x x? = Ix
~ @+Da—-—2) xe—7) _ x +4 Multiples of 7: 7N, 7N + 7, 7N + 14,
ee— TNE 3) Ga. x+3 IN + 21
4(7N + 2) = 3(7N + 14) + 15
28N + 8 = 21N + 57
IN
= 49
aa X of =i
N=7
The desired integers are 49, 56, 63, and 70.
BeOS 0 W
100
W = 1360 words 2x + 4) + 2(
=) + 2 +1)
+ 2x 31 in.
(a) 4Nr + 21N;, = 290 2x +8 +x+2x
+2 + 2x 31 in.
(b) Nr + N;, = 30 19 2) 1n.
Substitute N,; = 30 — Nr into (a) and get: x 3 in.
(a’) 4Np + 21(30 — Nr) = 290
-17Npr = -340 (x + 5)@ — 4) OAH OGTR 2)
Nr = 20 fats (x -—2)(x+1 (x + 8)(x'- 5)
E (x + 5)(x - 4)(x + 2)
(b) (20) + Nz; = 30
N, = 10 leans (x - 2)(x + I(x - 5)
(x — 6x +3) @ = 5S) + 4)
(aide == WNT=BD (x — 8)(x + 4) (x - 6)(x + 4)
LS
(b) N + D = 40
_ (& + 3a - 5)
(x — 8)(x + 4)
Substitute D = 40 — N into (a) and get:
(a’) 5N = 3(40 — N) | aeRO oe
10. = —3* = -9
8N = 120
N = 15
(b) (15) + D = 40 11. x) a oe 1 =
ee 2972/8 (27'Bp s 9
D = 25
1
The fraction was = 12. (8144) 297
8 13/4 =
gr 3/4 =
13. (27) a 2/3 1 a 1 s.
(2757/3 ((—27)""°)*
9
RpTp = Ry yy; Tp = 10; Ty =. 8: Ry = Rp + 10 1 + 4 y+ 4x xy
10Rp = 8(Rp + 10) 14, ee xy Say 1
ee 1 3xy + 1 xy
2Rp = 80
xy xy 3xy +1
Rp = 40
y + 4x
Dp = Dy = 40(10) = 400 miles
3xy +1
f
94
Algebra 2, Third Edition
Problem Set 36
<a 4 -— 3x x 24, Bee /O(xp?) ey let gucicd|
15. x = 28 § 490 ashes =4x(2 3 :5°)E a Ix
La (i>: x 1 +.2x 3 = 2x +2 — 13 = 8x — 4% ex
x Br 7+ 2x
—2x — 8 = -3x
ya 8
16.
5 Eo S0 lk 8 25. (a) Every point on this line is 2 units above the x-axis.
_ 3V15_— 2vi5 _15V15 615 ye2
Mails Mtr oo gS (b) The y-intercept is -2. The slope is positive and
_ 3V15 the rise over the run for any triangle drawn is +.
ake 1
y= 3° - 2.
17. (Gael?) tee Toe | is « 10%
18 x 10710 ~ 18 x 19710 26. A = 180 — 140 = 40
= 1x 10’ B = 180 — 40 - 100 = 40
C = 180 — 100 = 80
18. be ape ae NS DB 40
3 Me an eee’ ihr
E = 180 — 90 - 40 = 50
_ 2V21 321 14V21-— 9/21
M = 180 — 80 - 50 = 50
© Spade?
Reet Oy" 2354 oe
oe S21 re ty x-1-—
a 3G 27.. x + L)x* + Ox?) Ox Oe 2
K+ x
19. 4/12(3V2 - 4V3) = 42/2 /3(3V2 - 43) -x> + Ox?
= 24/6 - 96 -x3 x?
x? + Ox
20. 2428 - 3163 4.217% x7 + x
= BIT — 303/94 Fetal S547 —x —2
= 4/7 - 9/7 + 10/7 = 57 —x- 1
6 3 Check:
21.
(44 2)— x7 +3x+2 x(x +2) 2" (e +11) Wices)
6(x +d)5 4) 3x? een gan lor cok ft
x2 (x + 2)(x + 1) x2 (x + 2)(x + 1) I(x + 1) lig a Pee
-3x7 + 6x + 6 er eae ree eee ee
x7 (x + 2)(x + 1) 28.
Pp cua fmx + b
22.
ax ts ax 3 24
as
pax? * ax(cx + a) mx + b
axe ax a’ x"
pax* + ax(cx + a)+mx
+b
a’x4
epee ee
23.
7 4 2
129 Ses re er ee. = "28
—9x = -6
2
x=
3
Algebra 2, Third Edition 95
Problem Set 37
r ance y ? 344% 4 53 2. j
Nitrogen: : 1 x 14 = 14
ch ee y woe
ee ire oe Hydrogen: 3 x 1 = 3
Total: 144+3=17
= 3 - 3x%y?
308 302 ee) 23S ea Oy 2 ihe =
See ieens =: 261604 ee S16 17
17N = 14(850)
N = 700 grams
PRACTICE SET 37
a. Carbon: Ve Vig 3. Nitrogen: cal es
cam O Hydrogen: 4ex l= 4
Chlorine: 35x02
19°x 22=058 Chlorine: 103552 35
Fluorine:
12 + 70 + 38 = 120 Total: 14 +4 + 35 = 53
Total:
Cet 255 Cle
53 a 7195
Total 120
eG spe le 53Cl = 35(795)
L200 re Cl = 525 grams
120C = 14,400
C = 120 grams 4. me
b. mZABC + mZBCD = 180° Ds 100
(125°) + mZBCD = 180°
mZBCD = 55° Rplz + 100 = Rey; Tp = Tr = 4; Rr = 2Rp
mZDAB = mZBCD 4Rz + 100 = 4(2Rp)
mZDAB = 55° -4Rp = -100
Nor 2 =.6 Rp = 25
ae Rr = 2(25)
= 50
2y-4=8 = :
eG B = 25(4) = 100 miles
Dy = 50(4) = 200 miles
c. 8y + 20 + 10y + 70 = 180
18y = 90 ns Fe
ofp = 5
y=5 @
2— x GR = 550
x + 50 + 8y + 20 = 180 4
x + 8(5) + 70 = 180 GR 4
x + 110 = 180 = 550 x 1 = 200 gestures
x-=-70
6 (ot 3) 2) 4 OS LE vet a
x(x + 4)(x
+3) (x4
+ 4)(x - 2 “Sy +4
PROBLEM SET 37 m :
1. Hydrogen: 2 x 1 =2 7. ax(x —"4)Gst 3) OPA )* ae
Oxygen: 1 x 16 = 16 (X% 4) OOS) ax(x — 4)
Total: 2 +16 =18
g 948 . __} 1 1
Gr Oxygen ~ 94/3 (gl/3)4 16
18 5400
18(Oxygen)
(Oxygen) (
= 16(5400 ) 9. (27743 = ee eee 7 ne
Oxygen = 4800 grams (-27)43 ((-27)/3)4 ~ 81
96 Algebra 2, Third Edition
Problem Set 37
oy ast la Tay 1 1 Since the slopes of perpendicular lines are negative
10.
als ~~ (971354 ERT reciprocals of each other,
= = 1 m= 2
— A
ie
93/2 = (-3)(-932) = (-3)(-9!2)3
9
12. =x +d
mZPON = 180 — 85 = 95° 4
Xe Sel 9
oa
2= ie + b
—(-4)
y+4=9 ll=b
y=5 9
Since A and b= 1l,y= ax +
13; x = 2(90) — 80 = 100
yd = Oe eI Ss 19. 2x + 280 = 360
2 2 2 50
gee100 ee2
SY2 — 80 x = 40
y + (40 + 30) = 180
ae | e y = 110
14.
eee a BO ae
ae ae: x? Xi, 20. (a) 3x — 3y = -6
x2 Be2 x 4 =D
a he a oa ee
15. eees ety
ere Gx iby 5 ax
— my
x Xs Che
= ins
16.
3J2 V7 5V7 V2 _ 3V14 5/4
iE Gf a ON
242 7 2
6V14 _ 35V14 _ 29/14
a4 Me. 14
17:
211d WS. vo WES = 85/33. 54/33
S640 . J Vt< 03 11
_ 22433 _ 15¥33 _ 733
33 33 Khe
Substitute y = —3x + 6 into (a) and get:
18. Graph the line to find the slope.
(a’) 3x — 3(-3x + 6) = -
12x = 12
a3 = Il
(a) y=()D+2=3
(1,3)
ry, By
21.
x+5 x? + 4x -—5
x(x -— 1) ae
TRG 1) et Se
x” — 4x
(x + 5)(x = 1)
Algebra 2, Third Edition 97
Problem Set 38
4 ellie: 4 6(x
+ 2) 36x = 16x? — x?
22. — x — 16x* - 36x = 0
x(x + 2) Ey x(x + 2) x(x
+ 2)
_ 6x
+ 16
x — 16x2-436).= 0
~ x(x + 2) x(x = 18)@+4+'2) = 0
x = 0,18, -2
x+2 x -—3
23.
ee PROBLEM SET 38
21x + 14 — 5x + 15 = 70
16x
= 41 3Np = —4(-Np) - 15
Np = 15 protrusions
41
“elgg Zee ie
9 1440
24. anes 9Fe = 2(1440)
Fe = 320 kilograms
16x — 6x + 12x = 60
22x
= 60 Hydrogen: 2X = 2
30 Sulfur: b %32> 632
Sane Oxygen: 4x 16 = 64
Total: 2 + 32 + 64 = 98
25. 3./2(5V12 - 22)
32. 3S
= 3V2(5V2V2¥3 - 2V2) = 3076 - 12 98 196
26. 4./20(3V2 - 25) 98S = 32(196)
= 4,2/2-/5(3V2 — 2V5) = 24/10 - 80 S = 64 grams
ie (2B) ) alae ye 4 Gua
D 2
aN Ser
(x2 iy x4 x9 y? x2 x?
(b) 10D ="2NE
+ 84 Sea vN = SD 2
28. es 2y (ee Substitute (b) into (a) and get:
=-4+8+4+2-1-2=3 (a’) 2(5D — 42) = 7D
3D = 84
29.
(35-< 10 *)(4plo ae: 105s 105© D = 28
24 x 10529 2.1 x 1076
(b) N = 5(28) — 42 = 98
= 5x 109
De
-—.
4x? y [= 3x? y* z*
30. = — = *_| = gy} - 1277!
ze ieee p mn = Dr
RcTc = RrTy; Rc = 30; Rp = 20; Te + Tr = 10
30T¢ = 20(10 - T,)
PRACTICE SET 38
50Tc¢ = 200
Gy =" PD + 2H + 2) Tc =4
= (x + 2)(x* + 4x + 4) Dc = 30(4) = 120 miles
Koch Anh cd
(& + 5) Si@ + 5)Gnea Geet)
x +2
x + Ax? + dx (x? + 10x + 25)\(x + 5)
Dx Bers I x? + 10x? + 25x + 5x2 + 50x
+ 125
x? + 6x? + 12x + 8 x + 15x? + 75x 4+ 125
98 Algebra 2, Third Edition
Problem Set 38
(x + 4 = (x + 4x + 4x + 4)
= (x7 + 8x + 16)(x + 4) mp” F mp? — 20 4
x? + 8x? + 16x + 4x? + 32x 4 64 19. 4 4 l6p? — p?
} p*y - 2
16 p= 4
Pipl apsA agar
x + 12x? + 48x + 64 4 4 Pp p
2
Ketek a - mp” — 20
x? = ¥ Shipr=-O 15 p*
(x - 4a + 3) = 0
x = 4,-3 7 x 10778)(3. x 1077
20.
WSS AND
—48x = ~2x? _x
x + 2x? — 48x = 0 32 Vil Svil 2 _ 3922" 522
21.
x(x + 8) — 6) = 0 MILANI DF Ae 11 2
x = 0,-8,6
_ 6V22 , 55v22 _ 6122
10. 2 + 2x - 112 = 0 22 Sh By:
2(x + 8x - 7) =0
x = —337/ 2nd NA 3 Nl aN aa oe
22.
4303 SOY TT 3 11
(x + 5)(x + 2) ’ x(x + 7)(x + 4)
11. _ _ 2233 f 2133 _ 33
x(x + 7)(x + 2) (x + 5)(x - 3)
33 33 33
“ait 4
x -3 23. 3/24(2V6 — 312) |
ck eee a1 ee
re ee1
= 3/22 V2V3(2V2V3 - 3V¥2V273)
12. 16 a 16!/4 Ss 32
72 - 108/2
13. Ree ee er ee et 24. Graph the line to find the slope.
16°/4 (1649 8
32/3 = (3'?) =A
14.
(-8)!4 =.
1s.
16. (11? = # + [
gee ares4
35 _ yp
4
V35
_oy
Zz
ve =
= perec
in. X 5 renee
1
cal= 5 4 . y <lileees’5°Ae 1
phe)
320 —5 = -——(-2)
(-2) + b
ii. Length DEF = 360 - 2(7)(20 m) ~ 111.64m
23 _ »
2 a x -a iG 5
Pare oy aE ha Gn pte 4
18. 23 1 23
2 Baw pe igri l > Gace O Since m = —— and Db = a er
x x Guak
= a
Algebra 2, Third Edition
Practice Set 39
PRACTICE SET 39
pe pee ee
3 4
20x — 8 — 3x = 84
144s”
- 36 = 0
ley
= 92 (12s — 6)(12s
+ 6) = 0
_ 92 12s —On—a0)
rA7
12 se=a6
en ce See |
1
6 x4 1 Oe + Oxo Ot s= =
Wy,
xt + x?
-x3
+ Ox? 12s +6=0
_x3 x2
12s
= —6
x? + Ox
x? + XxX
—x-— l
bed
Zz
—x-1
0
e s Dik
27.
HEAD
121m
— 64 = 0
(1lm — 8)(11m
+ 8) = 0
llm
—8 =0
Limc= 8
8
m= —
Tt
llm +8 =0
llm
= —8
ap = ee8
La
1 By 2x +1
28. + + nT m= -—,—
WS x+2 x7 + 5x +6 11 11
* x+2 - 3x(x
+ 3)
(et Bix #2) eG Se) X + B = 360 — 306
2x +1
X+B=54
(x + 3)\(x
+ 2)
M2 + 3x Ox +07 + 1
pam pene
- (x + 3)(x
+ 2) D,
_ 3x7 + 12x + 3 Huse XG=—27)
WG HG 2)
X + B + 2A = 180
By eee ue Pe be (27) + (27) + 2A = 180
(ep py) Op) Pp op poy y opp
= p2yl3 2A
= 126
A = 63
1
30: Bee ar B37 = 37) Z = 90 because the diagonals of a rhombus are
perpendicular.
=8-— 4+ 36 = 40
100 Algebra 2, Third Edition
Problem Set 39
PROBLEM SET 39 6. 7(90 — A) = 2(180 — A) + 110
le ES of 630 - 7A = 360 - 2A + 110
3 -5A = -160
—xS$=4
16 se A = 32°
Ss= 420 x 16 = a9 ’
3 7. Area = — = 52 ft
Number with ulterior motives = 2240 — 420
2(52 ft?)
2
= 1820 sophomores Deb = ———
13 ft
ax = 18ift
P Areacircle = Zr? = (2)(8 ft)? ~ 200.96 ft
2: 100 xX of = Is
8. x? —9 =0
=e pe OSG (xi - 3)(x
100 - + 3) .=0
29 Oe
100 4 (Kt)
ead ay
28 9. 36x2 — 36 = 0
Ukases = 34,200 — 9576 = 24,624 36(x? 5) ai)
36 - 1)x + 1) =0
Sa Carpons sl e127 12
x = 1, =1
Oxyeen: 2 x 16 = 32
Total: 12 + 32 = 44 10. 24x = -11x” — x?
te x3 + 11x? + 24x = 0
44. 528 x(x + 8\(x + 3) = 0
44C = 12(528) 03-8
C = 144 grams 3
ll. @-1P =@-DE-DE- 1)
P = ete = 1h
ao a =x - 2x7 +x —-x74+2x-1
=
3 — 3x* 2 + 3x - 1
eer = Bee gape OT
100
100 v x(x + SX + 1) (x + 7)(x — 2) ose
Mae ORs re ee = (x 5x= 3) xe Dal) x= 3
Sr = 800 — 688 = 112 grams " 90 = 79432. _(412)3 = 8
‘ _4°3/2 S e
5, Dp
Pa 14. —
-3?~~ = -l|-27
2/3 = (27
1/3)2 =n
RrTr = RyTy; Rr = ys Rw =p SP Tr ae Tw = 6! ; 4 Pee!) eA a
5Tr = 3(8 — Tp) 15 a
ee ey oe ee x? + 6a
ee
: 2 2 4+ 6a a
elk io, S25.
a ie a x” 5 + 6a
eas A dsb
Dp = 5(3) = 15 miles x2 + 6a
Algebra 2, Third Edition 101
Problem Set 39
24. Graph the line to find the slope.
mp m* p — 6x as
—= 6 i as. 2 4 y
16. =
x ay es =
a ere Mm px —
mp — = m* px — 4 x
x m px -— 4
m? p — 6x
m? px - 4
17.
(359072) <0) 24s 1077)
2.4 x 10!4 2.4 x 10!4 i
Low
= 1x 190° i
ig, 293. V3 _ 5V13 V3 _ 2V39 _ 5v39 Ey Sede
W413 (13-4328 13 3 dae ar
= 6/39 = 654139 = _ 59/39 Since the slopes of perpendicular lines are negative
39 39 39 reciprocals of each other,
5
19, ONS 2 2 202 35 _.5/6 256 (agora
2 2 A 2 3 5
_ 15V6 4V6 _ 11V6 Vie ae
+16 6 FG
eens —=(2) +b
20. 545 - 2/75 + 2¥108 6
= 1534945 = 2954595 242/213 -133 = Bip
= 15/5 — 10/3 + 12V3 = 15/5 + 23
; 5 17
Since m, = —— and b = —,
21. (a) 2x + 4y = 120 6 3
b) 3x - 2y S © 17
(b) y = 60 y es
6 3
1(a) 2x + 4y = 120
2(b) 6x - 4y = 120 95 9! eagle
8x = 240 : —
v= 730 s. = ee + oe
16 16
(a) 2(30) + 4y = 120 ee)
4y = 60 6 OOF 36
Sig ES)
26. 0.002x — 0.02 = 6.6
22. mZWXY = mZYZW = 68° 2x — 20 = 6600
Oy 3 13 aa
See 10 x = 3310
w= 5 3 2
ore epee Ma ernie3 iy
2,
3y +4 = 11 Soothe RL) ee |
3y = 7 i FT ea) y(3x + 2)
, y(x+1P yx + 1h yx +1
as = Sy +3y+2xt ot2x +Sry Hy
y(x + 1)?
23. 3V12(4V3 -— 3V3) = 3V2V2V3(413 - 3/3) _ 2x* + 2x + Sy + 6xy
= 72 — 54 = 18 ye eel
102
Algebra 2, Third Edition
Problem Set 40
5S. A4xy + 4x? y2 7 4xy(1 + xy) _ ee 3. Iron: P< OG 56
4xy 4xy Bw ais Sulfur: 1 x 32 = 32
it ae Ss Se Total: 56 + 32 = 88m
29. le Axe oe oxy 56 448
2 4 zy 88 = Fes
= 3+ 84%“ 56FeS = 88(448)
FeS = 704 grams
30; 2° 2-29(99-5°) > aut x*y — xy
as: 4. (a) 20Ns + 60Ng = 8000
=-1+3 +44 + (2)7(3) - (2)3) (b) Ng = 3Ng
mh eS ee 12+ 6 924 Substitute (b) into (a) and get:
(a’) 20Ny + 60(3Ns) = 8000
200N, = 8000
PRACTICE SET 40 :
Ns = 40 pairs of shoes
es re ed ae (b) Nz = 3(40) = 120 pairs of boots
m ¥
=) Dp 5. Da
— > mx — ay = mx =f— -imx + Ss my 120
58
5x — amxy = mp + msx PK
5x = amxy + mp + msx RaT, + 120 = ReTxs Re = 15; Rg = 3;
5x = m(axy + p + Sx) T, = Tx
sie Ey, 3TK + 120 = I5Tx
axy+pt+sx 120 = 12T,
10 = Tx
b. tes
x m
Dx = 15(10) = 150 miles
Pe mx = iz et 6. te =u
s m x a
m + px = mxz ac + mx = acx
m> = mxz — px mx = acx — ac
m = x(mz - P) mx = a(cx — Cc)
m? ow. Li
Vs = +c= =
a
PROBLEM SET 40 m
ax + acm = my
1 poe x of = is a(x + cm) = my
100 ae my
ao x 1,200,000 = ME x + cm
100 .
ME = $1,968,000 8. Le nL ee
Pp My
Ve é 6y — apxy = mp + kpy
2. — x of = is
100 6y — apxy — kpy = mp
y(6 — apx — kp) = mp
380. 4800 = MS ‘4
100 — mp
MS = 18,240 words : 6 — apx — kp
Algebra 2, Third Edition 103
Problem Set 40
eg
Cc
ee
k
Bs ee ee
18. P Ra A apeee
ak — bck = cm
ide (eel gs teeoe
k(a — bc) = cm P Pp a’py— 1
ha om xy? - 4p
a — be ~ @tpy —1
10. (ce 33) ea 3) 3) 3)
=" (x7 eer 40 )\Q 3) 19. (4 x 1077 x 107) -_2.8 x 107
1.4 x 10716 1.4 x 10716
= ay = Gxt Ox — 3x° + 18 — 27
= 2x 108
= x° — 9x? + 27x - 27
20. 3/50 22/72
+ 34162
Le
360
-—318 _ 21
XKa=
2 3V2V5V5 — 2V2V66 + 3V¥2 133 V3V3
15V2 -— 122 + 27/2 = 30V2
Since diagonal bisectors of a rhombus are
perpendicular, Z = 90.
21.
33V7, 2V7 V3 _ 3V21 , 2v21
Y + 21 + 90 = 180
LI AT? 3 3 7 3
Y = 69
_ 9v21 , 14V21 _ 2321
12. 4x2 - 49 = 0 eel 21 et
(2x — 7)(2x + 7) = 0
22. A = 30
alae B + 30 = 180
Perse,
B = 150
13. x? + 3x2 = 18x
2C + 150 = 180
x? + 3x? — 18x = 0
2C = 30
x(x + 6)\(x - 3) = 0
C = 15
x = 0,3,-6
23. Volumec,):
Cylinder =7r-xh= 32 cm?
x(x — 5)(x - 2) Cat-6)(e = 3)5 _
14. 3
G+
6) 2). e#= Hie
s) - 2 32m7cem
r= ————_
87cm
15.
(-4°)?
rap = (1)(47/7) Ba au) 28
r= V4cm? = 2cm
24. (a) Every point on this line is 1 unit to the left of
16. ee a the y-axis.
x =-l
4x7 a she (b) The y-intercept is -2. The slope is positive and
y? a the rise over the run for any triangle drawn is +1.
17.
py)
2
<_ Se y=x =2
y a
4x2 a + ve ya’ 3 —- 2x x
25. + —=
i ax ya? 2 4 3 :
7 ya a 9 —- 6x + 4x = 60
ya 2y? a =i2 =2h
=a 51
4x2a3 + y?
51
x= =—_—
2y* a -2 7}
104 Algebra 2, Third Edition
Problem Set 41
26. 0.004x - 0.02 = 2.02 PROBLEM SET 41
4x — 20 = 2020
1. Odd integers: N, N + 2,N +4, N +6
4x = 2040
4(N + N + 6) = 3(N+2+4+N
+4 4) + 12
x = 510
8N + 24 = 6N + 30
2N =. 6
7 34 2
A ae 3x
N= 3
Ct De eee ee
The desired integers are 3, 5, 7, and 9.
Oe oe Le he) 2x(x + 1)
x(x + 1)(x + 2) x(x + 1)(x + 2)
Chromium: IP xX 52 R=) 52
3x2
al —————EEE—— EEE EEE
Chlorine: Br Se Sey = bs!
x(x + 1)(x + 2)
Total: 52 + 105 = 157
_ 3x7 + 9x +6 + 2x? + 2x + 3x?
x(x + 1)(x + 2) 105. Cl
[57 | 0256
_ 8x? + 11x + 6
~ x(x + 1)(x + 2) 157Cl = 105(1256)
Cl = 840 grams
eee x1
+ 4) =
2A RR 11136
100
100
5 {re2 22
2
ay Meee TR 1136 x —
284
= 400 tons
(a) 5Ny + 10Np = 700
30. Pa xy- 3 =[ (b) Ny + Np = 100
Substitute Np = 100 — Ny into (a) and get:
(a’) S5Ny + 10(100 — Ny) = 700
—SNy = —300
Nw = 60 nickels
PRACTICE SET 41
(b) Np = 100 — (60) = 40 dimes
1 ft _ 840 ~ 70K
OM SA tat eee Dr Dw
£54 eae OR SED [P+
76 |
12in. _ 12in. = 90(12)(12) in.
mee! RrTr + RwTw = 76; Rr = 16;
b.: 9042x 3 x £12 — (12)(12)
Ry= 4, Tp + Tw =7
16(7 — Tw) + 4Tw = 76
e 40 ai? x we sl * 52807 y 528 Of
1 mar 1 war | par
-12Ty Il | WwON
12m r ipa " 12 in. Tw = 3
1 & 1 lf
Tr = 4
= 30(5280)(5280)(5280)(12)(12)(12) in
Dy = 4(3) = 12 miles
Dr = 16(4) = 64 miles
thie lei
6. 87 ft? x x — = §7(12)(12) in.”
1 ft 1 ft
105
Algebra 2, Third Edition
Problem Set 41
3 ft 3 ft 12 in. 12 in. 14. 16x = —x? + 10x?
61 yd* x —— X —— x x —
x? — 10x + 16x = 0
lyd tlyd 1 ft 1 ft
= 61(3)(3)(12)(12) in.” x(x — 8)\(x- 2) = 0
x = 0,2,8
5280 ft . 5280 ft g 5280 ft
32 mi? x
1 mi 1 mi 1 mi 15. WeOx 0
an. Pie 82 N,
x x x x(4x
- 9) = 0
1 ft 1 ft 1 ft
= 32(5280)(5280)(5280)(12)(12)(12) in. x= 02
4
2x2
+ 4x + 8 + —32
16. (m2)
Ove OV Hee el
2x3 — 4x?
Ax? + Ox
4x? — 8x
8x = al
8x — 16
10. 15
2
17. 42 = H* + (3
D
Cae
16.= H? + Es
4
11. eee oy! cols H
4
i GC m
4cmp — x?km = cxy 5D otter,
4cmp — cxy = x?km 2
Ate x“km
4mp
—- xy
Area =
ax
5 +
n(2)° = 11.84 units”
12. 8 x SF = 20
ShiaDD 18.
x(x + 5)(x +3) (x + 4)(%— 2)
(x1 —2)(4—2) xe + 4)Geees3)
et)
fags wie
2 Seygeee>
Eas
ce
x 1
19.
32 = 322/5 aay ;
5
ily
Se SS
) Pp
a ates a x
Ce ern = a leet
Te
aes 20.
aes axe x
x x ax — 4
13. 2x + 40 = 180
De
= NAW - Cerra
+ ax? - 4
x = IAL
Since ADFF is equilateral, y = 60. (2.1 x 105
21. (5X 10°) 105 102
k. + 70 + 50 = 180 1.5 x 107! ie 10s
k = 60 =7x 10°
106 Algebra 2, Third Edition
Problem Set 42
22. 3V5V7 _ 6V7 V5 _ 3V35 _ 6/35 4x+8x?
Ax 4x(1+2
+ 8x" _ 4x(1 + 2x) et ey
fetes St 4x 4x
_ 15V35 _ 42V35 _ _ 27/35
35 35 35 29,
x2 x9 x 1-2 7 yx-3 x oy 2 Wa
(x2 yy? xyx2 x2 x7 ips x
23. 4/24(2V6 — 3V2)
= 4/2/22 /3(2V2 v3 - 32) 30.
= 96 -— 48/3
Write the equation of the given line in slope-
intercept form.
1 2
p= > Ss3 PRACTICE SET 42
Since the slopes of perpendicular lines are negative a. 13(5280)(5280)(12)(12)
reciprocals of each other, = (1 x 10!)(5 x 10°)(5 x 10°)
y=3x+b (1 x 10')(1 x 10!)
-5 = 3(-2) +b = 25 x 10? = 2.5 x 10!
| at
yo ir +i b.
_(4353)(933,216 x 107!)
(319,214)(0.01603 x 1073!)
25.
— =e x * ~ 109
= (4x x10%) _ 36 = x
1073
a: (3.10702 10972): » 6 10728
—3-x-x = 6 x 107
=—25
= 17
(0.013926 x 107!*)(27,153 x 107!)
6354 x 1077!
(gx 102)"(3, x 0a)
6 x 10°78
3 x 10"! gu
aah tl ees cel
x
PROBLEM SET 42
= 204d eae
17
396 fish = P
2 2X 2
27.
e(x-2) (2-4) 2. ates = 4800
& AA Xetkee) 100
Ss = 43000x< _, = 6000 students
x? (2x +2)
: x? (x — 2)(x + 2)
i ee
Pe 4 Abe ae — Dx" 3.
3600 43,200
x? (x — 2)(x + 2)
3600P = 600(43,200)
-2x3 - 2x7 + 2x +4
P = 7200 grams
Algebra 2, Third Edition
107
Problem Set 42
4. Potassium: 1 x 39 = 39 a Gd e pe
Chlorine: 1 x 35 = 35 : y d
Oxygen: 3 x 16 = 48 adx + dmy = pcy
Total: 39 + 35 + 48 = 122 adx = pcy — dmy
Ones
122-488
ea
ge
122K = 39(488) 4x? - 8x + 16 -
K = 156 grams 12. x +2)43 4 0x2 + Ox- 1
4x? + 8x2
5. N; = 4Ny + 2 2ox-
2
+ Or
CYNE 2 8x? — 16x
(b) N, = 8Ns
—6 1éx — 1
Substitute (b) into (a) and get: r+
(a’) (8Ng - 6) = 4Ns +2
4Ns = 8
Ns = 2 13. Area = = - (2m)? =~ 8.37 cm?
(a) Ny, = 4(2) + 2 = 10
14, 20x = x? — 9x?
6. (24)(5280)(5280)(5280)(12)(12)(12) pen aes eur
= (2 x 10!)(5 x 107)(5 x 10°)(5 x 103) Paes
(1 x 10!) x 10!)(1 x 10!) Ave
= 250 x 103 ~ 3 x 1015 eee
. ESE
(2472)(570,185 xx 107!2) ie Oe S\Oxete
pe5s) =
(243,195)(0.0003128 x 107°) \ oa ai
5 5
| OLeU 3 Ot Ney 7 Sa k= 2hes
(2 x 10°)(3 x 10719)
x(x =— 2)% =1) ‘Get sy +3)
=i\5 37 16... ———— |)
8. UIE ee zene) (x + 4) = 1) xe + 3) — 2)
x
be Geek}
-17 41 =
To 10-°
7 913% Ge) peek es
Me fe i an = “(gis * 27
lyd lyd lyd
x in. J 12 in. 12 in. ee m— 4x x
1 ft 1 ft 1 ft 18, +~—__—(= <= 6x -1
= 40(3)(3)(3)(12)(12)(12) in.3 6= Ox oul ae
x Ey 6x -— 1
m — 4x
100 =Se.0e3
m D =
- x 5 6x - 1
2 — =
Ne ie Ze 19. 2V3 Vil _ SVil V3 _ 2V33 533
m~ = cp + cmx A Aa, 3. “4B a eT re)
reheonea
2
_ 6V33 _ 5533 _ _ 4933
p+mx 33 33 33
108 Algebra 2, Third Edition
Problem Set 42
- 4,2)
20. 3\6(2V6 - 4/2) = 3¥2/3(2V2¥3 :
=2ei= 4 x
= 36 - 24/3 28, 5 ee
21. 3/20 + 245 — /245 -6x - 12 - 2x = 30
= 3¥2V2V5 + 2V3V3V5 - 5/77 (ee
= 6V5 + 6V5 - 7V5 = 5V5 -
22. 16 - SF = 2 E48
== 2
ie: 26 agro
stra
3 et a 2
45 =a 4 16 16
3 a4 5
ea 4 ig eens
51 =b
273 TE
3 4 eT
3
23. (a) 3x + 2y = 80
(b) 4x — 4y = 100 mee ;
2a) Gx + 4y = 160 Beary) | (ee Na
(b) 4x - 4y = 100 SOPs}
10x = 260 TG 2)
ame 4 - 32-6
Ox © 6 2)Gcet
2)
= 100
- 4y )
(b) 4(26 PERERA
yet 4 -16
(x — 2)(x
+ 2)
24. Graph the line to find the slope.
—2 —2
28. {2 > | = -1 + 3x4
y
29. ——j— =
30. -3x -x?% -x° = ->(3] yee: WE
Since the slopes of perpendicular lines are negative
reciprocals of each other,
m,
= -l
y=—x+b)
—2 = —-3)
+b
-5 =b
y=xu-S
Algebra 2, Third Edition 109
Problem Set 43
PRACTICE SET 43 5. | a Cc |
a. 4cos 75.8° = 0.98
RcT¢ + 20 = Ri TL;
b. 6 sin 37.42° = 3.65
To = T, = 5; Ry = 2Rc
; 3.61 1
c. sinC = 752 = 0.48 ryan + 20 = SR,
3
20 = —R L
d. cosD = ~ = 0.76 2
; 8 = Ry,
D, = 8(5) = 40 miles
e hy arpa ok = 0.77 y
1.30 i ,
6. sineA = _Oppestt =: cosine A = faecpeent
f. Inverse tangent of 0.405 ~ 22.05° hypotenuse ypotenuse
emgentAce opposite
g. Inverse sine of 0.794 =~ 52.56° adjacent
7. (a) 417 cos 51.5° = 259.59
PROBLEM SET 43 (b) 32.6 tan 86.3° ~ 504.12
1. 0.0032 x TP = 1280 Seana ests . *. Dare
ss 1280 = 400,000 people ypotenuse ;
0.0032 ¥ .
(b) cosB = OEM tse = 076
732 hypotenuse 9.2
ah (c) tanC = on = : = 1.33
LM = 9280 x —— = 4000 boys cota
piey
9. aS £2, in. é 12 in. ‘ 12 in.
3. Sodium: Ipxe03-= 23 1 ft 1 ft 1 ft
Oxygen: 1 x 16 = 16 = 4(12)(12)(12) in
Hydrogen: Lx le
Total: 23 + 16 + 1 = 40
| 10, -=-x=
b Z
23. Na. az — bxz = pb
40 320 az = pb + bxz
40Na = 23(320) a2
Na = 184 grams pt xz
XZ m
4. eC The ppb oe
100
NC = 2310 grams eR Mia ip
cxzZ = mp + ckp
Cw
m + ck i
110 Algebra 2, Third Edition
Problem Set 43
18.
1 Ax* +1 x
CXZ
19. ee a ee
=m ihe ay” — 4x x
cp
+ kz x x ay* — 4x
2
ix ¢ = #+(Z)
2
164 4 20.
give AS
4
113
a= —
*
4-SF=6
21. 34/9 (SV12/— 2) =93V2 (BV24243 = V2)
on.2 = 30/6 - 6
2
egies 22. Since the triangle is isosceles, A = B.
fe ae
AO Qeee A$
21 = sh
—— A2 = 4 + 16
4
= /20 = 25
Vi13 , PFJ SCA
= B= 2,5
SL hk 2
Baits side 424113
23. pee ae ys
4113
= 4c
113
gilS.2,,
4
£5 = -=(-2)
+b
14. x — 9x2 = 20x aie4,
7
x + 9x? + 20x = 0
Lah BAN:
x(x + 5)\(x + 4) = 0
soma bi Pees
x = 0,-5,-4
See Oe a a
24.
15. 4x? —~ 81 = 0 2 5
(2x — 9)(2x + 9) = 0 25x — 35 — 6x + 4 = 40
9 9 19x
= 71
Xia
oy)
Pai
19
16. (x — 3° = (x - 3)x = 3)x — 3)
= (x — 6x + 9)(x - 3) (47,816 x 10°)(4923 x 107!*)
x = 6x204 Ox — 3x* + 18x = 27 25.
403,000
2 ye Sy + 21 = 27 (5 x 10°)(5 x 107!) = 6.25 x 10°!
7 4 x 10°
x(%— 3x = 1) (x — 3)(x + 3) z
ut 6 x 107’
Cnr NG = 1) | xt = 5)(x — 3)
111
Algebra 2, Third Edition
Practice Set 44
26. (a) 2x -— 3y = -9 29.
-3y
= -2x - 9
2
y Say37 5
(by-x
+ y= 2
y=x+2
I-11!
why
anh
DS Ghats
D? = 36 + 1
D = 37
30.
NO
ll | | |
Re
Nl
~~ Se) lr|e N|[e Ile
N Sa aan
(= SS,
Substitute y = —x + 2 into (a) and get:
(a’) 2x 3 oS
Spe=
PRACTICE SET 44
tan bes aa
ye="6.4 tan ol
y = 2.46
27.
H? = y? + 6.4?
H? = (2.46)? + 6.42
3x* (x - 1) 2x. =a) = 147.0116
x(x + I(x —-1)_ x7 (4 + DG - 1) H ~ 6.86
x? (3x + 2)
x* (+ Dix) p-=i2? 418"
3x? — 3x7 — 2x +2 + 3x? + 2x2 p = 208
x? (x + I(x - 1)
p = 4V13
6x° - x7 - 2x + 2
x? (x + 1)(x - 1) Lane ae
8
28. +(e
—— - x = 56.31
p BY P
tany = 17
8
y = 33.69
112 Algebra 2, Third Edition
Problem Set 44
PROBLEM SET 44
cos 31° = hua
—2(N + 5) = 6(-N) + 18 14
4N = 28 m = 14cos 31° = 12.00
NG=a
sin 31° = is
4
Odd integers: N, N + 2,N +4,N +6
v= 14513) a7 20
S(N + N + 4) = 8(N +2 +N + 6) + 22
10N + 20 = 16N + 86 cosA = ft
—6N= 66 4
N = -ll A = 55.15
The desired integers are —11, —9, —7, and —5. B + 55.15 + 90 = 180
B = 34.85
270 i
700 ~ 2800 sin55.15%= —
1
700T = 270(2800) % = 7 sin 555° 05.74
T = 1080 tons
ets 5280 ft 5280 ft
Carbon: IP S62 12 = 4(5280)(5280) ft?
1 mi 1 mi
Chlorine: pies See 70
Flourine: DSS (= 38 10.
m d
Total: 12°70: 438 = 120
2c
120 550
dk
120C = 122350) =a
pt+cd
C = 55 grams
| Dr Dp ! 11. uae
ia ep c
200
RrTr RpTp = 200; Rr = 25:
Ry, = 50; Te Tp S03
25T + 505 — Tr) = 200
-25Tr = -50
Tp = 2 12.
Dp = 25(2) = 50 miles
B + 28 + 90 = 180
am
B = 62
x — dm
cos 28° = 20
H 13, If mZBAD = K°, then mZABC = (180 - K)°.
(a) 4x + 2y = 8
BPD = i) 1°
cos 28 : (b) 4x + y = 6
(ay 49-4
2y = 8
tan 28° = ——
9.5 -l(b) -4x - y = -6
y = 9.5 tan 28° ~ 5.05
7. B+ 31 + 90 = 180 (a) 4x + 2(2) ll oo
B= So Xx ll —_
Algebra 2, Third Edition
113
Problem Set 44
14. 24x
21. [2-5sit. W2 MA Sat
3 _ 10x? + 24x iT V2 V2
_ WM17 _ SNS _ 634 _ 85/34
ll =
x(x — 6)(x - 4)
i 34 34
0, 4, 6
Xx
_ 79/34
0 ties
15. pape ast i]
(Sp — 9)(Sp + 9) 0 22. 3V6(2V6 - v12)
36/2 /3(2V2V3 — V¥2V23)
p 2 = 36 - 18/2
Boge
23. (a) Every point on this line is 2 units above the
x 3s
4x?
+ 16x - 64+ 255
x+4
X-axis.
yates oe! Ox? + Ox- 1
y=2
16.
x? + 4x3
—4x3 + Ox? (b) The y-intercept is —4. The slope is positive and
-4x3 = 16x" the rise over the run for any triangle drawn is x,
16x2 + Ox
1
16x? + 64x y = —x-4
2*
eye = I
—64x% — 256
255 v7 2 _ as tix,
24.
3 uv
x(x + 7)(x + 2) (x + 5)(x + 5) WOke = 4S ye te ID BIE
17.
(x — 3)(x + 2) x(x + 7)(x + 5) Aes
Be Se S) 5¢ =A)
25. he eee
5 5 10
18.
Lees,
5 10 10
1 rine ig ae LaeARO 9
19. Height = 4 ft x in
= ain.
10-"16 S* 32
er x height) 26. (a). 2% yr =5
Veone = 3 Base
Vermek —o
i
= 3 m(4)? + 5(12)08) (b) 4x + 3y = 9
o\2
= = H(2) Jas)in.
5(6x + 48)(48) in? ~ 1069.44 in.?
20. ponnnwence =
114 Algebra 2, Third Edition
Problem Set 45
Solve the equations by elimination: b. (x +9)? = 11
—2(a) -4x + 2y = -10 V(x +9)? = +11
(b) 4x +3y= 9 x+9= +11
ye 7 x=-9+ 11
ae -9 +kJ11: (-9 + Vil + 9)? = 11
Chec
(J11)? = 11
(a) 2x - (-z) 5)
1 sul
Ph OMe: Check =9;—./11;(-9 + 11.4 9)2i=11
See & 5 C11) =F
(2 -1) (les ail
5. 5
7)‘
( Z
c 6G Pal) eats
71. es x3 a
x-+ x — 4 (:i =] Sra
IPE Te Shain F SP?
x74 x4 x2-4 4 2 Be
7
5g (51,463 x 107"*)( 748,600 x 1077!)
SHA —21
a Ae + 14D
7,861,523 7
(5x 107!9)(7 x 10716) 3)
= t——__ -.___«“=£4x 10 Check -—1 ; | ae
1 ia\ ces
esl
8 x 108 - 5 + 8:(5 +48 +2) i
; : (/8)? = 8
29. a - a? = (2) -2(-3) =4-5-4 8 =
2 2D 4uek 48
2
Cheek rs (-+ - vé+=] =
a0) 2" — 2? — (2) = |-2 46% atey? 7 7 7
a
Bes oe ie ag ee eS (-v8)*
=8
Ome
PRACTICE SET 45 PROBLEM SET 45
a 2 = 14
e Pepa tenet
vx? = +14 100
x = ty14 S = 248 x oe = 400 students
Check 14: (14)? = 14
14 = 14 2. Phosphorus: 1 x 31 =.31
PRE nh (oda se: 14 Hydrogen: 3 x 1 = 3
eae Total 31 +3 = 3 34
ea
34 if
= 272 grams
Algebra 2, Third Edition 115
Problem Set 45
a5 iat oT =" 536 Lacs = ze
100 Pp
4
% 100 _ = = 6.50
T = 5376 x Tae = 8400 Pp sin 38°
Bromides = 8400 — 5376 = 3024 bottles 8
12. cosA = D
4. (a) 12Np + 14Nc = 138 A = 48.19
(0) Ne + Ne = 11 B = 180 — 90 - 48.19 = 41.81
Substitute Nc = 11 — Np into (a) and get:
, 12Np + 14(11 — Np) a= 13. 100.000 mi2 x 5280 ft . 5280 ft
(a’) 138 : ; ee igi
Nr = 8 bunches of roses 12 in. : 12 in.
(b) Nc = 11 — (8) = 3 bunches of carnations 1 ft 1 ft
= 100,000(5280)(5280)(12)(12) in.”
5 Dr
(571,652)(40,316) (6 x 10°)(4 x 104)
214,000 x 10 DESK,
= 12x10! =1x 109"!
Rplz + 80 = R7Tr; Tr = 8;
Ty = 12; Ry = Rp + 30 15. 8 x SF = 12
12Rp + 80 = 8(Rp + 30) ry eee
4Rz = 160 2
Rp = 40 mph rxo=5
Rr= (40) + 30 = 70 mph 10
x= —
3
6. x7 =5 ;
2 2 10
x= +5 Co =a =f 3
1. Gey= 11 2 ees
x+7=+H/11
x=—7+/11 Saar e
SG 12)s 29)eee
x+12=+/2 32 3
x= —12 4.5/9 eee 2
2 39-26
9 a3 =13
(++) —-—=d
5
a 13 _G
+= +/13 tke
x= -—43 + 13 16, mx
eo
4 Pp
ae
. mxp + 4y = 4cp
10. sin,33°
= — 4yy == 4cp 5— mxp
10
dy
x = 10sin 33° = 5.45 4 =p
Cc —- mx
116
Algebra 2, Third Edition
Problem Set 45
7 (Sh _yp im 25, 24/5 (4a15 — 21)
P C . = 25(3V3V5 - 22/5)
Sxc — ckp = mp = 30/3 - 20/2
Sxc = mp + ckp
5xc =. 26. y= ~=x +
m + ck
18. x = 5x? + 50x oe rege Ae
x + 5x” — 50x =O 1930
x(x + 10x - 5) = 0 Soya
x = 0,-10,5 :
y =f2-1
19. 36x7 —- 25 = 0 3
(6x — 5)\(6x + 5) = 0 ze Sx-2_x-3_,
as =.- 7 on
25x - 10 — 7x + 21 = 140
20 x(x + 7(x-1) (x + 5x - 3) 18x = 129
"(x + 7) - 3) x(x + 20%-1) ee
_x+5 <2 6
sas ae
28. (a) 2x + 3y = 3
z 1 1
ea (2797/3 ((27)'/3)3 y
a St237 oj
oe SD (b) x — Sy = 20
243 1
y= =x -4
22. Perimeter = (24 + 477) cm
Circumferencesemicircle = 24 + 4a - 12 - 6
Am
= 47cm
: 27r
Circumferencesemicircle = oS 47 cm
r= 4cm
2
Al€asemicircle = ee = 25.12 cm?
f= 7979.91 x
3 eo aegie Xx 6px - p” (a) xa fy= 3
SOSA Opis er -— -2(b) -2x + 10y = -40
a x 6px — p se)
4 4x7p -1 37
6px - p” y 3
4 5/243/8 = $28,382 (o) x - {-22) = 20
5 v5 2 2
5V10 3V10 10V10 15V10 ga 2 21S.2D
Rie Gee, = Tiga} 10 13 sh18
_ 25/10 _ 510 & -7)
oy ee. : 13’ 13
Algebra 2, Third Edition 117
Problem Set 46
») 3 2% 3. Nitrogen: 1x 14 = 14
ee. ep Oh Hydrogen: 4x 1 =
i
_ Ax +2) 3(x - 2) Ix Chlorine: 1 x)35) =y35
An) ee, Bye Total: 35 = 53
14444
ae BE > See 140
2-4 53. NH,Cl
-3x + 10 35NH,Cl = 53(140)
yt = 4 NH,Cl = 212 grams
iver 2342
a, eston = (22Ly - (ey 1 1
Dae) mie tech 6
. Geleed EB
a) Ne 2
N 7 age,
Prmee
ee ee (b) Np = 3Ng + 11
16 16 16
Substitute (b) into (a) and get:
(a’) 2(3Ng + 11) = 7N
PRACTICE SET 46 - Nr = 22 elves
K
a. a [45/3 — vee (b) Nr = 3(22) + 11 = 77 fairies
Bok AS. pA? oa 5. De
V3 V3 V7 V7 ferme rea
Dy
eel) ON 2/21
3 sad RyTy + 16 = RpTg; Ty = Tra = 4; Ry = 16
n cee | Svar _ 14y21 _ —2./21 16(4) + 16 = 4Rp
u t : 80 = 4Rp
» 28-55 sv 20 mph = Rp
eeedo oS
AS Sasa
er rn a 6 5/3 +2/2-
2 3 600
5"
aewes Llead 8 _ 5V3 V2 , 2V2 V3 _ io
= EE MOD ola BS
| S16,sagennaaoeng
26 a0
_ 6VIS _ 25VIS | 45V15 _ 26V15 lle
15 15 1
: a _ 15V6 | 4V6 _ 60V6 _ _41V6
ce. ¥7V7 = [7)!2]!2 = 712714 = 73/4 6 6 6, 6
ep a3 " 5 [2+ 3/140
PROBLEM SET 46 :
240 = 3V5 V7 _ 5V7 V5 , 6/35
i pee = te aie Sess
100 =
SWihey = s
335 = ee a 6/35
T = 1440 x 540 7900 7
1535, 55s , 210V35 _ 38/35
1000 B 35 maser. EV358 te
4000 24,000
4000B = 24,000(1000)
‘ 8.
i5¥53=
5 =
[SG
1/2)]1/2
2 =
—
siks4 =
51/261/4 — 53/4
B = 6000 kilograms 9. \6V6 = [6(6!7)]!5= 61/366 — 612
118 Algebra 2, Third Edition
Problem Set 46
1= Vey3.3 4 day = (3y3)!2Qy)4
= x3/2y32,1/4y1/4 18.
= xAyi4
11. 4/ry 3]xy? us (Py) 4(¢y9)18
ee ybm
= my 3/4). 1/3 ys = x 19/12,29/12
x —- mp
2
12. («-2) =7 19. 6 SSP oe ee
3) 2 12
TN es od
> 12
2 7
2i — = AYV7Ff
Ns btasr
3
1
2 qj
5.—=
13.
(: a
—--—|] =
? 12 tai
35
CS
4
ae 9.007
1 20. —40x
= 13x? + x3
ee ete
or ew
V5Fs
x? + 13x? + 40x = 0
x(x + 8x + 5) =0
14. B+ 24 + 90 = 180
x = 0,-8,-5
(x + 9)(x + 5) xX = 2) x +4)
21.
x(x + 9)(x + 1) (x — 7)(x + 5)
x-2
cos 24°
1 1
22. glisy4 16
Seacos AL 4
if
23. ad Il 2(105) — 40 = 170
500-210
sin 55.15° = ba ,) = 75
—
C v= 7sin55.15° = 5.74 _ 360 — 2(65) = 115
(476,800)(9,016,423 x 10*)
16.
408 x 10!° — x2 x
foal Oe: 10°)(9 x 10!°) 24.
38 p- a 2x?
= 1x 104 + 2x7 ad
4 x 10!
x p- + 2x?
k
17. ES
fh Ee ee
m
25. 3/7(2V14 - 7) = 3¥7(2V2V7 - V7)
kb
= 42,/2 - 21
ax
= bc
Algebra 2, Third Edition 119
Practice Set 47
26. Graph the line to find the slope. (b) x + 2y = 2
=
aaa
Substitute y = x + 3 into (b) and get:
(b’) x + 2 + 3) = 2
Since the slopes of perpendicular lines are negative
reciprocals of each other, 3x = =8
vial
30.
Pep
=xt3 ix+4 PRACTICE SET 47
2 3 First method:
1644/4 2 143./4 = [4(412)]13 = 438416
—3x + 9 — 4x -— 8 = 30
= 46/641/6 _ 47/6
29
x= -—— Second method:
a
644 ahigoyaets [29(22) 12] 13
= 76/35-2/2
1/3 =_ 2 5251/3
2% GO
=e ome_ 2 47/3
28.
4 3 $9.3 = $132.3 = [32(3)1/2] 1/6
—6x - 15 - 4x + 8 = 84 = 32/63 1/12 = 35/12
aeol 85 mi 60 s 60 mitt mi
10 —
8
:' >=
| matt
- ——
1 hr
= §5(60)(60) —
on ar
29. (a)x-y=-3 207 wat 4. SURE _ lLmin _ 5280ft
hr 60 mtr 60 s 1 mr
Vrs Xo+03
2 207(5280) ft
(60)(60) s
120
Algebra 2, Third Edition
Problem Set 47
PROBLEM SET 47 805 mi _ 5280 ft 1 hr 1 min
—— x x x
hr 1 mi 60 min 60 s
i f x T = 2800
7 _ 805(5280) ft
(60)(60) s
e
T = 2800 x 7; = 4900 people
13 in. Dit 1 yd 60s _ 60 min
x x x x
2. Odd integers: N, N + 2,N + 4, N + 6 S 12eine ft 1 min 1 hr
_ 13(60)(60) yd
5(N+N+2)=71N+2+N +46) -10 (12)(3) hr
10N + 10 = 14N
+ 46
N =-9 Vasv5= [52/5 = [57(5!/7)] Wes 2 52/35 1/6
= 55/6
The desired integers are —9, -7, —5, and -3.
10. (93 = V3 V3 = (37(687)| WA = 32/731/14
<e ae x G = 1400 — 35/14
100
be2 y3] yx bs (x2y)!2(y2x)15 = Ky ye eats
G = 1400 - = 2000 girls 11.
70. ee: 7/6
By
4. Hydrogen: Dees Lae L 2372 = [2(2"3)]!2 = 21226 = 228
12.
Sulfur: Jexn32 3232
Oxygen: 3 x 16 = 48
13. 3/2 - 5/3 + 224
Total: 2 + 32 + 48 = 82
48 =—s192
SER ee ye
82 H, SO,
seas) 2 0
48H,SO3 = 82(192) 2V6 _ SV6 , 8V6 _ 5V6
2 5 We
H,SO; = 328 grams
14.
RrTr + Rwlw = 48
Rr = 8; Rw = 4 i iiaaa tread eRe
35 Sh) 35 35
Tr + Tw =7
8(7 — Tw) + 4Tw = 48
15.
-4Ty = -8 7
Tw = 2 x = 7sin 19° = 2.28
Tr = 7 - (2) =5
16.
Dy = 4(2) = 8 kilometers
Dr = 8&(5) = 40 kilometers
52.1ns 1 ft 1 mi 60 s 60 min
——__—_ X sx x — X ——— 17.
s 12 in. 5280 ft 1 min 1 hr
_ 52(60)(60) mi
~ (12)(5280) hr
Algebra 2, Third Edition 121
Problem Set 47
x*p xp — m* =
m a m x — pm
24. oie Hd m
x
ll +& m m x — pm
_ x2p =m?
S|
+ni
alm "ox = pm
I |
= (aT a Oe CE + 3)
a
x1|
Pays
Cx TGP Oe ae — S)Ge “F 3)
(36,421 x 10°)(493,025)
19. _x-2
40,2 16 x 10’
x -5
_ (4x 10°)(5 x 10°)
= 5x 10°
4 x 10!!
1 1 1
26. 97-43 = Sa SS ee ee
20. Zip
x |3
a
acy + cpx Pfs (a) x — 4y = 8
acy = mx — cpx 1
See
=-—-x-2
(b) 2x — 3y = 9
04Viel cca( Se y =
Ds,
—x37 -—3
22. B? = 42 + 32
oa
eh
| ah
BBEo
Solve the equations by elimination:
—2(a) —2x + 8y
= (b) 2x - 3y
Sy
N|o
Ww
ri[|m
Nw] y
23. x = 180 - 150 = 30
y = 180 -— 90 - 30 = 60
Zi y= 60
s = 180 - 90 - 60 = 30
p = 180 - 90 - 30 = 60
m = 180 - 90 - 60 = 30
122 Algebra 2, Third Edition
Problem Set 48
28 ee et), PROBLEM SET 48
5
O32) Fosr512
5 = —(-2)
+b
1? = LF = 1600 frontiersmen
0.32
ces
5
1400 _ NG
=-x
ond
+ —
cae 1540 6160
Ist s
1540NG = 1400(6160)
29. O0e =.-12r"
—x NG = 5600 grams
x? + 12x? + 20x = 0
x(x + 10)\~ + 2) =0 Iron: I Ne Ste = Sie)
x = 0,-2,-10 Bromine: 2 x 80 = 160
Total: 56 + 160 = 216
30. -a — a* —~a@ — ab
Soe
216
Fe.
972
1 1 1 216Fe = 56(972)
24 8 Bbw Fe = 252 grams
PRACTICE SET 48 (a) 5Ng = 3NG + 17
a VWx>5-3=7 (b) 6NG = Np +2 —" Npe= 6NG - 2
v¥x —-5 = 10
Substitute Np = 6NG — 2 into (a) and get:
x — 5 = 100
(a’) 5(6NG — 2) = 3NG + 17
x= 105
21NG = 27
Check: ,/(105)
-— 5 -— 3
NG = 1 girl
10 —3 =
A (b) Ng = 6(1) -— 2 = 4boys
b vx+5+12=
-ll1 Dr
jx
5 = —12
x+5= 144
aDe
xi= 139 RpTp = ReTx; Tr = 20;
Check: + 5 + 1 = -ll
4/(139) T,;= 10; Re = Rr + 30
4144 +1 = -l11
20Rp = 10(Re + 30)
12+ 1 # 11
10Rp = 300
No real number solution exists.
Rr = 30 mph
C axe sx 10 ee der
Rr = (30) + 30 = 60 mph
jx? +3x-10 =x-2
x2.+ 3x — 10 = x? - 4x 4x <3 = 3.5
Ix -14=0 Vx-3=7
ix
= 14
x-3=49
Kotz
ee 152
Check: ./(2)? + 3(2) - 10 + 2
V4+6-10+2 Check:
52 -3 -3 =4
) f= 3's
Algebra 2, Third Edition 123
Problem Set 48
1b vx+54+3=-4 Jw
N
Nn
pa = | HT] aS
vx+5=-7 SN ee
x+5
= 49 * | | tl S
x = 44
Check: 44 +5 +3 =-4
7+3=-4
The last statement is not true, so no real number
solution exists.
16. (x — 3)? = 16
x-3=74
axe + Bx + 15 BOS =
x= 3.1.4
x? + 8 + 15 = x? + 10x + 25 x = 7,-1
2x = 10
x i] -5
17. C = 180 - 90 - 63 = 27
Check'*4(—5)- + 8(—5) +.15 — 5 = —5 sin 63° =
-5 = -5
60 mi 5280 ft 1 hr 1 min
——_ x ——_ x x 18.
hr 1 mi 60 min 60 s
_ 60(5280) ft A = 48.19
~ (60)(60) s
19.
10. lo3/2 a te) = 1/29 1/6 = 22/3 200,000 x 107!9
_ 4x
4x 10°
Dee OOi se
|my 3 my
11. = (my) !2(m4y)!3 2 x 10
= my m3 y/3 a m7!3y5/6
20. LACIE
Ta PR:
12. (Pel SAGE BL [23(2/3)]12 = 23/2916 x Pp y
bpy -— cxy + kpxy mXp
= 25/3
bpy + kpxy -— mxp = cxy
Be
13. 3/2 - |B+ ava by + kxy — mx
3/2VS V5 V2 21. 2(180
— A)° = 4(90
— A)° + 40°
7g cero 360 -— 2A = 400 - 4A
_ 6V10 _ SV10 | 40V10 _ 41V10 2A= 40
10 10 iO S10. A = 20°
|7 }11
3x7’ - 3x + 3 = 8
14, 2,/—- — .J— + 2V308 22. x + 1)3x? + Ox? + Ox - 2
11 7
3x3 + 3x2
ONT es AL 7, ~3x7 + Ox
“Hag
wae =3x- ay
3x - 2
14V77 1177,+———
308V77 = 311/77 3x + 3
vl 19 77 77 -5
124 Algebra 2, Third Edition
Problem Set 48
a
wl eee
_ =
(a
a eo)
|e “
Be: _
as BH
XN = a
*
oS
Rla Rial
® + Ha +*
" | ‘
be + lla *N Ria
Daas reid |
a* N
ata R ae
Nl ~*~
5 Nhe a
ae i
(he na
nN _
cae ax o)s eee s 2 aS = 6 aS:= = o 3 Q,
wy + = #3 oO ;
ES +t=
Q
IS
a od tad<<| N ai i|!cl “IM ® + na
Pca wiOla
ee
i: 3 ®
3
® ot
|
|e
Bis eae
Elm”
a an\o « ” =o m=
i |
sr st
i] |
\ St om i] 1 |
00
ao
|
PP wits
n iS)2 ll r
y ~ Bl re
fy
go
—_<x
2 iN
ost
x oO
8
coin ss
a i | we
atten Sh © <™
_ alan
+t
——~
a
rs| a
TT | 7
M4
colin =
i |
a +
ae a wrcal
|
on
~~&
¢ &
erie
ke wt
I 00
u
—_ eat
vtisthin
it D
%
AQ is m=
>" 7 +
lun
N® veen wicGeoes
a H l MT
X=)se ae a ce
nou
alx
TT
Q ro} ++ =
D? = 145
D= V145
125
Algebra 2, Third Edition
Problem Set 49
PRACTICE SET 49 4, (a) 20NR + Ng = 205
a. y=mx+b ONS
Oe ae
Substitute Ng = 15 — Np into (a) and get:
5
es, (a’) 20Np + (15 — Np) = 205
19Np = 190
Tethenee
Use (2, —6) for (x, y).
>
BO) fp? (b) Ng = 15 — (10) = 5 qurush
at
eee 5. Da
y=
5
2 =i
ea Da ae
RpTp + 200 = RgTg
_ TR = Tg = 10; Rg = 240
100 N 10Rz + 200 = 10(240)
10Rzg = 2200
iH Rp = 220 yards per minute
Ia 6. y =-6x +b
Use the point (3, 0) for x and y.
114 + P = 180 4b
0 ==66)
P = 66 1S#=8b
= -6x + 18
sin 660 = ee:
100
7. B= 180 — 160 = 20
N = 100 sin 66° = 91.35
=~ Br="20
in 20° = N
PROBLEM SET 49 ee ap
L. SONe 13) AEN) ao N = 40sin 20° ~ 13.68
ON = -27
8. Vx = 4 — 3 = 5
hae x-4= 64
2. — xX CB = 1092
100 Check: (68 — 4 -3 =
CB = 100 5 =
1092 x —— = 420 celebrants
260
9. lx? — De Steer
3. Beryllium: 1 x 9 =9 x? = Dyhpeseetae
te ey
Fluoride: 2 x 19 = 38 4x = 4
Total: 9 + 38 = 47 he
38 95 Check 412-94 5. = 2
41 BeF, 2=2
Sia read to, 200in. 1 ft min 200 ft
BeF, = 117.5 grams Dante | cee ee iw (12)(60) s
126 Algebra 2, Third Edition
Problem Set 49
11. = [32(3!/4)]"3 = 32/33 1/6
= cept 21. Cpt eas
x ay P
12. q'/4g1/3 = 27/12 2py — 3px = mxy
mxy
13. [x23 ixy4 = (x2y3)/2(¢y4) 1/2 = xyr2yl2y2
ie ZY =:3x
= xi/2y7/2
Sate
22. (x+ 12 = + Dat DO + 1)
14. 3Ixy [xy = (xy)!Beay/? = gef3
= x76 5/6
= (x* + 2x + 1) + 1)
=a 28? 4 ho ett
=x° + 3x7 + 3x +1
15. {2 + 2/2 - 356
_ 3V7 V2, 2v2 V7 PER
(t= Sx = 1) Mos Oe 3)
5 ea BEape x(x —5)x-4) (x -2)(x% - 1)
21Vi4
, 4vi4 __ 84V14
8414 _ _ 5914 £3
14 14 14 “ie 2"2,
24.
16. 5 [2+3/2 + 162
9 2
5/2 V9 , 3V9 V2
“eo ae”
10Vi8 | 27V18 , 54
5418
18 18 18
91yi8
_ 91V2
6 Miles!
Sis)
OW
bohm
o>)
7.
b= oa
cad | | I Er wi D* = 81 +1
D* = 82
D =n82
18. 25. Use the graph in the solution for problem 24 to find
the slope as follows:
i=)ee 4 —|
Slope
= — = —- 2
x = 2,-6 +9 9
(517,832 x 107!4)(80,123)
19,
200,000 x 1074?
Bits: 10-°)(8 x 10*) = 2x 10° i) Il | |—
—& wa ae >
2x 107°”
20. | + S i | ©
360°
26. Number of sides = —— = 30
12
127
Algebra 2, Third Edition
Problem Set 50
PROBLEM SET 50
1.
(b) Nr = 10Nw
Substitute (b) into (a) and get:
(a’) 10Nw = SNw + 10
5Nw = 10
Nw = 2 Waltzers
28.
(b) Nr = 10(2) = 20 Two-Steppers
—3(x - 3) 3x +2
x%— 9 x? -9
eet eee ee
x2 —9 x7 -9
RsTs + RpTpz = 104; Ry = 8;
Rp S12: Ts cg 10
29. = =2
8T;, + 12(10 — Ts) = 104
Ge=el2) -4Ts = -16
eS 1 Ty =4
Tz = 10 - (4) = 6
30.
Ds = 8(4) = 32 miles
Dg = 12(6) = 72 miles
Sodium: “Tx 23°= 23
Chionme:” 16x 35 = 35
Total: 23 + 3) = 58
23a
58
Ne
348
PRACTICE SET 50
58Na = 23(348)
x? = 9x -7
Na = 138 grams
& | \O& + HT] | ~
aa Even integers: N, N + 2, N + 4
1(N + N + 4) = 10(N + 2) —- 48
aS SS
SS
14N + 28 = 10N— 28
NO
Lo)
% | I 4N = -56
ae |\o
N Nee =
N = -14
The desired integers are -14, -12, and -10.
sD x OS = 432
100 _
100
OS =
Il 432 x ahs = 180 inches per
day
128
Algebra 2, Third Edition
Problem Set 50
12. = -¥x+5+1
6=vx+5
36 = x7+15
(x + 4)? = 20 Bi se.y
x+4= +2,/5
Check: -5 = -y31+5+1
x= -4+2/5
-5
= -5
7. 1x ye = 5 =f)
400 yd Bh Alt 1 mi 60 s 60 min
13. ——— x — x uA
S lyd 5280ft 1min 1 hr
_ 400(3)(60)(60) mi
(x + 6)* = 41 (5280) hr
x+6= +/41
14. oe [2(2!8)]5 = 2'Sqiis _ 24s
x= -6+/41
15. [of3 = (32/3 = [32(3!2)]!2 = 3 . 34
8 x* = 7x — 3
_ 35/4
(2-74 )=3
|my 3] my" ek (my?) my’) !2
16.
Lae pe egg 49 = m2 yr m2 2/3 = me i6
4 4
(+) = a
a =
37
=>
2 4
17.
7 {37
x--—
= +—
4 2
Tay _ 15v33 , 2233 _ 99V33 __ 6233
=
2 2 33 33 33 33
a. Ve ar +
(746,800 x 10'*)(703,916 x 104)
18.
Use the point (—3, 0) for x and y. 500,000
0 = 3(-3)
+b C0 C7 10")
= 98 x 10%
Qumeh 5 x 10°
Yoo +9
v 1 x 1074
10. B = 180 — 120 = 60 19.
C = 180 — 90 - 60 = 30
= = ie) = | <2 5 I boNv
tan 60° = “as mxcr = 2p + kpr
10
; 2p + kpr
M = 10tan 60° = 17.32
mxr
1. x2 — 4x + 20 55 ew)
x? — 4x + 20 = x7 4+40 44 20. ce eS
a p om
Sx
= 16 4pm — 3x2m = cpx
a
4pm — cpx = 3x?m
Check: 4 — 8 + 20
p=
4
Algebra 2, Third Edition 129
Practice Set 51
21. a = 2474+77 Since the slopes of perpendicular lines are negative
reciprocals of each other,
a* = 576 + 49
2 = 625 daly =
Salis3
a= (625 = 25
>
24 x SF = 30 y = -—x+b
3
=tG0 > 5
=
Serre, 4 § = --(-2)
30? +b
Bs >4 207, 4395 ee:
6}
125
— =crt—
100
4 4
25
— =C
4 28.
2E 2 See FLO 23
+ os
22. Op 3 )ae ee hae eS
4x? — 6x2
6x2 + 3x
6x? — 9x
12x + 5
197) = 18
23
23. x? — 28x = 3x2
x? — 3x? — 28x = 0
x(x — 7)x + 4) =0
x = 0,-4,7
(x —7)(x —7) x(x -— 6)(x + 2) 29. 27Gos— xp BS B= 3) = x = 2622 4)
x(x —7)(x—6) (x —7)(x +5) —2(-x -2)-4=x+4+ 8
6 ey 2x+4-4=x+412
x +5 x = 12
25. 499 = _(491)3 = _343 a tie — . |
30. 22 p0 y= ep
26. Verism = ABpase * height
= 4° - 3p!
; Vp ;
height = —
Base
(600 — 507) cm? PRACTICE SET 51
|cloyc10) + (10)(10) - 55 |cm? Dit Ahi as
(600 — 507) cm? = (ii) — Tien Si °4a-nel 63
a Sr em = 261) = Tiel) aol 2s
(150~ en) cm? = 24 7i + 2i 54 5 43}
=-6+ (9 - /3)i
27. Write the equation of the given line in slope-
intercept form. BP 4 Sie +e Shia OP
Sy = -3x + 2 = 3i(ii) + 5(ii) + 5i + 7 + 2i(ii)(ii)
= 3i(-1) + SC-l) + 5i + 7 + 2(-1)C1)
Pee eee
2 aes —3i-5+51+7+2i=2+4
130 Algebra 2, Third Edition
Problem Set 51
PROBLEM SET 51 8, Sitevna? fei 4-274
lg ee = 5(ii)ii + 3(ii) + 7(ii)
(ii) + 7(ii) + 4 + 2d) GDI
(ii)(ii)(ii)i
17 = 5(-1)i + 3(-1) + 7-1) + 4 4 2C1)-DCDi
K = 72 x +1 = 408 knights Soi ee
3 = -6 - 7i
anil: Merge deat 3 9, -3i7 — 21+ P — 3 = -3(i) - 21 + iii) = 3
100 = -3(-1) = 27-1) — 3 eetiee Use 3
S = 528 x sy = 2400 students = -3i
; 10. el el
3. Sodium: 2 x123"=146
Sulfur: 1 x 32 = 32 (e+e4 )er
Oxygen: 4x 16 = 64 i 1
Total: 46 + 32 + 64 = 142 (Peret}=t+ 7
46 |
115 (
x + —
af aad: = —
4
a
142 Na,SO, Z
46Na,SO, = 115(142) gdh Bhs
Na»SO,4 = 355 grams 2 :
grey aD|
4. (a) 10Np + 25Ng = 650 NaS 2
(b) NpD + No Q = 35 11. e4° Se eax
Substitute No = 35 — Np into (a) and get: ;
(a’) 10Np + 25(35 — Np) = 650 (« re ie }at
-15Np = 225 9
Zz Zo\ = Al eas
Np = 15 dimes (: Benes | Hi
2
(b) Ng = 35 - (15) = 20 quarters (:+ >] ¥ 25
5. Dg
tr 2 = so
: 2 2
se x=——t—
Bot 55
TR = iy +2 2 2
RpTp = R,Ty; Rp = 40; Ry = 50;
X= i a4
40(T, t- 2 = 50T;
pti 12 9 =A D
8 = F;
Use the point (2, 2) for x and y.
D, = Dg = 50(8) = 400 miles ete ea
—2 = 343 10 =b
6. 5 + 61.43.
. y = -4x + 10
7. Si — iii + 2i- 4 - V-5
= 5(ii) — 8tii + 26= 4 —N5i 7
13> =" =
tos30°
: faa
Do Si ee 05yee ke:eo mer
fo
aie Rigg 4 Gi ge ASI
DS 03 SOR
=) 29, (10,84) 5)i
131
Algebra 2, Third Edition
Problem Set 51
14. Jx—-3+5=2
24. x= - =) 25)
Vx —-3 =-3
Yea y = 180 - 120 - 25 = 35
55 1 i =x = 25
Check:
J12 - 3 + 5 =2 z=y= 35
8 = 2
25. Write the equation of the given line in slope-
The statement is not true, so no real number intercept form.
solution exists.
3 5
= =x +>
20 in. 1 ft 1 mi 1 hr 2 2
15. SG eee OM x
hr 12in. 5280 ft 60 min Since the slopes of perpendicular lines are negative
20 mi reciprocals of each other,
~ (12)(280)(60) min
16. V3V3 = [3("2)]¥5 = 31/5310 . 33/10
17. lox@=* 3223 = [32(3'/3)]12
ee 31/6 ae 37/6
18. [<2 y2m Al ey? = (x2y*m)!/2(xym2)"4
= xym!2xM4y M42 = y5/4y5/4in
2 /13
19. So — a on Ht oy te
13 2
_5 3¥2 ey
V13 | 66
3v13 V2 26.
Nig vie ao A2
_ 6126 , 39V26 | 156V26 _ 201/26
206 26 66 + 26
20. 977-53 = 1 i 1 ae 1
oH (-27)°3 (273) 243
21.
0.00007142 x 10>
~ 10°)(7
(5x x108)
mene
x TX
22.
mx = kp + cmp 27.
28. Pe) De rae: 4
23. 2x3 + 4x2
Ax? — 2x
ig BS
ie)
wo
|
S5 i)
| s ka -4x? — 8
6x + 4
6x + 12
-8
132
Algebra 2, Third Edition
Problem Set 52
2, = at tx + iy 4 Substitute Dy = 1400 — Py into (a) and get:
4 3 3 (a’) 0.25Py + 0.05(1400 — Py) = 140
37 16 = Ay B= 26 0.2Py= 70
=) pe = 50 Py = 350
pauper) (b) (350) + Dy = 1400
7 Dy = 1050
BO Ga yi) 1 oben a0 ay ee 350 mL 25%, 1050 mL 5%
PCED = P80 Mer 3. Sodium: x 23 = 46
ay tae SOE 60 FS Sulfur: Ofxa82 364
ml BE Oxygen: 3 x 16 = 48
xe -3 Total: 46 + 64 + 48 = 158
Marea
158 1580
PRACTICE SET 52 158Na = 46(1580)
Py + Dy = 1600 —= Py = 1600 - Dy Na = 460 grams
0.75(Py) + 0.95(Dy) = 0.85(1600) 4. (a) 4Ny = 76 — 6Np
0.75(1600 — Dy) + 0.95Dy = 1360 (re Nee
ee Ne Substitute (b) into (a) and get:
Sihtor es (a!) 4(Np + 4) = 76 - ONp
a es ae 10Np = 60
Py = 1600 — 800 = 800 mL Np = 6 reds
800 mL 25%,
800 mL 5% (b) Ny = (6) + 4 = 10 yellows
5-1 th) D; De
PROBLEM SET 52 a
Gikestemcoaas
1. [Iodine Py + Iodine Dy = Iodine Total
R, Tr, oth RcT¢ = 28; Ry, =I
0.1(Py) + 0.4(Dy) = 0.25(100)
Rc = 8; 1a ar Tc =a8
(a) 0.1Py oF 0.4Dy wo)
OT; HERR eT) ats
Substitute Dy = 100 — Py, into (a) and get: Tins 6
(a’) 0.1Py.+ 0:4(100 — Py) =:25 Ta hee (Oa
TetN Foe! D,, = 2(6) = 12 kilometers
Py = 50 ;
Dc = 8(2) = 16 kilometers
(b) (50) + Dy = 100
Dy =°50 6. 4i7 - 3i + 2 = 4(ii) - 31 + 2
50 mL 10%,
50 mL 40% Sg le Okt atsir ae ae 2 emOe
2 Salt Py + Salt Dy = Salt Total Te Bi NESS OT 5 a
0.25(Py) + 0.05(Dy) = 0.1(1400) ee pate) eS
(a) 0.25Py + 0.05Dy = 140 S16, ise ey Gi)
(b) Py + Dy = 1400 20% 2S 234 By
au4pe
Algebra 2, Third Edition 133
Problem Set 52
ne enf{eOren 37 = 2(ii)(ii) + 3i — 3i(ii) 200 in. 1 ft 1 mi y 1 hr
16. oN E
=2+ 31+ 3i=2 + 61 hr 12in. 5280ft 60min
10. Ox 4.5.
= x” ___ 2000 mit
~ (12)(5280)(60) min
ate)
nae
a eee ee 17.
2242 = 2[2(2"4)]12 = 2 - 212 . 218
(x - 1% = 6 = 2713/8
3 943 = 3 3243 = 3[32(3'4)] 2
18.
= 3.3.
318 — 3178
11.
|4x3y° 3/ Sxy? a (22x39) !2(27 xy)
19.
cd pxe yey ca 4x11 19/6
y
20. 3 a He lie
\ 13 \2
©” 3 BNE AsO! Fe
alte ae « (WQecl2
_ 6V26 _ 6526 | 52V26 _ _7V26
"696 26 Maren - 026
12. y=2x+bD
Use the point (4, 0) for x and y. (987,612 x 10°)(413,280)
21.
0 = 244) + bd (74,630)(400)
-8=b
_ 10")(4
(x x 10°) _
~1x 10°
y=2x-8 (7 x 104)(4 x 107)
13. B = 180 — 165°=.15
22.
mise ote
P bx — bcp + ap = bmp
| l = 27.05
bx = bmp + bcp — ap
sin 15°
, = bmp _+ bep - ap
14. \x2 42x 410 =x+2 b
x7 +2x+10
=x? + 4y 44 x- + 3x — 28
23. = lk EFM A)
Dh=10 21x + 10x? + x3 x(x + 7)(x +°3)
ay 6)
_ «x -4
Check: 9 +6410 =5 (tS)
i)
24. _16>/4 = (iG) = 32
15.
xy ¥ x*y = p? p?
2 5)
64 =x - 3 25. P as Dp” mp 1
67)=x
=m - =] wp P :St p2
ae Pp mp — 1
Check: 3 = -5 + /67 -3 x’y = p>
3=-5
+ 8 mp -— 1
134
Algebra 2, Third Edition
Problem Set 53
26.
60 PRACTICE SET 53
a a
a. Carbon (C): 1x 128412
p=z=30 Chloride (Cl): 4 x 35 = 140
Total: 12+, 140-5 152
& Ul 180 — 100 - 30 = 50
Carbon: gees x 100% = 7.89%
152
1000 m , 100 cm
3 2x +4 0.073 km? x
27. + 1 km lm
x-3 x2-9 100 cm 1 in. 1 in. 1 ft
x —$<$<—— > X —_
—3(x +3) 2x+4 lm 2.54 cm 2.54 cm 12 in.
ett 1 mi 1 mi
x* -9 x- —9 x x
12eins 5280 ft 5280 ft
oe tae a ae 5 _ __ 0.073(1000)(1000)(100)(100) 2
x*-9 x29 (2.54)(2.54)(12)(12)(5280)(5280)
28. Write the equation of the given line in slope-
intercept form. PROBLEM SET 53
Sodium: 2X 23-2" 46
ee rel Sulfur: 2x? 32> =,64
a= 2 eK
Oxygen: 3 x 16 = 48
Since parallel lines have the same slope, Total: 46 + 64 + 48 = 158
y= -—-—x+b Sodium: ~~ x 100% = 29.11%
Alcohol Py + Alcohol Dy = Alcohol Total
Fe aS
3 0.2Py + 0.6Dy = 0.52(100)
(a) 0.2Py + 0.6Dy = 52
ak
3 (b) Py + Dy = 100
Substitute Dy = 100 — Py into (a) and get:
(a’) 0.2Py + 0.6(100 — Py) = 52
AS oe ae
—0.4Py = -8
3x7 + 12x + 47 + 8 Py = 20
4
29. x — 4)3x3 + Oc?= ox t+ 70 (b) (20) + Dy = 100
3x3 — 12x? Dn = 80
12x27 - x
20 mL 20%,80mL 60%
12x?
— 48x
47x + 0O Iodine Py + Iodine Dy = Iodine Total
AT 15s 0.4(Py) + 0.8(Dy) = 0.72(250)
188
(a) 0.4Py) + 0.8Dy = 180
(b) Py + Dy = 250
<2Tate Ape Eee ae
30. Substitute Dy = 250 — Py into (a) and get:
2 3
(a’) 0.4Py + 0.8(250 — Py) = 180
oF 22 4+ 10 = 4
—0.4Py = -20
ji Sie Py = 50
C= =
Zz (b) (50) + Dy = 250
5 Dy = 200
50 mL 40%, 200 mL 80%
Algebra 2, Third Edition 135
Problem Set 53
4. (a) Ng + Nz = 80 12. 4 + 2./-16 = —4i(ii)(ii) + 2(4i)
= -4j] + 8i = 4i
(b) Ng = 5Nz + 8
Substitute (b) into (a) and get: 13. 23 — i4 + 37 = 2i(ii) — (iii) + 37)
(a’) (SNg + 8) + Nz = 80 =] — | — 3 eeaAieels
6N3 = 72
14. x2 —5= 5x
Nz = 12 results
i) | Nn& Wl Nn
(b) Ng = 5(12) + 8 = 68 results & +
aN ey
ee Oe ae
4 4
( a 45
ia |S PF
> 4
RyTu + 200 = RpTp; Ryu = 50; Ty = Tp = 4
5 3/5
x-=—= t—
50(4) + 200 4Rp 2 p
400 bliSin, 12x15:
100 mph tt >a
Tie
v8 pay Sa:
Im 1km Use Be eyas
6. 9350cm x
100 cm
_-9350
~ (100)(1000)
100 cm , tain: e 1 ft 1 yd
Ts 32amy x<
lm 2.54 cm 12 in. 3 ft
_ 32(100) 16. A = 180 - 120 ll onoO
~ (2.54)(12)(3) > 10
cos 60°
5280 ft. * 5280 ft. y 12 in
m
8. 16,480,000 mi? x
1 mi 1 mi 1 ft m= —— = 0
12 in. 2.54 cm 2.54 cm
x x
ott 1 in. 1 in.
17. vx —11 -1= 16
= 16,480,000(5280)(5280)(12)(12)(2.54)(2.54) cm?
vx
=11= f7
9. W062 kane 1000 m x 1000 m - 100 cm x — 11 = 289
1 km 1 km lm x = 300
100 cm 1 in. 1 in. 1 ft
x x x Check:
/300 - 11 - 1 16
1m 2.54 cm 2.54 cm ein
1 ft 1 mi 1 mi 17- 1 16
x 4 x
VD ci, 5280 ft 5280 ft
18. en
_ __9.063(1000)(1000)(100)(100) 2
x7 +2x+5
5 x° 4 6x 4 0
(2.54)(2.54)(12)(12)(5280)(5280)
-4x
=4
10. —/24 oer e974. 2 + 2G x=-l
= Df BD) SE Ohies 9 Check]. as) 3eet
2-3=-1
11, 277 + 5i +44 /-9 = 2) + 5i+ 4 4 3i
=2+ 8 19. 2 yee [2(2'/3)]/5 = 2iSqi/s _ 2ans
136 Algebra 2, Third Edition
Problem Set 53
20. 81V3 = 34 4/3, = [34(3"4)]!2
80
= 3231/8 _ 3178 Ld 28. y = —9 = 40
21. 5 1%2, 3 [xy2 a (x2y)"/5(xy2) V3 x = 180 -— 90 -— 40 = 50
& PSS 20 xlW/15) 13/15 P=y= 40
O =tx = 50
oy eel 1 be 1
22. -4- = “G4DS = 732 R = 2(50) = 100
; > ee = YH 1
23. 3/2 - 2/2 — 21% 29. «+ 1)x? + Ox? 2x +2
Pi ee,
_ 3V2 V9 _ 2V9 V2 gg —x? — 2x
19 V9 2 2 =x? - x
= He Satine) ceed Po%. —X —,
3
24 (2,135,820)(4,913,562)
801,394,026 40) Gora ats 5
6 6
_ 2x10 Sx 10°) _ 4 x 104 —3y = -2x - 9
8 x 108
DZ
y= es + 3
25 ee
te ee (b) Sx + 3y = 3
cpx — cmy + kpy = 0 3y = -5x + 3
cpx + kpy = cmy ys -tx+1
“ gemay,
cx + ky y
p as
266 tc = 0 Te
x el
ptcx= a a
ap
p= =
dt _~ cx aes
aoe
> fel lee
a —-C 72 eka)
ei
27. 5 x SF = 10
SF = 2
Cx = 12 (a) 2x -— 3y = -9
ae (b) 5x + 3y = 3
a2 . y : el ex ="=6
6
AZ = 25 + 36 ey ae
A= V6l1 me i 6)eee
a S| -= =.
V6l x2 = vol +B PrePes| =7 7
2/61 = J61 +B 6 2)
J61 = B (-7 i)
137
Algebra 2, Third Edition
Problem Set 54
PRACTICE SET 54 Substitute Dy = 150 — Py into (a) and get:
(a’) 0.1Py + 0.4(150 - Py) = 45
oe oN -0.3Py = -15
(b) (50) + Dy = 150
28 Dy = 100
50 mL 10%, 100 mL 40%
sin 50PeSE = 3. Potassium: i x40 = 95
8 Chlorine: xaos = 35
28 sin 50° = B Oxygen: asx 46 = 48
Cis Total: 39 + 35 + 48 = 122
cos 50° = = Potassium: = x 100% = 31.97%
28 cos 50° = A
18.00 = A 4.8) eeNir ee 4Np =y 5Nc
18.00R — 21.45U ie ey
(b) 4Np = 3NG + 40
pes al -1(a) -4Np + 5Ng = 0
Sw (b) 4Np - 3NG = 40
RV = ST = 3 aN = 40
ae ST Ng = 20 geese
v (b) 4Np = 3(20) + 40
Np = 25 ducks
PROBLEM SET 54 5. De
1. Iodine Py + Iodine Dy = Iodine Total meee,
0.2Py + 0.7Dy = 0.575(400) ae
(a) 0.2Py + 0.7Dy = 230 RcT¢ = RyT yr; Rc = 2Rr
Tc = Tr = 3s Rr = 79()
(b) Py af Dn = 400
100(Tr = 3) = 50Tr
Substitute Dy = 400 — Py into (a) and get: 50T; = 300
(a’) 0.2Py + 0.7(400 - Py) = 230 Dias
—0.5Py = —50 Dr = Dc = 50(6) = 300 miles
Py = 100
(b) (100) + Dy = 400 *
Dy = 300
100 liters 20%, 300 liters 70%
2. Glycerine Py + Glycerine Dy = Glycerine Total
0.1(Py) + 0.4(Dy) = 0.3(150) A = 5cos 56° = 2.80
(a) 0.1Py + 0.4Dy = 45 B= 5sin56° = 4415
(b) Py + Dy = 150 2.80R + 4.15U
138 Algebra 2, Third Edition
Problem Set 54
17.
V4 Me eek oe cin
4 4
A = 8cos32° = 6.78
(ie iy.= 61
B = 8 sin 32° = 4,24
2 4
-6.78R — 4.24U
T £ sT=RvV— r= %¥
2 2
S s 7, v6
x=—+—
2 2
1000 m 1000 m 100 cm
100 km? x —— x x
1 km 1 km 1m 3
100 cm 18. =—x+b
x ——
= 100(1000)(1000)(100)(100) cm? “ #:
lm
Use the point (6, 0) for x and y.
12 in. 12 in. 2.54 cm
10. 100 ft? x —— x x 0 =) +b
Lit 1 ft 1 in.
2.54 cm
x = 100(12)(12)(2.54)(2.54) cm? 9=b
lin
60 mi 5280 ft 12 in. 2.54 cm
11. —— xX x x
hr 1 mi ft 1 in.
lm 1 km 1 hr x Lmin
xg x Be 19.
100 cm 1000 m 60 min 60 s
8x — 4 — 5x + 15 = 70
(100)(1000)(60)(60) s By
= ah)
12. ee, a DiGi) AMID x
=2i+2i+4=4
13. ap yg Of ay 4 [0 20. yo eS 4 Sey
= 3i(ii) + 2i — 4(ii) + 3i xe —-x-—-S5=x*-2x
4+1
= 31+ 21+4+ 3i=4+2i “T= 76
14. Sliet live 2 del Bis WG) ~ 2"+ i Check: 36 -6—5 +
ai iio ath iin tei 5 +
15, Vi + 2G — 2./-25 = ii) + 2 — 10i 21. v¥x—-2-11=1
sa TPE Ri je SD
cep ie
16. Si pees x = 146
(<2 - 3x \=7 Check: 146 — 2 - 11
9 9 12
- 11
So#4 die—-=7+=—vi
(O34 ZI
X ser
3
= Dray
4
37
22. | ] 9 = i
2 D
mk
x=
— +t —
2 2
Algebra 2, Third Edition 139
Problem Set 55
23. a :
—-x+—e=Fy ae
b. Xi Mie 2
p m
am — mpx + cp = mpy = x(m + Z) + ———— x(m + Z) = xs(m + Z)
De (m + Z)
cp = mpy + mpx — am aym + ayz + cx = xsm + XSZ
ee eee 2 Rage ayz + cx — xsz = xsm — aym
Py Pee ayz + cx — X82 _
: == xs - ay
24, {4/2 = 22/2 = [27(2'%)]!*
= 22/3916 ~ 25/6 c. Consecutive integers: N, N + 1, N + 2
a= ¢ NN + 2) = 5 + S(N + 1)
25. 1933 = ¥323 = [32(3"3)]"5 N24+2N=5+5N4+5
= 32/S3I/IS _ 47/5 NW BN 10 0
ae (N — 5)(N + 2) = 0
26. Sfxy xy? = (xy)(xy2)!8 N = 5,-2
= xl By S18 ee = ha ak The desired integers are 5, 6, 7 and -2, -1, 0.
27, -81!4 = -3
PROBLEM SET 55
28. 2/3 on
[7 + 2/84
3 om te 1.: Integers: : N, N + 1,N+2
+
ee NS NN + 1) = -6(N + 2)
eea3 f7
ee Is
my a Ne2 +N Be: 6N — 12
V7 V7 V3 V3 N? + 7N + 12 =0
_ 6V21 _ 3521 | 84/21 _ 5521 (N + 4)(N + 3) = 0
~ sal 21 21 21 N = -4,-3
‘ . The desired integers are —4, -3, -2 and —3, —2, -1.
Sk ee p
‘e sia p- _ p- x?y — xp” 2. Evenintegers: N, N + 2, N + 4
ete n'ont Sap afte.in® MN + 4) = 9(N + 2) - 24
p? p> xy = xp? N? + 4N = ON - 6
A 2 .
ce ep? = xy, ep NS SON 6 = 0
xy - xp? xy - p2 (N — 3)(N — 2) = 0
N.=,3, 2
30 4x?yp ( p7!m?y A 2x7yp = y? E gy2y2 Since the problem asks only for even integers, the
; my! l6x-? y m2 oe we desired integers are 2, 4, and 6.
3. Bromide Py + Bromide Dy = Bromide Total
0.05(Py) + 0.4(Dy) = 0.12(60)
PRACTICE SET 55 (a) 0.05Py + 0.4Dy = 7.2
x a (b) Py + Dy = 60
Pp x Substitute Dy = 60 — Py into (a) and get:
Oe re os (a’) 0.05Py + 0.4(60 - Py) = 7.2
P x —0.35Py = -16.8
xm + XS + pa = tpx Py = 48
xm + xs = tpx — pa (b) (48) + Dy = 60
xe * 8) Dy = 12
wag = 48 mL 5%,12mL 40%
140 Algebra 2, Third Edition
Problem Set 55
4. Carbon: Lexmli2s= 12 10.
Chlorine: 4 x 35 = 140
Total: 12 + 140 = 152
12-Lgc
152° 1368 A = 10 cos 20°
152C = 12(1368) A = 9.40
C = 108 grams B= 10 sin 20°
B = 3.42
5. Multiples of 3: 3N, 3N + 3, 3N + 6
-9.40R — 3.42U
6(3N) = 4(3N + 6) + 48
18N = 12N + 72 11. 4 yd? x 3 ft) 3ft | 3ft in
eine Zein: 2.54 cm 2.54 cm
N= 12 x x x x
1 ft 1 ft 1 in. 1 in.
The desired integers are 36, 39, and 42. 2.54 cm lm lm lm
x x x
b Tin: 100 cm 100 cm 100 cm
+
6. = aes = =k _ 4(3)(3)(3)(12)(12)(12)(2.54)(2.54)(2.54) ak
(100)(100)(100)
ma + mb + xy = kmx
ma + mb = kmx —- xy v. 1000 ft " 12 in. ¢ aoe cm 1m
ma + mb S dit 1 in. 100 cm
peas =x 1 km 60 s
m-y x x :
1000 m 1 min
r mp, d+e_ 5 _ 1000(12)(2.54)(600 km
k c pee (100)(1000) = min
mpx + cd + ce = cdx 13. 4 — 2-9 — 2i4 = 4i(ii)\ii) — 61 — 2Gi(ii)
ce)= .cdx, —mpx — cd = 2) — 2
BAL
_ cdx — mpx — cd
aa. se 14. 404 Or eB = At a) ie 2k
= Bhs DSS SY) Dh] p74 FE
mx d
8. x + Piaget 15) 97 + Jip e D-oedl = Dit) FQGGhen DD =r
Fe i ee maiiny cbpy =—2i+2+2-2=4-4
amx + dy — apy = bpy — bmx 16. 39° = oP = oe?
amx + dy - apy _ , = —3(ii)(ii)\(ii) — 2i — 2 — 2(ii)
py — mx 2°39 Pp 4 23 2
9. 325° 17. x? = 5x +5
A
<r. 7 (.?— 5x + =5
40 . Be
; 4 4
2
A = 40 cos 35° (x- =} pena
A = 32.77 D) 4
5 Pek!
B = 40sin 35° sia lice |
B = 22.94 ee 5 + NS
32.77R — 22.94U 9 i hay
141
Algebra 2, Third Edition
Practice Set 56
18. x* —-6 = 6x SF = 8
27. 4x
(x27 -6x + )=6 SF = 2
x? - 6x +9=64+9 Cea
G=sy= ee
ata A? = 4? + 3?
ary as
= Ge
A? = 25
19: Wee 4= 2
+) A=5
he aah 5a 8
5 Male=
x=6 10=5+B
Check: 6-2 +4=2 5=B8
Dp (sh 3D)
55 m
The statement is not true, so no real number 28. Ke = rs
solution exists.
px = ym
20. =x
Vx2 —-2x+14-12 om
x? — 2x + 14 = x? + 24x + 144 Pp
~26x = 130
x=-5 29. Verism = ABase x height
1 1
Check: 4/25.4.10 414. — 12°=-5 = |8x6) + Fe) eS sms
= hae Tenens ;
es sno)? | x 8
ft
21. MoV 2. = [a4 af = [atv yiie
SEI aD ya = (72Be * n\s) ft? ~ 1342.16 ft?
22, 2733 = 3343 = [33(34)]!4
4 4
= 33/43 1/12 _ 35/6 3p, aps + Sx = 2) _( + DC = 3)
(ee De = 2) ap(x + 5)(x - 4)
23, 4 detyeye, = Gay) ees
= xl/2ylM4,5/2y — 35/4 x-4
24, 81° = —(60!")? = 243
PRACTICE SET 56
25. 4/2 + 2|t — 4/198 80 +60 140
11 2) a x= Tue 8 = Zs = 70
_ 492
4
AT | 2vil aL 12/22
Vat tle ae v2 80-60 20
_ 822, 22v22 _ 26422 _ 117/22 Oh Ads Oe
oh: OD re 11
10:20" 56
9g, (4:183,256)(704,185 x 10-*) eR ERT Tag ee ee
; 802,164 x 1022
(4 x 10°)(7 x 10737) 66 280 - 80 200
jaf ee An 0 Gag ) SES SP ee ae
8 x 10° cae y2 g} 10
142 Algebra 2, Third Edition
Problem Set 56
PROBLEM SET 56
3= x D = 1440
1. Even integers: N, N + 2, N +4
N(N + 2) = -10(N + 4) + 8 D = 1440 x = = 400 were desired
N? + 2N = -10N - 32
N*$12N +°32'= (0 x + 10 = 46
(N + 8)(N + 4) = 0 X-=-36
N = -8,-4 y + 20 = 30
The desired integers are —8, —6, —4 and —4, -2, 0. y= 10
2. Carbon: is (2 12 (abe 120 + 100 Srri0
Hydrogen: 2x tos 2 2
Bromine: 2 x 80 = 160
oe, 180 - 60 _ 6%
Total: Lett LOO 174
2
160 320
174 CH,Br, x+1l_om
160 CH,Br> = 174(320) y Pp
xp + p=ym
CH Br) = 348 grams
xp = ym — p
Bromine: aol x 100% = 91.95% aoe ym — p
174
Dp
3. Dey
aDr a+
b
Xe ci Ep
m k
akm + kmx — bck = bmp
RywTw = RrTp kmx = bmp + bck — akm
Rw = 4: Rr ='20
_ bmp + bck - akm
Tw a= Tr =) 12
km
N S= i 240 10. as
Tw = 10
m* — ax — cx = amp + cmp
Dy = 4(10) = 40 miles m? — ax — amp = cmp + cx
4. Fluorine Py + Fluorine Dy = Fluorine Total C
0.2(Py) + 0.8(Dy) = 0.56(1000)
(a) 0.2Py + 0.8Dy = 560 Li
(b) Py + Dy = 1000
Substitute Dy = 1000 — Py into (a) and get:
(a’) 0.2Py + 0.8(1000 - Py) = 560
—0.6Py = —240 = 10 cos 30°
Py = 400 = 8.66
(b) (400) + Dy = 1000 ll 10 sin 30°
Dy = 600 D>
+nd 5
400 gallons 20%, 600 gallons 80% -8.66R -— 5U
Algebra 2, Third Edition 143
Problem Set 56
12. 21. > Il ~= + ~
=
Ne l x= a il x
a eS
eg ii ame i
Ay. %
4
2 4
= 20 cos 60°
10 a ee
2 2
ll 20 sin 60°
ge
S&S
>
aw = AS 2 2
10R + 17.32U 22. —& - 8 = -x
(x7 -8% + )=8
1ES. kn
hr
x? - & + 16 = 8 + 16
1 hr (x -'4)?
= 24
x
60 min x-4= +6
_ 60(1000)(100) in. x=
4+ 2/6
~ (2.54)(60)(60) _s
23. Vx+1+1=1
x ele
14. 400 yd? x
t= =]
Lam,
x Check: /-1+1+1=1
1 ft
Oe rerrN
2.54 cm
x
x .
1 in. 24. jx? -2x+21-1l=x
= 400(3)(3)(3)(12)(12)(12)(2.54)(2.54)(2.54) em? x7 — 2x +21 =x%7 42x41
—4x
= -20
15. 3i° + 2V-25 — 31% = 3i(i)(ii) + 107 — 3(ii)
Liss
= 3 + 10) + Sod ee
Check: OS: 10 + ate tT = 5
1632 = BP we OF ea 6-1=5
= 2(ii)(ii)
— 3i(ii) + 21 + 4
=2+31+2i+74=6+ 5 4x* + 12x?+ 36x + 108 + 3.
25. x — 3)4x4+ Ox? + Oe? - Ok- 1
VE 14312 = 192.579 = [22(2/5)] V6 4x* — 12x
= 91/39 1/30 = 211/30 12x3 + Ox?
12x? — 36x2
18. y4 xy? = (y4)!2(xy2)!2 = ye 36x7 + Ox
36x* — 108x
19, 2595 = 1524/5 = [52(5!)]'2 108x — 1
= 5. 51/6 108x — 324
323
ax — yp ax" —- yp xy
0, 4/2 -
\4 y) x " 7 9
Xy 2
yp-x
26.
_ 4/3 : ; y pa xy?
: xy y xy? y px
_ 123 ax? - y>p
exe y*p x
144 Algebra 2, Third Edition
Problem Set 57
27. Graph the line to find the slope. PRACTICE SET 57
é AY BY,
poe
(200)(8) _ P,(10)
(1000) ~— (800)
800 (200)(8) _
10 ~~ (1000)
128 = P,
P, = 128 newtons per square meter
b. P\V, = PyV2
ee (11)(44) = P(4.4)
+8 8 (11)(44) _ P
—
ia tee
Since the slopes of perpendicular lines are negative 4.4
reciprocals of each other, 110 = P»
m, = -8 P, = 110 atmospheres
y = -8x + b
7S = —8i-5).+ b PROBLEM SET 57
—
S=b
=
1. PY, > PiV,
y = -8& - 45 T, T,
100(4) _ P, (12)
if, eee ae 800 600
4 : P, = 25 N/m
2x -6-2+x = 48
a, = 56 Bie Pi Vy ova
56 7(42) = Pp(49)
oo P, = 6 atmospheres
3. Stee
Pe Yee
er
29 a= 3 3) a Bh
2 ar rae:
te - 10-3 +a3a = 417 M00 eas
1200 30: ;
P» = 100 N/m
a=+!3 4. Alcohol Py + Alcohol Dy = Alcohol Total
0.2(Py) + 0.4(Dy) = 0.352(1000)
Bi sn ceax?
Sate (a) 0.2Py + 0.4Dy = 352
S gty- oye — 1 theea
b) Py + Dy = 1000
x(x? -1) +1
(Bx + 2)(x Ne re ae
— Te MSL Gar SLOte ee Substitute Dy N = 1000 — PyN into (a) and g get:
a*y(x* - 1) a?y(x? — 1)
4ay (a’) 0.2Py + 0.4(1000 - Py) 952
~ azy(x? — 1) —0.2Py = —48
Py N = 240
Poy 9g = SF 12 — Aa*y
a*y(x?=1) (b) (240) + Dy = 1000
a
x? — 3x2 - 6x — 4a7y - 2 Dy = 760
* a*y(x? = 1) 240 gallons 20%, 760 gallons 40%
145
Algebra 2, Third Edition
Problem Set 57
5. Odd integers: N, N + 2,N +4, N+ 6 40 cm 1 in. . 1 ft 1 mi
11. x x
(N + 4)(N + 6) = 10N + 49 s 2.54 cm 12 in. 5280 ft
N2 + 10N + 24 = 10N + 49 60 s S 60 min
N2 — 25 =0 1 min 1 br
(N — 5(N + 5) = 0 40(60)(60) = mi
N = 5,-5 ~ (2.54)(12)(5280) hr
The desired integers are 5, 7, 9, 11 and —-5, -3, -1, 1.
1 in. 1 in.
12. 1000 cm? x SK ae ee Se
en ae 2.54 cm 2.54 cm
1 ft 1 ft 1 ft
x x
ee horieee es 12 in. 12 in. 1Qeine
3 1000 3
, (2.54)(2.54)(2.54)(12)(12)(12)
a
er, Cc
a4 oo
ap — cx + cy = mpx — mpy
13. AP Py nS
mpy + cy = mpx + cx — ap = 3i(ii) — 2(ii) + (iii) — 5
mpx + cx — ap = 3742 ¢ Parasz
mp +c
14. 20 /EQr = 37 a
xX
— @4
='-6)
— BG) 4 i= 2
dx — ad — cdp = kp =-6+3+2i-2=1-4
dx — cdp — kp = ad
dx -— cdp — kp 1S: 5x — 6 = -x?
(?- 5x4 }=6
2 sy eo eee
4 4
( 7 49
x-=—/] = —
A = 4cos 40°
2
A = 3.06
Ph Sok Fd
B= 4 sin 40° y 2
B ui PIR)
erese
3. 06R + 2.57U 2) D
10. x = 6,-1
330°
Sear
A
a
16.
A = 40 cos 30°
A v 34.64 (x - 3 =4
B 40 sin 30° is
Th Vln
B =e) pers
Ue
34.64R — 20U x = 5,1
146
Algebra 2, Third Edition
Problem Set 57
17. Use the graph to find the slope.
11 . 25. a2 + 7 ]3- ava0
Lege =P 5 2
Use the point (-1, 6) for x and y.
_ 3V2 V5, 75 V2 _ 4 ap
1 ASS) wen eyo,
6 SSR _ 6V10 , 35Vi0 _ 4010 _ V10
5 10 10 10 10
Se
4
26. 16°94=
1 2
1674
1 Pe 1
S16!Pe 332
ne Oa
y=—x+ —
5 5
Jpg) Kacey ee
18. Vx— 22°F2 = 27 pz x pz xz - 4p
x-2=1 wee xz - 4p? pz
Ppa P ‘ pez xz - 4p
2 5
Check: ¥3 -2 +2 =3 = 7J-P?
eho Rh gyi?
(See ie ee 28. 2 = 4°) 2312 = Gf)
et Ae 4 18 abt 2d = (x)(-2 - 37 - 2) - x(-2 - 2°)
me Pees ip) 4 — 6x + 3x = -13x + 3x
x=4 4 = -7x
Check: J16 —-4 +13 -1=4 is, OF
Sa 1% u
eee) Ce 29. 285igh max"
20. =— 2/5 A ;
4 3 x- = 3x — 28x = 0
3x — 6 — 4x = 60 xx — Tx + 4 =0
oo x = 0,7,-4
x = -66
ig x+2
30. = -
yg a lee oe ge Mata #2)
4 2
x-@-2= 12 Pepe 1ane DB uk AE Beli
Pay aa + 2) a*(a + 2)
ben eke OCh —2 | 2ee 2G
ofa.- 7 : + 2)
a“(a 2 at(a
a“(a +2 )
22. (93/3 = 43233 = [32(31)]?
ee OP 31/6 dt 37/6
23. ey
Nt Vx2 y = [x*(x2y)!P]?
= x2xN/3yl6 — y7/yV6
24. 3f4 2/2 = 2232 = 223245 = 215/15
147
Algebra 2, Third Edition
Problem Set 58
PRACTICE SET 58
3. Chlorine: 2) x 100% = 27.13%
a. 3x2 + 5x -6 = 0 129
|
S +
Nn
wf
f wv ll oS
4. Bromine Py + Bromine Dy = Bromine Total
ars 0.1(Py) + 0.4(Dy) = 0.16(50)
(: ts 37 = 2
(a) 0.1Py + 0.4Dy = 8
(2 + 31+ 3) Dee (b) Py + Dy = 50
Be 2-36 36
Dy
(:+ 2) = ei. Substitute Dy = 50 — Py into (a) and get:
6 36
. 2 (a’) 0.1Py + 0.4(50 - Py) = 8
ne re 36
-0.3Py = -12
oe 5 + 197 Pn =F4()
6 6
(b) (40) + Dy = 50
b 3x° D —-x-12=0
as
Dy = 10
oe hn
boat ue O 40 mL 10%, 10 mL 40%
(e-}e )e3
(2 -dr+
3 ie
cle sed
8 5.
— D
Dp
?
Bs) BG a 6
i,
(x- 2) -2 RyTy
+ 15 = RpTp; Tr =.Te = 3;
Rr = 20
ee arg ee
26 a6 3Ryp + 15 = 20(3)
Perce 13 3Rr = 45
= Rr = 15
Dr = 15(3) = 45 miles
PROBLEM SET 58
: PY _ PV, 6. 3Xe 2 Ee 4x oO=
fb1 ie2
x + ao = L=#0
4(6) _ 3(8) 3
- 2 (3?a 3 oat =2 |
T, = 600K
2. Carbon: 6x 12=72 a Fs ee
Hydrogen: 8x 1=8
Nitrogen: 1x 14=14 Pas
Chlorine: x35 = 35 (xe 3) =O:
Total: 72 +8 + 14 + 35 = 129 2 V3
x+— = +—
ph, 360 3 3
129 C,H,NCI a VB
72C¢HgNCl = 129(360) 2 Bont
C.6HgNCl = 645 grams
148 Algebra 2, Third Edition
Problem Set 58
i) |
= ca) | Nn ll (=)
11. go td
| | * | | I! x—p k
* oS
ak — ckx + ckp = xy — py
ckp + py = xy + ckx — ak
_ xy + ckx - ak
ck+y
Z|-
+
12.
|
Bilin
Om
oo}
é
[— coo|—
———
ie pS
pay
00
\o
|
~i) ll
IIl I+ A = 4cos 40°
|
00
\o
A = 3.06
| poh
cole
Blu B = 4sin 40°
13) = OLS]
-3.06R - 2.57U
x ERE
3 3
6 13. 315°
Cia
ae
A 10 cos 45°
A v 7.07
B i] 10 sin 45°
B u AO
= iH | 7.07R -— 7.07U
70m 100 cm 1 in. 1 ft
14. x x
x = 176 — 100 = 76 S 1m 2.54 cm 12 in.
1 mi 60s 60 min
ve
360 - 176 _ 92
x
5280 ft
x
1 min 1 hr
2
‘. 70(100)(60)(60) mi
ae 100 + 76 _ gg (2.54)(12)(5280) hr
HG x 12 in. $ [Zits2Z-in,
15.
1 ft 1 ft 1 ft
10. | 9 ll
al< 2.54cm _ 2.54cm _ 2.54cm
os x x
1 in. 1 in. 1 in.
kx — ak — ckp py
= 40(12)(12)(12)(2.54)(2.54)(2.54) em?
kx — ak = py + ckp
4195/3 = 432 213 = (32)"431/5
kx - ak _ 16.
= 31/231/5 = 37/10
y+ck
Algebra 2, Third Edition 149
Problem Set 58
17. Vav2 = {22J2 = [22(2'17 26. (a) 3x + 4y = -4
= 22/In1/14 _ 95/14 4y = -3x - 4
18. [x2 y3 3x = (x2y3)V2(xy5) 1/3 3
y =-—x-1
re
= xy 1395/3 a x43 y19/6
(b) x — Sy = 10
iS 604s ee
16/4 ive
(161/43 ares
“8
—5y
= -x + 10
ee
20. 5[2 = 2|2 + 3220
_ 5Vv5 Vil _ 2v11 V5 ¢ peg
Vil Vil V5 V5
ENS 2255 33055 _ 333/55
55 55 55 Se)
218 5) = Ci eos 45"
= Si(ii) — 6(ii)(ii\(ii)(ii) + 101 — 44)
= 5i-6+10i+4 = 2+ 5i
22. (40,621,857)(6,031,824)
19,610 x 10-74 Solve the equations by elimination:
7 6
is (4 x 10 )(6 x 10°) ~ 1x 1034
(a) 3x + 4y= -4
2 x 10°20
—3(b) -3x + 15y = -30
3 (x + 8)(x + 2) ; (x + 5)(x + 3) 19y = -34
(QeF 8) 3) (rie 5G 5) —h.34
Se ae Pe ‘ae i.
x -5
(b) x = s(-] fio Ct
24. Vcylinder = Apase x height 19 19
Apase X 3em = (72 — 62) cm?
(72 — 62) cm?
ABase = 3
cm
1 4x? = By 4 16 — x+2
7
(6)(4) - la = (24 — 22) cm 27. x
3
2)4 > Ore Or — 5
Ax?
+ 8x2
a)
—nmr~ = 27cm 2 8x2 + Ox
D
8x2 ~ 16x
r? = 4cm* lo = 5
r= 2cm 16x+ 32
—37
xa’ e 4
3
ae
25. o P 28.
8
-— 5 D 3
mp?
9 — 3x — &x = 42
xa’ p> — 4m mp?
-llx
= 33
ms mp? DIY cee Smp?
Xx = =3
xa — 5mp? mp?
mp? 50 5mp°
- x7a*p> - 4m
xa —- 5mp°
150
Algebra 2, Third Edition
Problem Set 59
29. oor = 5x oP) eat —7
Wee 2X
ce eee — 1.
—]x
= =15
b cx
30.
a(iy+l1) mea*(y +1)
ab cx ab
+ cx
en ercty ch trea
Gy ty
ee MT 8pce as
‘arly el). ra GeeDD
tanion= 2
3
@ = 59.04°
PRACTICE SET 59
Since @ is a third-quadrant angle,
@ = 59.04° + 180° = 239.04°
N = mF + b
H = 32 +5% = /34
Use the graph to find the slope.
(34 /239.04°
Slope = = = 83.33
N = 83.33F + b
Use the point (32,200) for F and N. PROBLEM SET 59
(200) = 83.33(32) + Db.
Integers: N, N+ 1,N +2, N+ a]
—2466.56 = b
NN + 3) = 10(-N — 2) - jig
N = 83.33F — 2466.56
N2 + 3N = -10N — 42
ersee |
a
3 N* + 13N + 42 =0
@ sta 7 (N + 7)(N + 6) = 0
(b) 0.01x — 0.6y = 3.03 N = -7,-6
21(a) = (a’) 7x + 6y = -9 The desired integers are —7, -6, -5, —4 and -6, -5,
—4, -3.
100(b) = (b’) x — 60y = 303
(a’) 7x + 6y= -9 Carbon: 2x12 = 24
~7(b’) -7x + 420y = -2121 Hydrogen: 6x 1=6
426y = —2130 Oxygen: 1 x 16 = 16
y=-5
Total: 24 + 6+ 16 = 46
(b’) x — 60y = 303
36 agen
x — 60(-5) = 303 46 460
x + 300 = 303 46Oxygen = 16(460)
x=3 Oxygen = 160 grams
151
Algebra 2, Third Edition
Problem Set 59
5 ae 2
—+— = -—
1
(Oa 4
(b) 0.04x -— 0.2y = 1.13
RrTr =e RyTw = 41; Rr = 8; Rw = 3;
Tr te Tw ial)
(a’) 4x + 9y = -3
8Tp + 3(7 — Tr) = 41
(b’) 4x — 20y = 113
STp.=-20 (a’) 4x + 9y iH} | Oo
Tr = 4hr -1(b’) —4x + 20y =
29y =
Tw = 7 - 4) = 3hr
Dr = 4(8) = 32 miles
Dy = 3(3) = 9 miles lye tae BN
Iodine Py + Iodine Dy = Iodine Total
0.9(Py) + 0.7(Dy) 0.78(100)
(a) 0.9Py + 0.7Dy = 78
Substitute Dy = 100 — Pa into (a) and get: :
(a’) 0.9P, + 0.71100 — Py) = 78
O2Pxy=e8
Py = 40
(b) (40) + Dy = 100
Dy = 60
40 liters 90%, 60 liters 70%
Ri Vie, Bov
tan@ = ze
(14)(10) = (20)V> 4
V> = 7liters Ol= 65a
Since @ is a third-quadrant angle,
Zr = mCa + b
@ = 2057 | 180° =.206.57°
Use the graph to find the slope.
H = AP IN
Slope = = = -1.43 2/5 /206.57°
Zr = -1.43Ca + b 10. 330°
Use the point (100, 18) for Ca and Zr. ibs A
18 = —1.43(100) + b
SSS, 7
30
161 =b
Zr = -1.43Ca + 161 30 cos 30°
v 25.98
78
+ 42
(a) x = 60 30 sin 30°
2
> =
nme 15
Zooi— 102
(b) x 5 = 78 25.98R -— 15U
152 Algebra 2, Third Edition
Problem Set 59
11. 2x? = s + Nn 2
14, 2 (2) + H?
&
nN | & + ll o4
———s Nile
a ae
4
Hy!
Paataa [4
Say,
=
a|—
+
2 16 — 25
a oT
esa)
4
to- = = |
NIM
&wplN
—
4 — on 5
—=f
2
] —
x ——— = H+
4 |S)
a Nee ree
—_
2
x= I+
|— o|3) vil xe=)
Area = ——2 cm? = 5Vi1 cm?
2 4
12. ax° 5 =
(2-204 = 15. ae t= 0
m
4p — 4s — cm + 4mx = 0
ees5 psn
9
+
Ole
4p — 4s = cm — 4mx
ApPp - lr ater
Fe) Re =
3
=
oe
©|5
WwilM
Wln
c - 4x
1
a 16. a pet)
3 Dae) x
mx — cp — cs + 4px + 4sx = 0
x=
4px -— cp = cs — mx — 4sx
cs — mx — 4sx
—— St ae
4x -c
40 in. 2.54 cm lm 60 s
13. Bn A= 17.
S
6
1 in.
x
100 cm
x
1 min
60 min
(2-04 B x —————EEE
3 1 hr
i 40(2.54)(60)(60) m
+ sueet.) Re (100) hr
3 36
(*-5)
ie
6
= 18.
18,000 cm? x
ai!
2.54 cm
1 ft 1 ft
1 in.
2.54 cm
& tn, ; 1 in.
2.54 cm
1 en SC x
52 =
12 in. 12 in. 12 in.
6
. 18,000 3
C= (2.54)(2.54)(2.54)(12)(12)(12)
19, it + 5 413/=9
— 2,/=4
x=
= (iii)
+ 5 + 9t -— 4
=14+54+9-4=6+5
20. 3° - 27 - 4 -i = 3)
- 2G) - 4- i
jeA ee
e e DT
Algebra 2, Third Edition 153
Problem Set 60
(2-41
+39 +1=x4+4 a Cone
30.
2_ ax + 39 = x7 + 6x + 9 te yy ek “Byp?
—10x = -30 ype
56 = 8! nap ype pe
poxy —2 a xty4
@heck: 4/9 = 12/4739. 4 1 i] 3 #44
= ca py” = pi!
6+ 1 3+4
22. Vx —-34+2=
x— 3
PRACTICE SET 60
ce ay Be) a. B = kR
Check: (39 —-3 +2 =- (12) = k(3)
6+2=-4 4=k
The statement is not true, so no real number 4R
solution exists.
4(6) = 24 bluebirds
23. eS [3(3!/7)]!6 = 31/53 1/10 = 33/10
24. MAd Dom N22
nl Dee 40) |1/4
ee qi/291/8 = 25/8
25. =I 6-3/4 = nee ee |
|xy 3/yxt =a, (x2y) 9x4) 13
26.
=visa = x7/3y13/6
PROBLEM SET 60
27. = a4 > + 3/234
1. B= kG
Pais
*a a
See oFeo (8) = k(2)
4=k
2V26 _ 52V26 , 234/26 _ 9226
26 26 26 13 B = 4(7) = 28 boys
28. REG
N,
oye 10 115
oom
ei So) \ 40
4000 = k
29. Aa x7 +2 x?
RPM = 4000 = 160
+ Se eens
87a y(a + y)
Y axy(ar y) _& + 2)(a + y)
a5
- yaa + ae ya(a + y) 2. C = kP
ax*y
40 = k(20,000)
y’a(a + y)
0.002 = k
axy(a + y) + (x? + 2)(a + y) + axy
y*a(a + y) C = 0.002(8000) = 16 clowns
154 Algebra 2, Third Edition
Problem Set 60
Fe 8.
2
(a) is* + ue = 34
TY T, 2
400 OCP (b) 0.02x + 0.3y = 6.2
500 1000
P(500) = 400(1000) (a’) 4x + 15y = 340
P> = 800 N/m? (b’) 2x + 30y 620
(a) Np = Nr + 50 ~2(a’) -8x - 30y = -680
(b’) 2x + 30y = 620
(b) 2Np = 3N7 + 60 —6x = -60
Substitute (a) into (b) and get: x= 10
(b’) 2(N7 + 50) = 3N7 + 60
(a’) 4(10) + IS5y = 340
Nr = 40 trepid students
y= 20
(a) Np = (40) + 50 = 90 pugnacious students
(10, 20)
B 135°
a Bach ES
A
> ee!
x“ -—--—-x+—e=
1 + —
1
A =14 cos 45°
4 64 4 64
A=1283
(
x--/|
Lip=—49
B= 45in45° 8 64
Boe 83
Fees!
8 8
—2.83R + 2.83U
Pe ee
8 8
ot, =
4
10. 3x2
= 2x + 1
(2 - dr 4 J=4
ss) 3
2 | 1 1
x°-=x+—-= -+-
9 Seed
(r-3) 25
Kae
3
=
9
6 = 63.43° Papas
3 )
Since @ is a second-quadrant angle,
O= [80° — 63.43" =MI6 ST
ped
AIT gd
3 SRh3
H= j27+4 = 25
era ee
2/5/116.57° 3
Algebra 2, Third Edition 155
Problem Set 60
11. Na=mC +b 1.400.000 ay ilvin, 1 in. b, 1 in.
18.
Use the graph to find the slope. een ie Oigiem Osa cut sd cel
1 ft Git . 1 ft youve lyd J {ya
y
m= cite 0.0375 ‘ 12 in. c 12in..” 12 in. gey3ift 3 ft 3 ft
80
Na = 0.0375C + b
G 1,400,000 ya?
~ (2,54)(2.54)(2.54)(12)(12)(12)(3)(3)(3)
By inspection b = 3.9.
Na = 0.0375C + 3.9 19. Va7V3 = ¥33V3 = [39(32)]!7
= 3. 316 — 3746
12. (a) 6x + 10y + 30 = 110
(b) 4x + 10y + 10 = 70
20. 4913/3 = 134 33 = (34)'4(314)
l(a) 6x + 10y = 80 -3. 313 — 343
—1(b) -—4x - 10y = -60
2X = 20
x = 10
21. 3 1-3/4 = oul
g 13/4
= oe
(31/*)
& = i
a7
(a) 6(10)
+ 10y = 80
lm? p 3 lm? p4 ca (mp)"2(m>p*) 2
22.
paar
= mpm p43 = me pe
13. ere
xy d
23.
ad = cdx + cdy + mx + my
ad — cdy — my = cdx + mx
BO ePORY NY ee
cd +m 5
10/2 , 27N2 60/2 23/2
6 6 6 6
Peat eb and FaePes
m Zz
xyz + cz + dmz = km (5,162,348)(0.0000165)
24.
cz = km — xyz — dmz
0.003217642
km — xyz -— dmz +A SHC 1002 Oe)
C=
re 3x 103 2 ee
1500 37. 4. 9 0" 2 iA 25.
= 3i(i) — 21 + 97 = 2 = (ii)
= 3) 0) 497-0 = a= Se
16.0372 07 3.2 3 oo x Il a
= 3(ii)(ii) — 2) + 2 — 15% wl
=i oi te Och = Sys 7/ IIGY | I ~
60 cm 1 in. 1 ft 1 mi
iy x x x
s 2.54cm 12in. 5280ft 26. & i) 180 — 90
— 40 50
60 s . 60 min
1 min 1 hr k = 180 - 90 — 50 = 40
_ __ 60(60)(60) mi a I 180 - 90 - 40 = 50
~ (2,54)(12)(5280) hr
156
Algebra 2, Third Edition
Problem Set 61
27. (a) 2x)- 3y = 3 PROBLEM SET 61
(b) 2x + y = 8 1. Rp = kS
Substitute y = —2x + 8 into (a) and get: 0.005 = k(5)
0.00L =k
(a’) 2x — 3(-2x + 8) = -3
Rp = 0.001(0.3) = 0.0003 kg per second
SV eal
5 21 3. eal = gy
8 (hah
21 11 P be ape ced
poe 300
(b) y= -( =) 18-= 7 1000 600
300(1000)
(B4)
8° 4 P= 600 -
P, = 500 N/m
3 = 2
ae, x + 50x = 15x 3. Water one — water out = water final
e150 + 50x = 0 W, - Wo = Wr
x(x — 10)~% - 5) = 0 (100) — (E) = (100 —- E)
x = 0,5,10 0.9(100) - (E) = 0.8(100 — E)
90 —- E = 80 —- 0.8E
29 2 b —0.2E = -10
- y=r=xt
5. E = 50 gallons
Use the point (—40, 2) forxand y. 4. Butterfat one + butterfat added = butterfat final
B, + Bz, =B
De S40) th wap teer
5 (900) + (Py) = (900 ao Py)
ich 0.02(900) + (Py) = 0.1(900 + Py)
18 + Py =,90 + 0.1Py
vis at $18 0.9Py = 72
Py = 80 pounds
-420. 9 Me 0,0 ,-2
30. cee aea [2? L ae | 5. Glycol, + glycol added = glycol final
te y cd (100) + (Py) = (100 + Py)
TOL Vat ok ePaire 0.2(100) + 0.3(Py) = 0.25(100 + Py)
ae a ie 204 03P, = 25 + 0.25Py
= 6x~> - 3x 1y-2y-2 0.05Py, = 5
Py = 100 kilograms
PRACTICE SET 61 6. abe 20 165°
Water one — water evaporated = water final A
(400) - (Ew) = (400 - Ey) ieee
0.85(400) — 1.00Ey = 0.60(400 - Ey) As 19:32
340 - Ew = 240 - 0.6Ew B = 20sin 15°
100 = 0.4Eyw B = 5.18
250 gallons = Ew | -19.32R + 5.18U
Algebra 2, Third Edition 157
Problem Set 61
10. y= Th tee
ee ee. Pe 7
(: 2” 2
Len 1 nepint. ee Bae ay
eae ilies mS st
ri) -2
2 ii |
—|—|—| 1 57
Te 4A 4
o}aL
Le) = == +—
|
Lise
o>)
1, 57
Ba
4. \4
tan@ = 2
6
11. Pore
pS ae
0 = 18.43°
Use the graph to find the slope.
Since @ is a fourth-quadrant angle,
m= ale. 42.6
@ = 360 — 18.43 = 341.57° Zo5
H = ¥6* + 27 = 210 Pb = (42.6)Sb + b
2/10/341.57° Use the point (0.75, 0) for Sb and Pb.
0 = 42.6(0.75) + b
-32 = b
(a) ot = =) = -4
Pb = 42.6Sb —- 32
(b) 0.2x + 0.9y = 19.2
12. x +3 68
(a’) 10x — 6y = -60 x = 65
(b’) 2x + 9y = 192 360 = y — 12 + 30 + 68 + 62 + 78 + 32 + 54
{(a)° 10x —"6y = -—60 48 =y
-5(b’) -10x - 45y = -960 (m+c+x)b
13.
Es 1020 k
Al) bdm + bcd + bdx + ka = dkp
(a’) 10x — 6(20) = —60 dkp — bdm — bdx — ka = bcd
ye dkp — bdm - bdx ~ ka _
bd
(6, 20)
4y m
14. + = d
-~4x — 4 = 5x? 2a+x C
4cy + 2am + mx = 2acd + cdx
(x? - dx 4 )=¢
5 5 2am — 2acd = cdx — 4cy — mx
sin4 ee ee Se cdx - 4cy — mx
x* —- —x + — = — + —
awa tiga 2d
oo
5 O5u =5p 295
(
x--
i ke
24 15. 297” 1187)_=4ha 9 SAU oi = 37 4 a
5 25 = (Siteae = —2-Ti
plage 216
5 5 16. BE er nO a te
8(ii)(ii) — 2i(ii) — 6 - 16i
2, 2V6
Me em 1 Sees 8 + 2i- 6 — 16i = 2 = 14:
5 5
158
Algebra 2, Third Edition
Problem Set 61
40 cm? r. 1 in. 1 in. 1 in.
17.
s 2.54 cm 2.54 cm 2.54 cm
26.
60 s i 60 min
ryt
v= st-—s
1 min 1 hr
_ ___40(60)(60) in?
(2.54)(2.54)(2.54) hr
A fe 12 in. . 12 in. é. 12 in. be 2.54 cm
18.
1 ft 1 ft 1 ft 1 in.
27.
2.54 cm 2.54 cm
SE x
1 in. 1 in.
= 4(12)(12)(12)(2.54)(2.54)(2.54) em?
=— ee
19. /32./2 = [25/2 = PQ?) |e = 95/29 1/4
= 214
3/3
iP bp eal 3CD [23(2!8)]18 = 2 . 219
20.
= 210/9
_g-5/3 +? 1 a 1 1
21.
ey 8t
22. Sy 4 LP ey v=) (xy) ¥2(y2x) 4 = x !2yl2yl/2 1/4
23. 0 2 oo Y eter
3 CY
3 eu
pp
ni V3 43
ts
Substitute y = —x — | into (a) and get:
(a jix we 2x — Ly = =6
8
bia
a
(41,685,231)(0.0012846 x 107!*)
8
0.001998 x 10710 b) |-~|]+y =-l
w) (-5) ,
(4 x 10") 10} =2x 10°
2 x 107}3 ye 2"
3
25. AZ = 5% + 12?
A* = 169
A= 13
qu) 3°3
5x
SF =4 28. ~15x
= —x° + 2x?
x? — 2x” — 15x = 0
we
5 K=O)
Fes) "= 0
3x Seen x = 0,5,-3
5
52 _¢
5
Algebra 2, Third Edition 159
Problem Set 62
29. Write the equation of the given line in slope- PROBLEM SET 62
intercept form.
k
2X te Sy = pe =
Ves
k
Bas 2
Since the slopes of perpendicular lines are negative 16=k
reciprocals of each other,
sages Ny = ~ = 2 victories
we
3
y =-—-x+b
5*
PV tN
T, T,
4= —(-2)
+d (600)(2) _ (400)(4)
(300) T,
T=D T, = 400K
3
y =—-x2 +7 S; + SA = Sr
(40) + (Py) = (40 + Py)
30. S23 (Oe) leet 5°)27 0.05(40) + 0.2(Py) = 0.1(40 + Py)
=4° = 9 ian erOx tex)4 2 + 0.2Py = 4 + 0.1Py
-15
= 12x
5
——— =X
Py = 20 liters
4
Ds Dr
PRACTICE SET 62 RsTs + RyTy = 152; Rs = 8;
eee Cee rae eatp Ry
= 20; Ts + Tr= 10
5x7 -x +3 =0 8T; + 20(10 — Ts) = 152
-12T; = -48
Oe ee)
5 5
He )-4
Ts = 4
5 5 Ds = 8(4) = 32 miles
(2-3 _ 3 1
x* — =x + —]= -— +— Carbon: bx Ire 12
5 100 5. 100
Hydrogen: 4x1=4
( t\" 59
Xe | ee
10 100 Oxygen: 1 xX. 16%= 16
Nitrogen: 2 a4 = 28
eee a ed
10 100 Total: 12+ 4+ 16 + 28 = 60
15
Fe entree |
10 +10 Nitrogen: = x 100% = 46.78%
160 Algebra 2, Third Edition
Problem Set 62
6. 9 -x + 2x7 + 30= 0
(3? or: {3 =
p) )
2 —
1
-xXx
1
—> «= ——
3 +
1
—
A = 20 cos20° 2 . 2 16
A = 18.79 (:e *] aoe
4 16
B = 20sin 20° 1 3
x-—-— = +I
B = 6.84 4 4
18.79R — 6.84U oe 123,
4 4
u 10. 5x + 6x2 = -3
(x? - 3x+ }see
6 5)
Mos
Pe 25
Ae. eePOGoe
ane5 = 47.
12 12
tan@ = =Zz
5 rear12% peel;
ar?
6 = 68.20° 11. Bi = mHg + b
Since @ is a second-quadrant angle, Use the graph to find the slope.
@ = 180 — 68.20 = 111.80° tae. - te
H = V5? +2? = 29
Bi = 16Hg + b
1 29/111.80°
Use the point (6, 24) for Hg and Bi.
1 2 24 = 16(6) + b
8. (a) A gael es
(b) 0.02x + O0.4y = 2.58 Bi = 16Hg - 72
(ix —2y = 3 12. Volume = rh = 11,5207 in
3
ee
(b’) 2x + 40y = 258 eee La
2070 in.
—2(a’) -2x + 4y= 6 r= /576in2 = 24in.
(b’) 2x + 40y = 258
44y = 264 ia. eee
y= 6 : G r
amr + bmr — ckr = cp
X= 2(6)s= =3
oor _ bmr = cp + ckr — amr
sae det che Shar
(9, 6) | ean <a ae
161
Algebra 2, Third Edition
Problem Set 62
5. C= Weer
6x Pp
14. ———— - ¢ C2
= 625
2y + 4a r
6xr — 2cry — 4acr 2py
+ 4ap Cle
6xr — 4acr — 4ap = 2py + 2cry 24 x SF = 30
3xr -— 2acr - 2ap _ —s 5
pce we Sia
4
Jeet, WE) = OB) Xe Vil
15. BP el Aea9 + 3
5
= 3i(ii) - 27 - 2 —3 + 3 B= 25 x — - 25
4
= 2-5
pa 128 _ 100
16. S34 01 — 4a Se OND 4
esi)" 21 — 4 — 3) — 6
= -7 + Si
Se 4
WE
600 cm?
—————. Xx
1 in.
x
1 in.
x
1 in. (a) x = B+ 58 ~ 63
26.
min 2.54 cm 2.54 cm 2.54 cm Mu
1 ft 1 ft iGtt 1 min
x x M x ee 160 — 52 _ 540
aro, 12 in. Dane 60 s
S 600 ft? 2
- (2.54)(2.54)(2.54)(12)(12)(12)(60) s
27. (a) 2x - 3y = -9
3 ft 3 ft Bvt 3y
= 2x - 9
18. 20 yd? x — x — x
1 yd 1 yd 1 yd @
=—x+3
x
12 in.
12 in.
———=
12 in. aes:
(sit 1 ft 1 ft (b) 2x + 3y = -3
2.54 cm . 2.54 cm . 2.54 cm
x ay ==2x
=3
1 in. 1 in. 1 in.
2
= 20(3)(3)(3)(12)(12)(42)(2.54)(2.54)(2.54) cm? a) ysl
a
Ao PAIN = [24(2¥2)] 2 = 22(2"/4)
19. 41642 =
= 29/4
4
20. May ee ee (22)'/4gU/5 _ 2l/2q1/
= 27/10
_4s/2 Es at 42)? ee) NA
ehee
El
21.
|xy! 3 wy = (xy)? @9)? = play 1278/31/83
22.
= x13/6, 23/6
ae
SERESNSSEE
23.
5 2 a : 2
Substitute y = aie — 1 into (a) and get:
_ V2 V5 SNS.V2 6 ip
V5 V5 V2 42 Gy 2 = a(-2 ~ 1 = -9
_ 6V10 25V10 —-60V10_ 79-10
10 10 10 10 4x
= -12
eS
(0.000618427 x 10!*)(7,891,642)
24.
3,728,196,842 b y =
(b)
2)
(23)
3 3) = 1 i —
_ (6 x 10!°)(8 x 10°)
aan? = ix 10°
x (-3, 1)
162 Algebra 2, Third Edition
Problem Set 63
28. 50x + x? = 15x?
PROBLEM SET 63
x? — 15x? + 50x = 0
x(x — 10)\(x - 5) = 0 T = kD
x = 0,5, 10 20 = k(400)
29. | eseth
se ll » + >
20
to
1]
] — il l . mn >
~ +
Ie $60) = 3 troubles
20
Le ges)
1 T,
fel aps
800,
30.
T) = 480K
Wi + Ws, = Wr
(50) + (Py) = (50 + Py)
0.96(50) + (Py) = 0.99(50 + Py)
48 + Py = 49.5 + 0.99Px
0.01Py = 1:5
Py = 150 gallons
Do
25 sin 20° 4000
u 25(0.34) = 8.5 DR
25 cos 20° RrTr + 4000 = RoTo; Rr =920?
2
Bev 25(O94) = 9235
Ro = 40; Tr = To
20T + 4000 = 40T
20T = 4000
T = 200 seconds
Length = Do = 40(200) = 8000 feet
Carbon: 3X 1[2-e36
S = 50 sin 35°
Hydrogen: 7k 1LS=7
So= S007 )e= 28.5
Chlorine: 1 X35 = 35
W = 50cos 35°
Total: 36 + 7 + 35 = 78
W = 50(0.82) = 41
23.5R + 85U 30 72 48
—41 R —- 28.5U 78 C,H,Cl
-17.5R + 20U 36C3H,Cl = 48(78)
Tracker was 17.5 miles west and 20 miles south of
C3H7Cl = 104 grams
the cabin.
Algebra 2, Third Edition 163
Problem Set 63
6.
N
a" ol
EB.
30 cos 30° = 25.98 (oe)&
SUSE
Led a it = |
pt}
hgte
N = 30sin 30° = 15.00 as mi
&
eRie)
220° e2u
ia
We \Yv
a
S
H = (7.3 + (26.34% = 27.33
tan@ = Ao
W = 50cos 40° = 38.30 Ws
@ = 74 51°
= 50 sin 40° = 32.14
27.33 /74.51°
25.98R + 15.00U
—38.30R — 32.14U
—12.32R — 17.14U 1 5
(a) —xFen — ~yTG = —48
Running Bear was 12.32 miles west and 17.14 miles
south of the village. (b) 0.4x + 0.05y = 5
(a) 2x = Sy ~480
(b’) 40x + Sy = 500
(a’) ~ 2x25 %= "480
30 5(b’) 200x + 25y = 2500
202x = 2020
Z\ (a’) 2(10) — 25y = —480
R y-=_20
os) iH 30 COS 45 -e= 212} (10, 20)
S Il 30 sin 45° = NO — i) —
10.
" 10 160°
R
z
20°
R 10 cos 20° = 9.40
UO enllsin 20-4 =3.42
21.21R + 21.21U
—9.40R + 3.42U
11.81R + 24.63U
164
Algebra 2, Third Edition
Problem Set 63
in 3x? = 4.42x (Dieay, a
15.
(2? - 2x + )--4 m. tawle-ap
3 3 bp + bzy — bem = am
emt
x“ - =x +— = -— +=
Aer bp + bzy = am + bcm
3 9 3 9 bp
+ bzy
( S 11 at+be
x--| = -——
3 9
16. Ap rey) te ian
ee ee Ai(ii) — i(ii)(ii) + 2(ii) — 4i = -2 - 93
$ 3
1, Vil, 17; 9 Fe Bl a et =f
ig RE U
3 3 = 3 — 2i(ii)@ii) — 3) ii) + -2i -— i = -i
12. Mo = mZr + b 400 cm 1 in. 1 ft 1 yd ¥ 1 min
18. x x <j
min DS Ancim 12ins 3 60s
Use the graph to find the slope.
a 400 yd
y= ets
=i
20
(2.54)(12)(3)(60) s
3 5280 ft 5280 ft 5280 ft
Mo = -7.5Zr + b 19. 4miv- X x x
1 mi 1 mi 1 mi
Use the point (116, 500) for Zr and Mo. 12 in. 12 in.
2am
i x
1 ft 1 ft 1 ft
500 —7.5(116) + b
2.54 cm - 2.54 cm ” 2.54 cm
1370 b 1 in. 1 in. 1 in.
Mo = -7.5Zr + 1370 1m 1m 1m
100 cm 100 cm 100 cm
1 km 1 km 1 km
13.
H 000Kn ~~ 10008. 100d
4(5280)° (12)° (2.54)? 3
3 3 (100)(100)(100)(1000)(1000)(1000)
5 343 4 [3(3'/4)]¥5 = 31/531/20 _ 314
LESS TIS 20.
H* = 25 21. 440 2i 192 419 = [22(2N/4)]U5= 22/591/20
ies _ 79/20
core see + 2(20 x 5) + (20 x 6) 22. er Gen = d((—27)282)) = AG aoT a)
(2723
= 24 + 200 + 120 = 344 m? = 36
Be te, i Og LORE E5449 000 cm? 23. 4 cy? |xy = (ey (ey = eye oe ie
m 1m
= xy
fie Wes Seed
m b 24. sf a [7 — sum
bp + bzy — bem = am
bzy = am + bcm — bp
3V2
V7 , SV7 V2 _ ong
i a2 AD
_ am + bem — bp _ 614 , 3514 _ 126V14 _ _ 85/14
by 4 14 14 14
Algebra 2, Third Edition 165
Problem Set 64
(476,158 x 1027)(79,318,642) PRACTICE SET 64
25.
(983,704)(514.0 x 107!*)
Dos
eeee RaIimnz
ae
orl (Sal ay 1
es
aG
8 x 107 Ke
(1 x 10°)(5 x 107!”)
(ga
1 + 2mz neh ot Ss2
=D +
1+ 2mz Ki 1+ 2mz
26. Volumecone = $(Ateagase) (height) = 127m
Z 1+ 2mz
—areh = 12am? (1+ 2meb b+ Rbmz tz
1+ 2mz 1+ 2mz 1 + 2mz
r2 = 6m?
3
=—+
S
pt
Sen
r= 6m XN
[a
+
Z Se
aet
z z
27. 12(90 — A) = (180 — A) + 20 HA,
1080 — 12A = 200 -A £8 glee SS Zi Byhint ra eae ee
Zz Sz + 25 _ be Zz 5z + 2s
-11A = —880
vd 5z + 25s
A = 80° (Sz + 2s)s . (z)3z
(Sz + 2s)z (z)\5z -& Zs)
28. Write the equation of the given line in slope-
Dz as
intercept form.
KZ D'S)
Sat aye]
2s? + 5sz + 3z2 (2s + 3z\(s + z)
aes, eee Z(5z + 2s) S z(5z + 2s)
arr as
PLP S34 Sees
Since parallel lines have the same slope,
= (ii)(ii) — S5i(ii) — 3V3iV5i
5
a) =--xre +b = (-1)-1) - Si(-1) - 3V3V5(ii)
153 = 3V (SGD 1 57 2 BAIS
5 (1 + 3/15) + Si
ibe + b
—7 = —-—(2)
(3 + 5i(4 — 5i) = 12 — 157 + 20: — 25:2
rey
Z 12+ 5i-— 25C I=. 37 5?
oo 2 PROBLEM SET 64
et OD) aay 4
CoeF
Peete
Ae 12x = 352 7 4 300 = is
Lse= 9x 2
k = 600
De fy FG
600
C=——
05 = 12 00 grams
30. PEt eX Si
3 7
28x + 35 — 3x = 42
ie
i 7,
D5 7) 700 _ 2800
7 400,
x= —
25 T, = 1600K
166
Algebra 2, Third Edition
Problem Set 64
3. Iodine Py + Iodine Dy Iodine Total Dre eop ait ie Daf4 Rae?
0.3(Py) + 0.8(Dy) 0.4(50) = 9i(ii) — 37) Gi) + 41 + 2i-/2i
J
(a) 0.3Py + 0.8Dy = 20
= =O — 3 + 4] 2-2 =) 518}
(b) Py + Dy = 50
Substitute Dy = 50 — Py into (a) and get: 10. V-4 + J-2/-2 - 4
(a’) 0.3Py + 0.8(50 — Py) = 20 = 21+ /2i/2i -— 64 = -66 + 2i
-0.5Py = -20
Py = 40 Was bcus irre EDL Ai
(b) (40) + Dy = 50 = (ii) = 3(a) — 2/213) = 4 + 2/6
Dy = 10
40 liters 30%, 10 liters 80%
12.—2.,/-9 area ae
S161 <e SG)Gi) + 24/34/31 ef = 3-* 13k
4. D,
coms OG)
D
“ = 10 - 6i + 15i — 977 = 19 + 93
RT; = RoTo: R; = 400;
Ro = 100; 7; + To = 40 14. (3i — 5)(2 + 4i)
400T; = 100(40 - 7;) =067 + 127 — 10 = 201 = 22 147
5007; = 4000
T; = 8hr 15. (i — 4)G + 2) = 277 +°4i — 47 — 8 = 410
D, = 400(8) = 3200 kilometers
16. 10 U
§. Carbon: SS Bo jel
Hydrogen: 3 ol =t3 R 10°
Iodine: ial) = 107
Total: 12 + 3 + 127 = 142 R =>10 cos 10° =. 9.85
lodine: 7" x 100% = 89.44% U = 10sin 10° = 1.74
142
2 %3
A saytip:
2 oes 2+ cs y 30
Cc Cc
2G 2m + cms + 2c 150°
— +- ——qK— =
4 2+ cs 2 ECS a see
R
eeoe: ee ee Se ey A.
& Ti As ,
R = 30 cos 30° = 25.98
¢ My U = 30sin 30° = 15.00
aes, 3a _ 2am + ms + 3a”
o> a+ a(2a + s) 9.85R + 1.74U
—25.98R + 15.00U
os 8 m 8 -16.13R + 16.74U
8. = + — He. ni
PY ep! «© —
Cc Cc
m 8c 2cem + m* + 8c?
= —
Cc 2c+m c(2c + m)
Algebra 2, Third Edition 167
Problem Set 64
17: 20. =3 2h =e
7
Ce }-3
(3 +g x-
a
—
3
ete
V47.
= +—Zi
0G
4 -
3M NAT
Se a
4 4
21. W = mir + b
Use the graph to find the slope.
D ul Sob: ih = = = —0.14
250
Since @ is a third-quadrant angle,
W = -0.14Ir + b
6 = 56.31 + 180'=9236.312
Use the point (1850, 50) for Ir and W.
2/13 /236.31°
50 = —0.14(1850) + b
3 1 309 = b
18. (a) ey
5% 3) = 8
W = -0.14Ir + 309
(b) 0.02x + 0.4y = 4.4
22. Area = Areatyiangle + Afeasemicircle
(a’) 15x — 2y = 280
(b’) 2x + 40y = 440 = yi + Ley
e-0-(9
2) 2
20(a’) 300x — 40y = 5600 2
(b’) 2x + 40y = 440
Dy
302x = 6040
x = 20 Bey
4
(b’) 2(20) + 40y = 440 ye 119
40y = 400 2
y= 10
1 119 1
(20, 10) Area = |Fo
—(5 5 + —1(3)*
5 (3)
19. =
2 119 +
9 )
—7jm
5
4 2
és
iw)
+
| & + ll |
—N <7 n
Wl]
23.
400 cm? lin. 1 in. 1 in.
5
teeee
je] ————_
Ss
x
2.54 cm
x
2.54 cm 2.54 cm
9 3 9 60 s f 60 min
( al 14 1 min 1 hr
SE Sp ae
3) 9
_ __ 400(60)(60) in
poh ee (2.54)(2.54)(2.54) hr
3 3
24. 08/3) 82218 oo [32(3¥3)]¥3 = 328419
eee
x= —-— + ——— _ 37/9
3 3
168
Algebra 2, Third Edition
Problem Set 65
56x = -15x? — x?
25. 2|2 = a + 2140
x? + 15x? + 56x =10)
_ 27V5 __ 3V5 V7
"Ss ys rae“ee
mse 7)(x: +28), =A
x = 0,-7,-8
_ 14V35 _ 15V35 | 140/35 _ 139/35
35 35 a 35 (146,842 x 107)(0.0007892)
30.
a(b + c) d (96,478 x 10'*)(0.000712 x 104?)
26. See
— (1x 107)(8 x 1074) 0754
= ——— = 1 xX ]1
abf + acf — fmx = dx (1 x 10!°)(7 x 1078)
abf + acf = dx + fmx
bye acl.
d+fm PRACTICE SET 65
RpTp = 288
27. eae
a(b+ec d
(R12 — Te) = 288
abf + acf — fmx = dx 36Rc — 3RcTc = 288
acf = dx + fmx — abf 36Rc — 3(108) = 288
o = et fmx - abf 36Rc — 324 = 288
aera
36Rc = 612
(a) x -— 3y = -6 Rc = 17
—3y
= -—x - 6
Rp = 3(17)
1
y =<"3" + 2 Rp = 51
(Bb) 2x + Sy = 15
(51)Tp = 288
Sy "= "2x 4+ 15 288
ji es
et a)
96
‘pe
Pi y17
96
— +T,= 12
17 q
96
Te =. [homme
- 17
Ta
108
C7 17
PROBLEM SET 65
Substitute x = 3y — 6 into (b) and get: Si = kP
(b)) -2y = &) + Sy = 15 400 = k(100)
lly Il NX
"NN A=k
y Si = 4(12) = 48 kilograms
@ x= 3[2)-6= 5 2. P\Vi = PyV>
11
800(2) = P(0.1)
ea
11°11
P = 16,000 N/m?
Algebra 2, Third Edition 169
Problem Set 65
3. 0.84(500) -— (E) = 0.8(500 — E) (a) RT; = 120
420 —- E = 400 - 0.8E (b) RyT> = 120
—0.2E = -20
E = 100 milliliters (c) Ry Il bdoa wo
4. (a) Ny; Ne + 400
(d) T; + Tp = 6
(b) N; = 4Nz + 100 Substitute (c) and (d) into (a) and get:
Substitute (b) into (a) and get: (a’) 2R,(6 — T>) = 120
(a’) (4Nz + 100) = Ng + 400
3Ne = 300 Substitute (b) into (a’) and get:
Ne = 100 (a’”) 12R, -— 2(120) = 120
(a) N; = (100) + 400 = 500 12R, = 360
100 were erudite while 500 were ignorant.
Ry = 30
5. Be x T = 1920 R, = 60; T, = 4; T; = 2
100
= 1920 x ~~ = 2000
96 + 4 be ce :
x aS oem iea-tyD
at+—
Phosgenecombined = 2000 — 1920 Cc Cc
80 kilograms
Cc acx+ bx +c
= ie db = ee
6. (a) RyTy = 8 ac +b ac +b
(b) RgTz = 8
(c) Rp = 4Rw 10. Fs it dey Soe ae ee
leona 3 ab +1
at+—
(d) Tg = 2 - Ty b b
Substitute (c) and (d) into (b) and get:
ser b _ 4ab+4+be
(b’) 4Ry(2 - Ty) = 8 € ab + 1 c(ab
+ 1)
Substitute (a) into (b’) and get:
(b”) 8Ry — 4(8) = 8 2
11. BG op g =xt g = Sy
8Rw = 40 oe a+-1 a+l
Rwy = 5 a a
_ axtxta?
a+1
12. 4 = [35-3 ta
= 2i — J3iV3i + 2i(ii)(ii) — 4
=2i+3.+2i-4=-1+4
Gets Te = 15
Substitute (c) and (d) into (a) and get:
13. (Si 2)Q7 33)
(a’) 5Rc(15 = Tc) = 600
Substitute (b) into (a’) and get:
= 107- 153 - 49 + 6 = ~4 — 19i
(a”) 75Rc - 5(105) = 600
14. (= B)G2r a4)
TSRe-= 1125
Rc = 15 = Di eae i6) we ae i
Rp = 75; Tc = 7; Tp = 8
170 Algebra 2, Third Edition
Problem Set 65
is 5 '
Ws (a)
a) 5x
=x - 6
~y= 1
20 U (b) 0.3x + 0.07y = 0.84
(a’).12x —Ty = 42
a (b’) 30x + 7y = 84
4 rit See vies
a 42x ie 126
x=3
R = 20cos 45° = 14.14 (a’) 12(3) - Ty = 42
-Ty =
U = 20sin 45° = 14.14 6
Y= i=
210° d
6
(2-3)
18. =) = 3x2 = 7x
De PL 2
R = 10 cos 30° = 8.66 aay a oe mes
U = 10sin 30° =~5.00 ch aed
3 36. -371.66
14.14R + 14.14U (é a 3
++OAT o/s oe
Xx = ==
“SA8R
é 9.
x + a = PME:
6 6
16. 1 JB
a
6 6
19 2x7 = 4 Fe =5x
fi + Sue + =2
p
xt 2 + i =2 =
2 16 16
(BE a
) a 57
16
ef
Sudo =
v7.
H = 3? +9 nae
4
x= 2) ia V7
H = 34 4°
5 20. K te= mRa
+ b
tare =: —
i Use the graph to find the slope.
@ = 59.04°
m = ee 0.89
} ; 560
Since @ is a fourth-quadrant angle,
K = 0.89Ra + b
a aan sa Use the point (2100, 500) for Ra and K.
134 /300.96° 500 = 0.89(2100) + b
-1369 = b
K = 0.89Ra — 1369
Algebra 2, Third Edition 171
Problem Set 66
21. Since the measure of an inscribed angle equals half 4 3 4(x — 2) 3
30.
the measure of the intercepted arc, Pnen ee aOa
mAB _ _ 4x
- 11
2 ~ PP4
mAB = (2y)°
Since the measure of an arc of a circle is the same as
PRACTICE SET 66
the measure of the central angle,
x° = mAB 1 * 3m 2 1 Fe 3m
x° = (2y)° z-2 -z+2 z-2 z-2
1+ 3m
600 12m 12 in. 12 in. 1 hr ‘ Caan
22. x x x -
hr 1 ft 1ft lft © 60min
— 600(12)(12)(12) in? Shit), Ae 2 eet ee
(60) min —-x -—4 x+4 x+4 x+4
23. eae [x2y32]12 = xy3/4 oo
a le ee
r x+4
24. 42/2 a “92 519 = [22(21/5)]u6 = 21/39/30 _ 2x -14 _2x+14
— 211/30 — yt 4” x+4
2 9 Since this is a 30°-60°-90° triangle,
25. ee Spee4 50
9 2 v50 \3SF =7
v2 Cae. ar aint Nee Fa) Fi3
ies ae SF = —= = ——
NO 3
_ 22 27/2 60V2 _ 852
6 6 On ae 6 ANB
aa CE (1)
i d
26.
abe Oa ae oa8
3
fx — abfm — acfm = abd + acd
fx = abd + acd + abfm + acfm b = D3
abd + acd + abfm + acfm 3
=
f pean
3
5 d
27. ———_ -m
ZO ec) ih
fx -— abfm — acfm = abd + acd
PROBLEM SET 66
fx — acfm — acd = abd + abfm
Co ue
fx - acfm — acd U
=b
ad + afm
soe
20
28. (x — 8)(x
- 4) x(x + 5)(x + 9)
100 = k
x(x + 5)(x
- 4) (x — 8)\(x + 7)
£ ED 100
Co = + oe 50 grams
LEES:
29.
(47,123 x 10°)(980)(476) AY _ BY,
(0.00134)(576 x 10°)
T, qT,
_ (5 x 10’)(1 x 10°)(5 x 102) (400)(6) _ (800)(60)
~ (1x 10)(6 x 107) 200 L
v 4x 101°
T> = 4000K
172
Algebra 2, Third Edition
Problem Set 66
3. Arsenic Py + Arsenic Dy = Arsenic Total 9, Ag = mr 9x in.”
0.36(200) r
3 in.
(a) 0.6Py + 0.2Dy = 72
Since the radius of circle Q is 3 in., the radius of each
(b) Py + Dy = 200 semicircle is 1 in.
Substitute Dy = 200 — Py into (a) and get: mr 2nr2
Ashaded = Boo ————=
(a’) 0.6P, + 0.2(200 — Py) e472 2 2
0.4Py) = 32 1 mr
ey ater5
Pr, -=-80 93 @ D
(b) (80) + Dy = 200
_ 9nin? = xin.’
+0 2
Dy = 120
80 liters 60%, 120 liters 20% = Spine = yomin? = 4r in.”
D, 2
4.
|10 — Mt (a) R7Tr = 300
Dp (b) RpTp = 1200
RcTc + 10 = RpTp; Rc = 4; Rps= 6; (c) Rp = 8Rr
Tc = Tp
(d) Tr = Tp + 3
4Tp + 10 = 67p
Substitute (c) and (d) into (b) and get:
10 = 2Tp
(byaBRe(Tp 3) et 200
Shr = Tp
Substitute (a) into (b’) and get:
5. Carbon: a= 2 (b”) 8(300) — 24R, = 1200
Hydrogen: 3x 1=%3 ARTF 1900
Bromine: 1 x 80 = 80
Re=es0
Total: 12 + 3 + 80 = 95
Rp &400; Ty= 6; Tr=3
80
_ Br
95 950 11. (a) RpTp = 624
Br = 800 grams
(b) R7Tr = 364
4 1 “x Ta - 1 A Ta (c) Tp = Tr - 4
x + 3 —-x -3 x+3 x +3
(d) Rp = 4Rr
sae Heel |
= os
Substitute (c) and (d) into (a) and get:
(a’) 4R7(Tr — 4) = 624
7 x+5 DXe— 3S KES 2x=5
> 843 72-3 =x > B43 x +3 Substitute (b) into (a’) and get:
Soe (a”) 4(364) — 16R; = 624
xt 3 -16R7 = —832
Rr = 52
8. is 2 XG
Rp = 208; Tr = 7; Tp = 3
Yio 42 42-2
Z 4 2% +2) i
Area = n(3r)" — nr?
eee x -4 ;
= Or? — ar
_ -2x% - 4x +4
7 x 24 = 82r* m2
Algebra 2, Third Edition 173
Problem Set 66
1 18.
13. ax + ie ea
a+ —
Xx Xx
x atx? +ax +x
= ax + =
ax + 1 ax +1
1
ie : pe oe 5)
i abs x x +1 eee
(rd
laa
akon
—fo»)
x By
m x mx” +mt x
3 x“ +1 x(x" + 1) 8 442
15. —/-3V-2 + J-4 - J-3J-3 - 28
Sea Oh Die 2 A iN 3 D1)
= (6 + 21+ 3+ 2i = (3+ V6) + 4i
16. (47 — 2)(33 + Si)
= 12i + 20i7 - 6 - 10i = -26 + 2i 19.
17. D (b) 0.012x + 0.02y = 0.496
as Pery] : (a’')"3m— 2y = —16
R
(b’) 12x + 20y = 496
R = 4cos 28° = 3.53 10(a’) 30x — 20y
U = 4sin 28° = 1.88 (b’) 12x + 20y =
42x =
baie
(a’) 3(8) - 2y = -16
10 U
2y =
y =
a a
R
20.
R = 10cos 35° = 8.19
NO
s + wwca) + i
U = 10sin 35° = 5.74 =, SS
3.53R + 1.88U = ae WwWS = | i]
8.19R + 5.74U
11.72R + 7.62U
% + ll
TN
Te WN
Ww gs
oN
| he
174
Algebra 2, Third Edition
Problem Set 67
21. 5x7 = otic
OR tee, k
= Cc
(P4x4 )-4 m d
5 dpx — dy — cdm =
Pht hd eee gad dpx — cdm — km = dy
E+ 5. ta dpx — cdm — km
( Ly 21 d
DO aaa =.=
2 20
m k
Pe spear 29. =-c
+ —
2 25 preay d
aed V105 dm + kpx — ky = —cdpx + cdy
2 10 kpx + cdpx = cdy + ky
— dm
_ cdy + ky — dm
22. Since this is a 30°-60°-90° triangle, kp + cdp
2SF
=7
(40,213 x 10°)(748,609 x 1073°)
wrea 30.
(0.164289)(506,217 x 107)
2
_ (4 10)(7 eas)
7 7/3 = 3x 10°72
P = 503)
—(//3) = —— , (2 x 10°')(5 x 107)
7 7
y =
5)
(1) =
5 PRACTICE SET 67
23. Mg = mCa
+b
3 +7.
1 meee Sig
RITA
awl 3et
Yaiay 2
Use the graph to find the slope.
3/2
- ¥3 15/2 — 53
= 1000 = 66.67
15 + 73 3V2
3/2 - V3 92) —3
Mg = 66.67Ca + b
_ 15¥2
- 573 _ 3/2 a3
15 3
Use the point (36, 2400) for Ca and Mg.
2400 = 36(66.67) + b
PROBLEM SET 67
0=b
1. R=KkA
Mg = 66.67Ca
2000 = k(12,400)
10in2 2.54cm 2.54cm _ 2.54cm
ee x —— x ——
hr 1 in. ire tn:
Lhr _ 10(2.54)(2.54)(2.54) cm?
3000 = (= 4
60min (60) min 31
18,600 Danes = A
as, Va/2 42 = [x!/2(x)] 4 = 1/6,1/3
xl/6, = 1/2
Te es le
a P\V\ =
PA 6d |bce zk xl/I5 1/12 os 3/20
(800)(300) = (1200)V5
V5 = 200 cm?
0 ae ~ 6/40
3. 0.2(400) + 0.8(Dy) = 0.32(400 + Dy)
V5, 15
32 as 2 4 — 1210 80 + 0.8Dy = 128 + 0.32Dy
3510_ 120710 _ _ 79/10 0.48Dy = 48
S
Dy = 100 pounds
— i=) 10 Sige 10
Algebra 2, Third Edition 175
Problem Set 67
(a) Ny = Np + 75 Substitute (b) into (a’) and get:
(b) 8Ny = 10Np + 140
(a) 6(295) — 12Rp = 1062
-12Rp = -708
Substitute (a) into (b) and get:
Rr = 59
(b’) 8(Np + 75) = 10Np + 140
Rp = 354; Ty = 5; Tp = 3
460 = 2Np
230 doctors = Np
12. i GCithannen bE aca
(a) Ny = (230) + 75 = 305 nurses Sate
b b
ab 4bx”
+ 4ax + ab
hulk <a) = 4x + =
100 bx +a bx +a
Pos = 400 xX = 2.5
60 13. ON (eo
me ed
ihe ee
NR = 400 — 240 = 160 grams x x
= x? amx + a + x?
-42- V5 Gs m+1 mx +1
-4+ 1/5 -4-5
S825 =8 = 2N5 14. (5 = NG +. 20)
1Ox=55 11 = 30 + 10i — 6i — 27% = 32 + 4
1 2/2
- V3 15. Aa 0 ae
2V2+ V3 2V2 - V3 =)25237 FA Dial BAG)
_ 2N2= V3 _ 2V2 - V3 = Dp = By) SS Oy a Ghee A
oe, 5
16. 27s
2. 3V5+3_
65 +6 a
eS
\W
35 ~3 BAS 3 — 4554.9
5 td S 10
6
Ae Se he 3 R = 10cos 37° = 7.99
x —2 22 =X x —2 x —2
10 sin 37° = 6.02
. der 5
~ x -2 227°
4 2
10. +
x2 — 9 x -3
4 , 2% + 3) _ 2x + 10
x7 —9 x2 —9 x2 -9
R = 8cos47° =
uy 5.46
11. (a) RpTp = 1062
U = 8 sin 47° u 5.85
(b) R7Ty = 295
-7.99R — 6.02U
(c) Tp = Tyr - 2
(d) Rp = 6Rr -13.45R - 11.87U
Substitute (c) and (d) into (a) and get:
(a’) 6R7(Tr — 2) = 1062
176 Algebra 2, Third Edition
Problem Set 67
17. 20 3x2 + 5 = 2x
(2 +Sr4 )--
3 3
ee) 1 5 1
xe + =—X + — = —— + =
3 9 3° 29
( i 14
x+—-] = -—
2 9
1 IAS
og oe ey
3 .
5 = 1, 914,
3 3
21. Since this is a 30°-60°-90° triangle,
tan@ = - V3 x SF = 7 af
6 = 36.87° SF = = 73
Since @ is a third quadrant angle, 5A. 5/3
@ = 36.87 + 180 = 216.87° oe 0 =macs
5 /216.87° ye a ov we
1) = = .) = oo, eAO0IGiieeS Lined fanpop tein
(b) 0.2x — 0.04y = 2.4 eet 2.54 cm sae cm a cm
(a’) 10x - 14y = 0 * 60min (2.54)(2.54)(2.54)(60) min
(b’) 20x — 4y = 240
23. Ag = mAu + b
2a’) —20ceF 2ky = 10
Use the graph to find the slope.
(b’). 20x —. 4y.= 240 8
24y = 240 ih org = 0.2
p= We Ag = -0.2Au + b
(b’) 20x — 4(10) = 240 Use the point (120, 2) for Au and Ag.
20x = 280 2a SC120Ee E
x= 14 Die hh
(14, 10) —= -0.2Au + 26
19. -x = -1 - 4x? 24. = ¥2%2?= [2(2'5)]!2 = 2'2Qu6
(ares, ic \ ~ioe
x* —-—x+ = ——
: : | 95. pe =a «nie eect
x2 7" it a = Fi at mn = xADyl5p'5 1323 — Dis ie a
(x-4) --8 26. a[2 + 2] - 2097
8 64
wre oF SA et | 2v11 V3 6 is
Secu tat 8 Vil vil V3. V3
jee 15" 12V33 . 22V33 198-33 _ _ 164/33
fot alga ge | aie alk | ee ee 88
Algebra 2, Third Edition 177
Problem Set 68
' y
a(b + c) m 2
27. ——— - =p
x y Ela
Beveatitd
aby + acy — mx = pxy Giainkaa
7+
aby + acy = pxy + mx att tt
aby + acy _ * ae een x
py +m
38. pee NAS, Ss hs =p
a(b + c) m
mx — aby — acy = abmp + acmp
mx — aby — abmp = acmp + acy
mx — aby — abmp =
amp + ay
PRACTICE SET 68
29. Graph the line to find the slope. 0.00042 x 10717 4x 1072!
a. Estimate: —~oregit ae © aRIO
568,425 x 10 6 x 10
Md
= 7x 10 —32
RERENPZean Calculator answer = 7.3939 x 107?
0
ep ageeneoe
REZSR aot eld b. Estimate: *4/156 —= 34 = 81 and 44 = 256
so the answer should be between 3 and 4.
Calculator answer =~ 3.33
c. Estimate: (0.00042 x 107!’)(568,425 x 10°)
uv (4 x 10°7!)(6 x 10!°)
2 <i’
Slope = S Calculator answer ~ 2.39 x 10710
f)
d. Estimate: (1.86) *°© —= 2-5 = 0.03125 and
| Eg b 1-* = 1 so the answer should be
eee between 0.03125 and 1.
Use the point (2, 4) for x and y. Calculator answer ~ 0.049
3
Ae D4
7?) PROBLEM SET 68
34 = 1. Dike
ye 800 = k(9600)
a
Bi a
ane
1
De po) = 2000 dastards
Dd PiV, = PxV>
(1200)(200) = (1600)V>
V, = 150mL
178
Algebra 2, Third Edition
Problem Set 68
0.4(800) + 0.2(Dy) = 0.36(800 + Dy) 10. x + (2x + 10) + 54 + 89 = 360
320 + 0.2Dy = 288 + 0.36Dy 3% = 207
—0.16Dy = -32 Xi= 69°
Dy = 200 pounds
x = mMN
= 69°
(a) Nr = 2Ng + 14
1 2+ J2 24+ 2
11.
(b) 2Ne = 20Nzr — 100 2-42 2452. Gee
Substitute (a) into (b) and get: _ 24+2
(b’) 2(2Nz + 14) = 20Np — 100 a 2
-16Ne = -128
Ng = 8 12.
4 342 +1_ 12V2 +4
2a 8 | ee
(a) Nr = 2(8) + 14 = 30
_-—
12/2 +4
8 were ephemeral while 30 were fugacious.
80. x T7="6 720 13.
2 3/3 +5 _ 63 +10
100 ASCEea es hah
T 6700 2 = 3/3 +5
80
T = 8400 14.
—7 2x
et eo
(a) 9:000418 x ‘Aiaege 2. ah(age es hae) 26 Gye
Ge PAL
501,635 x 10° 5 x 10!! Wage g, Py 9. FPG
= 8 x 10-2
Calculator answer ~ 8.33 x 107° 15.
2 3
Poy 4 16
(b) (0.00037 x 107!7)(7231 x 104) -2(x + 4) 3-11-22
= (4 x 10°!7)(7 x 10’) = 2.8 x 10° ont lb eadasl—
16 Vere 16
Calculator answer ~ 2.68 x 107°
16. (a) RgTg = 140
(a) *8192 —= 34 = 81 and 44 = 256 (b) RgTz = 140
Answer should be between 3 and 4.
(c) Rg = 2Rg
Calculator answer = 3.99
(d) Ip = Te =)
LG) oe eet anes =. 0,125 Substitute (c) and (d) into (b) and get:
Answer should be between 1 and 0.125. (b’) 2R¢(Tg — 7) = 140
Calculator answer ~ 0.14
Substitute (a) into (b’) and get:
Since this is a 30°-60°-90° triangle,
(b”) 2(140) - 14Rg = 140
1x SF = sae
-14Rg = -140
SF = ome)
Rg = 10
x = 2/3(2)
= 4V3 Rg
= 20; Tg
= 14; Tg = 7
y= 2/3(/3) = 6 m m
17. 4r + = 4r + 5
Ce Par, MLL.
Since the area of a triangle is equal to half the
all triangles x i
product of the base and height and since
in the parallelogram whose base is AD share the mx 4rx* + 4mr + mx
same height, all have equal areas. seis oes veD) BS ye ke Be a, s/) Oe,
Algebra 2, Third Edition 179
Problem Set 68
re 23. 2x = =1)\- 4x?
18. 3a + ———
= 3a + Se oui tl f ae
I
A ie CaS: (« oe
a a
2 3 Pye} 1
2) See 1 ae41 1
= 3eee cm: _ 3a ae ax x aa 16 16
+
Gmaet G a x ( i 3
19. (Si — 2)(3i + 4) 4) Wer
B
; go es oe
PGR une 6) 8 = 223 Melal 4 4
20. 22 - 3i7 - J-9 + 24 +4 aos 13,
= /2i/2 — 3(ii) — 3i + 2(ii)(ii) + 4 4 :
Oy 3 tet ad = 9 24. 3x2
+ 5 = -2x
21. 315 ° x2 —
3
2. + )—
3
3
= ») ice Sila
EN, ~ x? = ot SS See
3 ees
Se)1 ee16
2
40
U (: 7] 9
3 3
R = 40cos 45° = 28.28 1 4
1 = —hie—
U = 40sin 45° ~ 28.28 Bmulds
5
x= -,-l
10 3
U
aaa a 700 cm? 1 in. 1 in. 1 in.
R 25. ———— xX x x
min 2.54 cm 2.54 cm 2.54 cm
R = 10cos 24° ~ 9.14 , 60min _ 700(60) in?
PER sis apn 1 hr (2.54)(2.54)(2.54) hr
28.28R — 28.28U 26. 11,35
CuWie) = eG ee = 31/73 1/21 a 34/21
9.14R +— 24.210
37.42R
4.07U og h
: Ixy? [x>y a Gy)!4 ery)? = aye yea
= x17/6 13/6
= y
b} 6)
22. (a) =x —= =y = —2 D
Ory = 3) 28. 3/2 — 53 + 224
(b) —0.06x — 0.4y = -7.2 : 2) V3 SO Rie
(a) 6x — 10y = -30 V3 V3 2 2
(b’) -6x — 40y =.=720 _ 6V6 _ 15V6 | 246 _ 56
—50y = -750 6 6 6 2
y = 15
29. 8
ES elo
(a’) 6x — 10(15) = -30 1 (ae) a
6x = 120 my — apx — bpx = acxy + bexy
Poe, my — apx — acxy = bexy + bpx
my — apx — acxy _ b
(20, 15) cxy + px P
180 Algebra 2, Third Edition
Problem Set 69
30. Graph the line to find the slope.
Length of side = ——
Ty = —
782
ynits
V2
2
Area = s? = 7N2 = a ernites
2) 2
ome
it
400 = ‘ss
5
2000 = k
= hy = 80 tons
25
(a) 0.8(Py) + 0.4(Dy) = 0.64(2000)
Use the point (—8, 2) for x and y. (b) Py + Dy = 2000
9
Ds m8
13 )+b Substitute Dy = 2000 — Py, into (a) and get:
(a’) 0.8(Py) + 0.4(2000 — Py) = 1280
46
=F 0.4Py = 480
13
Bx = 1200
13 13 (b) (1200) + Dy = 2000
Dy = 800
PRACTICE SET 69 1200 mL 80%, 800 mL 40%
FY, _ 4%
Dg
T, T,
ToP\V, = T\P2V2
eee e|
Dr
Ty = TiFV)
R7Tr + 40 = RpTp; Rp = 50; Rr = 70;
FY,
_ (300 x 10°)(0.00005 x 10!*) TR = Tr +2
(50 x 102)(0.008 x 10+) 10Tr + 40 = 50(Tr th 2)
(0.000014 x 1078)
(50 x 107)(0.008 x 1074) 20Tr = 60
= 525,000K Tr = Sh lev
Since the train started at 2 p.m. and took 3 hours to
get within 40 miles of the bus, it was 5 p.m.
PROBLEM SET 69
L didi 4 ios l _ 2x-3 _ 1 ee
T, T, x-5 5-x x-5 x-5
Ty PLV,T.
e224 _ 2x -2
PY, gees,
_ (0.0008 x 10'°)(0.013 x 107'7)(800 x 10°)
(80 x 10°)(0.0005 x 107’) 2x +3 DX " 3
80s 40 8.x 107) x-2 B=" hp —-x+2
Fy (8 x 10°)(5 x 107!!) ZL AMSY ULE 3 6
T> = 2.08 x 10'°K
Algebra 2, Third Edition 181
Problem Set 69
- 16i - 4i + 8 = -20i
3x45 +
o) 15. (4i — 2)(2i — 4) = 8i?
x—-5 x?
— 25
16. 220 2h
ns (3x + 5)x+5) Z
= Ji) + eLGD
a5 1 25
3x2 4+15x + 5x+25+2 = 2-2)
+ lipid aet
ae 05
17.
3x? + 20x + 27
x? — 25
4 efit 252 12 Aa
R = 20cos 30° =
SFnOe Bae 2 Cee)
12 + 4/2 U 20 sin 30° =
7
10.
2 ane Sit 32 10 + 62
et oe? 25
=18
10+ 6V2
7
11.
2. 3+2V8
_ 6+ 8V2
BS 33+ 28) - 9— 32
_ -6- 82 60 cos 30° = 51.96
23 U ll 60 sin 30° = 30
12. (a) RoTg = 171 17.32R + 10U
-51.96R — 30U
(b) RrTp = 171
-34.64R - 20U
(c) Rr = 3Rg
18.
(d) Tr = Teo
Substitute (c) and (d) into (b) and get:
(b’) 3R¢(Tg - 6) = 171
Substitute (a) into (b’) and get:
(i) 307) 18h na wd
EIShp = 340
Rg = 19
Rp = 57; Tg = 9; Tr = 3
1 1
13. m+ =m+ —;
ieee m + 1
m m
m m>
+ 2m
=m t+ 5) — ,)
m~ + 1 m +1
a a a D v 56,316
14.
as pin me AOe: aa ax +1 Since 6 is a fourth-quadrant angle,
a a
6 = (56.31) + 360 = 303.69°
Siar
x as
ax +1 113 /303.69°
182 Algebra 2, Third Edition
Problem Set 69
2, 2 3 : : ; ;
4 ft . 12 in. y 12 in. ’ 12 in. 60 min
19, a) —x + —y = 28
23.
a) get 5? min 1 ft 1 ft 1 ft
x
1 hr
(b) -0.05x — 0.2y = 5.5 in?
= 4(12)(12)(12)(60) ——
(a’) 10x + 6y = 420 hr
(b’) —5x -— 20y = —550 24. Cr = mV + b
(a’) 10x + 6y= 420 Use the graph to find the slope.
2(b’) -10x -— 40y = -1100 ns zt nf ee
24
—34y = -680
Cr = 37.5V +b
w=220
Use the point (62, 200) for V and Cr.
(a’) 10x + 6(20) = 420
200 = 37.5(62) + b
TOx-=4300
—2125 =b
ce Shi)
Cr = 37.5V — 2125
(30, 20)
25. 12753 = 1348 = [QM]!
20 3x7 + 8 = 5x as 33/231/8 = 3153/8
i = sit 25 }= eas 3 Pees) 4 ipne: es (xm>)!3(xm?)!4
3 3 26.
= 4/3 5/34 2 ~ x 7/12, 13/6
ono.
xe x
325
tS See
emerges:
3 36 3) 36
( a 71 21.
i oS ===> 9
6 36
eee Loe ale)
x--=t
5 V71,I 42 o Ee SS
6 6
_ 18V2 | 10V2 _ 150V2 _ _ 61v2
53.4 11. J Xs 6 ou x 3
XS — l
6 6
28. eG Ore
21 3x2 + 8 = —5x m Zz
(# ot aie = al
axz + ayz — cm = kmz
3 3 axz = kmz + cm — ayz
kmz + cm — ayz
——————S——
az
x+2 x 3x—3 =
29.
5 4
4x + 8 — 15x + 15 = 40
~llx = 17
aes
5)
22. Since AABC is equilateral,
472.2 x 10726
mBC = mAB = mAC = 120° and 30. Re
1658.27 x 10AO,
BC = AB = AC = 10 -24
Use the Pythagorean theorem to find the height.
ee ae Sa 077
OSG 10)
10% = HH? +5? Calculator answer =~ 2.85 X 10-°’
H? = 100 — 25
H = 5V3 (b) (1.24)-2-73
Answer should be between 1 and 0.125.
Atrear= 10 x 5v3 = 25./3 units” Calculator answer = 0.56
Algebra 2, Third Edition 183
Problem Set 70
SET 70 5. Evenintegers: N, N + 2,N + 4
PRACTICE
Z 25 A(N\(N + 4) = -10(N + 2 + N + 4) + 28
m= of +2] 4n2 + 16N = -20N - 32
4N* + 36N + 32 = 0
mx = pz +2 ps
a Seam N+ 9N+8=0
ya + mx = |(x— + k) Pe e ek) (N + 8(N + 1) = 0
N= -8-1
y
kmx = +2 + 2k j
ymx TONGS EDS Pe PS Since the problem asked for only the even integers,
= 2
ykmx — 2kps = pyz + 2psx — ymx the desired numbers are —8, —6, —4.
k(ymx — 2ps) = pyz + 2psx — ymx?
=
Wa Dist PiNas 3ET
2
6 a y| ee
ymx — 2ps 2p b
x 3%
PROBLEM SET 70
paar H mean
2bmp = bx + 6pxz
eS
RY _
Se
Pa, 2bmp — bx = 6pxz
F T,
b= _ 6pxz
PLV5T,
We 2mp —- x
FY,
e004 x 10°)(500)(4 x Ee 103) : A t
= ar aT Fe = +=
* ~ (0.001 x 107!3)(0.04 x 10!) * {- +5 *)
(4 x 103)(5 x 107)(4 x 10°) a ny
Daa x0 eS
Ge 1°)5)4, & 10°) ac = . ieseee
13
T, =2%x 10~K ACER ACSZE= NXZ EP TLE SUX
2. 0.2(400) + 0.5(Dy) = 0.26(400 + Dy) DOSE: —~ SERS INE Tee ere
80 + O.SDy = 104 + 0.26Dy 5 = me bare er
0.24Dy = 24 acz — tx
Dy = 100 milliliters
8 SN By, pe py ete By
3. (a) 50N; + 100N¢ = 5000 “De ee eee ee
Nr = 60 _ 6x -2
(b) Nr +
Substitute N- = 60 — N7 into (a) and get: Mee
(a’) 50N; + 100(60 — N;) = 5000 5 A _ ana
-S0N7 = —1000 x7 +8x+12 x+2
pea _ atts*ayGade
(x + 6)(x + 2) (x + 6)(x + 2)
(b) Nr = 60 — (20) = 40 people flew first-class : 4 fsyawn 76 pag
4. Calcium: 1 x 40 = 40 ree Cenk Ny x” + 8x +12
Sulfur: 1 x 32 = 32 _ 4x* + 19x - 26
Oxygen: 4x 16 = 64 x? + 8x + 12
Total: 40 + 32 + 64 = 136
10 ait ; V2 +4 = 22 +8
Sep 8) 24) (OA SEG
136 ~—-680 vy as
S = 160 grams a ae
184 Algebra 2, Third Edition
Problem Set 70
2 3V12
+2 _ 12¥3 +4
3.12 2,40) 32 +0 PlOSEAE
_ 3v3 +1
26
12. 2 —2V2
+2 4/244
240
= oo Sig
= /2+1
R = 6cos 40° = 4.60
U = 6sin 40° = 3.86
13. (a) RpTz = 65
3.76R — 1.37U
(b) RyTy = 104 4.60R — 3.86U
8.36R — 5.23U
(c) Ry = 2Rz
(d) Ty = Tz - 1 19.
Substitute (c) and (d) into (b) and get:
(b’) 2Rp(Tz - 1) =
Substitute (a) into (b’) and get:
(b”) 2(65) — 2Rg = 104
-2Rz = -26
Rp =/13
Ry = 26; Tz = 5;
14.
>
ab xyb + xya + ab tan@ = 2
bt+a 4
@ = 51.34°
15. ll Since @ is a second-quadrant angle,
ay +1
+ +
|
< |
<
6 = —(51.34) + 180 = 128.66°
y bf
m xy © amy + m+ xy? (41 /128.66°
— eeSe
y ay +1 — -y(ay
+1)
20. (a) st = > =.-2
16. (2 — 3i\(5 — 6i)
= 10 — 177-— 151.4 18% =-8 —1271 (b) 0.07x + 0.3y = 5.5
(a’) 4x - 5y = —40
M7, —,/-4 - J-2-J-3
+ J-9 (b’) 7x + 30y = 550
at = 4)Zig 34. + St
6(a’) 24x — 30y = -240
= 446 + 31 = 46 +i
(Bb): 7x + 30y = 550
18. 340° SiG = 310
i Halt)
(a’) 4(10) — S5y = -40
Sy = -80
R = 4cos 20° = 3.76 yr= lb
U = 4sin 20° = Ie (10, 16)
Algebra 2, Third Edition
185
Problem Set 70
a0 3x2 + 1 = 4x 25. (a) 46,831 x 10-4 EBX i
een 1 9140.26 x 103 9 x 10°79
(: a Je ~ 6x 10°
ip oe - 4p. 1 4 4 Calculator answer ~ 5.12 x 10°
3 RON aye 9 (b) 74146
ae 1 Answer should be between 5 and 7.
(«3 2) ‘.9 Calculator answer ~ 6.33
ee 2 2 re 26. 5/ xy 4 xy = (x4y3)/5(x2y) 14
3 3 = xM/Sy3/5, 1214 _ 13/10y17/20
pee mea
3 27. {2 + 2,|2— 2126
1
r= 1,5 a MT M2 IT ee
(212 AT AF
oh 4x + 7 = -3x _ Wi4 | 4Vi4 — 84Vi4 _ _73V14
14 14 14 14
2 eM!
Oe, wir 4x* + 4x +442
28. x — 1)4x3 + 087= Ox = 2
Pang ater SA LARC 4x? — 4x?
3 9 3 9 Ax? + Ox
2 er tale Ane = Ax
eS IS a TG” ay 9
4x — 4
ta 2 = Nts YD
3 3 9. : alee :
2 V7. 2! Write the equation of the given line in slope
os = + ae i intercept form.
ax + Syr=e
23. Since this is a 30°-60°-90° triangle, y= ad + 5
= é| 3
1x EsSF =4
Since the slopes of perpendicular lines are negative
SF = 4 reciprocals of each other,
a = 4(J3) = 4/3 my = =
b = 422) = 8 ;
Ve ae 0
24. Co = mNi + b 4
Use the graph to find the slope. 5 = (2) +b
ree13 tele say
S =
Co = -1.54Ni + b yoireS
Use the point (138, 50) for Ni and Co.
50 = -1.54(138) + b 30; 22 eae
5
262.52 = b .
l6x
— 8 —- 15x + 10= 200
Co = -1.54Ni + 262.52 weeoton
186 Algebra 2, Third Edition
Problem Set 71
PRACTICE SET 71 3. (a) 0.1Py + 0.3Dy = 0.17(200)
(b) Py + Dy = 200
a. ax* + bx +c=0
Substitute Dy = 200 — Py into (a) and get:
ee ert 20 (a’) 0.1Py + 0.3(200 - Py) = 34
c
-0.2Py = -26
(2 +2r+ )-- Py = 130
(b) (130) + Dy = 200
a a
= — - — Dn = 70
130 mL 10%,70 mL 30%
ee: 14 4 § b? — 4ac
2a io Aa 4. Dr
Finn A b* — 4ac | D; |
2a 4a
: RrTpr = R,Ty; Rr = 8; Ry; = 24; Tr + Tj = 8
ee ah ree BT pe —-24(Ri—
Tp)
he 2a 227 =r 192
ya TO tafAe Tr= 6hr
2a Dr = Dy; = 6(8) = 48 miles
b. 2x7 - 6x +4=0 5. Hydrogen: pee lene
Bi Bob 6 eS? Carbon: Lote ake
- Oxygen: 3 x 16 = 48
ee TO) # (6)? =42004) Total, i 2 12-448
=62
2(2) A
u 6 + 36 — 32 7 642 ae Hydrogen: x 100% = 3.23%
4 4
6. See Lesson 71.
2. “hee = Ix GU
PROBLEM SET 71
2) £2)? = 4046)
1. PV, = P2V> ape: 2(4)
PY, 2a" we
_ 2391S
Les Te 8 8 2
ee (40,000 x 10~°)(4000 x 10-7) 8. eS |
8000 x 10 wr+x4+4=0
eet
gy NBER) = 10)) = 20 Nim? eo hE VO? = 44) _ = + 431
x 10°) (8 2(2)
2 PF = kP x= oe = V31,
; ‘ 4 4
240 = k(4800)
1 ane 360 — 280 = 40
— =k 9
20
. ee 80 _ 50
600 = (=)
au Since the diagonals of a rhombus are perpendicular
12,000 people = P bisectors of each other, z = 90.
Algebra 2, Third Edition 187
Problem Set 71
17. (-i — 1)-3i + 2) =°3i? —)2i #°3i - 2
10. =-5 +i
m 3m
r= =
x+c y 18.
rxy + rcy = my + 3mx + 3cm
rxy — 3mx = my + 3cm — rcy
20 U
_ my + 3cm — rey
4 ry — 3m
pe aes
11. R
= 20cos 70° = 6.84
U = 20sin 70° = 18.79
4 ea I2 oe 8 ANS
12.
jeniO oeNE
pt 73.3
13
13. (a) RyTy = 160 10 cos 40° = 7.66
(b) RpTp = 400 U =
i) 10sin 40° = 6.43
(c) Rp Ri 6.84R + 18.79U
(d) Tp = Ty + 1 7.66R + 6.43U
14.50R + 25.22U
Substitute (c) and (d) into (b) and get:
19.
Substitute (a) into (b’) and get:
(b”) 2(160) + 2Ry = 400
2Ry = 80
Ry = 40
Rp = 80; Ty = 4; Tp = 5
14.
x+1
b b
15.
a ab+a
b b
b @ = 45°
ab+a abt+a
Since @ is a fourth-quadrant angle,
16. —3 Fane aco ioeb oie 0) 6 = -(45) + 360 = 315°
= (EDEN OR Ol, mae
4,/2 /315°
3i + 4i - 3i = 4
188 Algebra 2, Third Edition
Problem Set 71
20.
3 1 24. f81V3 Heastal3 = [3473 42)] 1/2
ianh=xXee. - 5)
—y = 2
= 3231/4 = 39/4
(b) 0.06x — 0.2y = -0.64
25. Vx5y6 Jay? = (x5y)/3(,y3)1/2
(a’) 3x - 4y = 16
(b’) 6x — 20y = -64 = yy Ma, SP 13/6 7/2
y
-2(a’) -6x + 8y = -32
(b’) 6x — 20y = -64 26. 3/2 +3/3 - 6/9
1296
ane a5 375 V2 12/10
y=8
VERSE DAD
(a’) 3x -— 4(8) =
3x = 48
6VI0 |15V10 _ 12010_ 99/10
10 10 10 10
x= 16
27. =2) hes ey
21. C= An +4)= 2)
= x° — 2x” — 4x2 + 8x + 4x — 8
(# SSS FS }= =
3 = x> — 6x? + 12x - 8
: Sager 1 5 1
x*—-=x+—-= -—-+-—
3 9 3) sD Ping2 i ye Be
(x--| ih=-—14 ps Mh 5 i DN A Ue 2 Heel(SY
28.
3 9 a eS mel (Oy
eS 5) y
1
x-- = +—i
J14. y y x“ — 6y
3 3
» See ya 1
he ay
x=—+—i
4
x“ 93
- 6y
3 3
22. Since this is a 30°-60°-90° triangle, ~471,635 x 10°
29. a et
2x SF =11 > 0.0071893 x 107!4
Tae 2
mt
= —(1) = —11 Calculator answer =~ —6.56 x 10°
eo vs 2
11 11/3
b= 5 (3)3 . Answer should be between 0.125 and 0.037.
Calculator answer ~ 6.86 X 1072
23. Pb = mB + D
Use the graph to find the slope.
30. ee
i = te = Ae z y
2
by
= Xz
Pb = 36B + b
Use the point (4, 0) for B and Pb. b==
Vy
0 = 36(4) + b
—144 = b
Pb = 36B - 144
Algebra 2, Third Edition 189
Problem Set 72
PRACTICE SET 72 PROBLEM SET 72
. 0.52(500) — (Wp) = 0.4(500 — Wp)
260 — Wp = 200 - 0.4Wp
A 5
—0.6Wp = —60
a (A Wr = 100 mL
ay
-230° . 3.07°x AP = 31,314
AP = 10,200 things
sin 50° = a
5 sin 50° = A Of at5410
100
3.83 =A M = 8200 people
. (a) Np = Ny + 160
(b) 6N; = Ne + 40
Substitute (a) into (b) and get:
(b’) 6N, = (N; + 160) + 40
5N, = 200
N, = 40 pieces of jetsam
(a) Nr = (40) + 160 = 200 pieces of flotsam
sigs Seo
Yr
Py =
PT.
1-2
T,
|fol ||| re (500 x 10°)(0.002 x 10°)
C7 Ha alolWa ae 0.0004 x 107
100 102 104 106 108
ae (510524104)
7 2
Carbon in grams
4x 10°
Use the points (100, 80) and (106, 130). Py = 2.5 x 10° N/m?
(a) -80 = -100m — b
. There is never an exact solution to these problems.
(b) 130 = 106m + b
One possible solution is given here.
50-= 6m
S = mC + b
m= = = 8.33 Use the graph to find the slope.
3
3
m= — = 0.075
80 = 100 - ee +b 40
3
S = 0.075C + b
i) = a = —753.33
Use the point (40, 1) for C and S.
1 = 0.075(40) + b
H = 8.33C — 753.33
2=b
S = 0.075C - 2
190 Algebra 2, Third Edition
Problem Set 72
7.
12. Q ll =
|-
3 Sak
+
eS
tA: <_
——
a=—+-—
amymy = MX + M4xXy
amym), — MxXy = Myx
mix
2
my = ——
am,
— xy
R = 3cos 70° = 1.03
2x +3 3x —2
U = 3 sin 70° = 2.82 13. = Saar Ce
xX —2x-8 x-4
-1.03R + 2.82U
2x +3 4 Gx = 2) + 2)
8. (x — 4)(x + 2) (x — 4)(x
+ 2)
2x +3 + 3x°.+ 4x -—4
x - dees
= 3x7 + 6x -1
7 2y ees
14.
3-2-3420
_ 6 - 18V5
es ND = 34 08 eet 190
3+ 9/5
88
15. (a) RpTp = 693
(b) RcTc = 165
(c) Rp = 3Rc
Gil, Te 2
D v 18.43°
Substitute (c) and (d) into (a) and get:
Since @ is a third-quadrant angle, (a’) 3RATc + 2) = 693
6 = 18.43 + 180 = 198.43°
Substitute (b) into (a’) and get:
20/10 /198.43° (a) 3(165) + 6Re = 693
9, See Lesson 71. 6Rc
= 198
Rc = 33
10. 5x27 = -7 ~ 2
Rp = 99; Tc Si5j/Tp = 7
5x2 + 2x +7=0
_ 2+ 42? - 467) 16. ab + e = ab al:
2(5)
72 2+ -136 1, V34.
Li Vet
b b
10 5 5 b* ab> + ab + B*
magus br2 +1 ne Lie ks
Ged |
11. 3x2 + 7x = 3
3x2 + 7x + 3 = 0 ~u8n a 4 =
Li
y+ cyte tet
xy xy
2(3)
- ay xdy? + x2 + xy?
i a Pea) ie be
= 6 6 6 xy +1 xy +1
191
Algebra 2, Third Edition
Problem Set 72
18. (—3i — 5)(i + 5) 6 2/3 2 2 — 2135
S38 = Spe sae 22 0h
4
19a ay? eo 3S _ 23 V5, 4V5 V3 _ 6 ag
a
= —4i(ii) — 3(ii) + 3i —
ae P
V3 V3 (ii)
Asie Sit AE el) I siz
EAM 34 Or 4+3 26 + Ti _ 615 , 20VI5 _ 90VIS _ _64V15
iS 15 15 15
1 1
ce gaa ar 27, A? = 6 + #
(b) 0.03x - 0.4y = -1.64 A = 2413
(a’) 5x — 4y = 40 6 x SF = 10
(b’) Bx = 40y = =164 oF = e)
3
-10(a’) -50x + 40y = —400
ies aetna et 23 x 2 = B+ 213
-47x = 564 3
x= 12 Seren
6V13 _
1ovi3 _ eee
(a’) 5(12) - 4y = 40 3
x4. = 5p 4/13 _
3
a)
(12, 5) 28. 2)" 2? 2 ee ae
21. x7 -x+4=0 Set)
' l-2.4-0
= f= 4298 See
x? — =x + = -2
Z =) = 4%
5pte Racers ee) Be ee elle
Dee ey 16 2
ge
et4 et
= 16 29. Write the equation of the given line in slope-
elee 7 31 intercept form.
* aig ube 4y - 3x = 1
4: ies Gite Falag
1350 17 . Since the slopes of perpendicular lines are negative
Z2a(arx = oy 7 reciprocals of each other,
272 ee
— 40 m, foe
= Z
(b) x = Pie. = 116° 3
2
23, 34/2 = {22/2 = [22(2'2)]"5 y= Se 8
= 22/5910 _ 41/2
Okeee——(—7) +5
= 1/3)4 ee i:
a 74/33 -[-(8'°)*]= -[-(16)]=
3
25. 4 la5 yjay* = (aye Gy) 4 ; iF
us Syl? q7/4y9/4 y= ae ake
192 Algebra 2, Third Edition
Problem Set 73
30, (a)? 35123 x 104-4 xno? a a) a Nate
-798 x10 -gxi0l3 (a’) Np = 2Np + 4
=5 “ao”
(b) 7Np = 3Np -— 4
= 20
Calculate spewer 4 AO aLt Substitute (a’) into (b) and get:
(b) °9/243 (b’) 7INp = 3(2Np + 4) - 4
Answer should be between 3 and 5. Np = 8daisies
Calculator answer ~ 4.24 (a’) Np = 2(8) + 4 = 20 pansies
PRACTICE SET 73
5+ V3 2V3 _ 10V3
+ 2(3) R, Ty, a8 R7Tr = 256; Ry, = 16; Rr = 12
AER ee 4(3) Lye= Lyra
_ 10V3 +6 3 +53 16(ip +2) + 127 = 256
12 6 28T, = 224
bp, 212 = 3v3 2v3 + 3V2
Tr = 8 hr
’ 245. = B49. 231, # AG) Dy = 8(12) = 96 miles
ABOVE k
4. Nr SSS
2a — 63 Se.
2/5 Foy) tot:
A = 18 50 = =
— 9/6 +126
64436 2500 = k
BELOw Nr = ae = 100 frangibles
213 — 3/2
D9 Fad v @
12 -"6./6 5: rae es
- 6 — 18 1 2
2 ons = 6 ae Fe
6 + 3V6 _ Bakn1| V6Gi = meni M6& eet(0.08)(700 x 10°)
—6 —2 2 i
0.0004
es 107 )(7. x 10°)
|) a Sire;
a ae
PROBLEM SET 73 4 x 10r*
T, = 14 x 10'°°K
1. Carbon: Xe S24
Hydrogen: Sox 1 = 5 BS Dh
ee eA
Bromine: i x<7505=a80 = 5 Beals
Total: 24 + 5 + 80 = 109 _ 6+6V5 +2V5 +10 -2- V5
7 A =926 2
24 48
109 C,H.Br : V8 7 2/2 . 0 +
C,HsBr = 218 grams Mh phere) 5] tet 28
Algebra 2, Third Edition 193
Problem Set 73
H= 9% +4
8. Length of side = “all = 7N2 m H = V97
42 2
2,
4 tan@ = 2
Area = 2 = (=) = =m? 4
@ = 66.04°
9. There is never an exact solution for these problems. 197 /66.04°
One possible solution is given here.
12. See Lesson 71.
H=mC +b
13. 2x? +5 = 5x
Use the graph to find the slope.
2x? + 5x +5 =0
eh aa
5
a _ 5 £45 — 425) _ -S.t V=15
a 2(2) 4
H = -8C
+b
5, wis.
x= —— 2
Use the point (105, 80) for C and H. 4 4
80 = -8(105) + b Dx? — A= 55x
14.
920 = b ee Se 0
H = -8C + 920
SS
Ra
Be ee Oe ae
2(2) 4
10. 8 5. /57
U x=-—-—t
eet i 4 4
R = 8cos 20° = 7.52
15. re. ates : 2|
km, m,
U = 8 sin 20° =~ 2.74 xc = —— — = —
km, mM,
ckmympx = mypx — kmpx
ckmjmyx + kmypx = mypx
Bila ss
ckm, + kp
R = 6cos 40° = 4.60 3x2 4x3
16.
pao. ep eae
U = 6sin 40° = 3.86
22g et 3 eS
7.52R + 2.74U x= 2 x-—2 Bg 2
4.60R — 3.86U
12.12R - 1.12U 17. (3) Ral, = 160
(b) RpgTpz = 240
11. (c) Rp = 2R,
(dd) Ty +737
Substitute (c) and (d) into (b) and get:
(bo QR bak Se2a0
Substitute (a) into (b’) and get:
(b”) 14R, — 2(160) = 240
14R, = 560
R, = 40
Rp
= 80; T, = 4; Tp = 3
194
Algebra 2, Third Edition
Problem Set 73
18. gress 2 2 2. 80
feed
a= BS a-x = | ee
ax ax
ax? J ax
80 40
wines
fe a;
= ax2 _ a*x a>x3 =
gs Ted a*x —1
19. (3i + 2)Gi — 4) -— ¥-9 die aes
= 3ricel
28 eR lic
13 Sell r
os. 13; 2
20. (a) 2x + 3y = 6 pees
2 2
y = 3%
=-—x+ 2
c = 180 — 40 — 25 = 115
(b) x - 2y = 4
1 3 23. 3./913 = 3 32 4/3 = 3/37(i2) |?
=-x -
AE ae 3G!) = 39/4
-2? 1/4)3
24. Seis mele las
25. ae x9 pe aye i (a2x° yx ud 2)U2
= ayl2zli4y = gyl/4y3/2
26. 2,[2+ 3/8 - 20
Dao8)+ ——
ONS AS
2(a) 4x + 6y = 12
3(b) 3x -— 6y = 12
=
Roe
——— — -
ges caera 2,40
fe = 24
_ 10/40 , 24V40 _ 80/40
40 40 40
24
oS He
7 _ _46V40
_ _23/10
. 40) « 10
24
w) (|—]-2y
b) (+) =4
x(t Bix 12) (x — 10Ke4 3S)
4 27.
I ee (x —10)(x
+ 2) x(x — 8)(x
+ 3)
er
— sites
ae
alll,
oF
weap6!
ae 28. ya arth
21. x* +6= 3x
— 34.4
(7 = "-6 2 = —-—(-5)
+b
og I ae a ed Ol=5b
4 4
( Al 15
Nea | =
p: 4 1) ee
x-—--—-=tHt
3 45--l
2 2
ee Sen IE
2 2
Algebra 2, Third Edition 195
Problem Set 74
ke 4,168,214 x 10% 4 x 10° 2. Dp,
NT ) oo
74.612 x10" 79 x 10-4 -—
80
=~6x 10%
34 De
Calculator answer = 1.22 X 10 eet
os 30
(b) (4.01)>4*
be between 9 x 107* and
Answer should RrTr = 80; RcTc = 30;
ioe | a Rp = 4Rc; Tr = Tc - 2
Calculator
~
answer
¢ OU?
= 6.01 X 10 -4
=
4RATc - 2) = 80
30. 3x -—5 a = ay 3 4RCT¢ = 8Rco = 80
? 4(30) - 8Rc = 80
9x -— 15 -— 2x = 36 -8Rc = —40
ix sr R Rink
ey a
— 7 Rr = 20 mph; 7c = 6hr; Tr = 4hr
PVs PV
PRACTICE SET 74 3 pe = aa
Ds 1 2
———— PLVoT
120 LS ee
PY,
Dp
SS 7, we= (400
SOE x 10°)(500
ax 10*)(0.06 x 10°)
105 A (0.004 x 10°)(0.02 x 104)
RgTg = 120; RpTp = 105 7, = (4X 107)(5 x 10°)(6 x 10%)
Ry 2 Ril = Ty — 3 Z (4 x 102)(2 x 102)
RzTz = 120 T, = 15 x 104K
(2Rp Tp a 3) =a
2RpIp = 6Rp — nD 4. 0.92(2000) = Ex = 0.8(2000 = Ey)
2(105) — 6Rp = 120 1840 — E, = 1600 — 0.8,
—6Rp = -90 -0.2E, = -240
ee Ae E, = 1200 liters
(15)Tp = lis
Tp = 7hours —> Tz = 4hours 5. Odd integers: N, N + 2, N + 4
4(N + 2)(N + 4) = 20(N + N + 2) + 12
PROBLEM SET 74 4N? + 24N + 32 = 40N + 52
1 De 4N? — 16N — 20 = 0
a= 4(N* — 4N - 5) = 0
RyTr = 28; RpTz = 28: OE ne
Rp = 2Rp Ty + Tg = 12 N = -1,5
2Ry(12 — Ty) = 28 The desired integers are -1, 1, 3 and 5, 7, 9.
24Ry = 2R7Ty = 28
24Rp — 2(28) = 28 5 A ee
24R; = 84 " =V/3
= 2 30
Ry; = 3.5 mph _ -2V3 +44+3-2y3 _ ouaG
Rg = 7mph; Ty = 8hr; Tp = 4hr i. aa we
196 Algebra 2, Third Edition
Problem Set 74
3V2-4 V2+3 | 11.
J2-3 243
_ 6+ 9V2 ~ 4V2 - 12 _ 6 ~ 5V2
2-9 FE
4/2 -5 2+ 38
pete,his Bae
by3)
R42 + 48 10 30,/2
rae
_ -19
+ 11/2
34
9. There is never an exact solution for these problems. 5
One possible solution is given here. tan@ = rine !
N = mF + Db ra)
Use the graph to find the slope. Since @ is a third-quadrant angle,
se @ = 45 + 180 = 225°
m= st = 70
5/2 /225°
N = 70F + b
12. See Lesson 71.
Use the point (30, 50) for F and N.
50 = 70(30) + b 13. aa eas
~2050 = b ay + 4
N = 70F — 2050 ee lt jl? — 4034) _ =1 +4647
2(3) 6
10. = >
1, ¥47.1
6 6
14. xb 3x° = =4
3x7 +x -4=0
R = 4cos 45° = 2.83 epee {2 a 4(3)(-4) ei {49
x Sooo =
2(3) 6
U = 4sin 45° = 2.83
ol es
6m 6G 3
Leer
15. a= nf ~ “|
pe x
m km
SS
pe x
R = 6cos 20° = 5.64
acpx = mx — ckmp
U = 6sin 20° = 2.05
ckmp = mx — acpx
2 83 Rae) SOL :
-5.64R — 2.05U Be ey
m — acp
-8.47R — 4.88U
197
Algebra 2, Third Edition
Problem Set 74
Bie as Fe ES ay
a 1
16.
anaes.
m x, ih
a he be
1 1 A 1
oe
m x, te o) Soe ee a) a
ee OG. at ae
ax \X_ = cmXx, + bemx,
axX\X_ — bemx, = cmxy 6 36
cmx x=
1 Set
83.
38 SSS 6 6
: ax, — bem
pa eeRa Se
6 6
17. gees = Gee %
peakoe 22. (a) 4x - y = 45
a
Ae Bites ghee ae (b) 3x + 3y = 90
95 GR =
ax — x ax -x 3(a) 12x -— 3y = 135
(b) 3x + 3y = 90
18. Ci —3\@es) eso 15x =1225
= Or — Gi— 3i + 9sse8i it ili) per aket
0 6 Sie Ot 3h 31 eae SF (b) 3(15) + 3y = 90
3yr= 45
3x —2 a 1 y= 15
19.
x7 +7x+10 «+5 2 AW. ="2 9? 2 a OY ed We:
23.
- 3x —2 - x+2 = 202") = 29/4
eeTy P10) x? 4°7910
30 (30,2
2x -4 24. a = _[-(9")3] = 27
x? + 7x + 10
25. x*y 3V8y> x = (4x2y9)1/2(gy5,)18
3 2 = Ixy lay yk = 4x43) 25/6
20. (a) ities
a Sry
)
26.
6
planar 7 eG)
enc
(b)'3x + O0.5y = 17 7 6
(a’) 15x - 4y = 20 _ 2N6 V7 , 3V7 V6
iT ERE ci
(b’) 30x + S5y = 170
_ 12V42 | 2142 _ 12642 _ _31/42
-2(a’) -30x + 8y = -40 Soa SAD, 42 es
(b’) 30x + Sy = 170 27. Write the equation of the given line in slope-
intercept form.
4x -y=7
i) nn= | KRLae S — Il NOi=)
y=4cx-7
—~ —
— —
15x = 60 Since parallel lines have the same slopes,
y=4x+b
0 = 4-2)
+b
(4, 10) Si
=b
y=4x +8
198 Algebra 2, Third Edition
Problem Set 75
28. Vx -5-2=7 P. P.
oie Score
x—) = Sl if 1
x = 86 ee PT,
— 5 425.7
Checks «(86 T,
9 = a7 py = (0.0036
0.0036x 1072)(40
x50
1077)(40
x 10
x 104)
x104
oo tad i ee p, _ (3.6 x 10™)(4 x 10°)
7 3 5 5 x 108
ee 9 91x 4414=. 105 P> = 2.88x 10-8 Nim?
5x = 100
x = 20 3 peat
pat Sooo
Nu
k
30. Lengthof side= 222 = 9m aes 500
: =k
5000
Area = 5 = 9m)* = 81m
T = TO = 25 days
PRACTICE SET 75
4. Dr
oe
x 8x4 4%
Leet
eee
Dp
See onl
tail i allan aa ae RgTp + 20 = RyTr: Rr = 60;
(x — 4)(x + 1) (x —4) (* 41)
Rp = 40; fp = Typ +2
eae = fae + Gee 5) 2 x Sr = 7
fd ix A)(x rf 1) (x a 4)(x ri 1) 40(Tr + 2) + 20 = 60TT
40Tr + 100 = 60T r
100 = 207;
PROBLEM SET 75 Shr = Tr
1 Dp
ister, 5. (a) 4N, + 6Np = 192
(b) Nr. = Np oF P)
D-
aes Sl Substitute (b) into (a) and get:
(a’) 4(Np — 2) + 6Np = 192
RpTp = 120; ReTrE = 360; 10Np = 200
Rp = 2Rp; Tg = Tp + } Np = 20 poinsettias
2Rp(Tp + 1) = 360 (b) N;, = (20) — 2 = 18 lilies
2RpTp = 2Rp =5360
x+4 2,
120) + 2Rp = 360 [ig Se
coat it
ie ¢ x4+2x-3 I1-x
=e
2Rp 7 x+4 a 20Gb ae
Rp = 60 mph TANG 1 -Ce 3-1)
Re = 120 mph; Tp = 2hr; Tr = 3 hr pees 10
A vay m3
199
Algebra 2, Third Edition
Problem Set 75
ie — a5 x+2
Fae que
Roe ie ore
Z x —3 , + 2) = 4)
(x -—4)(x+3) (x - 4)(x
+ 3)
i
Ree 2gee tae 8 R=" 10 cos 20%=
9.40
x* —-x-12 U = 10sin20°
= 3.42
x7-x-11
Stee pants 8.66R + 5.00U
x" -x - 9.40R — 3.42U
18.06R + 1.58U
3 Ne
SA iat ap) 14,
Wier Or ee Be,
7 TD 5 ow
; 2-3V3 _ 2-3V3 3+ 43
Be 4 3 aS
_ 6 + 8V3 - 9V3 - 36 _ 30+ V3
9 — 48 39
10. Acircle = ar? = 9ncm*
r= 3cm
Length of diagonal = 2r = 2(3 cm) = 6cm
2
Asquare = Ce (se en = 18cm?
a
11. There is never an exact solution for these problems.
One possible solution is given here.
ViesonK oop Since @ is a third-quadrant angle,
Use the graph to find the slope. iach eS ree
5 2,17 /255.96°
m= ar = Ss
15. 2x + 4 = —5x?
V = -1.25K +b Gs aes
Use the point (95, 27) for K and V.
27 : 1.25(95) + b 7 EG EEO
145.75
= : ;| my
_ Bt Neer
= ae
V = -1.25K + 145.75 ie ds a
12. See Lesson 71. m a Uh
16. ——— en(©+ °)
oe x (ep
13.
my = acmy + bemx
my — acmy = bcmx
a Il 10 cos 30° u oe D fon
ebone 2) ex
iH 10 sin 30° I Nn SS I
m — acm 1 -ac
200
Algebra 2, Third Edition
Problem Set 75
17. Since this is a 30°-60°-90° triangle, i
22. Atriangle = ou Il asoO 5
NO
3 -SF =8
Seta
3
en ON, . 163
3 s) ae’
Asemicircle = Beh 20
y = 83, - 83
r? = 4m?
3 oS
r=2m
2 a2 a
18. a“y + = a*y+ zt
23. A232 = 22[2(2/3)]¥2 = 2292916 — 28/3
ane a
y y
2
ee
3a 2 3 2
Y 24. ae tas) 1/2)3)
ae 42 = (4 = -8
ay+a ay+a
a*y?
+ a*y + ay 25. Alesmp Srey
m pt we = (mp?)'! 4 (mp*)'9
yr = m4 4B p¥3 a pe
19. (240)
— 4) — -/-16
S284 4 —4 26. at + 2/8 - 256
== S21 = 4 4 = 9 -
3/7 8 +
2487
—— — -
2/56
20.
1 1
(a) =e7 =yred = 115
AS ATT
_ 21V56 , 16V56 _ 112/56
(b) 0.02x + 0.2y = 6.4 56 56 56
(a’) 3x + 4y = 180 _ 75V56 _ 7514
va O56 es
(b’) 2¢°4 20y = 640
ay 215 = 209 = =900 fw) 4813 x 10714 bees io
27.
(b’) 2x + 20y = 640 C00 Me 0
S1\2% =O) ~3 x 10)
eo) Calculator answer v 2.53 x 109}3
(a’) 3(20) + 4y = 180
(b) *°198
4y = 120
Answer should be between 3 and 5.
y = 30
Calculator answer ~ 4.34
(20, 30)
ie 3x7 +2=—%
2
es ai st + = “3
Sp Ch SEES ae
3 36 3 36
ee Pees
6 6
Ine 23),
x=-—t l
6 6
Algebra 2, Third Edition
Practice Set 76
28. Graph the line to find the slope. PRACTICE SET 76
x = 3y (a)
a. x+y+z= 56 (b)
x —-2y —-3z = -25 ()
We can substitute (a) into (b) and (c).
(b) Gy) + y + z = 56
yet 2 =50" (0)
(c) Gy) - 2y — 3z = -25
—— y =937
= =25 “(e)
(d) 4y+ z= 56
—4(e) —4y + 12z = 100
Since the slopes of perpendicular lines are negative 133.2056
reciprocals of each other, z=12
m,= 1 (d) 4y + (12) = 56
y=extb 4y
= 44
(7) = (5) + b y=Il11
2=b (a) x = 311)
y=x+2 x = 33
seB
20. Se5 -— 2
Nid~ | Nn iH] om
il ao pes
~~ s5 4 =15°
A-= 2siniS” = "052
B =.2 cos 15° .= 1,93
Check: Pe) aoe 2 I
= ~
‘Oo | i) i] ~
30. i SRS lO
2 3
12x + 21 -— 10x iT] i)as
2x
Ww
Ni[w C = 5sin 70° = 4.70
D = 5cos 70° = 1,71
A + C = 0.52 + 4.70 = 5.22 up
B + D = 1.93 + 1.71 = 3.64 to the left
The solution is -3.64R + 5.22U.
202 Algebra 2, Third Edition
Problem Set 76
PROBLEM SET 76 (a) x = 2y
(b) x+y
+z = -198
(c) x =—3y = 22 = 16
Substitute (a) into (b) and (c) and get:
(b’) 3y + z = -198
RcTc¢ = 32; RrTr = 72;
Rr
= 3Rc; Tr = Tc - 2 (c’) -y — 2z = 16
3RC(Tc — 2) = 72 2(b’) 6y + 2z = -396
3RcTc - 6Rc = 72 (c’) -y-2z= 16
3(32) — 6Rc = 72 5y = -380
—6Rc = -24 y = -76
Rc = 4mph
Ry = 12 mph; T¢ = 8hr; 7; = 6hr (b’) 3(-76) + z = -198
26 i Wwi=)
ta (a) x = 2-76) = -152
T, T,
ae PVT, (-152, -76, 30)
PT,
ue (700 x 10°)(700 x10~’)(8000 x 10°) (a) 2a 2y 2 =i 81
(3500 x 10*)(56 x 10*)
(D)e3ieny
te, ae,
ee (7 x 10’)(7 x 107>)(8 x 108)
(3.5 x 10’)(5.6 x 10°) (c)
x —3z=0
V> = 2 x 10"! liter x = 32
3. 0.64(500) - E, = 0.4(500 — E,) Substitute (c) into (a) and (b) and get:
320 — E, = 200 - 0.4E, (a’) 2y + 5z = 81
-0.6E, = —120
E, = 200 mL (b’) -y + 1lz = 27
4. Sodium: L.x-23 =) 25 (a’) Oy ¥95z = 81
Chlorine: Lx S5.ne. 3D
2(b’) -2y + 22z = 54
27z = 135
Oxygen: 1 x 16 = 16
Total NaO: 23 + 16 = 39
ee Rimes. (a’) 2y + 5(5) = 81
39 NaO 2y = 56
NaO = 312 grams
y = 28
5. (a) 4Ng = 3Nez + 14
(c) x = 3(5) = 15
(b) 6Np = Ng +7
Substitute Ng = 6Ner — 7 into (a) and get: (15, 28, 5)
(a’) 4(6Np — 7) = 3Nr + 14
21Npr = 42
Nr = 2
(b) Ng = 6(2) - 7 = 5
A total of 7 chips were down.
203
Algebra 2, Third Edition
Problem Set 76
11.
2V2-1
1+ +2
(eae THAD
ee
fo
R = 5cos 20° =
emcomel} Zo 4-400
U = 5sin20° = 1.71 12. 44 2\8° 44 492.
A 4d2
_ 12 -12V2 -4V¥2+8
_ -5 + 4v2
re 16 — 32 4
13. There is never an exact solution to these problems.
1S. One possible solution is given here.
Al = mB + b
R = 4cos 65° = 1.69 Use the graph to find the slope.
U = 4sin 65° = 3.63
Pe ale in
—4.70R + 1.71U 4
—1.69R + 3.63U
-6.39R + 5.34U Al = 10.5B + b
Use the point (4, 180) for B and Al.
180 = 10.5(4) + b
138.270
Al = 10.5B + 138
14. See Lesson 71.
15.
a m m
=—_—-
+ —
Cae as r t
art = cmt + mtx + cmr + mrx
art — cmt — cmr = mtx + mrx
14.04° art — cmt -— cmr
Since @ is a third-quadrant angle, mt
+ mr
6 = 14.04 + 180 = 194.04°
17 /194.04° a (- 1
16. =m—-+-
(Gare 3. i? t
5x +2 iG
10. a m m
x 437-10 2— x =—-
+ —
(Car BF r t
= 5x + 2 2x(x
+ 5)
Ga 2). 1G £520) art = cmt + mtx + cmr + mrx
Sx ie 247 & 10% art — cmt — mtx = cmr + mrx
Re ee10 cmr
+ mrx
= ae eae ar — cm — mx
x? + 3x - 10
204 Algebra 2, Third Edition
Problem Set 76
17, WES weak =O 24. ony? Mm y - (my?) (my) 1/4
(=?a Od eal = ml3/4U4 _ pq112,3/4
7 a
oe
Yep7
a 1 1 1 3 8
ng
196 Ree7 roa
196 25. = + 38
2/3 ro 2216
2
ee ere a 2/3 J8 + 378
= SA DS
V3 - 604
14 196
Ts V8 V8 V3 V3
29
Samean _ 624 | 2424 144J24 _ 114-24
ve a5 ‘ 29 24 24 24 24
“fla 44 _ _19V6
2
18. x + ng “ 18
ae! ax +a 0.00842 x 10 Sal 15
a 26. (a) FE.
SPP aCR 68 ey
nea Ds
x ; x A NO Ses axe Om A Se KOT
ae: ax 4 OX wyat AX ct AX = 9S 1022
ax + ou ax +a
7 x? + 2x Calculator answer ~ 2.01 x 1072
ee
"a (b) (4.63)!
19. pe Sy 237? = /-9 Answer should be between 1024 and 3125.
i i pe > 23;
; ‘ : ’ Calculator answer ~ 2557.26
=6—3i = 23-—-1+2-3i=7
-8
27
1 2 r
20. (a)
(a) —xAgere
+ —y = 31
(b) 0.02x + 0.7y = 27.6
(a’) x + 2y = 93
(b’) 2x + 70y = 2760
—2(a’) -2x - 4y = -186
(b’) 2x + 70y = 2760
66y = 2574
yen 39
Graph
(a’) x + 2(39) = 93 ie the line to find the slope.
P
agi Slope 7=. —
(15, 39)
eile b
21. The radius of the circle is equal to the diagonal of Vee ay Ae
the rectangle. a
r= + 5°
(Vil)? 4= (2)+
yr =-6 .
£a =
92, 512575 = 5¥5°+V5 =95[525"/7]!2
= 5(5(5"*)) 215 9/4 Poae
y= Tr + —
4) 7
perry (41/2)
23. 7) = “36
205
Algebra 2, Third Edition
Problem Set 77
28. v2x-2+7=9 PROBLEM SET 77
2x -2=4
2xt= 6
oe =13
Check? 4/2(3)
= 2. + 7 =<9
2+
7 =-9
RyTy = 320; RsTs = 240;
29. I5x-2 2x-5_3
Rs = Ry + 20; Ty = 2Ts
3 4
60x — 8 — 6x + 15 = 36 (Ry - 20)2Ts = 320
54x
= 29 2RgTs — 40Ts = 320
29 2(240) - 4075 = 320
** 54 ~40Ts = -16 0
Ts = 4hr
1 dts
30. Ag;Figure = —bH + —ar
2 Ty = 8hr; R; = 60 mph; Ry = 40 mph
2
H = y(V3)?
-P E==
H=4=2 E
2
k
Argue = ow ' a(S) |n
in2 500)
(500) = (10)
——
5000
=k
(2+ an}in.? ~ 3.96 in.2
FE = 22 ~ sooo
PRACTICE SET 77 0.7(50) + Wy = 0.97(50 + Wy)
a. x3 +3x2-8 —x-1=0 35 + Wy = 48.5 + 0.97W,
¥x343x2-8 = x41 0.03W, = 13.5
xo + 3x7 -8 = (x + 1) W, = 450 liters
x? + 3x2 — 8 = x? + 3x2 + 3x + 1
Multiples of 7: 7N, 7N + 7, 7N + 14
—9
= 3x
—3=x IN + 4(7N + 14) = 3(7N + 7) + 133
Check: 3/(-3)3 + 3(-3)? - 8 - (3) - 1 35N + 56 = 21N + 154
Se ee 14N = 98
—2+2
N= am
The desired integers are 49, 56, 63.
b. vVs—16
+ Vs =4
Ss
WME 6 4. Ss
(Vs - 16)? = (4 - Vs)?
S 16 =816 — 85 4-8 Da
8s = 32 RcT¢ = RoTo; Tc = 6;
Vs =4 Tg = 12; Ro = Rg + 4
s = 16
Rc(6) = (Re - 4)(12 )
Check: ,/(16) — 16 + ./(16) = 4 6Rc = 48
4=4 Rc 8 mph
206 Algebra 2, Third Edition
Problem Set 77
6. ix + 9x2 —27 =x +3 10. Since this is a 30°-60°-90° triangle,
x74 9x7 — 27 = x3 + 9x2 + 27 + 27 2x SF=8
—27x = 54 SF 24
28 = Ey 5 5
Check: 4 (-2)3 Oe 27 SS ite D r= 4(v3) = 4V3
eee) t= 4(1) =4
De v¥m-—1 sr eS en 11.
Am 12 ="/m 2
m—-12=m-— 4/m+4
—16 = —4./m ia
ab Jim ache
16 =m
R = 4cos 30° = 3.46
Check: 416 -12 - 416 +2 =0
2—-44+2=0 = 4 sin 30° = 2.00
8. (a) 2x+y-z=7
(b) x — 2y + z = -2
(co) 2¥ + Z= 0
Z=-2y
Substitute (c) into (a) and (b) and get:
(a’) 2x + 3y =7 R = 6cos 20° = 5.64
(b).x—4y = 2 U = 6sin 20° = 2.05
(a) 2x+ 3y= 7 -3.46R + 2.00U
—2(b')+2x 4+ By.) -4 —5.64R + 2.05U
+ rae ia We TD -9.10R + 4.05U
Flynea 1)
yas 12. y
(b’) x.— 40) =.-2
=2 2
‘. il
(c) z = -2(1) = -2 ha
- x
(2, 1, -2) a
Bi
9% (axtyrz=7 -
(b) 2x -y-z=-4 Y
(c) z ="2Zy
Substitute (c) into (a) and (b) and get:
(a’) x + 3y = 7 Hom }72 +4 22
(Di) 2k = Sy = 4 H = V53
= 4 =o r
x =] * tan @=-—5
ee 3y = 6
9 ~ 74.05°
y=2 @ is a third-quadrant angle,
Since
(c) z= 2(2) =n4 0 = 74.05 + 180 = 254.05°
(1, 2, 4) /53 /254.05°
Algebra 2, Third Edition 207
Problem Set 77
fh Sw 2 20.
13.
==
15. Se x
rz Tx +2 , 2 +3)
(=
5S) +3)" “Aa SGeeS)
tet 2x Om 9x+ 8
oe en ee x? —~2x - 15
Wee AO. 2v2 +1
14.
ONae MOO art
AN? eo2 3/2
Sent ) I I+
& ll
al|_N
34+2V5 1+-5
15.
[eee en gS 21. -~8x
— 1 = 2x?
3+3V/5+2V5+10
-13 - 5/5
2x? + 8x +1=0
oe) 4
Ja -8 §? — 4(2)(1) _ -8 + /56
16. a+ =at ‘ 2(2) sali
Bp real
2
22. mZAOC 180 — 30 = 150°
ZOOL SOC ; hil 15°
17.
Mey ae Gy =
360
23. Agquare a = = 4m?
aR,Ry = amxR, + bmxR,
aR Ry — amxRy = bmxR, s = 2cm
bmxR, Length of diagonal = \2? + 22 cm = 2/2 cm
ee
aR, - amx
18, =/29-= 37 — tae
= —3i — 3i(ii) — 2(ii)(ii) + 2
= -3i + 3i-2+2=0
4000 cm? Meallan' 1 in. gy Kin.
24.
S 2.54 cm 2.54 cm 2.54 cm
1 ft 1 ft i ite 60 s
19. (a) =x + =) ='10 — xX x x
125ink 1Dine 12am: 1 min
(b) 0.03x — 0.2y = -1.58 : 4000(60) ft?
(AQ) Sse ay =e 350) (2.54)(2.54)(2.54)(12)(12)(12) min
(b’) 3x -— 20y = -158
25. VA = 192219 = [2221/3]1/4 = gl/gi2
(a) 15x 4-4 14y%= «3560
_ 27/12
=5(b’) —15%-4 100ve= 790
114y = 1140 (-3)9 (30
(b’) 3x = 2002-158
Ves LO 26.
Oo = 40%] = 27
x= 14 27. (2V5 + 5)(5./20 - 1) = (2V5 + 5)(10V5 - 1)
(14, 10) = 100 — 2/5 + 50/5 — 5 = 95 + 48,/5
208 Algebra 2, Third Edition
Practice Set 78
28. Write the equation of the given line in slope-
intercept form.
PRACTICE SET 78
Diet
Ay = 3
5 3
ys=-—x+ =
4 4 A 40
Since the slopes of perpendicular lines are negative
reciprocals of each other,
a (\
my)
ih
= =
B
5
A = 40sin 58° = 33.92
y ="sE-+-d .
B= 40\cos58 = 21.20
Q ll 30 sin 22° = 11.24
a iSor
==) COS 2 a= 27.52
9 33.92 + 11.24 = 45.16 upward
—21.20 + 27.82 = 6.62 to the right
The resultant force is 6.62R + 45.16U.
45.16
0
D? = 10748?
De, jose
cae ee 6.62
D = 2V41
In polar coordinates,
3x -1 —_
x -—5
—— = t eé = —
45.16
nt 4 7 ore Co.62
21x] = 4x +.20 = 28 0 = 81.66"
I7x = 15 F = (6.62)? + (45.16)? ~ 45.64
xs Lal The resultant force is 45.64/81.66°.
17
Algebra 2, Third Edition 209
Problem Set 78
PROBLEM SET 78 6. (a) N = 5D
(b) 2N = 7D + 6
1. DR
Substitute (a) into (b) and get:
as! (b') 26D) = 7D" +6
3D= 6
eS
D,
945 D=2
(a) N = 5(2)= 10
RrITpr = 135: R,Ty = 945;
Fraction = aw = i
Tr = T; - 4; R,; = 3Rpr D2
3Rr(Tr + 4) = 945 7 Vx-9
+ Vx =3
3RrTr + 12Rp = 945 x-9 =3-vx
3(135) + 12Rp = 945 4-929-— 6Vxe +
-18= -6x
12Rp = 540
3= Vx
Rr = 45 mph Te 6s
R,; = 135 mph; Tr = 3hr; 7; = 7hr Check: 9 -9 + /9 =3
0+3=3
Even integers: N, N + 2, N +4, N+ 6
8. Vx-8
+ Vx =4
(N + 2)(N + 6) = -3(N + N + 4) —- 16 Vvx—-8 =4- Vx
No BN 19 = —6N’
— 28 x-8= 16-8 /x +x
N? + 14N + 40 = 0 24
= -8Vx
(N + 10)(N + 4) = 0
3 = Vx
= Xe
N = -10,-4
Check: 9-8
+ J/9 =4
The desired integers are —10, —8, —6, —4 ere)
= 4
and —4, —2, 0, 2. 9.
Jk — 24 = 6 LHe
kim 2A = 36 inet Bahondcta
sis x Tc = 3440 -60 = -l2vk
100
SME
Tc = 4000 grams 2 =k
Check: (25 — 24 = 6 — 4/25
se C= 232 1=6-5
100
C = 400 chords 10. There is never an exact solution to these problems.
One possible solution is given here.
S = mFe + b
(a) Nr = 2Ns5
Use the graph to find the slope.
(b) Nr - Ns = 200
20
aa 1
Substitute (a) into (b) and get:
(b’) 2Ns - Ns = 200 S = Fe+bD
Use the point (140, 50) for Fe and S.
Ns? = 100
50 = (140) + b
Ns = +10
—-90 = b
The numbers were either 10, 20 or -10, -20. S = Fe - 90
210 Algebra 2, Third Edition
Problem Set 78
M1. (a) x- 2y z= +9 a=5 ; “V5%12
14.
(b) 2x —y + 2z=7 V5+2 V5 -2
en eS ae
y = 3x
2 rez 253 aD
Substitute (c) into (a) and (b) and get: 13.0 ees ee
, 48) $00.00) abaP 3223/2 5 ay?
(a) Sk — z= -9 _ 6+ 6V2 +62 +12 -6 - 4/2
(b’) x +2z=7 O18 3
2(a’) -10x — 2z = -18 igs BND. 1+ ala:
(b’) -x+2z= 7 1S wiioe es(eho
—llx
sa
;
= =1]
_ 4442-32-64 5 1-2
= =
(b’) -(1) + 2z = 17, xe =—x-1
z=4 x>+¢x+1=0
(c) y = 3(1) = 3 eronc: ia)
(A,3, 4) 2)
-1+ /-3 ee eC Perel
fe Ae
12. ) ph
8
10 C 18. mc = a ee
may x R,
B a a
me = — + —
° x R
B = 10cos27° = 8.91
cmR,x
= aR, + ax
G= 10'sin pg = 4.54 cmR x — aR, = Whe
0.00R + 8.00U | Pye
8.91R + 4.54U | cmx - a
8.91R + 12.54U
1254 19 ee 2 + ;|
tan = oT ane
@ = 54.61°
:
me = oy
+ =
R,
F = \(8.91)2 + (12.54 ~ 15.38 cmRyx = aR, + ax
15.38 /54.61 ° R, ni
Pila + heed
Vet
mR, x
B. =% + 3 ‘ ,
cE 5x% — 14 2 ex 20. a+ =a+ >
nee x a+x
ee xr=3 nj? 3x(x + 7) a
(x + 7(x-2) (x + T(x - 2) " a a +axt+a?
_ Sea = 2k on oa a? +x
x? + 5x- 14
es hy, 3 a. -5i? — J-9 + J-3V-3 = -SiGii) - 31 + 3()
LE =-3 + 2i
= 5i- 31-3
211
Algebra 2, Third Edition
Problem Set 79
22. Since this is a 30°-60°-90° triangle, 29. Solve for v:
2-SF=5 im
bs (500)? (100 x 107!*)
ve 4 x 10’
ves 6.25% 10"
30.
400in2 2.54cm 2.54cm 2.54 cm
23. ———- xX ———_ *K ——_ Xx -
Ss 1 in. tein: 1 in.
x 60 cm?
= 400(2.54)(2.54)(2.54)(60) ——
min
25/46 J2 = 2[22°62y)"2]5 = 2[ 21252110] PRACTICE SET 79
= 735/10 _ 272 1000 mL
12,000 E x 12,000(1000) mL
1k
12,000,000 mL
0 (5-2 es 2 LH WESS
25; sole Ca) yl ak
4-3/2
3 4 a od
4
Se
10 £2 - 12 ix Pree 12 ie
1f& lf 1£
26. (2V4 - 2)(3V9 - 2) = (2)(7) = 14 2.54 car ; 2.54 cr 2.54 crr 1 liter
1 ie la lie 1000 ert®
Ble [2 - a]8 + 20 3 10(12)(12)(12)(2.54)(2.54)(2.54) liters
(1000)
_ Ne
4V5V8 ee— ee
3v8 V5 AO), Since this is a 45°-45°-90° triangle,
V8 V8 5 HS —_—
1 > SFt=3
_ 20V40 24/40 - 80V40 _ 194/10
~ 40 40 40 ~ = ae5 SF= 3
y=1- SF
T0213 < 10 7 ee10" ya=us
28. (0) ee 1 x 108
5062 x 10 5 x 10
Z=
2 or
Calculator answer = 1.39 x 1078 Zee Sa)
(b) >/263
Answer should be between 2 and 4. PROBLEM SET 79
Tc = KNp
Calculator answer = 2.81
2142 = k(3)
714 =k
Tc = 714(10) = $7140
212 Algebra 2, Third Edition
Problem Set 79
: Dp Fake, coon ee ee I, son
900 : thie 1 ft Wait 1] shah.
2.54cm 2.54 cm 1 liter
De
arr
< Tin,
In. Lin.: ~ 1000cm>
cm
ey 20(12)(12)(12)(2.54)(2.54)(2.54) fiters
RpTp = 900; Tp = S5Tx; (1000)
ReTe = 120; Rp = Rr + 10
8. x’ = 4x +4 =x $2
x74
t+4x7 + 4x +4
RR Toe OT p= 900
5(120) + 507, = 900 —8x = 0
50Tz = 300 a0
Ty = 6hr Check: 0? - 4(0) + 4 =0+2
Tp = 30hr; Re = 20 kph; Rp = 30 kph 2=2
A Carbon: i244 [Qu 12 9. e 28
4a alg
Nec
Oxygen: 2 x 16 = 32 Ag ay Gann dea
Total: 12 + 32 = 44 eRe: (ores eee
Carbon: 12 x 100% = 27.27% 8 = 8s
44
lg= 4s
4. (a) 0.6Py + 0.9Dy = 0.78(50) Weete
(b) Py + Dy = 30 Check: V1 = 4 - +8
V1
Substitute Dy = 50 — Py into (a) and get: Reawee:
(a’) 0.6Py + 0.9(50 — Py) = 39
-0.3Py = -6 10 yer yt 22 =3
Py = 20 (b) x — y —- 2z = -6
(b) (20)
2 + Dy Dy = 50 (rennet
Dy = 30
View
20 mL 60%, 30 mL 90%
(ay 2x =) pee 22 = 3
a: (a) Newords _ 2 (6b) x= y — 227 =-6
rea (d) 3x — 2y aa
(a’) 7Newords = 2Nspears Substitute (c) into (d) and get:
(b) SNgwords = Nspears + 120 (te Tones
(b’) Ngpears = SNewords — 120 Ry ees
Substitute (b’) into (a’) and get: 36 =
(a”) 7Nswords = 2SNewords — 120)
(c) y = 3(1) = 3
3Nswords = 240
Newords = 80 swords (b) (1) - (3) - 2z = -6
a4
5(80) = 120 = 280 spears
(b’) Nspears =
= 7
1 liter
; ,000 mL. x ————— = 50 liters } (13,2)
ae 1000 mL
213
Algebra 2, Third Edition
Problem Set 79
11.
Sip iat 16.
Gon ee
EX, R, R,
aR, R> ns amxR> + bmxR,
aR R> _ amxR> = bmxR,
A = 2 cos 40° u 1.53 bmxR,
oh =
B = 2sin 40° = 1.29 R,R, - mxR,
a |a b |
17.
5e R, R,
Cee ne
x R, R,
CSri"cos 45 7 NI=" 071 aR R> = amxR» + bmxR,
D =1sin45° ul= 0:71 aRR> =_ amxR> = bmxR,
1.53R — 1.29U aR,R, - amxR, ae
0.71R + 0.71U mxR,
2.24R —- 0.58U
0.58 18. ax ax
ax — = ax —
a — —e
ax —a
6 = 345.48° x 3%
ax? a? x? — a*x — ax?
F = |(2.24" + (058) ~ 2.31 =) 165
ax —a
—
ax —a
2.31 /345.48° is ax” — ax — x?
x-1
4x +3 23%
12.
Te dee 19. a{ SATS) SI Dp ne 20-9
4x +3 2x(x
+ 3) =2i+3-24+42+4+2=5
+ 2
x*
2 —9
eer x”
G: —9
Ax +3 + 2x? + 6x 1 1
So 20. ee
a) —x re =
-— —y -6
x7 9
2x* + 10x #3 (b) O2nc 029 eee
<9
(a) Ax = Sy-= 120
13.
-2-3 2v3-2 (b’) 25 4 e120
PEE phe OMe Pe,
(a) 45 =15y = 120
_ 443 +4-6+ 243 | -1- V3
-2(b’) -4x — 4y = -240
a ad ~~ Pay 4,
—9y = -360
-1-J2 -5+ 2 y = 40
14.
5 -J2 -5+2 (69) 20% 2(40) = 120
_ 5- V2 +5V2-2
93+ 4y2 2xXC= 40
ae) 23
x = 20
15. See Lesson 71. (20, 40)
214 Algebra 2, Third Edition
Practice Set 80
21. 3x7 — lox Ze (b) (194)! .09
3x7s~
Ome lero Answer should be between 5 x 10°? and
3 x 10°,
2(3) Calculator answer ~ 3.21 x 10°
= 2+ V16 = iL + 2 = 1, a}
6 3 3 3 29.
22. Since this is a 30°-60°-90° triangle,
V3 x SF =3
SF= V3
x = ¥3(1)
= V3
y = ¥3(2)
= 23
30.
23. Since this is a 45°-45°-90° triangle,
1 x SF = V3
SF
= 3
a V¥3(1)
= V3
b V3(/2)= V6
40 mL 1 cm? nt n. iain:
x x
Ss Lil 2.54 cm 25476
1 in. ¥ 60s 60 min
x
2.54 cm 1 min 1 hr
40(60)(60) = in?
~ (2.54)(2.54)(2.54) hr
25, 3/9243 = 313443 = 3[343!2]!2
= 3[373!4] = 3135/4
PRACTICE SET 80
wi,
Carlw ey
wi
pee
LTO
= On 14
14/14 F 6V14 84V14 32-14
14 14 14
27. (4V2 + 3)(5V2 - 4)
Mee 6a + 152 1
ll 92, on
W1,e5a%« 107° 4 x 10°2
28. (a)
0.00492 x 107!4 os i’
= x 10°
Calculator answer = 8.51 x 1048
Algebra 2, Third Edition
215
Problem Set 80
PROBLEM SET 80 8. Vs - 18 + Vs - 36 = 0
1.. S@ei s — 36 = 324 = 36+/s-+s
119 = k(14) 36Vs = 360
k = 8.5 Vs = 10
C = 8.5(32) = $272 s = 100
Check: 100 - 18 + 100 — 36 =
D
i ; ee Pa: D
Bn lOMaNBe 4 8 s=
SOR 0);
B, (5) 9. (abxtyt+z=8
100B, = 1250
By = 12.5 blues (b) "2x a Sy st oe eG
(c) 2x —-z=0
NB 4
ah (a) a SNR = 4Np z= ax
Np 5
(b) Ng = 2Np — 1200 Substitute (c) into (a) and (b) and get:
Substitute (b) into (a) and get: (a’) 3x + yoe ts
(a’) 5(2Np = 1200) = 4Np (b’) -3y 2g
6Np = 6000
Np = 1000 acrobats in pink y=2
(b) Ng = 2(1000) — 1200 = 800 acrobats in blue (a’) 3x + (2) = 8
1 SX =—6
4- 3— =< P=, 650
4 x sz
P= 650° X < = 200 people (CZ = 20) et
(2, 2, 4)
5. (a) 40Np + 2Nc = 820
(b) Np ate Ne = 310) 10.
Substitute (b) into (a) and get:
(a’) 40Np + 2(30 — Np) = 820 af am
38Np = 760
Np = 20 dogs adeesi
A
(b) Nc = 30 - (20) = 10 cats
6. er ee = A = 4cos 60° = 2.00
x*—-x-Q2=x* = 4,44 B = 4sin 60° = 3.46
3x6 y
ap = oe
Check? 4/273] 0°=) = 4) a a ss
—200°
20./p = 80 C = 6cos20° = 5.64
VP 7 D = 6sin 20° = 2.05
iia+ 16 = 10 pee 3.46U
2.00R +s,
Check: /16 + 20
Oot oe 18 7.6+4R
1.41U
216
Algebra 2, Third Edition
Problem Set 80
|W wo = re) ok tO Il | Ww=
(2-24 )-4
3
9 (Peas |
xo -x+—
SS t+ —
4 3 4
( y 11
x-=|] =—
2) 12
1 433.
x--
= +—
2 6
x=—-—-t—
i, Aik!
94 6
17. a= n(2 +4)
tng = 2- =3 pee:
y
mqx
Cc
@ = Ji 57° y Cc
acy = cmpx + m@xy
Since @ is a second-quadrant angle,
acy — mqxy = cmpx
6 = -(71.57) + 180 = 108.43°
_ empx
2/10 /108.43° Z ac — mqx
12.
5 ere 18. 3a — = 3a - 5 2
x 366-16 x«-8 a- 3 a~7—"3
YS 4x2 B(x 4:2) a a
BiB) x 2) BG-= 82 +2) = Bye 5
3a =
3a 5
9a — 3a
a“ —3 a“ —3
eet es
i Ea!
a3; 19. (a) Since this is a 45°-45°-90° triangle,
V2 x SF = 5
eS
clas
14.
2-3V2
3+ 22 _ 5V2 (4) _ 5v2
a 2 Se 7 2 2
7 eth
_ey.Bay.SB
2
15. A? = 37 + 47 (b) Since this is a 30°-60°-90° triangle,
A=
4 x SF =
SF =2
im) ll | ay T
(5)Z 2.8 +20 ~~ —
Mm
mi]
10.
= B +5
Q ll | il ll
S="B
Cae we
Algebra 2, Third Edition
217
Problem Set 80
20. —f=4 + RiaOuee ? ae Hopah ee 4i4 (b) 2x + y = -l
237 Bi — ii). + 2G) 4 AG)
23 434+ — 2-84 =\-6 + 2
1
21. (@)Sx - bare
me
(b) 0.07x + 0.06y = 1.32
(a’) 8x — 3y = 72
(b’) 7x + 6y = 132
2(a’) 16x -— 6y = 144
(b’) 7x + 6y = 132
23x S216
Substitute y = —2x — 1 into (a) and get:
xt=i2
(a’) x — 3@2x = 1) = 6
(b’) 7(12) + 6y = 132 Adie S
6y = 48 3
eS S
res 7
(12, 8) i3 Bon
b) yyo=. —2( (>
(b) = beeis
7
22. ee es
x7 -x-5=0
_ -(-1) + ¥en? = 40-5)
can
eee |
3 3X 7x
2(1) 28. pest
la 2 a1 ax a“x x+a
2 pa eo . 3a +a) 3x(x + a) Ta? x2
a® x(x + a) a> x(x + a) a” x(x + a)
600 cm? 1 in. 1 in. 1 in. 3ax + 3a? + 3x2 + 3ax + Tax?
23. ———— xX
min 2.54 cm 2.54 cm 2.54 cm
a? x(x + a)
1 ft 1 ft itt 60 min
x x 3a” + 3x” + 6ax + Ta?x?
12 in. 12 an 12 in. 1 hr
a? x(x + a)
600(60) ft?
. (2.54)(2.54)(2.54)(12)(12)(12) hr 29. x? = 4x2 4 32x
x? = dy 2 390, = 0
24. Volume = Apase X height
Ds Ue — 3G 4) = 0
[25] xh x = 0,-4,8
2
12 =| 30. Write the equation of the given line in slope-
1(6 in? (6ft x
1 ft intercept form.
q= $138.88 1-inch sugar cubes x-y-Tl=0
4 [xy Ixy? = (x>y)!4(xy3) 2 y=x-1
pay
a sa a lla = x AyiA Since the slopes of perpendicular lines are negative
reciprocals of each other,
26.
an
= (eo
ee |= 8 m,
= +1
Y= Xe b
3=2+b
27. (a) x — 3y = 6
-l=b
y= 3 —2Z
y=x-1
218 Algebra 2, Third Edition
Problem Set 81
PRACTICE SET 81 Equal ratio method:
2
2+3i 343i ehigae 2
3-31 343i Mo Ely
100 _2»
ABOVE
M, (5)
2 + 3i 4M, = 2500
3 + 3
M> = 625 monkeys
6 + 9
6i + 9:2
k
6 + 15% -— 9 = 34+ 15
eae
BELOW
4= E;
Bet (10)
3+ 31 400 =k
9 — 9j
9i — 9i7 Ves = = 100 macaws
Dr
re
Dy
2-21 2-21 -2-2i
2i-2 -2+2i -2-2i RrTpr = SIS RT; = Bil:
Rr = 3Rj; Tr = 1; — 10
ABOVE
3R(T, = 10) = Bi)
BO
Ok
3R,T,; — 30R, = 375
—4 + 4i 3(375) — 30R; = 375
~ A -30R,; = —-750
—4 —A = 8
R,; = 25 mph
BELOW Rr = 75 mph; 7, = 15hr; Tr = Shr
Dy
=) —E2I (a) 0.1Py + 0.4Dy = 38
Asay
(b) Py + Dy = 200
4i — 4i2
4 +4 =8 Substitute Dy = 200 — Py into (a) and get:
(a’) 0.1Py + 0.4(200 — Py) = 38
ge me —0.3Py = —42
Py = 140
(b) (140) + Dy = 200
PROBLEM SET 81 Dyn = 60
Variation method: 140 liters 10%; 60 liters 40%
M = kT? 40
oS, = 1120
100 = k(2)" Tir
2 =k Ny = 1120 - Boe
40
M = 25(5*) = 625 monkeys Nr = 2800
Algebra 2, Third Edition
219
Problem Set 81
5 Sa alES 12.
Dees oS 5
6-15 -21+5% 1 _ 17,
a 4 — 2572 “39 29°
32), Died
Di, SA Di FRA taba ink
Gi: 2 difie A
= 4i2 — 16 Seo 10
8 cos 71° = 2.60
(180
— A) = 4(90
—- A) + 30
B = 8sin71° U 7.56
180 — A = 360 — 4A + 30
—2.60R + 7.56U
0.00R + 5.00U
—2.60R + 12.56U
ipa es
2.60
~TIx
= -21 @ = 101.70°
so = 8}
F = (2.60) + (12.56 ~ 12.83
Check: ./32 — 3 + 30 -3 = 3
12.83 /101.70°
6=3 23
1x = 2 Bex
10. ip — 48 =12.-4/p 13. +
to9) (3x
p — 48 = 144
- 24,/p + p
2 Tx —2 SxS)
24,/p = 192
(Kee 3) 3) Be 3G
P = 8
_ 3x4 - 2x -2
p = 64
x2 -9
Check: 64
— 48 = 12 - /64
4.= 108.8 V2-5 ¥2+2
14.
nD 22) Oeste?
11.
(b) 2x -y-z=0
_ 2+ 22 - 5vy2 -10 _8 +32
Dora 2
(c). y.— 3z = 0
y = 3z
15.
2/3 -1 14+ 3V3
Substitute (c) into (a) and (b) and get:
(23.3) 15379
(ain jn = 95 RN Sich 18 carl SalSee kde Ee
(Db!) 24 — 47 = 0
1-27 26
—2(a’) -2x — 6z = -10 1+ V2 342
16.
(BON 2.42 I i=) 3.2 Tot Ne
~ =10z = =10 _34+V24+3V2+2
5+ 4,2
z Hf aR 9-2 ye
(ix 3(1)-=' 5
17.
40
go UG ie
x= 2
2
(c) y = 3(1) = 3
140
—1
(2, 3, 1) y= = 00
220 Algebra 2, Third Edition
Problem Set 81
18. See Lesson 71. 25. 3943 = 3/3243 = 3[32(3"")]!2
ee ees 31/8 = 3178
eee 3a ae 3a
19.
at at a #4
a a 26. 2,| — 35 + 34020
2 ay
ee ne 2at
- a® Bos
a~ +1 a+1 afg fae P aoe
20. V-9 — J-2J-2 + J-242 - 313 - 27? 205 eae= ae 17/5
= 3i — 2(ii) + 2i — 3i(ii) — 2(ii) 5 5 5
=3i+2+2i+3i+2=4+8
Zi
2 1 x+y
21. (a)
a) -xa = oe
--yo= 6
wy rs SOY ty) + 2X(xX + y)
(b) 0.15x + 0.0ly = 0.84 2,
X-YW(X
th VW wah W(x ty)
2
x y(x + y)
(a) 2x-y= 18 xd y + 3x + 3y + 2x? + 2xy
(ob) 15x + y = 84 x*y(x
+ y)
ix = ike
= 0 28. 4 ee) a Oe) ll
-16-1-x+1+2x+5
(a’) 2(6) - y = 18
y=-6 x 18
(6, -6)
29.
22. 2 = -2x* - 3x
2x2 + 3x +2=0
x=
3+ /3? —4(2)(2), _= —
-3 + V7
POD) 4
meer ir
4 4
4x? + lox + 63 + 4
23. x — 4)4x3 + ne —- x FAS
4x? — 16x?
16x27 - x
16x? — 64x
63x + 2
63x
— 252
254
0.5061 x 10° 5 x 104
30. (a) 0.0071643 x 10 ase
Since this is a 45°-45°-90° triangle, 7x 10 31
V2. SF = WW = 7x 10-*
oi - LIND Calculator answer ~ 7.06 X 1074
2
_ 1124) _a5Wv2 (b) °2/594
= 5
Answer should be between 2 and 3.
17/2 17/2 Calculator answer = 2.80
ya= mune mat Ay
221
Algebra 2, Third Edition
Problem Set 82
PRACTICE SET 82
a. = =
ae x ax RrTr + RwTw = 60; Rr = 10;
3 lla =
omen a ap
rae 3 BO Ee 3a +3 Rwy
= 5; Tr + Tw = 8
a a
. 1 a 3a + 3 10Tp + 5(8 — Tr) = 60
— 9a+9+ax ~ 90+94ax STr = 20
3a + 3 Tr = 4hr
a 3a + 3
Tw = 4hr
~ 9a + ax +9
Dr = 10(4) = 40 miles
b. = sae al =H - Dy = 5(4) = 20 miles
ate xe AB ae
Vie E qz + 1 qz +1 Sodium: 2x 23 = 46
re Z Hydrogen: Lxl=,/
‘ m = mqz F 1) Phosphorous: 1 x 31 = 31
Gxt
+ X+ PZ guztxtpz
Oxygen: 4 x 16 = 64
qz +1
Total: 46 + 1+ 31 + 64 = 142
46 Na
PROBLEM SET 82 142 852
2
1 Bh 22) 142Na = 46(852)
a 5; Na = 276 grams
300 _ (10)?
05 6) Sodium: s x 100% = 32.39%
100Q, = 25(300)
Q> = 75 m z m
m m
2 + 2
2 Dp 742 2p+2
nl
1920 P
reek LSet wos mt
Dp 2 4 —mP 4p +4+ mp
320 2pit 2 2p.+2
RpTp = 1920; RrTr = 320;
_ _m(2p
+ 2)
Tp = Tr + 4; Rp = 3Rp
4p + 4+ mp
3Rp(Tr + 4) = 1920 P # p
3RrTR oe 12Rp = 1920
a+ ? at to
3(320) + 12Rp = 1920 oe ym +
12Rp = 960 y Y
Rr = 80 mph
Rp = 240 mph; Tr = 4hr; Tp = 8hr apy yb aym + 3a + yb
ym + 3 ym +3
5 AM By _ __Plym
+ 3)
T, T, aym
+ 3a + yb
(eo PLVoT, 2
ar“h
TORY Veons = ala
Ty 2 (50)(200)(540) ~ 4320K
12(3) m>
(5)(250) i= ee = 4m
9m
222
Algebra 2, Third Edition
Problem Set 82
9 223i 4-i _ 8-2i-121-3 14,
a ey et a 16 +1
5M;
17-17
10, ee 24 Bh 10 d5iGe21 403
72-2 ar oo +o: 4+9
= 1 +i A = 45cos 70° = 1.71
; B = 5 sin 70°. = 4.70
me x —-x+47-5= x
xe —-x4+47 = x2 + 10x + 25
llx = 22
Se eo)
Check: ¥4—-2+ 47-5 =2 Cie 30 cogn0° = 2819
: hse? a AT — =
°
LEA =e D = 30sin 20° =~ 10.26
12, Vx +24+ wie 1 7iRere A0U
= vx -28.19R + 10.26U
x+24 = 144- 24/x +x -26.48R + 14.96U
24./x = 120
15.
Vx =5
se AD
Check: 25 + 24 + J@5-= 12
71+5 = 12
13. (a) 2x -— 2y — z = 16
(b) 3x — y + 2z7 = 5
(c) -y + 3z = 0
y. = 3z
Substitute (c) into (a) and (b) and get:
(a’) 2x — 7z = 16
(b’) 3x —-z=5 6 = 60.26°
(a’) 2x-7z= 16 Since @ is a second-quadrant angle,
—7(b’). 21x +292*= 1-35 6 = -(60.26) + 180 = 119.74°
~19x = -19 65 /119.74°
Ze Al
ax = 2 x
hy aay = = 5 1G.” Sry
IsTo
e= —2 ‘: 3x2 I xix 3)
(6) y = 3¢-2) = -6 Es
»
BS ieiaaamnniaeee
3X
=
a Xk
3
Oy ae
oO
oe
= 3
= 2
(1, -6, -2) re x2 =) Ls x? -— 9
223
Algebra 2, Third Edition
Problem Set 82
2 1
Gp OB Lee 23. (a) oe
57 3) |
ee ned (b) 0.07x + 0.2y = 2.15
(a’) 6x -— Sy = -15
18:
2-75
Se
2-2V5 (b’) 7x + 20y = 215
DES e245
4(a’) 24x — 20y = -60
_ -44+ 4/5 - 2V5 +10 _ -3 - V5
4 — 20 8
(b’) 7x + 2y = 215
Sky Up)
19. 5? = A? + 3? x=
25 = A7 +9 (a’) 6(5) -— Sy = -15
16 = A’ Sy = -45
4=A
y=9
5 x SF = 9
(5, 9)
24. See Lesson 71.
25. 3x2
— 2 = 5x
3x* + 5x42= 0
5-4
5 Sian Se) scant
5-385 2(3) 6
5 2
6) «6s ae
20. +=
Zz
2(2+))
x
m m\a [55%
LVN Sie oes
RY = Gy) l2G4y)!s
26.
ee BEE a x72 yll2 2/3513 = x iby Isle
m am m
27.
az — ampy = mpx
se Pe
Z — mpy = 24 22 - 0
z
6/2 ¥ AF 9072 29,/2
aie +=F (ys i y) 6 6 6 2
m m\a
ean eg 2° 1/2)5
m am sm 28. Spare
= 1-4'2)"],=.22
az = mpx + ampy
az - 29. (a) 4x -— 3y = -3
mx + amy
oe Ae
di 3
22. Length of side = suagonal = om = 3/2m (b) 4x + 3y = 6
4
Area = s* = (3/2 m)? = 18m? y Sree
5h
224 Algebra 2, Third Edition
Problem Set 83
PROBLEM SET 83
1. Equal ratio method:
upsur a
R, C,
1200 _ 300
R, 100
300R) = 100(1200)
R», = 400 were resentful
Variation method:
R= ke
(a) 4x -— 3y = -3 1200 = k(300)
(b) 4x + 3y = 6
4=k
8x ="3
R = 4(100) = 400 were resentful
RyTy = 420; RrTr = 270;
Ty = 2Tr; Rr = Ry + 20
wlo
wl]wo (Rr — 20)2Ty = 420
2R7Tr — 40Tr = 420
(F) 2(270) — 40T; = 420
-40T; = -120
400 cm? 1 in. 1 in. 1 in. Ty = 3hr
30. ————_ x ——_ X x ——
S 2.54 cm 2.54 cm 2.54 cm Ty = 6hr; Rr = 90 kph; Ry = 70 kph
ett ett lett 60s
x <x Du
ars)
1210. Lan: 12 in. 1 min
_ 400(60) 5 pal Ds
(2.54)(2.54)(2.54)(12 )(12)(1 2) min RsTs + 400 = RyTy;
Ry
= 5; Ty = Ts = 400
R,(400) + 400 = 5(400)
PRACTICE SET 83
400R; = 1600
a* = x/2,.m +m/5 _ gq X!26mI5
m 73x,ml5 qx!2 2 Ry = 4yd/s
Be
a—-2_a+1 (a) 4N, = 8Ns + 80
m Pp = m! A. 2m staid na + lyre 2a
m3 p24 (a’) N; = 2Ns + 20
= men epee = m3 -2y1-4 (b) 10Ns = N;, + 140
Substitute (a’) into (b) and get:
Zz +3 v4 42 3)
Epp ren 3
(b’) 10Ns = (2Ns + 20) + 140
(m*) (ye bs mz"
m * m
8N, = 160
Ny = 20 sedimentary rocks
= m= +*zm + 3
(a’) N, = 2(20) + 20 = 60 igneous rocks
225
Algebra 2, Third Edition
Problem Set 83
5. (a) 0.05Py + 0.2Dy = 64
11.
2-41 1-i _2-21-4i1-4
——-
(b) Py + Dy = 800 ete as 1 +41
Substitute Dy = 800 — Py into (a) and get:
= -] — 3
(a’) 0.05Py + 0.2(800 — Py) = 64
5 345i 2+2i 6+ 6i + 10i - 10
—0.15Py = —96
OF Oe ees 4+4
Py = 640
(b) (640) + Dy = 800
Dn = 160
13. (a) 2x + 3y -z=-3
640 mL 5%, 160 mL 20%
(b) x + 2y = 0
6. (a) x + 6y = 40
x = -2y
(b) 5x + 6y = 80
(c) x —-2y+z=-2
—l(a) -x -— 6y = -40
Substitute (b) into (a) and (c) and get:
l(b) 5x + 6y = 80
4x = 40 Oa he ee)
ca eLo (c’) -4y + z = -2
—Sy = -5
(a) (10) + 6y = 40
6y = 30 yea t
y=5 (a’) -G) - z = -3
z= 2
aps a
3G (ca ae = a* = 9S x m* yee Xx
(b) x= —2(1) = =),
= q2*!3-3y_-2x+2
(-2, 1, 2)
8. Cae Oe ae = gmx t+mpmx + 3x 14, Vie + lies 32 = 8
Gee
k +32 = 64-16Vk
+k
a a 16k
= 32
= [eee 1 Vk =2
at vr
ax +1
at+— k=4
x 5
Be a z a Check: (4 + /4
+ 32 =8
ran x axx + x4+x 2+6=8
ax +1 ax +1
_ a(ax + 1)
15.
ax” + 2x A
b b
10. Se 5
a+ a+
a ab+a
at—
b b
I 4 cos 20° = 3.76
< b 2 b
b ab+a*+b B 4 sin 20° = 1.37
Gat
ab+a ab+a
—3.76R — 1.37U
~ b(ab + a) _0.00R + 5.00U
-3.76R + 3.63U
ab + a* +b?
226 Algebra 2, Third Edition
Problem Set 83
epi _ 3.63 21. Since this is a 30°-60°-90° triangle,
3.76 /3 x SF =4
6 =~ 136.01° =
SF = mci=
F = \(3.76) + (3.63)? = 5.23 : :
5.23/136.01° oe { 2) . a4
ree: cee yen aa,
oe ed 2X 3 3
eee
ee Sie
Ge aa) ats
ee) 22. : Sinceage
this is a45°-45°-90°
-45°-90" triangl
triangle,
1x SF =4
— 5x? = 20 + 6x + 3x3 + 6x? _
x(x? — 4) tr DO
_ 3x3 + 11x? + 6x - 20 m = \2(4) = 4V2
xx =A) p= 1(4)=4
V7. =22 = 2) 42 23..
a} 2) A 2 5
_ -8¥2+4-8+2V2 2-32 (b) -0.05x — 0.2y = 1.65
16 — 2 7 (a’) x —82V = 35
nt oe eee (b’) 5x — 20y = 165
18, ———— . —=
2-J3 2+3 Solve the equations by elimination:
arts Spe 6 3303, gr Si; (a) 5x- 2 = 55
hog (b’) —5x — 20y = 165
-22y = 220
19. p= 2-4-5] Dame
(a’) Sx — 2(-10) = 55
pao 5X =o
x. m
ge = 7
mpx = am — acx + cmxy
(7, -10)
am + cmxy — mpx = acx
a acx 24. Qx* — 5x = 5
~ at cxy — px 5 5
be eX te ==>
2
a a
20. p=4-2-5] 2 ~ yp eee
thn 1 ge ee pa ade
pee ae ey 5 eae
x m i 4 Lg 16
mpx = am — acx + cmxy 5 65
Pie! ete
mpx — am = cmxy — acx 4 4
ae Oe
mpx — am _ .
mxy — ax = :
227
Algebra 2, Third Edition
Problem Set 84
7 4 Use the point (—3, 2) for x and y.
25. + 2[4 - 5/63 9
4 d y= <a +b
= a a i v7 ae Sa
2D Ain ALT i ——(-3) +b
oa a Pag 1a
14 14 1 8 8
11
6. -2° (-3°) = -{-[-(27'/7)
a 1/32 11 == -9
? ]} raha
8
9 11
27. 3]
(3/x2y)4 = [(x2y)"3]4 = (x23 y!)4 = 28y3.4/3 eo a
28 4 liters .,1000 cm? . 60s . 60 min PRACTICE SET 84
aks liter min 1hr (b) 64 — y=
3
= 4(1000)(60)(60) 2(a) 6x - 2y = -4
hr 0 =-9 False
Inconsistent and independent
29.
PROBLEM SET 84
1. Equal ratio method:
Ay B,
A, B
200, B10
A, 80
10A, = 1600
A> = 160 were admired
D2 = 92 + 82 Variation method:
D? = 145 ee
as :
(uses80
30. Graph the line to find the slope. 16002 k
4 = 1600
10
A = 160 were admired
2. 50 + 273 = 323 kelvins
ies
i 7,
4 gots
4T)
= 2584
T, = 646 kelvins
646. —.273 = 373°C
228 Algebra 2, Third Edition
Problem Set 84
3. Lithium: 2X) S14 B SAL Bek Bie yt 121 OF = 12
Calcium: 2 x 40 = 80 8 31 + 3 oro
Oxygen: 7 * 16 =112 ee 7,
Total: 14 + 80 + 112 = 206 18 6
14. 356 14. Vs - 48 + Vs = 8
B06—" 'sL.i,Ca;05 s- 48 = 64-16/s +s
14LiyCa,O7 = 206(56) 16Vs = 112
LiyCa,07 = 824 grams Vs =7
aon 14
ithium: 706 - 100% = 6.80% Checkiei4o 246
ie 49 one
1+7=
4. 0.1(160) + 0.3P, = 0.22(160 + Py)
16 + 0.3Py = 35.2 + 0.22Py NS seAAG 5 dee cei
0.08Py = 19.2 DLP a8
Py = 240 mL Ces ae
Ri=ay
5. 250 — 140,000 = By Substitute (c) into (a) and (b) and get:
(ax 3) =) =6
350,000 tons = Br (b’) a)
6. yPxl
+2yPB,-2-P = xo+2y-2—Pybyb3 — ye —pyAbl3 x= 1
(49.30) 2.3y = =6
a aa yet 32+ PybyBIS= ye+p + 1y2b/3 —3y = -9
yb/3 .2-—p
Park y=Si. 3
8. (x4 +3)by-2b+4 _ yah +3b,-2+4 _ yhtab+4 (by21) =") S'a= "2
-l+z=2
9 ivy vo E yf) + 2ay-aby2—a z ytt? 2 =a
ae
dg, eye 3)
16.
10. ——P__ 2 —_?
ee he
= am + |
am
es __ Plam +1) peal ~150°
am Fat am? p28
1 | A= 60s 30° = 5.20
fi; Bee 3 bx Dae Oe) es
|e — vi ine
bx — 1 / .
is i" 5
is 1h eon | C
bpx — p — bx C = 4cos 20° = 3.76
He 2 cig, | du em Seis 3 D = 4sin 20° = 1.37
Sih ico ilkaea 1+1 5.20R
+ 3.00U
(es Tas Warr 3.76R + 1.37U
= 5 = “> a 3" 8.96R + 4.37U
229
Algebra 2, Third Edition
Problem Set 84
17. 22. J-3V3 - \22 — J=4 + 3i? - 27°
=3i-2-2i-3-2=-5-i
23. ~2x*
x =3
eeeie foe
1a) —4
24. Agquare ieee mn
s=6m
3
’Circle = ee=3m
Acie = r’- = 1G m)* 2 = 97m za
6 = 69.44°
25. There is never an exact solution for these problems.
(73 /69.44°
One possible solution is given here.
4 2 DN Y = mB + b
18.
(ey ay Use the graph to find the slope.
4(x — 3) iS ee
pe aA sg
Bn) x a 3) ee EB) 6
_ 4x -12 + 2x - 5x3 Vee 50: aah
i ee (ye 3)
Use the point (99, 100) for B and Y.
-5x>+ 6x - 12 100 = -50(99) + b
x” (x - 3)
100 = —4950 + b
4/2 -5 3/2 -2 5050 = b
19.
21),2. BD= 2 Y = -50B + 5050
_ 24 - 15V2 - 8V2 + 10
. 18— 4 26. @ ~2.065
—
500
x 104
x 10
.
=) x04
-5 x 108
=4~x 10>
34 93,12
14 Calculator answer ~ 4.13 x 1075
(b) (84.9)-4-9!
20. (2 + 3V20)(4 — 5/45)
Answer should be between 3 x 107!° and
= (2 + 6V5)(4 - 155)
4x 10710.
8+ 24/5 — 3015 — 450 = -442 - 6/5
Calculator answer ~ 3.38 x 109719
21 Bean ofirs
iG Se y 27. 33 +22 — ava
ee ee Pe. _ 35v3, 2W3V5
Cc x y
mxy + xcy = cpy + cbpx
eae ee
mxy = cpy + cbpx — x*cy Siig ees
mxy
J _- =
—- 25 _oJ8 | sng
py + bpx - x”y 5 5
230 Algebra 2, Third Edition
Problem Set 85
og, 40em , 60s | 60min Lin. PROBLEM SET 85
‘- :- mee aii 1. Equal ratio method:
-
Pane. ys : Rn Gan ie ii, 2b
I, T,
29. 7 — 2(2x + 2) 800 _ 2400
Eee [2 3x = 2")| — SOx es) Sea 4
T= 4% — 47> --2 ~ar Qe Ox + 3 I, = 3 were improvident
—4x + 3 = 3x — 10 — 6x + 3 Variation method:
x= 10 1 = kT
30. -3{[(-2 - 3) - 2] - 2[-4(-3 - 2°)]} pd crated
2°53) :=k
= pfs 2) <= ali6)} = 3
iJ o= 30) =< 3 = 114 Ny; = 5(9)
N, = 3 were improvident
PRACTICE SET 85
2 ee
a. (a) BTp + 9Tp = 36 /———_—_—__+]
a ae = = = 18 ‘ aD
2BT = 54 B
(c) : BT = 27 ee
(a) (27) + 9Tp = 36 RpTp = 640; RpTz = 280;
OT) ="9 Tp = 2Tp; Rp = Rg + 20
Ip Sih (Rp + 20)2Tp = 640
(b) B(1) = 27 2RyT y+ 40Tp = 640
B=27 2(280) + 407g = 640
b. x + y? = 20. (2) eee
Tz = 2hr
2x -y=5 b
4 He Tp = 4hr; Rp = 160 mph; Rz = 140 mph
(b) y = 2x — 5
3. (a) SNy = Np + 90
(v’) y? = 4x? ~ 90x +025
(b) 3Np == Ny + 10
Substitute (b’)into (a) to get:
(a) x2 + (4x? — 20x + 25) = 25 (b’) Ny = 3Np — 10
5x7 — 20x 4 252°25= Substitute (b’) into (a) and get:
x(5x — 20) = 0 (a’) 5(3Np - 10) = Np + 90
x = 0,4 15Np -—50 = Np + 90
14Np = 140
(b) 200) -y = 5 2(4) -y=5
y=3 Np = 10 were demure
y=
(b’) Ny = 3(10) — 10 = 20 were hoydens
There are 2 solutions: (0, -5) and (4, 3)
231
Algebra 2, Third Edition
Problem Set 85
4. (a) 0.1Py + 0.5Dy = 136 Substitute these values of x into (b) and solve for y.
b) y = 2(0) - 4
(b) Py + Dy = 400 lan e
y=-4
Substitute Dy = 400 — Py into (a) and get: 16
(a’) 0.1Py + 0.5(400 — Py) = 136 () y sy (ee5 Ss ee
-0.4Py = —64 7 12
Py = 160 5
16 12
(b) (160) + Dy = 400 (0, —4) and (=. 2)
Dy = 240
- —2x
Guba), nag
160 ft?3 10%, 240 ft?3 50% 8. Tey ae
_-xXpxy
_ a“b
N + 2, N + 4
— b y-2x+2x
=a ,x-2xpx
5. Odd integers: N,
x p,x/3 1-2
NN + 4) == 10CN - 2) - 25 9. ties = gt H2pe3-2 = gryri3-2
N* + 4N = -10N —- 20 — 25 a
N? + 14N + 45 = 0 fe 3 é .
(N + 5)(N + 9) = 0 Bok ae Pees 0
Ni 5-8 ea eaXx cm + x
m
The desired integers are —9, -7, -5 and —5, -3, -1. + x(cm + x)
acm + ax + bm
6. (a) BTp rf 6Tp = 24
(b) Bip Oly = 12 11. —_ = .
2BT = 36 2a. gr want
: rant co +b
BTp= 18 rs
(a) (18) + 6Tp = 24 NOR GC Ry) ee OE)
;
Tp = 1
2c? +2b+c? + 2b
3c?
(b) B(1) — 6(1) = 12 19, 2 Pies chiP = Cuzz GinOr 10
B= 18 SP Sinn 3 teat 9 + 25
16 + 4i 8 ;
= = — + —]
34 ty Se Wo
7. (a) x7-+ y* = 16
(b) 2x -y = 4 ig fm hey Sein
Seay reee 3+ /2i 3-2i
12 — 3V2i — 4/2 — 24 10.0 92
(by = 4° = 16x +16 = Was BAedcetN2t ns2 il Ogeifaiaa
042 tit
Substitute (b’ )into (a) and get:
pees ie : = 14. Jk -32 + Vk =8
(a’) x * ik Ate Se te eg ee EN Ao
Spe s IG)
SS 1)
Ses 16. k= "96
Svar loxten Nie6
xOx = 16) = 0 k = 36
16 Check: 436 — 32 + 136 =
a =) ail se SS =
5 2+6=8
232 Algebra 2, Third Edition
Problem Set 85
AGS 24+ 2N3
(b) 2x -y 432 =6 DO Fas 2423
ic) ¥ ~ zemG _ 8+ 2V3 +8V3 +6 _ 144+ 10V3
y=z A=r(2 ~8
Substitute (c) into (a) and (b) and get: -7 - 5/3
(a’) x + 4z = 6 * 4
(b’) 2x + 2z = 6
d
(a) x+4z= 6 20. c= m2 —p)
~2(b’) -4x - 4z = -12 '
—3x ~6 c="(G _ mp
x= 2
c~2 = dm — cmp
)= 6 ; ;
+ 4z
(a’) (2 cmp |p = dm — .
et,
d —
aes oe m-c
aa
cm
(c)y
=1
WA, =fald — 3) a ee 9
(2, 1, 1)
So for lt = 2 = 3h Se
16
B a a ae ES
‘ J - 2
ele 140 = 120 + x
20°
= 3
(b) 360 — 120 = 240
B = 4cos 32° = 3.39
240 — 120
C = 4sin 32° = 2.12 eee 8)
0.00R + 6.00U
3.39R — 2.12U 1 1
3.39R + 3.88U 2B. ADS Rea mee
ee 3.88
es Gn (b) 0.3x — 0.04y = 2.46
E 3.39
@ = 48.86° (a’) x + 3y = 27
F = (3.39) + (3.88)? ~ 5.15 (b’) 30x — 4y = 246
5.15 /48.86° 30(a’) 30x + 90y = 810
-1(b’) -30x + 4y = -246
94y = 564
17. Tide GeV xin 2 7%
vis 6
tas J 0 ee of , 9, 6a!
sila a oa|
(a’) x + 3(6) = 27
- yx -a
a)
ig BN ee
Soma 2: 54 ae (9, 6)
% 040-4 342 —4 ~ 11-72
2: 15=1002 +902 —4 3 24. See Lesson 71.
233
Algebra 2, Third Edition
Practice Set 86
25. There is never an exact solution to these problems. 100 £2 . 60s . 1? in. * Din TOT.
One possible solution is given here. 28. 3 tani ft ift a
en 2.54cm _ 2.54cm _ 2.540m
Use the graph to find the slope. ‘ > ae % Ga. es iiaa,
—-10 cm?
ion we ees = 100(60)(12)(12)(12)(2.54)(2.54)(2.54) —
Na = -5Mg + b
Use the point (54, 82) for Mg and Na. aha I 3 2 ap
82 = -5(54) + b 12 5
Na = -5Mg + 352 2V3 V3 (V5 V5
5J15 12V15-120V15 _ 137/15
a al as “19 AO Oe ee
1
y= ie a
_90
30, 2
47312 = 452 = (412)5 = 32
PRACTICE SET 86
a. -x £ 4; D = {Integers}
-x > 4; D = {Integers}
x < -4; D = {Integers}
=61- 7-6-5) -4
Substitute y = —3x + 6 into (a) and get:
(a’) x — 4(-3x + 6) = -8 , x74 y%= 11 (a)
Tee 2 5e= 8 "Wawel (b)
13x = 16 (bo) y=x+1
_ 16
pacer
ie
(b’) y S097 ae Peers||
Substitute (b’) into (a) to get:
w y= aE) +6 - (a) tog
13 (a) 07: $8 9 et anid
y
30
3
,
= — 2x + 2x _-- 10
a) = =0
2 =) = Lt YQ) = 4(2)-10)
13’ 13 2(2)
_ 2, 1 P21
Ae 4 ee:
ia S
rt
Xt
Ss
234
Algebra 2, Third Edition
Problem Set 86
y= (46B).
4. Hydrogen: 20x? lvek?
Ohyextl
1 Carbon: X12 22
2 2 Oxygen: 3 x 16 = 48
lr Tet Total: 24 12.4 48-= 62
ya eth
~*~ ie 372
62
y=x+l
62GE=aI2G72)
1 21 C = 72 grams
ae pee enn ot: |
2 a
Lo eat 5. RN = 0.14 - 120,000
"orl iim RN = 16,800 ducks
6. Acircle = mr? = 251 m?
(e + N21 1 + 1) ng
Beistene. 2 Bk ee Paty)
(
fyDla azt, ie a) r=5m
2 Length of diagonal = 27 = 2(5m) = 10m
Length of side = ae . 4 = 5/2m
PROBLEM SET 86 yess
Area of square = s2 = (5/2 m)? = 50m?
Waa
1 7 a i. 7. x £—2; D = {Negative integers}
jc SS a?
400
BS
Vy
aa IEE SE dan ee
1273 273 =3020 S404
% = 400(2273) _ 714.22 liters 8. x + 3 4 2; D = {Reals}
1273
aap 8 By SD)
3 Dr =e S Sil
braee ir ee
D ——|——9—$_|
—}—
P OG 1, 2%ae4
RrTr = RpTp; Rp = 20; 9. -x — 6 2 -3; D = {Negative integers}
Rp = 2; Tr + Tp = 11 aia< 3
—x < 3
(20)Tp = 2(11 — Tr)
Te =922 — 27h oe =
22Tr = 22 Lakegecat.t* OF 4
Tr =-1 10. (a) BTp + 3Tp = 60
Dr = RrTg = 20(1) = 20 miles (b) BTp — 3Tp = 36
; 2BTp = 96
ae
Since Dp is only halfway to the swamp, the total
(c) BTp =
distance to the swamp is 2Dp or 40 miles.
(a) (48) + 3Tp = 60
eT es le
3. 0.5(100) + 0.2(Dy) = 9.23(100 + Dy)
50 02D y= 23 POULIN Tp = 4
27-= O.03Dy (c) B(4) = 48
= Dy | B= 12
900 liters
235
Algebra 2, Third Edition
Problem Set 86
ll. @) x’ +y =4 pg aaa ae
(b) y-x=1 , 5
y=x+1 N + 22 = 45
(b’) y2 = (x + I? = 2x7 + 2x41 N = 23
Substitute (b’) into (a) and get:
(Qe ao + 2x + 1) 4 18. Jp + 48 = 8- yp
2x? + 2x -3 = 0 +p
p+ 48 = 64-16/p
Solve this equation by using the quadratic formula. 16,/p eat
x =
SAO),
eoCaE OT
wale, VI Jp =1
2(2) 2 2 p=1
Substitute these values of x into (b) and solve for y. Check: +
4/1 Aas Ge fit
1 si) ee ll
if =e
ab =}|—-—
we +—]+1
ep:
Pigg 19. + 2y-—-z=0
(a)x
2 2
1 7 (b) 3x + y — 2z = 3
wy=(t-) a1
2 4 (c) 2x —-z=0
pee a Sk z= 2x
2 Sipconeg
Substitute (c) into (a) and (b) and get:
Yo i: aa ean 7 - _ j
2 a7? 4 2 (a’) x + 2y =0
(ees (@) eae
2 Ppp oP 2
(a) =x + 2y= 0
baal GL Weaea mea yo? 42 y e7ai5 Sabi3 -1(b’) gee yells
Peg
a or. er ia es
ab/3 ee >
ye uy y=-3
13. eid = Lae — y3at2 (a’) —x + 2-3) = 0
x27 4 2-—a
x = -6
14. —_+— Se : ~,
Sa=eeSS = -12
(c) z = 2(-6)
bee i ax” +x+a fae a5
x+=— ea ax +1 Sore?
a
o ax +1 20.
+x +a?
ax?
15 Pt) iege eet é -10 — 2i + 15i — 3
—5+i -5-i 25 4-4
=;
—13 + 131 = SS
1 + —I
26 2
i, 2+ 2! Sti _ 15 + 3i + 10: - 2 A = 10cos 40° = 7.66
Sb DSA 2p + : 4
B = 10sin 40° = 6.43
a=
13 + 13% =
| Ls
—+-—1
26 De ok
236 Algebra 2, Third Edition
Problem Set 86
2A! Eo16 = 3y 53+ en -s ef
= 3-T7i
: ] 7
—220 25. 5,|2+ 3 [2SIG
Gr=s10\cos402 >= 47.66 seal Deb gig AMES re
D = 10sin 40° ~ 6.43 Th AT ey Dean)
-1.66R + 6.43U _ lovi4 | 21V14 _ S6vi4 _ _25V14
-7.66R + 6.43U ay (do lA ee 14
-15.32R + 12.86U
21. 26. 3x? — 1 = 6x
5 1
K+ 2x4 |= =70
3
, 1
(x° + 2x + ) = -=
3
Crt Ox leeeel
5)
(oe ) owes ee3
H = N10?
+ 3? = 109 PR
x he
s
pete eae
3 x= Syn
@ = 73.30° 3
Since @ is a third-quadrant angle,
40 in.? pelt, piaieties) tydi eld
rth 180 = 253.30° 27. eee ee
an eatin Fe aie a ake ft
v109 £253.30" ,, Omin _ _ 40(60) yd?
Paesepiaahea + Ww 1 hr (12)(12)(3)(3) hr
2, ——___ - ——_=
Merels. 14 34/3
2
eget 943 — 213 218 15-21 1173 28. — 1 = 6x
-2x°
[97 —26 x? + 6x +:1=0
A+ 11435
pie TE -6 + /6% - NT
26 ia ey Oe ee a
2(2) huge ey)
x+2 (¢
23. = (S| = Ges
y b
Bt eon
y b
bx + 2b — bcy = amy + bmxy
bx + 2b — bcy — bmxy = amy
amy
=b
x +2 -—cy — mxy
237
Algebra 2, Third Edition
Problem Set 87
29. (a) 2x — Sy = -15
pore
2 9
Saket S
Ges
ee
(b) 3x + 4y = -4
3 2 = -—(7)
+b
oy = -—x-1
re:
2 ep
lA 2
3 25
= —-—x + —
Pee
PRACTICE SET 87
J)
~\
on
=
xj = —3; Vi = 96; Xo = —11; v7 = 49
|
[||
aon
eS
pes
———————— SS
(49) — (96) _ 49 - 96
X,-xX, (-11)-(-3) -11+3
Solve the equations by elimination:
ae!
—3(a) -6x + 1I5y = 45 . 8
2(b) 6x + 8y = -8
23y = 537
eel PROBLEM SET 87
23
D,
{-————+]
(b) 3x + (2) = -4 1440
23
Dg
ape eel, 120
23
x=
80
-—
R,T, = 1440; RgTp = 120;
23
Ry = 4Rp; T; = Tz + 4
4Rp(Tp + 4) = 1440
Gye
23° 23 4(120) + 16Rz = 1440
16Rp = 960
30. Write the equation of the given line in slope-
intercept form. Rp = 60 mph
4x -— 6y = 25 R; = 240 mph; Tz = 2hr; 7; = 6hr
—6y = —4x + 25
Equal ratio method:
ee HS; SRA
Gs are
BS, opti,
10 _ 200
Since the slopes of perpendicular lines are negative
reciprocals of each other,
HS, 400
HS, = 20 hours
238 Algebra 2, Third Edition
Problem Set 87
Variation method: si (x1, 91) = (-2, 108); (x, yp) = (—21, 47)
ne m=
Yy- My _= 47-108 ==_ -61_=
61
H
X,-x, -21-(-2) -19 19
10 = HE.
400 x £ —4; D = {Integers}
k = 4000 x 2-4
4000
H ——— 500 = 2 0 hours =5"=4 -3 -2,-1
-x + 2 < -3; D = {Positive integers}
3. (a) 0.8Py + 0.3Dy = 240
x < -5
(b) Py + Dy = 600 x IV 5
Substitute Dy = 600 — Py into (a) and get:
Ce pl5) alse Th ols)
(a’) 0.8Py + 0.3(600 — Py) = 240
0.5Py = 60 10. (a) BTp + 4Tp = 36
Pye 120 (b) BT = 4Tp =. 12
(b) (120) + Dy = 600 ORT em ees,
Dn = 480
(c) BTp = 24
(a) (24) + 4Tp = 36
120 in.> 80%, 480 in.? 30%
4Tp = 12
4 (a) 10Ny = Np = 102 Ty = 3
(b) 100Ny = Np + 168 (c) B(3) = 24
(b’) Np = 100Ny — 168 Bra: 8
Substitute (b’) into (a) and get: 11. (a) x7 + y =4
(a’) 10Ny = (100Nqy — 168) — 102 (b) x - 2y = 1
-90Ny = -270 y= 2y-+ 1
Ny = 3 holographs
(b’) x? = 4y? + 4y + 1
(b’) Np = 100(3) — 168 = 132 printed ones
Substitute (b’) into (a) and get:
5. Potassium (K): exes 9e=— 239 (a’) (4° + 4y + I) + y=| aS
Chlorine: exes) =)55 5y* + 4y -3 = 0
Oxygen: 3 x 16 = 48 Solve this equation by using the quadratic formula.
Total: 39 + 35 + 48 =1122 edit f4® = A(S)(—Bm =a2 19.
(sae 2(5) Kaean ae
Percent n Of Rea ae
= — eX 100%
ie ioefa Total Substitute these values of y into (b) and solve for x.
= 18 100% = 39.34% Oi pee
5
ae
5 5
122
48 © O76 xe (-2- 49) .1 -12- 48
122 KCIO, 5 5 5 5
48KCIO3 = 122(576)
iB+ 2V19 2 + 19 )ang
KCIO; = 1464 grams 5 5 5 5
er oe |
6. Refer to Lesson 87. 5 5° § 5
Algebra 2, Third Edition 239
Problem Set 87
a,,2a+2..-b 19. Vz —-V¥z-45 =5
12. Le, = mott2y-2
y dz —5 = Vz— 45
+25 =z— 45
z—-10Vz
13 (m2) -y = m24~ %y = m% ~ &y-5
10/z= 70
oo my? +1
Vz =7
z= 49
a a a(a* + b)
14,2 =.
—— =
ee a a at+abt+a Check: 49 — 49 - 45 =
b GateG
Ze = a’ +b 7-2=5=
a
be a> +b 20.
a*>+bta
B 5
15 a eee oe Ee ol et eet
ee ee ees 121 72°
a aes
2 2
A.= 5 cos 72> = 1:55
16 Bot 2 ele Osi ai
B = 5sin72° = 4.76
oe Laat
4 The -1.55R + 4.76U
=—- —i 8.00R + 0.00U
5 5
6.45R + 4.76U
17. (a) (9315 x 10°)(-2.065 x 10%) 4.76
tan 0
Estimate: (9 x 10°)(-2 x 10%) 6.45
St se 10" 6 = 36.43°
Calculator answer ~ -1.92 x 1011 F = (4.76)? + (6.45)? =~ 8.02
(b) 74/1001.94 8.02/36.43°
Answer should be between 10 and 31.
21.
2-3 ¥3-2 _ 2v3 -4-3+42v3
Calculator answer ~ 12.92 Va 42, \ao 2 34
= 443
=i)
18. (a) x+y-2z=7
(b) 3x -y-z=3
22. -3i(ii) — /-2J-3 = 6 + 3i
KC) 2607 =
23. a2 = 32 — 4 189
Z = -2¢
NE eek eesali)
Substitute (c) into (a) and (b) and get: = a eee) 1901
(a) Sx 49 = 7 3 NORE Aa)
Tam
(bh) 5x. y=, 3 _ 21V21 | 921 _ 252V21_ _74/21
10x east)
mil 21 Re oe
x=1
24.
(a) S@)P Fay =-7
a bcy + 4bc acmp + bmp
bey + 4bc — acmp bmp
(c) z = -2(1) = -2
bmp
(1, 2, -2)
by + 4b - amp
240 Algebra 2, Third Edition
Problem Set 88
25. y = 2(70) — 80 = 60 2. Tin (Sn): tex 71119. =41 19
2x) =2 360 = 2(75) Chromate: lx 52 = 52
2x = 210 Oxygen: 4x 16 = 64
x = 105 Total: 119 + 52 + 64 = 235
26. aay hae St elec
423 Poy ¢ ao 239 SnCrO,
F SnCrO, = 1175 grams
er caiss a OG a Pre
2(3) 3 an 3 3. DG
(oe
OH.
10km ~
1000m x
100¢m__lin.
a
ee
hr 1 km lm 2.54 cm Dp
1 hr Imin — 10(1000)(100) in. 400
x x =
60 min 60 s (2.54)(60)(60) s
RpTp = 400; RgTg = 1120;
28. See Lesson 71. Ripe R20 aT ore
OO ao45 933256 = 3G)Gl2) = 3178 (Rp + 20)2Tp = 1120
2(400) + 407, = 1120
_9-3/2 mt(ep) =O 1
30. (27) = LPR Cl gL oy en
~(9!/2) gp meee 40Tp = 320
Tp = 8 hr
Tg = 16hr; Rp = 50 mph; Ro = 70 mph
PRACTICE SET 88 : 2 S
a x = 4,9, = 23%. = 3392 = 1 4 FY,
_ tok
ney: [
D= \e, ~ ay + (y, a yy)? l 2
Siar 2
Pie 0) Z
See PVT,
ib)
so Ba?
= saa at y, < (740)(10)01473)
= SS 2 (573)(1480)
b. ipy HRT we ~ 12.85 litters
=< PY = _ Aa moles = 0.61 mole 5, PV = nRT
RT (0.0821)(159)
PV
—}$ =n
RT
PROBLEM SET 88 n= J OES mole = 0.24 mole
(0.0821)(251)
1. Evenintegers: N, N + 2, N+ 4
NWN + 4) = 6-N — 2) + 12 6. Refer to Section 88.A.
N2 + 4N = -6N Es 2 sy
N2-4-10N =-0 7. DCP) + (y — 2)
NIN + 10) = 0 (5 8)? + (2) = 3)
The desired integers are 0, 2, 4 and -10, -8, -6. | - /64 +25 = /89
241
Algebra 2, Third Edition
Problem Set 88
8. —x —- 2 < 4; D = {Negative integers} (Cie ta caotea Z 4a +4 +(a/2) -apl 2a
=x 6 13. y4p24 me
Xx IV
IA -6 = y(7a/2)
+4,1-2a
27-654 —3ee-1 0
r r _ rd + rm)
14. z 5) a D 2
a —x + 2 % 3; D = {Negative integers} ae i a Me r mt+rm” +r
x+2< 3 ;
x<l
x 2-1
15. pe 3 cp
hh
or) C
aren err na ee
3-2-1 0 1 Gay
| Os 1 Dez pe a
ig ee
pe
rie
10. (a) BTp + 6Tp = 22
(b) BTp bce 6Tp = 10
9 SPC BOAOO _op(cp*
= 90.
2BTp = 32 cp?-p-cp plep? - c? - 1)
(c) Bly, = 16
_ _e(cp* - 1)
(a) (16) + 67) = 22 cp” |
6Tp = 6
to eee | Aga! Oh ae -12 —- 31+ 8i —2
16.
3i-2 31+2 9-4
(c) B = 16
=i Sua 1d | 3,
11. —13 13 13
x=yrt+l 5712, <0 ~
lige Fae
0.0416 x 10
(Dees
pl) aay ety al
1
Substitute (b’) into (a) and get: Estimate: aes) SKS
G)i67 + Wy 4 1) ey =o (4 x 10°)
yt yet = 0) Calculator answer ~ 1.37
Solve this equation by using the quadratic formula.
(b) (184.3)-}-62
2427-401) 1 mad
Answer should be between 5 X 107 and
2) 2
25.310 ©,
Substitute these values of y into (b) and solve for x.
Calculator answer ~ 2.14 x 1074
(b) x See Pe Fe Naas.
2 2 2 2
18. Vz-35 +z =7
ten[ ek~ 2s] 5 aE
y 2 % y) z-35
= 49-14/z +z
(4eed: 7 oe 14/z
= 84
pe 2 2 2 Vz= 6
(4LAs gi “3 zZ = 36
oe vie? 2
Check: +36 — 35. + 36 =7
12. Cae eee see y24+2b-2a+b,a—ald
yi = » 1+ 6
in x3) y3a/4 7 iH] sw
ll
242
Algebra 2, Third Edition
Problem Set 88
19. (a) 2x + 2y-z=14 Hite y12?e+ 4% =) i/1600= 4.10
(by "3x + 3y + z = 16
tan 0
12 =
(ce)
x= Qyp=8 4
x = 2y 0 u Ta
Substitute (c) into (a) and (b) and get: Since @ is a fourth-quadrant angle,
(a) 6y — z ="14 6 = 360 — 71.57 = 288.43°
(b’) 9y + z = 16
4/10 /288.43°
1Sy "30
ye ad 22. Ai? = 5/29 4= 33
(c) x = 202) = 4
9
4 ae
3
9 7 5N27 = 43 + V3 - 1503
(a) 2(4) + 222) -z= 14 23.
12-z= 14
= -10/3
z= 2
(4, 2, —2) 24.
1-V2 34+2V2
_ 3-3/2 +22 -4
3 0s) 2g a3? 9-8
20. Re = =1 40
s A
SJ Pe
an ~ noe B 4+ V3 1+ 3 _ 443+ 4v3 +3
DRS,
—200°
20
[aa 1 +3 ieee
=i SNS
A = 20 cos 20° =918.79
2
rtd
= 20 sin 20° = 6.84
Loe 1 1
C 26.
=e, B
c Cees 2
Aone
2g aye Py
D Cc ue r
30
x?r + 2xyr = cyr — cxy
cxy = cyr — 2xyr - x?r
C = 30cos 30° = 25.98
cxy
D =*30 sin 30° = 15
cy — 2xy =e
18.79R — 6.84U
25.98R — 15.00U 27. ah er2 =O
44.77R -— 21.84U
x? + 2x = 2
21. ¥ Gr 22s 441) = 21
te File oa
x +1 tr
toe <li
& b 5)
10 mL 60 s lcm 1 in. 1 in.
28. x x x x
S 1 min 1 mL DSA Ci) 2o4,6em
Dalton es 10(60) in?
OE (2.54)(2.54)(2.54) min
243
Algebra 2, Third Edition
Problem Set 89
PROBLEM SET 89
29.
1. Equal ratio method:
en
A, 8B,
~ Il | nA5) 500
_10
An. gas
A, = 2100 altercations
Variation method:
A = kB
500 = k(10)
y + 83 = 180 50 =k
SA
Il 97
A = 50(42)
ote lan = 4 A = 2100 altercations
30.
pe 3
-9x —-6 — 4x + 8
—13x
RrTr ss RywTw = 65; Rr = 10;
Il
Rw = 5; Tr = 9 - Tw
109 494) + 5ky a>
PRACTICE SET 89 90 - 10Ty + 5Ty = 65
25 2857
—x-4<« —1 and x - 1 < 1; D = {Integers}
5 = Ty
—x <3
Dp a), aed
x = —3 and 36 Cae
Dp = 40 km; Dw = 25km
VLaQ aij, OF 2
0.79(800) — E = 0.30(800 — E)
=1 <x = [sy5: 2D =f Reats} 632 — E = 240 — 03E
=< ~ — leandesx
— I lA 392 t= OE
On<x and x lA a 560 liters = E
Sey (a) 10Np = 2Ng — 140
7x=3-4 (b) Ns = 3Np + 10
ths
= M2
(b’) Ny = 6Np + 20
12
eS = Substitute (b’) into (a) and get:
7
(a’) 10Np = 2(6Nz + 20) — 140
4(4
+y)= 10Nr = 12Np + 40 — 140
16 + 4y = 100 = 2N;
4y = 50 were fast = Np
y=
(b’) Ns = 6(50) + 20 = 320 were slow
244 Algebra 2, Third Edition
Problem Set 89
740 Substitute (b’) into (a) and get:
—— x § = 592
100
(a’) (y? + 4y + 4) + y = 3
S = 592 x a ay + 4y+1=0
740
S = 80 spotted ones Solve this equation by using the quadratic formula.
-4 + 4? — 4(2)(1) 2
x-S5< -3 and x - 3 < 1; D = {Integers}
2(2) 2
-x $2 and ee
x = -—2 and Substitute these values of y into (b) and solve for x.
ae cel
(b) x = [seen pel
mae ca He The aie2tiaehs rk Z 2
-4<x-4<
1; D = {Reals}
wx=(1-2)+2- je
Oreo 55
Ct Dia
St kG
wee OF -X < [3 De= {Reals}
2b/..a + 2\1/2
Syrce?, BOM tem 1 13. hi ail 0 tite di ee 14+ .4/2 +1,2b
~3b
ee = y
>) 0G a> —|
= x otal 6
—-3 2-10 1
14. Cee wees: a a* +2a+t+a* -2-a
7x = 2(14) I+a iat!
y
i 25 = y2a?+a-2
x= 4
k - k
10. 15.
m aa ))
m + 1 m + by
a. — am +1
m
a k(am
+ 1)
am2 + m2 +m
11. (a) BTp Sie 5Tp =
(b) BTp — 5Tp = m 7 m m(be
+ da)
16.
2BTp = ee ea ox abc + ad + cx
peed, Ae be+ d
(c) BTp = a
(a) (42) + 5Tp =
ate (ean MepOt 2) 400 BE
M7.
i-4 i+4 Liat 6
(c) BB) = 42 why ty SE, atl Se
B=14 ae ee, yan
12. (a) x” + y = 3 i= Shai A Bi 2 te 2
18.
41-1 471+1 -16 -1
(b) x -y=2
x=
y +Z ee ol ae es
(b’) x2 =(y + 22 =y* + 4y +4
OTA, Viel 7,
Algebra 2, Third Edition 245
Problem Set 89
19. (a) 3x + 2yv+z=9 F = (3.13)? + (-7.47)?
(b) x — 2y - 2z = -3 F = 8.10
(Cc) ~+z=0 8.10 /—67.27°
z = -2x rs
Substitute (c) into (a) and (b) and get: 22. 4 + 23 ; 3/3 +2 - 12/3 +18 + 8 + 4V3
Rigae BB! Pee
(a) x+2y= 9
(b’) 5x - 2y = -3 _ 26+ 16/3
6x = 6 pe
ie = 1 ae
S
(a’) (1) + 2y = 9 23. ate et
xy = amx + cmx + acmy
(c) z =-2(1)
Si es
am xy — acmy — c’my = amx + cmx
aay = amx + cmx
20. Vs = 3+-+5 - 21 x — acm —c*m
SAG NAO Sts 21
30 = 6Vs 24. x _ mx + mcy
Beis atc y
23=" 5 xy = ric cmx + acmy
+ c’m
Check: (25 = 3 + /25 — 21 : 2
5 nae xy — cmx — c*my = amx + acmy
xy — cmx - c’my
2A. A —#—_@\———— =a
mx + cmy
/ 6
~ oe
25, 37° 45) = ./20
J2 = =3p.
ne ores
B 8
26. AcsB am
B35 aldem “Sian
= ee ee
A = 8cos 69° = 2.87
PNeDN 1 7 v 3
B = 8sin 69° ~ 7.47 3.16
=-+ : ate
~2.87R — 7.47U
6.00R + 0.00U 4 906. ae _ 48V6 _ 76
3.13R -— 7.47U 12 12 12 12
tan@ = ciel pg am ig he joy
3.13 ;
@ =~ -67.27° O = 7x* —-x — 5
2.13 Pee LC)
ER ae Feel
Veo 2(7) 14
3 5 :
m 7.47 28. 10 ft . 60 s & 12 in. e 12 in. . 12 in.
s lmin 1 ft 1 ft 1 ft
in.?
= 10(60)(12)(12)(12) ——
min
246
Algebra 2, Third Edition
Problem Set 90
29. (6x + 20)° + (4x + 10)° = 180° PROBLEM SET 90
6x + 20 + 4x + 10 = 180
10x = 150 : f
x= 15 300
(4x + 10)° = A° De
AGS) 410 =e Beets
7 =A
mesh Shes s ; eeae “B RpTp = 160;
6(15) + 20 = B ra
110 = B (Rr — 50)(5Tp) = 300
5RrTr — 250Tp = 300
30. (xy, 1) = (-2, 3); (x9, y) = (8, 4) 800 — 2507p = 300
Do) Oey 500 = 2507p
Dews [f(y + (4 = 3y 2 = Tr
D = ¥100 +1 = 101 Rp = 80 mph; Rp = 30 mph;
Tr = 2 hr; Tp = 10hr
PRACTICE SET 90
2s PV SARE
x-—-2y+2z=2 “(a) ie
Byes 5 (b) V= cpa
ax — 294-22
= 2. (Cc)
_ (0.832)(0.0821)(400) . | :
(a) x -—2y+2z=2 V= Ss ore liters ~ 9.11 liters
(c) 3x — 2y - 2z = 2
4x -— 4y =/4 3. (a). Nr = 3Nz - 5
(d) x-y=l (b) 6Ng = 10Np — 70
(a) x -—2y+2z=2 -2(a) -2Np = —6Ng + 10
2(b) 2x + 2y — 2z = 6 (b) 10Np = 6Nz + 70
a i? 8Np = 80
pies B Np = 10 red boats
3
(a) 10 = 3Nz — 5
(d) ($)-y=1 15 = 3Npg
8 Nz = 5 blue boats
y Shore
4. (a) 0.1Py + 0.4Dy = 104
Peery (b) Py + Dy = 800
8 5 Substitute Dy = 800 — Py into (a) and get:
w (2)+(3)-<=3 ‘ F;
3 3 (a’) 0.1Py + 0.4(800 — Py) = 104
G4
ae = WE
—0.3Py = -216
3 Py = 720
4
qin € (b) (720) + Dy = 800
Dyn = 80
85 4
3°3°3 720 liters 10%, 80 liters 40%
247
Algebra 2, Third Edition
Problem Set 90
(c) 2(3) —2y +-C2yp= 0
— x SP = 672
6-2y-2=0
16 ae=Dy
~Ae, iH 672 x ane 1536 spotted ones
De ay.
6. (a) x4 y— 7 =93
(3,
2,-2)
(b) -x - 2y - 2z2 = 0 -1 <x -1<
4; D = {Integers}
ONS <e5
(c) x — 2y - 27 = 4
(a) x + y= z=3 =i sO ol? e2aes eas
(b) —x - 2y - 2z = 0
x-220
0x—2 >2; D = {Reals}
(d) -y - 3z =3
56 4 0) Ot se SS Al
(b) -x - 2y - 2z = 0 ee Ole 5g etl
(c) x-2y-2z7=4
(e) aAy ag 4 ehwel
4(d) —4y - 12z = 12 10. (a) BTp Gi 3Tp = 728
-l(e) 4y+ 4z=-4 (wills = cis, Sale
-8z = 8 2BT, OM wat4e
z= -l (c) BTp = 22
@) = = 36)=3 (a) (22) + 3Tp = 28
3Tp = 6
=O Tp =2
(a) (Orel) (c) B(2) = 22
x+1 B=11
52
11. (a) ee y = hs
(2,
0, -1)
(by) ya =4
(a) 2x—y+7= 2
x=y=4
(b) x + 2y + 2z = 3
(b’) x° = (y - 4 = y? — By + 16
©) 24 = 2y + we ©
Substitute (b’) into (a) and get:
2(a) 4x — 2y + 2z = 4 (a’) (y* - 8y + 16) + y? = 18
(b) x + 2y + 2z = 3 Dy? — Syr— Des 0
(d) 5x + 4z=7
yy
(b) x st Qyet
27 = 3 Solve this equation by using the quadratic formula.
(c) 2x —-2y + z=0 4+ 16+ 4
y= oa es = 5
(e) 3x + OZ 3
3(d) 15x + 12z= 21 Substitute these values of y into (b) and solve for x.
-4(e) -12x — 12z = -12 (bye = 2 445 ea
3% = = 2) ob BS
i=-3 (b) x = (2- V5) - 4
(ce) 3G) 4532= eS = 5
94-37 = (<204 5.24 5/5) and 2= 5, aes)
3z
= -6
4 = J) 12. oe wig? F302 425 — 34 2 mays
248 Algebra 2, Third Edition
Problem Set 90
—b)\2
13. y°(m~”)
— yom=2b 4 + b/2 = yom=3b/2 C = 6cos 60° = 3
m
D = 6sin 60° = 5.20
x a xX 3.06R + 2.57U
14.
as _ me 3.00R + 5.20U
inh mx — | 6.06R + 7.77U
Di
ie x(mx — 1) _ mx - 1 20. tan0 =10
a4
mx? — 2mx m?x
— 2m
S uv 68.20°
(be P a P p(x
- p)
a Sap 2 px? x? — px — px?
i-= x-p
x
ha 2k. it i, ese 10Pa4i-12
16.
-S$-i -S+i 25 +1
aa eee
26 48626
17. Since the mAC = 120°, the mABC = 240°.
Length of 240° arc = = x 107cm = a) pate Since @ is a third-quadrant angle,
360 3
6 = 68.20 + 180 = 248.20°
18. 45 == 15 = 3 H = (-4)? + (-10)? = 2/29
Sa 6s 29 = F57— "15
229 /248.20°
24 = 6s
4=s 21. —2-/3
jee?
24+2V3 _ -4- 23 - 4V3 - 6
t 2a A 12
16=s
Check: /16 — /16 — 15 = 3 _ -10 - 6V3 _ 5 + 3V3
Ale
Ka = ee
19, 4+V3 34+2V3
_ 12 + 3V3 + 8V3 +6
v9Fe
enki ee eee Cent
_ 18+11/3 _ -18
- 11/3
*e =3 rd 3
A = 4cos 40° = 3.06 23: -2i?7 + J-4/4 - J-3V-3 - 27°
B = 4sin 40° = 2.57 =24+4+3-21=5
+ 2
24. |x | 3 Il v |
aeels Sle
3
Lagan pegnys aa
c b
bcex — bemy = bpry — cpy
bcx — bcemy + cpy = bpry
te bpry
bx -— bmy + py
25.
3194/3 = (3231/43 = Gy 3 33/4
249
Algebra 2, Third Edition
Problem Set 91
26. (a) 3(12) = 4(44 + x) PRACTICE SET 91
36 = 16 + 4x
Vereee
Q0N=—4%
1
< —-x -2
QS Sa
ple
The first step is to graph the two lines.
a7. -~6x7 -x-5=0
ite)
Nn
& + + Il
aN
A. O|& Sas 6
(x7 eae
+= + —
al ) —-———
120, + —4
6 144 144 144
hc ee,
12 12
The region we wish to find is on or above the line
y = -x — 2 and on or below the line y = }x — 2.
1 V119. This region is shaded in the figure.
x= —-— tH 1
12 12
y Sex
=3
1Dsine 12". 12 in.
28. 42 ft? x x x ysx+3
1 ft eft 1 ft
The first step is to graph the two lines.
2.54 cm 2.54 cm y 2.54 cm
x x
1 in. 1 in. 1 in.
= 42(12)(12)(12)(2.54)(2.54)(2.54) em?
29. A = y13* —122 = 25 =5
13 x SF = 19
Sie
13
Rs eee eee
Be 43
30. Write the equation of the given line in slope- The region we wish to find is above the dashed line
intercept form. y =x-3 and on or below the solid line
Lee ODY =a, y = x + 3. This region is shaded in the figure.
1 4
sey 2S
ye hag &
PROBLEM SET 91
Since the slopes of perpendicular lines are negative
reciprocals of each other,
nae
m, =5 i, 7,
y = 5x + b P, wa PT,
7 = 5(-2)
+b i,
17 =%B ae (740)(1600)
, = +
400
y = 5x + 17
P, = 2960 mm Hg
250 Algebra 2, Third Edition
Problem Set 91
2. (a) O.6Py 4503Dy = 126 Equal ratio method:
(b) Py + Dy = 300
oe ee
Substitute Dy = 300 — Py into (a) and get: G, A,
(a’) 0.6Py + 0.3(300 — Py 27126 35 50
0.3Py = 36
G, 70
Py'= 120 Gy = 49 were glabrous
(b) (120) + Dy 300 6.
(a) y2—x-1
Dye 180
180 mL 30%, 120 mL 60% ) y< 3x41
Arsenic: 4 xX 75 = 300 The first step is to graph each of these lines.
Oxygen: 6 X 16 = 96
Total: 300 + 96 = 396
Arsenic: an x 100% = 75.76%
396
D,
e—— ++]
3000
<>
Ds
800
The region we wish to find is on or above the solid
T, = Ts + 1; R, Tr = 3000; Ry, = 3Rs; line and below the dashed line. This region is shaded
RsTs = 800 in the figure.
3Rs(Ts + 1) = 3000 7.
(a) Vie eel
3RsTs + 3Rs = 3000
(b)y
Ss x\+ 2
2400 + 3Rs; = 3000
3Rs = 600 The first step is to graph each of these lines.
Rs = 200
R,, = 600 mph; Ry, = 200 mph;
Le = 5 hr; Ts = 4hr
Variation method:
Ge
The region we wish to find is on or below the solid
line and above the dashed line. This region is shaded
in the figure.
~ G = 49 were glabrous
251
Algebra 2, Third Edition
Problem Set 91
8 (a) x + 2y-z=1 12. (a) x2 + y? = 12
(b) 2x — yor 2 =. 9 (b)x+y=4
(c) x -— 2y - 3z = -9 x=4-y
@Qax2y- z=) I (b’) x7 = (4 - y? = 16 - By +”
2(b) 4x - 2y + 4z = 18 Substitute (b’) into (a) and get:
(d) 5x + 3z = 19 (a’) (16 - 8y + y*) + y? = 12
(ayer yee ol 2y* - 8y +4 =0
(c) x - 2y —- 3z = -9 ee Ay th2 0
(e) 2x — 4z = -8 Solve this equation by using the quadratic formula.
2(d) 10x + 6z = 38 4+ 16 — 4(2)(1)
iT Baan
|)
—5(e) -10x + 20z = 40 2(1)
26z = 78
=2+ a = 2a 2
a= 3
(e) 2x - 4(3) = -8 Substitute these values of y into (b) and solve for x.
Ase
=a a (b)x = 4—(2
+ 2)
cae tu Aa
BS RS AD
(a) (2) + 2y - (3) = 1
2y
=2 (b)x = 4- (2 - V2)
x= 4-242
Sa
Mage te
(2, 1, 3)
(2 - (2,2 + /2) and (2 + /2,2
- 2)
9. 4<x+4< 6; D = {Integers}
b\2-a ab
Oe
52 eH 13. (x ) x s 2b — ab + ab — ab/2
x abl2
~ x 2b - abl2
=| @ 7. 2 2
10. t+ 2 ED Or x + 3) <.6; D. = {Reals} a is a
1 ‘ x
38 Soe DSS sornay eiesh GS b+ i b+ 5
5° SS) oe Kiel x
a(cx + 1)
Ome e253 4
bex* +b +x
11. (a) BTp + 2Tp = 51 m = m
(b) BTp - 2Tp = 39 15.
ma 2 © D
— a a+ —
2BTp
= =«90 CC Qa aint
(c) BTp = 45 a
m(a?
+ m)
(a) (45) + 2Tp = 51
a> + am + a2m
2Tp = 6
Tp = 3 16
: Absce Te ee —2+7i - 4i-14
(c) BB) = 45
i+2'°4@6e “a2 ae
-16 + 3i 16 3
B= 15 25 = ig” “wus
252
Algebra 2, Third Edition
Problem Set 91
17.
3i- 6 -2i-1 tan@=
207
—28+1 21-1 2.94
6 + 12i - 31 + 6 6 uv= 41.16°
eS P= 4057) + @94)- = 3.90
=
12+91 12 9 = ol= = C—'! 3.90741.16°
—5 5
23. There is never an exact solution to these problems.
1-2 4452 One possible solution is given here.
18.
4-5/2 4452 O = ml + b
4-4/2 +5/2-10 -6+-2
Use the graph to find the slope.
16-30 ne 34
LOO sO Bie / 0.5
6 - /2
30 — 20 10
34
O 71 +b
19. Use the point (20, 30) for J and O.
Sear oie =; 30 = 720) + b
XG
-110 = b
20. iH} Q >
my
+
Ql
——~ Si O = 7I - 110
x
iH] | a = oe oe (oa). mg T 3
my a 24. 72/3 ~ (273) Te0. ee
ax = dmry + abdmy
ax y(dmr + abdm) 25. 54 3i3 — J-9 + J-4 - J-2/-2
=f = 31 — 31+ 2b 2 = 2 — OL
7 ek
~ dmr + abdm
9 5 6
26. STB
[2 3/3 +8 =5 Not BS et OND
21.
3+ HT} Q
oNSoi
—
+ S
ee _ V5 , SVS | 45V5 _ 56v5
| oor.) 5 5 5
T =
27. See Lesson 71.
ax = dmry + abdmy
ax — abdmy = dmry 400 mL 60s — 1cm? 1 in.
28. x x x
a(x — bdmy) = dmry Ss 1 min 1 mL 2.54 cm
1 in. 1 in.
dmry x <———
2.54 cm 2.54 cm
x — bdmy
_ 400(60) in.
22. ~ (2,54)(2.54)(2.54) min
4
3x7- 6x + 10 - —8,
aera
? . E 29. v4 2)oe Or re 2
3x3 + 6x2
A = 4cos 40° = 3.06 ca mh
6x? — 12x
B = 4sin 40° = 2.57 10x + 2
10x + 20
—3.06R + 2.57U
6.00R + 0.00U
-18
2.94R + 2.57U
253
Algebra 2, Third Edition
MM @x-
x = 390 +3D PROBLEM SET 92
x= 36 ii (fa) (8 + WTp = Dp
x=6& (10 + W)T> = 70
() 22 + 4 = 1+ ®) (8 = WTy = Duy
W=1+4x
Tp = Ty, so we use T in both equations.
l=2z
(a’) 107 + Wr= 70
(b’) 107 - WF = 30
20T = 100
PRACTICE SET 92 T=S ip
a (8 + WT» = Dp (a’) 105) + WG) = 70
(4+ W)Tp = SW = 20
4Tp + WTp =N (a
W = 4mph
(8 _ Wy = Dy
(4 - W)Tp = 30 (6 + weH = €
(a’) 48 + 4W = 00
(>) (8 — WTy = Dy
(a) 4Tp + WTp =
(B — WS = 55
(bd) 4Tp —_ WTp =30
i z (b’) SB -— SW = 55
8Tp
Tp = 10 hours = T,, S(a’) 20B + 20W = 300
4(b’) 20B — 20W = 220
(a) 4410) + WUQ = 3
408 =3 20
10
10w =
B = 13mph
wz=1 mph
(a’) 4(13) + 4W = 60
4w=8
bh (6+ WI = Dp W = 2mph
{8B + WG) = @
3B + 3W=33
(a) (8 + W)Tp = Dp
(B + 7)Tp = 3S
B+We=ll
(bo) (8 - WTy = Dy
(B - W)Ty = Dy (B -— 7)Ty = 21
(B — WX4) = (12) TD = Ty, so we use T in both equations.
4B —- 4W = 12
(a’) BT + 7T = 35
B-W=3 &® (b’) BT - 7T = > 21
(a) B+ We tl 2BT = 5
(o) B- We= 3 (c) BT = 28
(a’) 28) + 77 = 35
8 = 7 mph WM=7
T= bie
(a) (7) + W= 1)
(c) BU) = 28
W =
" 4mph
B = 28 kph
234
Algebra
2, Third Edition
Problem Set 92
4. Equal ratio method: 1
(a)
a y <—-x
a +2
Seek
G, C, (b) aber 2
500 _ 20
17502 oe The first step is to graph each of these lines.
Cy = 70 delicious comestibles
Variation method:
G=kC
500 = k(20)
25 =k
1750.= 25(C)
70 delicious comestibles = C
The region we wish to find is on or below the solid
line and to the left of the dashed line. This region is
shaded in the figure.
Dy = RyTw = 6Ty; Dy = Dw + 50;
(a) x+y+z=6
Dy = RyTy = 8Ty; Tw = Ty so use T
(b) 3x -y +z= 8
8T = 6T + 50
+ Zz ll
(c) x —-2y i=)
2T
= 50
(a) x+y+t+ Z=
lS
(b)
3x ~y+ z= 8
(dj) 4x + 2z
= 14
2a) 2 Dy 22-= 12
(c) x- 2y+ 2= O
(e) 3a + 3z = 12
Std) l2vett6g=" 42
—2(e) —6x — 6z = -24
6x =
yas
(e) 3(3)
+ 3z = 12
ae
cl
(3) GO) yee(hy) ll ON
y=2
The region we wish to find is on or above the solid
line and above the dashed line. This region is shaded (3, 2, 1)
in the figure.
255
Algebra 2, Third Edition
Problem
Set 92
9% (x+y= 10 ig MHZ, WHT | -SHHFUVES
(b)x+y=s4 2-3 22+3 —4+-°®
ys4-x . Wee
(b’) FY = 4-w = 16 - eX? —
2x? — && +6 = 0 pi acsiiaingisites
Solve this equation by using the quadratic formula. sail pes
_ 82 (4-402)
4
og pea) ce”
=
Substitute these values of x into (6) and solve for». Check: \36 - 3 = J36 -—27
(> y=4-Qe=! 6-3=9
) y=4-(=3
(3, 1) and (4, 3)
10. -4 <x - 2 < 2; D = {Reals}
—2<x<4
S-2-1012345
Wl. x-122arx+
2 & 2; D = {Integers}
x-le2Qarxt+2<2
x23a x <Q
2101283 4
(y***)? x 2/3 m Rees =A
Lae = yatta, BB -6 [ = S$ cos 30° = 693
12. aap)
ey
= ytd 08 = §sin3Q° = 4
P : —4.70R -— 1L71U
is = SS —6.93R — 4.00U
ye = ye -11.63R - S.71U
a any L + xy
x © 19 y
oe . eet xy) ;
yt xy? + xy
14. m : m
a am
at ; at rtriTtiVintiit
“Pe. mta Buna —
me Lid it
_ m(m + a) x
2am + a
15 PS EE - eeoe ee
TVHti 49 +] adhe ee
7-i
-36+2i_ 18 1
30:té<CUS:tS os 7206
256 Algebra2, Third Edition
Practice Set 93
Since @ is a second-quadrant angle, 27. 3x7
x =5
6 = -75.96 + 180 = 104.04° <3x- ale a fee 0
H = (20)? + (-5)?
H = SV17 = MEAT TA) wal apse.
-6 6 66
5/17 /104.04°
20.
3-5/2 2+-2 28.
15 cm
x
1 in.
x
1 ft
OSS
1 yd 60 s
OSS
PExewarcK? Ss Do aIcm 12 in. ft 1 min
_ 6+3V2 -10V2 -10 _ -4- 7/2 15(60) yd
4-2 a 2 “A: 54)(12)(3) min
21.
8- V2. 8-2 »44 2/2 4 x ope
Mien dich 4ee2N2: AE 25/2 29;
x2-9 —34+%x
_ 32:4 16V2 -4V2-4 28 +122
7 4 = 3x(x
+ 3)
16 — 8 a 8
(x + 3)(x —3) (x — 3)@it3)
_ 74+ 32
melee 4 = 3x?
= 9x ~3uti
tx H4
5xy? 3 [xy? = xM553/5,1/3)2/3 _ y 17/15,,19/15 (C3 x="3) x7-9
22. Bs
23.
a 1 1 2 Mt
20% 6
by R, R, 2
2 i
aR, R> = bxyR> + bxyR,
21
= 4y
bxyR> = aR,R> — bxyR,
21_
ia?
Ri(aRy — bxy)
= R;
hs *a | PRACTICE SET 93
by R, R,
ceo = =ehe
dh
a X Xx
—[S = Ss oa =
by R, R, x? + 3x-2=0
aR 1Ro = byxR) TF byxR 1
aR R> = y(bxR te bxR,)
aR,R, b* — 4ac = (3)? = 4(DC2) = 9 + 8= 17
ey
bxR, + bxR, The discriminant is a positive number, so there are
two real number solutions.
25. 3 _ 9j4 + f-9 - J-2V-2
cine nde ee) ae ag ~Ax = 3x 43
26.
AS RD
= 3,7) SA ee S
tS nite ae GaltS
ks. adn b? — dac = (-4)* — 4(3)(3) = 16 — 36 = -20
boys Ov IS | ae The discriminant is a negative number, so there are
6 5 two complex solutions that are conjugates.
5/15 3615 | 120V15 _ 89/15
ps 20 30 30
257
Algebra 2, Third Edition
Problem Set 93
PROBLEM SET 93 4. De
+1
1. Downstream: (B + W)Tp = Dp (a) 40
Upstream: (B- W)Ty = Dy _ (b) Hier te ele
48
Since Tp = Ty, we use Tp in both equations.
R-Tp = 40:Tr = Tc 4;
(a’) BTp + 5Tp = 45 os; Pate
(b’) BTp - 5Tp = 15 RrTr = 48; Re = 2Rc
2BTp = = 60 R(T - 4) = 48
() Bi paare’ WRT - 8Rc = 48
(a’) (30) + STp = 45 2(40) - 8Rc = 48
STp = 15 32 = 8Rc
Tp = 3 4=Rc
(c) B(3) = 30 a Q |= 4mph; Rp = 8 mph;
B = 10 mph
Tc = 10hr; Trp = 6hr
2. Downstream: (B + W)Tp = Dp (a)
(b) 5: ge = T,
T, We:
Upstream: (B —- W)Ty = Dy
(a’) 4B + 4W = 48 PT,
T> =
P
SS
(b’) 8B — 8W = 64
1)(4000 y= 300K
T> = a
2(a’) 8B + 8SW= 96
(b’) 8B - 8W= 64
eo Sa isn
16B = 160
B = 10 mph a
x* +.5x —1,= 0
(a’) 410) + 4W = 48
b* — 4ac = (5)? — 4(1)(-1) = 25 + 4 = 29
4W=8 2
b= —-4ac >20
W = 2mph
There are two real number solutions.
3. Downstream: (B + W)Tp = Dp (a)
cf 3x = -2x% -
Upstream: (B- W)Ty = Dy _(b) io maak
2x? —- 3x +5=0
Since Ty = 2Tp, we substitute and get:
b* — 4ac = (—3)2 = 4(2)(5) = 9 — 40 = -31
(a’) 40Tp + WTp = 210
b* — 4ac < 0
(b’) 80Tp — 2WTp = 380
There are two complex number solutions that are
2(a’) 80Tp + 2WTp = 420 conjugates.
(b’) 80Tp — 2WTp = 380
1607p = 800 8. (a) x +2y <2
Tp = 5 d2y<—x+2
(a’)") 40(5
40(5)+ W(5)W(S5) == 2 210 wie ole,
1
5W = 10 2
W = 2mph (b)
y = -l
258 Algebra 2, Third Edition
Problem Set 93
The next step is to graph each of these lines. Solve this equation by using the quadratic formula.
y SS)
-§Sa+ {82 — So
4(3)(5) Ae =e
25) 7 ONS
3
= —-—,1
bin oS
- Substitute these values of y into (b) and solve for x.
tees) 5
eae
5) 5
(bo) FS 2EP) + 2 0
(0;=1)-and (<.-2)
The region we wish to find is on or above the solid
line and below the dashed line. This region is shaded
in the figure. 12. (yon? Pye? _f ytd ease ie yias+6
os = X — 4. < 2: P= {intevers} B x > x
(= <5 ett! 4 ae
ick 1 be +1
c
HF 3) ae Se B
x(be + 1)
10. (a) x + y — 2z = -3
~ be +1+ac
(b) 2x +y+2=7
id 3x = pit ze= ft) See i
ae 9 Bee
Dae te = ata! 3m + 1
#
esa yz =143
Sx = 20 _ __m@+m1)
x=4 6mx + 2x + 3m
ie = iS a? Fes) wail a SEE
ae aia Sd ” peer Spat
Le
Fiat _ 38 +16: 19 8,
Ms
(d) (4) + 3z = 10 50 055995
se 1 2i+4 3i-2 -6+12i- 41-8
~ 3. hte 20 Sx
oe ue ee
Wig
(a) (4) + y - 22) = -3
i A A aco ewan tk!
Aes? jp = 5+ p- 35
oie 17,
[ile ayer ey = 1 Vi = = ea|p = 35
b)x-2y=2 > p -10jpi+ 25 = p — 35
x=2dy+2 -10,/p = -60
Rp ro
(b’) x2 = (2y + 2)? = 4y? + By + 4
= 36
Substitute (b’) into (a) and get:
Check: 36 = 5 + V¥36 — 35
(a’) (4y? + 8y +4) + y= 1
Sys + 3 = 0 | 6=5+1
259
Algebra 2, Third Edition
Problem Set 93
18. A 15 25. i> — 2i2 — .[-9 + J-4./-4
+2
B eZ
= esi 4 0 el
12
26. a2 + [2 - sve = 42+ 3? - 9/2
A = 12cos 32° = 10.18
B 12 sin 32° = 6.36
_ 2v2 , 3v2 _ 18V2 _ _13V2
-10.18R — 6.36U D 2 jen ers)
15.00R + 0.00U
4.82R -— 6.36U
rah an Ser
(x2 42+ }=-2
Gee, 5064
H = J(-6.36)? + (4.82)? = 7.98 griigieg
hT Ae Beta oath
7.98 /-52.84°
Sera). Buu tad
19.
EO eae Nol er’ 1 123°
_ 92 + 24 +12 + 16V2 x+—
6
= +—I
6
a 18— 16
1 23
_ 36 + 252 XY =) —Sst 1
6 6
" 2
20. S45 5-2N5 28.
10m 60 s
x ——
100cm
xX ——
1 in.
xk ——_
1 ft
xk ——
Sabi2Wit (Sea
Dal Sh Ss 1 min 1m 2.54 cm 12 in.
_ 25+ 5V5-10V5
-10 15 - 55 _ 10(60)(100) ft
i D5 20 a 5 (2.54)(12) min
= 3-4/5
21. 3/45./2 = (07042) Ue = 92/39 1/15 = gis x7 — 6x + 18 - 6
29; <it 3)ak eee ONE
4
22. xy? [x3 = x4) 3/2,3/2) 5/2 — x7Ay4 x? + 3x2
—~6x2 + Ox
cae,
23.
g 9
I mi
— + —
=6x7.= 18%
5 d (Ese =. 2D
Gi m mb 18x + 54
+ —
—56
m 58 d
adx + cdx = dm* + m’bx
adx = dm* + m*bx — cdx 30; ees
2)
_ dm* + m*bx - cdx
90 — x = 50
x = 40
Q io)
|
+
24. TH} 3 +
[>
Sa
25 + 5y = 44
Sy =F19
adx + cdx = dm? + m*bx
adx + cdx — m*bx = dm?
a
_w7
dm?
ad + cd — m7b
260 Algebra 2, Third Edition
Problem Set 94
PRACTICE SET 94 Sets (a), (b), and (d) are functions.
a. Function Set (c) is not a function, because 6 has two images.
b. Function
p(x) = x? — 2x
c. Function p(-2) = (-2)? - 2-2)
p=4+4
d. Nota function
p=8
e. p-5)="5)7 = 55)25 +125 = 50
=x" = 4x +4
Oma + 4x 404
PROBLEM SET 94
b* — 4ac = 42 - 4(1)(4) = 16 - 16 = 0
Eve an
PV There is one real number solution.
n= —
RT
(a) 2x + 3y > -6
ee ph
sohmole
(0.0821)(473) 3y > -2x
-6
D2
2. 0.2(40) + 0.6A = 0.44(40 + A) >--x —-2
ae
8 + 0.6A = 17.6 + 0.44A
0.16A = 9.6 (b) x - 3y = -6
A = 60 gallons —3y
> -x - 6
3. Downstream: (B + W)Tp = Dp (a) SS lA eee
3
Upstream: (B — W)Ty = Dy (b)
The next step is to graph each of these lines.
Since Tp = Ty, we use Tp in both equations.
(a’) 10Tp + WTp = 48
(b’) 107p — WTp = 32
20Tp =)
Tp = 4
(a’) 10(4) + W(4) = 48
4W = 8
W = 2mph
4. P,V; = PoV2
(700)(1400) = (2800)V>
The region we wish to find is on or below the solid
Vous (700)(1400) _ 3659 mI. line and above the dashed line. This region is shaded
2800 in the figure.
k 10. —x +3 <2 or «+3 < -5; D = {Integers}
5 P= a
x+32>2o0rx+3<-5
gg k —x =-—5 or —x
< —8
10?
XS Or 8
400 =k
P= ae = 16 purples AeeOme
Gn) = Ome om 1
Algebra 2, Third Edition 261
Problem Set 94
11. (a)x
+ 2y-—-z=2 Pp 3 P
m 2
(b) 2x -yrz=2 m+ 1 mt me
(c) 3x —-y + 2z=4
mp
(a) x + 2y-z2=2
p(mp?
- 1)
(b) 2x - yrz=2 m? p? m+ mp
(d) 3x + y 4
x i x
2(a) 2x + 4y - 2z = 1S. b a 2
(c) 3x - yr2z= at a+ :
oo. =— = ab— 1
(e) 5x + 3y = §
3(d) 9x + 3y = 12 _ _x(ab?
- 1)
—l(e) -Sx - 3y = -8 a*b? -a +b?
[a Ae
x= 1 16. 247 31 9-6 HRMS _2 gd
—3i 0 3i o Baa
(d) 31)
+y = 4
y=] Si—-2 1-215
ikA
(b) 21) -() +z ~2i +5, -% <5
1 + a” i —10 + 41% 25810
= 20 | |
lal WN
Nw
— ¢ —4 = 25 Lae 4
(1, 1, 1)
18.
12. (a) x? + y = 3 vk -21 = 7- Jk
(b) x --y=2 k-21 = 49-14/k +k
x=yrt2 14Vk = 70)
(b’) x2 = (y + Way + oye 4 Vk =5
Substitute (b’) into (a) and get: k= 25
(a’) (y? + 4y + 4) + y? =3 Check: 425 + 425 -— 21 =7
2y7 + 4y +1 =0
OS ay ey
Solve this equation by using the quadratic formula.
+4o+ «/16 see [2 19, 20
‘ 4 2 ; ee 8
6 A
y=-l+— = ~160°
A = 20cos 20° = 18.
D a| oO
[> >
wx=(a+2] i) =1 += B = 20sin 20° = 6.84
| |
ro] j
t
Il iw) {I
a
ie
oe— - +
/
a
C = 20 cos 20° = 18.79
D = 20sin 20° = 6.84
18.79R + 6.84U
yal?2 y?9
ya3 - 3a/2)2 2a x 7aloy2 -2a -18.79R — 6.84U
13. yial2 (y*)? 0.00R + 0.00U
262 Algebra 2, Third Edition
Problem Set 94
20. y 25, 317 + 21° — 21+ [-9 - ¥-2V2
2-3 + 2+ 24+ 31-2) =-34+i
eae
am
. 26. (a) Since this is a 30°-60°-90° triangle,
a J3 x SF =4
coo
we eeea3
an < °
nie
3
+ 4/3 4V3
ae Vie woleex = —
3
tan@ = —-—
: Ne Ten
3 A
4
@ = —63.43° (b) Since this is a 45°-45°-90° triangle
Since @ is a second-quadrant angle, 1x SF =4
@ = -63.43 + 180 = 116.57° SF = 4
oe etsy C=1 xe4ged
H = 4V5
ss D= 2x4 = 4/2
4/5 /116.57°
27, ee ee
my, 2a 5V2 32 +2 epic ey
Ogee}
oat 3.2 +22
_ 6¥2 — 30 + 4 — 10V2 ee ee Oe ere
ce 4 Sete
=«
. gels -
x13
7
22 28.
20 liters x
1000cm3
——— x
Lin. x
Lin.
S 1 liter 2.54cm 2.54cm
% 1 in. \ 60s
yaa 3 = — 2240 2.54¢m, =) i min
= 6Ne Hg A _ __20(1000)(60) _ in.”
- Ce (2.54)(2.54)(2.54) min
215 <2 eagal 15 29. There is never an exact solution to these problems.
3 One possible solution is given here.
ods wis this = oe et are
15 15 15 15
Use the graph to find the slope.
+ D ] -
23. or
Xx = ae
aaa pega
120ices
— 180We,
aR R> i bR{R> = XR, of cxR NS (-1.17)R ad
aR,Ry + bDR\Ry — xRz = cxR, pe ORE Rind
aR, R, ¥ bR,R, 3 xR, Usethe point (120, 70) for R and N.
pote ee A (el 120)
xR,
210.4 = b
24. Ss acreage emgage N = -1.17R + 210.4
x R, R,
aR Ro + bR Ro = xR) a5 cxR, 3 4 3%
30. airs 7
Baa ct = On eS ex
AR,Ry + bR\Ry — cxR, = xRy
xR, fe 3(x? - 9) + 4atx> ~ 3a4x4(x + 3)
iy aR, + bR, - cx a*x> (x? = 9)
263
Algebra 2, Third Edition
—
Problem Set 95
PRACTICE SET 95 PROBLEM SET 95
3x —y = 4 (a) (straight line) I. | ey |
LY =a, (b) (hyperbola) Dr
(b) a= ii RywTw = Dy; Dy = Dr; Tw ='10\- Tr;
pect RrTp = Dr; Rw = 9; Rr = 24
y 6(10 — Tr) = 24Tp
60 — 6Tp = 24T
(a) Ble y=4
y
-
60 = 307.;
21 ‘a Fy De Tr
ee Dp = 2(24) = 48km
Ope
; cine ale es 2. 0.6A + 0.2(200) = 0.52(A + 200)
yo aye pelle. 0) 0.6A + 40 = 0.52A + 104
Gat Dy — 3) = 0 0.08A = 64
eo pes) shoes 64
(). x7) <7 0.08
A = 800 mL
x = =1
(b) x(3)=7 ete
7 Z, 2,
Xe
3 7 ee
cas
There are two solutions for this system: (—1, -7) and 1
400
13), Vy) = EUSP 73827 mL
473
2 Dg ;
ff ' an 10 a) tere) 4. RrTp = 360; RjT; = 480;
3x” — y" = —2 (b) (hyperbola) T, = 4Tp; Ry = Rp — 60
(a) x? + y? = 10 (Rr — 60)(4Tp) = 480
(stay = 4(360) — 240Tp = 480
ltPi retake -240Tp = -960
iis Tr = 4
x = +/2 Tr = 4hr; T, = 16hr;
(a) (/2)? + y? = IG Rr = 90 mph; R; = 30 mph
2+y = 10 5. Downstream: (B + W)Tp = Dp (a)
y = Upstream: (B — W)Ty = Dy (b)
y = +/8 = +22 (a’) 4B + 4W = 60
+ y? = 10
(a) (-V2)? (b’) 8B - 8W = 72
er y- am) 2(a’) 8B + 8W = 120
lee (b’) 8B - 8W= 72
16B = yn
= +,/8 = +22
, 2 B = 12mph
There are four solutions for this system: (a’) 4(12) + 4W = 60
(V2, 2/2), (V2, -2V2), (-V2, 22), and 4W = 12
(-V2, -2/2). W = 3mph
264 Algebra 2, Third Edition
Problem Set 95
6. (a) 4x - y = 3 8. (a)x+2y-z=4
(b)
xy = 6 (b) 2x -y+z=-3
6 (Cc) ty
+ giee4
x=
y (a) x+2y-z= 4
Substitute (b) into (a) and get: (b) 2x - y+tz=-3
(d) 3x + y = I
(a’) (£)-y=3 (a) x+2y-z= 4
y
4 (Cc) x- yrr=-4
ct SS
y (ec) 2k
eae+ 4
Vo Sa
24 — yy = 3y
(d) 3x + (-2x) = 1
y + 3y -24=0
x=
Solve this equation by using the quadratic formula.
() y = 21) = 2
ars Pete 3105
(b)24+2+z=-3
2 2 2
z=-7
Substitute these values of y into (a) and solve for x.
(1, -2, -7)
(a) 4x
(-2FE108 a
ee Systems (a) and (d) are functions. System (b) is not
a function, because 5 has two images. System (c) is
not a function because m has no image.
10. 3x7 -x +5=0
fs)— aNa iT | |
tw
| b? — 4ac = 1 — 60 = -59
The quadratic has two complex number solutions
that are conjugates.
11. (a) x > —2
(b) 2x + 3y > -3
3y > -2x - 3
y> -=x
z - 1
: 3
The next step is to graph each of these lines.
(a) (V5)? + y? = 16
y=11
y= +tyll
(a) (-V5)? + y* = 16
i aia Lb
The region we wish to find is to the right of the
y= tyll vertical dashed line and above the slanted dashed
(V5, +V11) and (-/5, +11) line. This region is shaded in the figure.
265
Algebra 2, Third Edition
Problem Set 95
12. 2 < -x —-2 < 0; D = {Integers}
O0<-x«<2
O>x>-—2
=3 -2 -1 0 1
ete 2 Ni 2
(a ) x Ee qi2t! — X,2a-a
13.
a x*
= qri2ztilya
EY P x
14.
4+ a 4+ ug
(ees 4+x Reet We
4 23.83
x(4+x) _ x(4+x) @ = -16.10°
16+ 4x + 16 32
+ 4x
H = .|(-6.88)2 + (23.83)? = 24.80
15. ee
ay oe 24.80 /-16.10°
—— fo ae
oR me ee
38
2x +52
20.
2/2 -5 22 -3
gn DNF IZ) 2k Fon) = 10(x + 1)
2/2 +3 22-3
Aga dee x «(Sxet4. | Sy 4A 8-10V2-6V/2 +15 23 - 16/2
Sot 6) a4
Ute FL ES ie
16. =23741642
3-Si 3+ 5i
Wie Al 25 ig 0 av 163
21.
3-2 2-4
‘ 9 + 25 ” 34 J2+4 2-4
sped Caat 3/2 - 2-12 + 4/2
TAT
2 - 16
17.
et AAI ga Pees ree -14+7¥2 2-2
-14 2
18. Acco OL. eee mor (242)
\x2 + 2x +34 2x44 22.
x? + 2x + 34 = x7 + Bx + 16 x pa
18 = 6x mapq + mbpq = qx + prx
x=x mapq + mbpq — qx prx
Check: 37 + 2(3) + 34 -3 =4 ampq + bmpq —- qx x
N49.= = 4 px
L=3.= 4
19, 14 6 23. mer). (242)
ee
x Pp q
at
mapq + mbpq = qx + prx
12 mapq + mbpq — prx qx
C = 12 cos 35° = 9.83 qx
D = 12sin 35° = 6.88 amq + bmq — rx
14.00R + 0.00U
9.83R — 6.88U 24. 4 — 317+ [0 4 e353
23.83R — 6.88U =44+3+31-3=443
266 Algebra 2, Third Edition
Practice Set 96
25.
7
2,|2- 3/3— av
3 30 mi —-5280 ft 2c 24 Cin
30. x x x
hr 1 mi 1 ft 1 in.
_ w3i3 _37 x
1 hr
x
1 min
60 min 60s
_ 14V21 _ 9V21 _ 84v21 _ 79/21 _ 30(5280)(12)(2.54) cm
21 21 21 21
(60)(60) s
26.
PRACTICE SET 96
x - 3y = 5° (a)
xy = 3 (b)
(b) xy" ="3
3
27. x=
y
| w
fat)—
a. — I Nn
x (1.32)(0.0821)(600) Sw
|wieee,
TE
V = 13.00 liters
5
28. Write the equation of the given line in slope-
intercept form.
3y — 2x = 5
3y = 2x +5
ee
ate ae
See
Since the slopes of perpendicular lines are negative 6
reciprocals of each other,
@ x-f-3+ 4) -s
Se +
| iH] Wn
N{[n
eth
(a) x | ee) | | ll Wn
HAln |
5a eee”
* + oo
Il wn
nN
Nl
29. 4(4 + x) = 5(12)
16 + 4x = 60 = iH] |
Nl]n
4x
= 44
re TY
6(6)
= 4(4 + y) Gas 5 4 16 nd
36 = 16 + 4y
Desde 46.26
20
= 4y ( - 4-8 - 4)
say
267
Algebra 2, Third Edition
Problem Set 96
b. Variation method: Equal ratio method:
me FA Sy 3 RM,
D
Sy RM,
(75)
k(15
k(5)
(85)
100 _ 4(5)
20 R(2)
63/5) =nlok
R = 2 ratchets
ADK:
(20) = 425030) Downstream: (B + W)Tp = Dp (a)
D Upstream: (B — W)Ty = Dy (b)
20D = 12,750
Since Ty = 2Tp, we substitute and get:
D = 637 deer
(a’) BTp ie 6Tp = 80
Note that half a deer is not considered as part of
the population. (b) 2BTp — 12Tp = 40
Equal ratio method: 2(a’) 2BTp + 12Tp = 160
ier 12 (b’) 2BTp — 12Tp = 40
A,D, 4BTp = 200
(iD): & (15)D, (c) BTp = 50
(20) — (30)(85) (a’) (50) + 6Tp = 80
S N T 637 deer
6Tp = 30
Tp = 5
PROBLEM SET 96 (c) BOS) 50
1. Variation method: B 10 kph
Me Se RgTg = RpTp = 300;
Tg = TR + 5; Rp = 2RG
65 = ——
2RG(Tc = 5) = 300
k = 13
2RoTg = 10Rg = 300
13(100) 300 = 10Rg
NB 30 = Rg
Nz = 260 boys
Rg = 30 mph; Rp = 60 mph;
Equal ratio method:
aa ll 10hr; Tz, = Shr
(a) 0.2Py + 0.6Dy = 160
(b) Py + Dyn = 500
5 100(3)
7 iH] 260 boys Substitute Dy = 500 — Py into (a) and get:
(a’) 0.2Py + 0.6(500 — Py) = 160
2. Variation method:
S = kRM
~0.4Py = -140
k(4)(S)
Py = 350
=~ Il
© 5 (b) (350) + Dy = 500
i)
— 5(2)R Dy = 150
aoO
S iH]
ll 2 ratchets 350 mL 20%, 150 mL 60%
268 Algebra 2, Third Edition
Problem Set 96
6. (a) x =3y= 2
Oa)" ox"+ saye=* 50
(b)
xy = 4 (-3)(e) -15x - 15y = -30
4 20y, = 20
ea —
y y=)
ofi)-s
Substitute (b) into (a) and get:
(ce) Sx eo(l =210
2 oe = 5
X= 1
Aa 2y
0 = 3y° + 2y—4 (b) (1) + (1) + 2z = 6
22 = 4
Solve this equation by using the quadratic formula.
z=2
ee ee
6 aye (1, 1, 2)
Substitute these values of y into (a) and solve for x.
Sets (a) and (b) are functions. Set (c) is not a
a) x — 3} -— +
i as function, because 4 has two images.
(a) -; 2) —|]=2
x + 12-13
=" 10. g(x) = x — 4; D = {Integers}
x=
17 413 g(-2) = (2 -4=4-4=0
a is (ete LED =P 11. (a) x —-y < -2
3 3
-y<—-x+-2
x+1+-
13 =2
See
ae
ie = all3
(b)
y >-2
[1+ 3,- i ae
+ 8) and
a-bs-3 The next step is to graph each of these lines.
fe Bae 1-13
tS 3-4-3
7 (@) axe = 4
(b) 4x? - y? = - 4
Ses = 0
x2 =
y=
(a) (0)? + y? l=
y=
y = Teg
Ee The region we wish to find is on or above the solid
(0, 2) and (0, -2) line and above the dashed line. This region is shaded
in the figure.
8. (a) x + 3y -Z Ul i)
(b)x+y+2z=6 12. x+432 or 4 > -1; D = {Reals}
(c) 2x + 2y-z=2 x+4<2 orx-4>-l
x = —2 or es
2(a) 2x + 6y — 2z
(b) x+ yt2z= 6
BSS SerO 1 2) etal
(d) 3x + Ty =uld
2
(b) x + yt2z= 6 io:
Ca ee J Hd+2-3oppee x
a?! b*
2(c) 4x + 4y - 2z = 4
at + V/2px
(e) 5x + Sy = 10
Algebra 2, Third Edition 269
Problem Set 96
SE
H = (3.52)? + (2.34)?
x ay
14.
“4
= ab 2 H = 4.23
a+ b i
a? + — ab +1
ab
x(a°b + 1)
a*h + a + ab?
15. 2\33/2 = 2(23(2"3))¥2 = 2(2103)12
= 2(253) = 28/3
16. 2i-8 446i _ 8i - 32 - 12 ~ 48:
4-61 4+ 6i 16 + 36
peore
er 11 10, 4.23 /56.39°
ae ee 1s
a? Or ee ee ee
17. = cObieey? 1-i e
—!
I l
2 AS Bilt ral 14425.
=4
= LS)
(s==39
13 — Ws
18.
s — 39 = 169 - 26Vs +5
si AE
2
26s
= 208
Js= 8 91, ee eT
WTS a tesa
s = 64
op Riel = 27 Be 9S Se
Check: 64 — 39 = 13 —- (64 el) a6
(25
= 13 =8 _ 3+
5 = 5
ere
19. 7 c= n(2te_,|
p
am
+ dm
C= ir
6 B Pp
2
cp = am + dm — mpr
beer cp + mpr = am + dm
A
gia asc
a c + mr
AS 600s 67 = 2:34
B= 6sin67°°= 5.52
23. c= nf2* _,)
2.34R + 5.52U
Pp
0.00R ~2.000
2.34R + 3.52U
=
am
+ dm
SSSnh
P
cp = am + dm — mpr
aoe
tan@=
2.34 cp — dm + mpr = am
6 u 56.39° cp — dm + mpr _
m
270
Algebra 2, Third Edition
Problem Set 97
2, 2i* - i? - J-16 - |-4./-4
=2+1-41+4=7-4
8x SF =.13
25. PV = nRT aoe
PV 5
RT
E97 5020)
= 1.81 moles
(0.0821)(673)
30. ACicle = nr = 1007 m2
26. 4x? + 6 =
ress 10 m
4x7 +x4+6=0
Side of square = 2r = 20m
_-12Vi-4@ _ 1, 95.
=
2(4)
= =——
8
i —!
8 Asquare = (20m)? = 400 m?
Zi. (a) x 4+ 2y = 6
2y=-x+6 PRACTICE SET 97
Me 3%
= Sys—ell (a)
2x
— 4y = -6 (b)
(aysx Sy =11
y= 11 — 3x
ete
5
dl!
5
(b) 2x - (Ss - 2) = -6
5 5
3 11
2x() = a(S = =\6 = —6(5)
SRSee4aan
SY 288 2088 x 10x — 12x + 44 = -30
ae
—2x = -74
veo
(b) 2(37) -— 4y = -6
—4y = -80
2(a) 2x + 4y = 12 ye= 20
(-1)(b) -2x + 5y = 10 The solution is the ordered pair (37, 20).
Qy = 22
22
o PROBLEM SET 97
(a) 2Nr = 3Ny + 12
(b) 4Ny = 3Nr — 48
3(a) 6Nr = INy + 36
(-2)(b) -6Nr = -8Ny - 96
Care|
9’ 9
0= Ny - 60
Ny = 60 were untalented
28. 4O in. . a5 L 1m - 1 km
(a) 2N7 = 3(60) + 12
hr 1 in. 100 cm 1000 m
2Nr = 192
1 hr 40(2.54) km
Nr = 96 were talented
* 60 min (100)(1000)(60) min
271
Algebra 2, Third Edition
Problem Set 97
2. 0.3A + 0.2(300) = 0.24(300 + A) 6. (a) 2x + 3y = -7
0.3A + 60 = 72 + 0.24A (b) 3x - 3y = 12
y=x-4
0.06A
= 12
A = 200
mL Substitute (b) into (a) and get:
(a’) 2x + 3@ - 4) = -7
os 2x + 3x — 12 = -7
Svi=3
Fig rly
|
Ear8
t= a
Ty = 21 = SON =150K
_
() y=()-4=-3
(1, -3)
4. Variation method:
ee (a) 4x -— 2y = 8
—2y
= -4x + 8
,-8 =
y=2x
(b) 3x — 2y = -4
-4
k= 1= Substitute (a) into (b) and get:
(b) 3x — 2(2x - 4) = -4
ek 3x — 4x + 8 = -4
ep)
| pinata
w|Rae, —x
= -12
to
4 x= 12
32 °= i (a) y = 2(12) - 4 = 20
3
(12, 20)
G = 6 greens
(a) y — 2x = 3
Equal ratio method:
2
y=
2x 83
BM,
(b) xy=4
B, GW?
Substitute (a) into (b) and get:
come 0 (b’) xOK +3) Hee
a) Ju,(2°)
2x? + 3x = 4
G>) = 6 greens
2x? + 3x ='4 B10
5. Variation method: Solve this equation by using the quadratic formula.
C = kFF tee V3" = 4GAR. ad _ v4
1000 = k(100)(4)* 2(2) Arta
k = 0.625 Substitute these values of x into (a) and solve for y.
C = (0.625)(10)(20)* = 2500 cheers @y-f-2+H)3-3.4
4 ee)
Equal ratio method:
3° Bla 3 41
2 (a) y Se) (-3
ee at)43 wees
5 5
S s Fi
C, Bao 3 41 3 741
1000 _ 100(47) [$+ G3 SE
& 10(207)
C, = 2500 cheers
34 a3
ding 2
_Va
2
272 Algebra 2, Third Edition
Problem Set 97
The next step is to graph each of these lines.
(a) (V3)? + y?
3+ y?
|—|—|—|—1
WP
Oaaf
(a) (-V3)? -y 2=1]
The region we wish to find is on or above the x-axis
3+y 2= 11 and above the dashed line. This region is shaded in
a oak the figure.
M4
y +2,/2 13. AP xd <= 1. DD = (Integers!
ARS Seba
a |
(V3, #22) and (-V3, +22) il S xe ah
10. (a) 3x +y +z
Oi
eae2eee:
(b)-2%
—y =z (x? + ye x3al2 b
14. YT’ GIT Sake et BVO
(cl 2 + 2p - —b/2 aa ary
y
= x24 1,36/2
EVR
oe oe oe ged
3) PS ge
Bod
13; Pp if P
5x
m A m-x
X
ft 4¢b== mx +1
MX
lay
ak ae Os ZS 2
p(mx?
+ 1)
(c) x + 2y - Zz 8 m2x> + mx — mx
(d) 4x + 3y
16.
5¢ 14 (= De Seat
(d) 4(1) + 3y = 2 i) -i0i er) 2i= 1
By.= DOLEes 7 an ie, eee
ve nar 5
(a) 301) + (2) + z= 2 17. SiaJ2 = (23(2'2))¥6 = 2'2gil2 — 27/12
Siar he 15 — 21), 8 = 2i 5 + 2i
18.
(1, 2, -3) fey a(-Doe Gl
19; Jz? — 33 + 4z = 11
11. g(x) = x2 — 2x + 2; D = {Reals}
az = 33¢=" liye
(5) 25" 26) 4 2 = 17 28 = 12122242 +2
222 = 154
12. (a) x - 2y <2
Vz =7
—2y <-x +2 7 -= 49
1 Check: (49 — 33 + /49 = 11
aes
> —-x-1
NIG + 7geadl
4+7=11
(b) y2 0
Algebra 2, Third Edition 273
Practice Set 98
20. 26. Il S
io |e
lea |e
a ey,
4 Ze 2 [rR
+a
els
[DS
& a é
C = 4cos 34° = 3152 pac = cmxy + amxy
D = 4sin 34° R 2.24 pac — amxy = cmxy
-4.00R + 0.00U cmx y
3.32R + 2.24U cp -— mxy
-0.68R + 2.24U
2.24 27. ACirele = mr = 497 cm?
tang =
—0.68 Pe IGN
@ ul= -73.11°
Diagonal of square = 2r = 14cm
Since @ is a second-quadrant angle,
@ = -73.11 + 180 = 106.89° 14
Side of square = —cm = qe. cm
H = J(-0.68)2 + (2.24)? 2
H = 2.34 Asquare = (7/2 cm)* = 98 cm?
28. See Lesson 71.
400 mL » lcm 2 1 in. 1 in.
29.
S 1 mL 2.54 cm 2.54 cm
1 in. 60s 60 min
2.54cm I1min 1 hr
400(60)(60) _ in.>
~ (2.54)(2.54)(2.54) hr
x —4 bead ee
2.34/106.89° 30. _ =a)
2 3
ay. eNews = 2y ga aie oe + 8 3x — 12 —- 2x 4 1D = 2
C2 2S ae 16-8 Be PA
_ 20-14V2
_ 10 - 72
8 4
3-V7 2 se
V7 _Vi3V7-7 PRACTICE SET 98
22; le _GAL7 -3V7
aie Le
-J7 V7 7 7
23. fo +242 - N27 = 3 + 83 153 a
os
LVS | 2N30 45y3- ) 40V3
3 3 3 3 7
=-+-)—- —
(2 =}
2 fa866 6
2. 247" 40 = Gid-0) esi
SA = Bye Dees; a4 = 10; a
Zz
25.
a
Feyts e tar
1
= at + —
5
Xa oy r 2 24
ar* — crx — crey = x+y
ar“
2
— xX — crx
a)
= y + cry
2; ae She
24 «24
ar?
p>
— x —cr’x
> a
1 + cr?
ao
_ 89
24
274 Algebra 2, Third Edition
Problem Set 98
PROBLEM SET 98 2(a’) 8B + 8W = 56
(b’) 8B — 8W = 40
16B = 96
B = 6mph
96 = Rwlw + RrTr: Tw = 12 - Tr; (a’) 4(6) + 4W = 28
Rr = 2: Rw = 6
4w=4
96 = 612 —T,) + 127, W = I mph
967=272 = 6TR = 12Tp
Variation method:
24 = 6Tp R = rabbits; S = squirrels; r = raccoons
R= %:
4 = Tr
Tw = 8
Dr = 12(4) = 48 miles Oz k(40)
2
Dy = 6(8) = 48 miles k = 05
2. (a) 0.1Py + 0.6Dy = 90
(0.5)(20)
-
(b) Py + Dy = 600 r = 2 raccoons
Substitute Dy = 600 — Py, into (a) and get: Equal ratio method:
(a’) 0.1Py + 0.6(600 — Py) = 90
LS
—0.5Py = -270 R, Son
Py = 540 LOTS (40)r,
5 20(2)
(b) (540) + Dy = 600
ry = 2 raccoons
Dy = 60
540 mL 10%,60mL 60% vets -4]
4,2
1 {2 >|
=-—+-)—-—
4 PNG 12
1200
1 {=}
=—+-—]—
4 PNG
360 a
— 2
12007 RcT¢; Ry = Rc = 10;
860) = RyTy: Tc = 4Ty
ao
Rc _ Ry = 10; 12
(Ry - 10)(4Ty) = 1200
4RyTy - 40Ty = 1200
—40T y= ~240
e
Ty = 6
5
Tc = 24hr; Ty = 6hr; _ 66, 2/85
dee _ 36
es
Rc = 50mph; Ry = 60 mph 5
4. Downstream: (B + W)Tp = Dp (a)
Upstream: (B — W)Ty = Dy (b)
(a’) (B + W)4 = 28
(b’) (B — W)8 40
275
Algebra 2, Third Edition
Problem Set 98
8. (a) 3x + 2y = 5 (a) (-2)? + y? = 16
(b) 5x + 6y = 7
y> = 12
6y = -Sx
+77
y = +2V3
yy Borel. (2, +2\/3) and (-2, +23)
a Gao 11. Sets (a), (b), and (c) are all functions.
Substitute (b) into (a) and get:
12. (a)x + 2y >4
(a’) 3x + (2+ 7)=a 1
>-—-x +2
Ox + (-Sx + 7) = 15
ap
Ale
<= fe) (b)
y21
The next step is to graph each of these lines.
6 B65 3
(2-5) iz
(aie 3x°="5
y= srr
6) xy= 6
Substitute (a) into (b) and get:
(b’) x(3x + 5) = 6
The region we wish to find is on or above the solid
of 5n,— 6. =10 line and above the dashed line. This region is shaded
Solve this equation by using the quadratic formula. in the figure.
x = -
-5 sooo
+ [25 — 4(-18)
_ =
a££ 97
-—— 13. £—-x+ 2 or x + 3 < -1; D = {Integers}
6 6 6 42—x+20rx+3<-l
Substitute these values of x into (a) and solve for y. < eG or a4,
sale) Spe eera cl =6'-§ =4 =3' 2-4
6 6 2 2
eae 6ee 6 er read
D 2
14.
(Gea
yal 2
x24
= 24/3 + 2a—a/2
— xfa/6 + 12a/6 -3a/6 _ 5134/6
el56 SE6.97 SEL
5
@
97
2
ee
15. xy bs xy
35). x97es aa i
xy
pet ED:
2
6 6 °2 4
esp = ty a
y
10. (a) x* + y? = 16
(by) 26 y = d =)OD
et 7 eS
x*y+x+ xy? xy + 1+ y?
3x2 = 12
x2
=4
16.
ip = a
ET; I I N + io)=.
(a) (2)? + y? = 16
y? = 12
y = +23 18.
276 Algebra 2, Third Edition
Problem Set 98
19. There is never an exact solution to these problems.
Sane
22, ——~— eee ON
One possible solution is given here. .-- “=
See ep
3 243. —3243
_ 9 - 3¥3 + 63 - 6
23
=11S, 7150. _ sen * 9-12
D278 Sen 3
enae aed
Se="—5)0P 4b =3
Use the point (4, 175) for P and S.
175 = -S0(4) + b PARK, = -{t+1)
vectmey) m n
a7) 312
GE R RAIN
S = -50P + 375 te ny. m on
20. 12 amn = nxz + nyzZ + mxz + myz
oo?
amn — mxz — myzZ = nxz + nyz
nxz
+ nyZ
an — xz — yz
is
24. a = (— + *)
Nagteey m n
a Zz 26
CG 10 cos 64° = 4.38 =—+—
5oii a) m n
D 10 sin 64° = 8.99
(BBE oF Va am BGG oe myz = amn
12.00R + 0.00U amn — nxz — mxz = nyz + myz
4.38R — 8.99U
amn — MXZ — NXZ
16.38R — 8.99U —l]
i ay
mz + nz
25, 9-9 A i> =
=-3i+2-3+2i+4=3-i
42 EB
ae BRE
16.38 keds 26. fo - 242 + sven S S21 221 Liga
| *
[20S
_ 721 621 , 210V21 _ 211/21
21 21 21 21
H = \(-8.99)* + (16.38)? 5
+
100
— —_—_—_—
100 ~=100
H ~= 18.68
18.68 /-28.76° 10 100
oF elon 3 ,
21. S=v2_ 2+ 52 10 10
PAD, 2 52
L311.
10 — 2/2 + 25V2 - 10 x=-—t
10 10
E
pe 4-150. ts
s pe
23.24 a2 oo ages ge ee SOUS 60(2.54)(60)
—~
ew P
dG 5 1 in. 1 min eet
Algebra 2, Third Edition 277
Problem Set 99
a «x ay =x 3. Downstream: (B + W)Tp = Dp (a)
hy BA ye xy? _ay-x Upstream: (B — W)Ty = Dy (b)
aS 845 Seay 8 Since Tp = 2Ty, we substitute and get:
x xy? xy?
(a’) (B + 8)2Ty = 280
47,162 x 107!2 (b’) (B - 8)Ty = 60
30. (a) ———————
®) 50,132 x 10° (a’) 2BTy + 16Ty = 280
= 2(b’) 2BTy — 16Ty = 120
Estimate: eal a = 1x 10°!’
5x 10 4BTy = 400
(c) BTy = 100
Calculator answer ~ 9.41 x 10738
(a’) 2(100) + 16Ty = 280
(> 311 Ty = 5
Estimate: 310!° = 6.77 (c) B(S5) = 100
Calculator answer: SAVE = 5.41 B = 20 mph
Equal ratio method:
PRACTICE SET 99 Ligon ins
2 PLP,
—|x| + 3 < -2; D = {Integers}
8,000 — 100(20)
=|7| aes 16,000 P, (2)
yl ?>~5 P> = 2000 people
Variation method:
SH 8} Say SHS)
SPS) Oi 2G} Gy
W = kPF
8000 = k(100)(20)
4 S06
PROBLEM SET 99
16,000 (4)(P)(2)
1. Silver (Ag): 2 x 108 = 216
2000 people EF
Sulfur: Lx 32 po
(a) 0.05Py + 0.1Dy = 96
Total: 216, -32°="248
(b) Py + Dy = 1200
Silver: NS x 100% = 87.10% Substitute Dy = 1200 — Py into (a) and get:
248
(a’) 0.05Py + 0.1(1200 - Py) = 96
2. (a) 2Np = 10Ny + 16 —0.05Py ll | i)&
Py = 480
(b) 13Ny = Np + 8
(b) (480) + Dy = 1200
(Wi) NpecaNye 8 Dy = 720
Substitute (b’) into (a) and get: 480 mL 5%,720mL 10%
(a’) 20BNy, — ee 10N ee 16 -|x| + 5 > 0; D = {Integers}
26Ny — 16 = 10Ny + 16 -|x| > -5
16Ny=-32 |x| <D)
Ny = 2 maunderers x >-—5 and
x < 5
(b’) Np = 18 purposeful walkers 5 -4-3-2-1 012 3 4 5
278 Algebra 2, Third Edition
Problem Set 99
7. -|x| + 1 < -3; D = {Reals}
of)
Substitute (b) into (a) and get:
|x| < —4
il 2
|x| > 4
x? —- 24 = 2x
ea
OL 4 4
Peer At FO)
(x — 6)\(x + 4) = 0
—§ —41=3
—2 E10, V1 208 84) 5
x = 6,-4
8. (a) Since this is a 30°-60°-90° triangle, Substitute these values of x into (a) and solve for y.
1 2 4
2 x SF= 124 (a) y = 3° )
—(6)-—-= ae —
SF= 62
1 (— v) (=)
m = ¥3(62) = 62/3 (a) y = —
rave =
3 ot
n = 1(62) = 62
(b) Since this is a 45°-45°-90° triangle,
2 x SF = Sle
12.
Se Se
3x? = 15
p=16)=S5
x2
=5
g= 16)=5 x = +5
(a) y? = 8 - (v5)?
ae Nw | yoiet3
I]t
aieWile ee
wl<ae,
ye ays
Il ++ (a) y? = 8 — (-V5)?
iae
SI}a
to
y? =3
NIE
NiI[F
mw _ NN = ee) y = +V3
- i) 42. 4 (V5, +,/3) and (-V5, +,/3)
10> (a) 37 4Sy7 = 9 13. (a) 3x + 2y-z=1
Ss! x ES (b)x+y-z=-l
(c) 5x + 2y + 2z = 8
(b) 4x — 6y = -8
(a) 3x + 2y-z=1
Substitute (a) into (b) and get:
-l(b)
x - y+z=l
(b’) 4x — 6(-x + 3) = -8 (d)2x+
y =2
4x + 6x — 18 = -8 2(b) 2x + 2y — 2z = 2
10=-10 (c) 5x + 2y + 2z = 8
Va (e) 7x + 4y =
(a) y=-(1I) +3 =2
—4(d) -8x - 4y = -8
(ec) 7x+4y= 6
(1,2) —x = —2
x= 2
11. (a) x -3y=2
(d) 222) + y= 2
neyg y =2
. (a? sean 3
(b) @) + 2) &z=Fi
(b)
xy = 8 Lan
8 (2, -2, 1)
y= x
279
Algebra 2, Third Edition
Problem Set 99
14. The solution is © or { } because 4 is not in the 22. O= ml +b
ne nO: Use the graph to find the slope.
200 4
15. (a)x-y<-2
= ae n=—— 50 =
UE apie O=41+b
(b) 3x + 5y < -5 Use the point (45, 0) for Jand O.
3 0 = 4(45) + b
Hi ales “30.20
The next step is to graph each of these lines. O = 41 — 180
23.
ae B
Pe Sa
A
A = 4cos 65° = 1.69
The region we wish to find is on or below the solid
line and above the dashed line. This region is shaded B = 4sin 65° ~ 3.63
in the figure. 0.00R + 6.00U
1.69R + 3.63U
16. x +4<3 orx+1> 2; D = {Reals} 1.69R + 9.63U
x < -l or caer | an Doe
1.69
SE or Wes Mie eects
=2°=1 40 132 6 = 80.05°
= 2 2
17. (x2-4)2y4/4 = y4-2aya/4 _ 4 -7ald F = (1.69)? + (9.63)? ~ 9.78
9.78 /80.05°
18. = e oe ng Cees A 2s Sumy Sx eats
a+ - ree i 2 8 =
ea oe b’ +a _ili- 6/2
; 7
ab(b* + a)
ab? + a? +b? os, 422V3 V3 _ 4V3-6 _ -4V3 +6
5 5 3
19. 216V2 = 25242 = 2(245)(2!15) = 228/15 oe v3 a ss
26. wig gsop
oy 22S° 9
ete ~y, ri
ee
= A =
3
)
tek :
apm = Cpa: acp
abm
sae ie e ee bp + ap .
21. 2i s! Betis ak 2 Dt
=f aan 1 We =O Bi a ee ee
= -31 + 31 + 2i - 41 + 6i = 4i
280 Algebra 2, Third Edition
Problem Set 100
9>,
/3
Plage ipEAs
[10
Saieaan£
From this we see:
10 3 “ (a) Opens downward
ae eS ansg (b) Axis of symmetry is x = —2
410 3
(c) y-coordinate of vertex is +1
_ ¥30
3
, 4V30 _ 4 135
10 3
_ _3V30 , 40V30 _ 60V30 _ _ 23/30
30 30 30 30
29. PV = nRT
Yee nRT
P
ie (0.0163)(0.0821)(870) ~ 0.12 liter
10
30. 40 cm Z lin. . 1 ft .
S 2.54 cm 12 in. 5280 ft
60s 60 min _ 40(60)(60) mi
x x SS ——————
1 min 1 hr (2.54)(12)(5280) hr
PRACTICE SET 100
ay SaOr + 3
Rplz = 320; RsTs = 240;
wax Ox y+ 3
Rp = 2Rs; Tp = Ts - 2
eG ee 1s — 1
Y =X 19h 422 2Rs(Ts ad 2) = 2()
2(240) — 4Ry = 320
From this we see:
4Rs = 160
(a) Opens upward
Rs = 40 mph
(b) Axis of symmetry is x = 1
Rp = 80 mph; 7, = 6hr; 7g = 4hr
(c) y-coordinate of vertex is +2
0.68(600) — 1(Dy) = 0.2(600 — Dy)
408 - Dy = 120 - 0.2Dy
O.8Dy = 288
Dyn = 360 mL
(a) 10Ny = 6Nc + 40
(b) 4Nc = 2Ng + 160
(b’) Ny = 2Nc - 80
Substitute (b’) into (a) and get:
(a’) 10(2Nc — 80) = 6Nc + 40
b. y=Ax — 413
20Nc - 800 = 6Nc + 40
14Nc = 840
-y = (x? + 4x ) +3
Nc = 60 coruscated
-y = (x7 + 4x +4) +3-4
-y=(+2-1 (b’) Ns = 2(60) - 80
y=—(x+2)? +1 Ny = 40 sparkled
Algebra 2, Third Edition
281
Problem Set 100
2 From this we see:
R | 4S
Res od NG, Hy (a) Opens downward
100 _ 40(5)? (b) Axis of symmetry is x = 2
R, — 20(10)? (c) y-coordinate of vertex is +8
100(20)(100)
Ry = ————
(40)(25)
R, = 200 reds
Consecutive integers: N, N + 1, N + 2
NN + 2) = 8(N + 1) + 32
N? + 2N = 8N + 40
N? - 6N - 40 = 0
(N + 4)(N- 10)
=0
8. (a) Since this is a 30°-60°-90° triangle,
N = -4,10
/3 x SF =5
The desired integers are —4, —3, -2 and 10, 11, 12.
raped
3
yas x = 6x 4-1
We (tee Ce 459) 1 9
m = 2939)
= 10N3
3 3
y=(x- 3-8 _ 534) _ 5V3
From this we see: 3 3
(a) Opens upward (b) Since this is a 45°-45°-90° triangle,
(b) Axis of symmetry is x = 3
1x SF =7
SF =7
(c) y-coordinate of vertex is —8
p=T01)=7
q = V2)
= 7/2
—|x| + 5 £ 3; D = {Reals}
=x) Sie
lel tag 2)
w= 2 and xy
10.
11 3
y=—-x*t+4r44
=sto ee
aye (xe = 45°44)
24 4 Ss ENG 6
~y = (x - 2)? -8 yPe Ss
=-+—
3 6
Y=—-@ =2)7% 48
282 Algebra 2, Third Edition
Problem Set 100
M. (a) 2x =<2)-<-1 7 13. (a) x27+y2=5
mi eae (b) 2x7 - y? = 4
‘pinoes 3x? =9
(b) 4x + 3y = 5
2
= 3
:
x = +3
Substitute (a) into (b) and get:
; (a) (/3)+?y? = 5
() a(y- 2)+a =5 y2=2
AyEsoie: ay 5 paca
7 (a) (-V3)’
+ y? =5
y=1 Filet
= +2
@x=()->=2 cairn
(/3, V2) and (-V3, +2)
(3) i 14. (a) 2x + 2y-z=0
(b) x + y — 2z = -12
12. fa)x—y=d
(cy 2x = ys+-z
= 10
(oxy. =. 2
2 (a) ean 2 oe 0
ive —2(b) ~2x — 2y + 4z = 24
3z°= 24
Substitute (b) into (a) and get: Bae
. 2
(a’) r-(2)=5 (b) x+y
—- 2z = -12
x
2 (jar yz =e 10
x" -2= 35x Ais merce ae)
i? 2 CSL Ei0
(d) 3x — (8) = -2
Solve this equation by using the quadratic formula. Beef
2(1)
5 as ( + 2y
(a) 2(2) v= — (8)
(8) = 0
6 ah erry
2 H2 eam 2
Substitute these values of x into (a) and solve for y. ya
(yl
$, +=434
— a =
eee) (2, 2, 8)
2 2 2 2
15. None of the sets are functions.
5 38 eae
(a) y= 2 = a < o) = Ty = eRe Set (a) is not a function, because 2 has two images.
Set (b) is not a function, because 2 has two images.
Soo ie aay 8 tae, i
2 vs eae) 2 - Set (c) is not a function, because —3 has two images.
5 33 5 a 1 33
2 2" se 2
Algebra 2, Third Edition 283
Problem Set 100
16. (a)
y < -3 23.
8
(b) 4x + y < -2 B
y<-4x
-2 el a
A
The next step is to graph each of these lines.
= 8cos 28° = 7.06
= 8 sin 28° = 3.76
6 D
es Ta
C
C = 6cos 62° = 2.82
The region we wish to find is on or below the solid D = 6sin 62° = 5.30
line and below the dashed line. This region is shaded
-7.06R + 3.76U
in the figure.
2.82R + 5.30U
-4.24R + 9.06U
li. ©4225 ox +3 -< 3; D = {integers}
9.06
tan 0
Xo On x <0 —4.24
0 v —64.92°
es} 0) Sl BE My ZEB
Since @ is a second-quadrant angle,
6 = (-64.92) + 180 = 115.08°
2x+4_b/2
18. a — = ght!
= gt tse
a
F = (4.24)? + (9.06)? ~ 10.00
10.00/115.08°
Oe
pe ates
ee 24.
5-2/2
V2 _ 4+ 5y2
ee xy + 1 f2 —2 2
xy
3-4/5
V5 +1
= ew ore Dae, bY t doy 25.
4 eal 5 +1
xy? ai x?y = 2Y xy? +xy-—y
_ 3V5+3-20-4/5
-17- V5
20. 3/9V3 = 3/37/73 =3-3.- 3/4 = 394 5] 4 4
Paes Sabres a
26.
2 13
[hee Od
41. 2 u 5 CANE cee Le nari te
13 2
—i —1 U —(-1)
~82 4B WEARao.
Sg = i. ay
13 Ailey a YD.
22 = :
RTT Oh Seas, RR, 4 2V26 | 39V26 _ 104/26 _ 63/26
26 26 26 26
ESO
tage ed
1+ 9 i, 27. —/-7/-7 + 2J-16 - 31> + 2i?
=7+8-3i1-2=5
+ 5i
284
Algebra 2, Third Edition
Problem Set 101
b+ ©
I |y + 3&
4. 0.6(400) + 0.8(Dy) 0.72(400 + Dy)
240 + 0.8Dy = 288 + 0.72Dy
ay = kmx + cmx + mdky + mdcy
0.08Dy = 48
ay — mdky — mdcy = kmx + cmx
Dy = 600 liters
kmx + cmx
a — mdk - mdc
5. Downstream: (B + W)Tp = Dp (a)
400fC * 412 in. = Prin. 12in- Llimin
29. - x x x Upstream: (B —- W)Ty = Dy (b)
min 1 ft 1 ft 1 ft 60s
_ 400(12)(12)(12) in.? Since Tp = Ty, we use T in both equations.
60 Ss (a’) (20 + W)T = 104
a e10-° (b’) (20 - W)T = 56
(a) Estimate:
ane
—) X wd sae (a’) 207 + WT = 104
Calculator answer =~ —3.92 x 19713 (b’) 207 — Wr =056
40T = 160
(b) Estimate: some number between
(4)> = 10 x 10-4 and (5) = 3.2 x 10-4
Calculator answer ~ 7.46 x 1074 (a’) 20(4) + 4W = 104
4W = 24
W = 6mph
PRACTICE SET 101
Selling price = purchase price + markup y= 2 opRE, ee
16,295 = Pp + 0.25Pp yee
ax feed Jared
16,295 = 1.25Pp
enter 2) eral
$13,036 = Pp
R=EG 4 2) 21
From this we see:
PROBLEM SET 101
(a) Opens downward
Selling price = purchase price + markup
78 = Pp + 0.3Pp (b) Axis of symmetry is x = —2
1oe= 1.3Pp (c) y-coordinate of vertex is —1
$60 = Pp
Markup = 0.3($60) = $18
Selling price = purchase price + markup
Sp = 2100 + 0.6Sp
0.4Sp = 2100
Sp = $5250
Selling price = purchase price + markup
16,5352= Pp + 0.25Pp
16,535 = 1.25Pp
$13,228 = Pp
Algebra 2, Third Edition 285
Problem Set 101
2. y= x? 42x +2 Substitute (b) into (a) and get:
>
3 = (x7 +
(x 2x +1) + 2-1 (a) ex+ {2x+3)=8
Ve ae 124 4+9 2
From this we see: 6p eer ~ = 8
(a) Opens upward
(b) Axis of symmetry is x = -1 8x + 15 = 16
—8x = 1
(c) y-coordinate of vertex is +1 +
oes 8
1 e:
(b) )7 =?)
( q 5
——|+—
ae om
ner wale tae
Fa)
8° 4
12. (a) 3x -y=4
l
8. Veone = yer h = 602 em> (b) xy = 5
, = 60%(3) em ae
ecm) x . x
3
h= er = 20cm Substitute (b) into (a) and get:
9ncm
(a’) 3x - (=) = 4
9. -|x| + 3 < 2; D = {Integers} x
ine —1 3x2 — § = 4x
|x| e723 2ay 5, 9
Solve this equation by using the quadratic formula.
———_9— 09 at _4)2 —* =
2(3)
=
10. N= 334 242-32] caren.
3 S\ 6 3 ra es
_ 10 2(29 20 7
ns as slg =e Substitute these values of x into (a) and solve for y.
ee @y= 2+) 4-240
xs sei at
50. 5. bS9 2 19
=—
is +
5
— i
is (a) i y= 3—_ —- 3
—— = Ao— os Deen — 4/19
11. (a) 6x + Sy Sy~ iH} (o-2) , v9
SS: 19 and
Ww)
Bie
ae
(b) 4x + 2y = 3
(32
- {19
P.-2- vo)
vo
|tN
286 Algebra 2, Third Edition
Problem Set 101
13. (a) x7
y* =-16 16. (a) 3y — 2x > -3
(b) 2x7 — y* = 2
Z
3x2 = 18 >=-x -1
z 3
x?
=6
(b) x = —2
x = +6
The next step is to graph each of these lines.
(a) y? = 16 — (V6)?
y* = 10
y = +v10
(a) y? = 16 - (-V6)?
y? = 10
y = +10
(./6, +10) and (-6, +/10)
14. (a) 3x +y+z2=7
The region we wish to find is on or to the right of the
(b) x - 2y —-z=2 solid line and above the dashed line. This region is
shaded in the figure.
(c) x +y-z=-5
17. -10 <x + 2 < -4; D = {Reals}
(a) 3x + ytrz=T7
(b) x -2y-z=2 a 2X <6
(d) 4x - y =9
(a) 3+ y+z= 7
(c)
x + y-z=-5
x+2\1/2 x/2 +1
18. Cte ain ea Y _ gxl2+11-20
(y7)4 $ yr J
(e) 2x + 2y = 2
2(d) 8x — 2y = 18
m a m
(e) 2x 4 2y = 2 19,
my — - i = my
10x = 20 al my — 1
ys sD my
m(my — 1) my - 1
(e) 2(2) + 2y = Z
y=-l
20. 7493/7 = WAT = 7(7)'27") = 71516
(c) -(2) + (-1l) -z=-5
z=2 2-31 3-21 -6+4i-9i-6
21.
se a Shwe OT on 4
(2, -1, 2)
= tS.mf
15. p(x) = x? — 4; D = {Reals} ee ee
i Zi) grog tot
22.
he ee) oe ee ee ee
ro Cage eae
-=43? —13
287
Algebra 2, Third Edition
Problem Set 102
23. 2x0y?
pix
pAypim _ 6xlyp
xp
tay”
B 8 =i ae
_ 2y ue * a B = 2mpty! — 6x-ly?
70s
p- x“ p
12
eo
A
PRACTICE SET 102
A Sicos 70 = 2.74
KX) =x +1
Ba=esisind 02s ee
g(x) = x? - 1
—2.74R + 7.52U (h + g)(x) = x2 + Xx
12.00R + 0.00U
9.26R + 7.52U (h + g)(5) = (5)? + (5) = 30
LS2 h@) =x 42
tang —
9.26 h(-2) = (-2) +2 =0
@ = 39.08° g(x) = x7 — 7
F = (9.26 + (7.52)" =111.93 g(-2) = (2) - 7 =-3
11.93 /39.08° hg(-2) = 0 - 3 = 0
fm =x+6;
gx) =x-4
24.
242
= 43 46 4/98 1342
V6 V6 6 fa(x) = & + 6) - 4)
fa(x) = x? + 2x - 24; D = {Positive integers}
25.
Br? 2 eee
6 2 2D
pO A228 ae
WES =150 PROBLEM SET 102
2 Ds
Ler
Br RG
26. Pic2D wea5 DAO Dp
RsTs + 40 = RyrT jp; Rs = 46;
_ IND V5 aN ing
Ryr = 50; Ts = Typ
TS Ser eee.
46(T;) + 40 = 50(Ts)
=
4V10
_ 20V10 +| 40V10
_ 1210
4Ts = 40
10 10 10 5
Ts = 10hr
ihe Ty
a ee ae eee Ds = RsTs = 46(10) = 460 miles
=3+2i+4-i+3i=7+4
Downstream: (B + W)Tp = Dp (a)
28. ee eee ee Upstream: (B - W)Ty = Dy (b)
x R, R, (a’) (10 + W)T = 78
mR,R> = CR, Rox = Rox “tr Rix
(b’) (10 - W)T = 42
mR,R> = CR, Rox = Rix = Rox
(a’) 107+ WT = 78
R,x
R, = 2
(b’) 107 - WT = 42
mR, - cR,x -Xx
OOF 28120.
29. 2x7
=x +4 T=.6
x7 4+x4+4=0 (a’) 10(6) + 6W = 78
x=
_ 1+ yay? = 424) |,
-—
V3.
+ ——j
6W = 18
W = 3mph
2(2) PS 4
288 Algebra 2, Third Edition
Problem Set 102
Ay Giay From this we see:
H, G,(RY (a) Opens upward
5” =e Gy (b) Axis of symmetry is x = —2
10 G4) (c) y-coordinate of vertex is —2
2(36)(10)
G, 2 = ———_
5(16) = 9 goats
Selling price = purchase price + markup
1666 = Pp + 0.7Pp
1666 = 1.7Pp
$980 = Pp
Selling price = purchase price + markup
1680 = Pp + 0.4Pp
1680 = 1.4Pp 10. yo = aK AD
$1200 = Pp Hy ="? 4x + 4) 4 DR
yes (4 422) te?
gees
Rt From this we see:
al ee me (a) Opens downward
716-1016 (b) Axis of symmetry is x = —2
V> = 589.18 mL
(c) y-coordinate of vertex is +2
hx) =x + 1; D = {Reals}
g(x) = x* — 6; D = {Negative integers}
h(-5) = -5 + 1 = -4
g(-5) = (-5)° - 6 = 19
SEAMS
hge(-5) = —4(19) = -76 $d tt
f(x) =x + 4; D = {Reals}
g(x) = x — 1; D = {Positive integers}
fg) = & + NO - I)
fg@) = ne One A
We cannot use this function to find fg(—3), because
~3 is not a member of the domain of g(x), so it is not
a member of the domain of fg(x). Therefore, the
answer is either © or { }.
pax? + 4x42
y= (x7 +44 4+4)4+2-4
ying
By — 2
289
Algebra 2, Third Edition
Problem Set 102
12. (a) 2x + 3y = -3 (a) y? = 12 - (2)
(b) 4x — 2y = 18 y= 8
y=
2x -9
y = £202
(2, +22) and (-2, +22)
Substitute (b) into (a) and get:
(a’) 2x + 3(2x - 9) = -3 15. (a) xX +y+Z ll oo
2x + 6x —- 27 = -3 (b)x+y-z=0
8x
= 24
(c) 2x -—yrzZ
8
(a)x+ yrZ=
(b) y = 23) -9 =-3 (bt) x+ y-Zz=
(3, -3) 2x + 2y =
13. (a) 2x -y = 6 (d) x+y=
(b)
xy = 4 (bt) x+y-Z=
4 (c) 2x -y+z
yes x
3x =
Substitute (b) into (a) and get: IS
(OS
(On|
oe)
colle
=
(d) () + y
(a) 2x. = (=) 6
Xx y= 3
2x?
— 4 = 6x (a) U). +.G) + z=|
x7 -3x-2=0 Z= -
oC
Solve this equation by using the quadratic formula. (1,
3, 4)
~(-3) + (3 - 4@(-2) 16. (a) 3x — 5y < 10
2(1)
3
i Oieastl y > =x5 -2
2 D
(b); yigeiaee?
Substitute these values of x into (a) and solve for y.
The next step is to graph each of these lines.
@ y= te
23 ae
2. 6 -3
+ 17
y
@y= 5S 3Sees
Z) 6 3- JI17
14,
4x? = 16 The region we wish to find is on or above the solid
line and above the dashed line. This region is shaded
x2 =4
in the figure.
Bey
i. x+2<0 or x +3 £ 3; D = {Integers}
(a) y? = 12 - 2°
x < —2 or xO
y? =8
y = +2/2 4-32-1012
290 Algebra 2, Third Edition
Problem Set 102
18. tan0
10
4
6 X= 68.20°
2/29 /68.20°
a b
25.
D5 TU ERC.
am + ac + bx = amx? eaex,
2 2
ac — acx” = amx = am — bx
19.
amx 2 — am — bx
(ay =
a—- ax
20.
2-2/2 BAO
26.
EID) an) EY aes ee)
6V2+4-12-4/2 _ RR Ale
= -k(kx - 1) =k - Wx 18— 4 14
ae Ware |)
3/5 4) xy as
= x 3yl/3, oy 2 x23/ 12,,5/6
21. y 7
31+4 oe F Pap Cy ee ae) ee) 3i
22.
1-1 1l+i
28. 4 [> + 3/= - 360
eal tkB — 3/60
= ia) atoe
4/60 , 3.160
See : — 3/60
=!Same
23. 12
—k—)2
20/60 é 36/60 _ 180/60
60 60 60
_124V60 _ _ 6215
GF aia 15
24. 29. See Lesson 71.
30. ZAED and ZCEA are supplementary angles and
need not be equal.
Choice (c) is correct.
Algebra 2, Third Edition
291
Problem Set 103
PRACTICE SET 103 (Ry + 40)2Ty = 720
2RyTy + 80Ty = 720
‘—— Sxy + 16y?
2(200) + 807 = 720
a. 2x + 4y) 8x3 ce 64y°
8x> + 16x7y 80Ty = 320
-16x7y Ty = 4hr
—16x7y - 32xy*
Tp = 8hr; Ry = 50mph; Rp = 90 mph
32xy? + 64y3
32xy* + 64y°
0 sean
Hy et,
4x7 + 8xy + 16)’
ae
b. 2x - 4y) 8x3 = 64y° 300% 47s
8x td 16x7y
T, = 450K
16x7y
16xy - 32xy"
Multiples of 11: 11N, 11(N + 1), 110@V + 2)
4(11N + 11N + 22) = 10(11N + 11) — 66
88N + 88 = LION+ 44
44
= 22N
PROBLEM SET 103 2=N
1. Selling price = purchase price + markup The desired integers are 22, 33, and 44.
1424 = Pp + 0.6Pp
ae 6xy + dy?
1424 = 1.6Pp
3x + 2y)27x3 a
$890 = Pp ee aes 18x7y
~18x*y
Downstream: (B + W)Tp = Dp (a)
~18x*y — 12xy?
Ld
Upstream: (B -— W)Ty = Dy (b) 12xy? + 8y°
(a’) (18 + W)T = 132 12xy? + 8y
0
(b’) (18 — W)T = 84
(a’) 187 + WT = 132 Since the lengths of tangent segments drawn to a
circle from a point outside the circle are equal,
(b’) 18T - Wr = 84
y+ = 5
36T HT] i) petON
r Il y=
(a’) (18 + W)6 = 132 y +8 =x + 15
6W = 24 12: $8 ll & + — Nn
W = 4mph Sax
Perimeter of triangle
3 Dn
a el =x+15+y+84+x4+x4+y+3415
720
3x + 2y + 41
Dy 3(5) + 2(12) + 41 = 80
200
Since 2 is not a member of the domain of b(x), it is
RpITp = (20: RyTy = 200:
not a member of the domain of ab(x). Therefore, the
Tp = 2Ty: Rp = Ry + 40 answer is either © or { }.
Algebra 2, Third Edition
Problem Set 103
9, y=x? + 4x 4+ 6
12. wea + 325-4)
¥ =a" 4) + 6 Ao SN 82) ame
v= Gur oy + 2 1
=—-+-]—- —
(2 *)
4 3\.4 4
From this we see:
1 3
(a) Opens upward = — + —
4 2
(b) Axis of symmetry is x = —2 1 6 af
=—
+ —- = —
(c) y-coordinate of vertex is +2
Awied, aid
13. (a) 4x + 3y = 17
(b) 2x - 3y = -5
2 5
- aes 3
Substitute (b) into (a) and get:
(a’) 4x + (3s + >|eely
3
Ce es Trem Sas lg
Ox 712)
10. yeh 4 4x 2% Xe=
y= (4x44)
+ 6 =4 D)
(b) = —@)=
ya Gs 22 3
y= -G—2y -2 2INn
4
=—+4+—2=+3
38
5
U6
From this we see:
(2,3)
(a) Opens downward
14.
(b) Axis of symmetry is x = 2
(b) xP ay" = 2
(c) y-coordinate of vertex is —2
y=ux {2
Substitute (b) into (a) and get:
(a’) x7 + (x- 2% =6
2
x = a 4 = 6
Ont dys 2a) 0)
Neer est)
Solve this equation by using the quadratic formula.
pe DEVRY 2
= 4OCD _ a 5
2(1)
Substitute these values of x into (b) and solve for y.
(b+) y= (1+ V¥2)-2=-1+V2
(b) yrs
Il (1 =+/2)
-2 ll24 S42
1° <2) 3) eA <5
(1 + J2,-1+ V2) and (1 - /2,-1 - V2)
Algebra 2, Third Edition 293
Problem Set 103
15. (a) x* + y? = 10 The next step is to graph each of these lines.
y= 10." ,,
(b)-2aciep2y? = 5
Substitute (a) into (b) and get:
(6 2x? O00 =x) an5
2 20 es ;
ee OES
a 4
x= po
2
; 1 fxi eees The region we wish to find is on or above the solid
pups penne ai Gham oa) and-solve Tory. line and to the left of the dashed line. This region is
(a) y? et Oe (> - qs shaded in the figure.
2 2
ben See 15 18. -3 < x - 3 2 4; D = {Integers}
pe eo Oma <a
5 ,vi5
Aamo Py pacers
ee SN = Ob haat es iad aie ae
0 ft 23°46 6. °7
x2 ab — 2b
16. (a) x + 2y+z=-1 19, ——— - 5b/2
= 2b
x
(b) 3x -y + z= 6
2 m m2
(b)3r— yy += 6 m° + <7 Ae i
(c) 2x — 3y — z=. 8 sa Rss m +1
(d) 5x - 4y 4. me +1) Pe te
(a) x+2y+z=-l wD +m +m m+ 2m
(c) 2x -3y —z= 8
—
3x — ay a
ae 21. 5) xy ‘ey = x2/5 3/5 1/4 1/4 Ee. x¢13/20,17/20
3}
€ = (3x —*7
i. ; 22 ae A ae hy
Substitute (e) into (d) and get: er a eet ea err anny tee:
(d’) "5x = 46x ars. 14 _ Bape see
SX = 124 2 28n=, 14 ° aeare et
ixe= 14
x=2 23. 41-23. Welt
© y= 30) 47 sal ie oreas
(a) (2) + 2¢1) +z =-1 = 210i 4 2
—10i — 4 + 25 —- 10: ale
25+4 29 29
z=-l
(2, -1, -1) 24 Ba 2/5) , SiS
17. - (a) Ady < -4
(a)x-4ys- Bas 5S ds
I - 15+ 65 +10/5 + 20
ats ge 25 — 20
_Riera
35 + 16/5
yy 758
294
Algebra 2, Third Edition
Problem Set 104
25. 30. Since there are 180° in a triangle,
ZABD + ZBDA + ZBAD = 180°
ZABD + 90° + 40° = 180°
ZABD = 50°
> ll 6 cos 29° = 5.25
Since AABC is isosceles,
& Il Qs 29° = 2.91
ZDBC + ZABD = 70°
-5.25R — 2.91U
ZDBC + 50° = 70°
0.00R + 4.00U
-5.25R + 1.09U ZDBC =-20°
PRACTICE SET 104
6 = -11.73°
8
Since @ is a second-quadrant angle, 0.00000513 x 22. = ——»/3
108 100,000,000
@ = (-11.73) + 180 = 168.27°
100N = 1.524 24 24 24 ...
F = \(-5.25)? + (1.09)* = 5.36 N = 0.015 24 24 24 ...
99N = 1.509
5.36 /168.27°
noe 1.509 _ 1509
he, 99,000
(? - | m
26. aj- ee
C x 2)
ab a_m PROBLEM SET 104
c x Pp
abpx — acp = cmx (a) 2Np = 24NG - 6
abpx — cmx = acp (a’) Np = 12NG - 3
(b) 10NG = Np - 5
Steareiae
Substitute (a’) into (b) and get:
27. Rte Je + 33 = 11 (b’) 10Ng = (12Ng - 3) - 5
(z+ 33 = 11 ~ar 2Nc = 8
z + 33 = 12> 22Vz +2 Ng = 4 geese
22/z = 88 (a’) Np = 12(4) - 3 = 45 ducks
Jz =4
z = 16
Check: /16 + 16 + 33 11
4+7 1] RyTy = RpTps Ry = 240;
Rp
= 360; Ty = Ty -—4
4 3 Fert [3 +548
28. a4 — 2/2 + sa = 240T, = 360(Ty — 4)
120Ty = 1440
- 93 _ Bs m3 = 2143
Ty = 12
29. i = Al + Oi -1 = 91
Pepi Dy = RyTy = 240(12) = 2880 miles
Algebra 2, Third Edition
Problem Set 104
3. (a) 0.1Py + 0.2Dy = 44 9. (a) Since this is a 45°-45°-90° triangle,
(b) Py + Dy = 400 1 x SF =3
Substitute Dy = 400 — Py into (a) and get: SF= 3
(a’) 0.1Py + 0.2(400 - Py) = 44 m = \2(3) = 32
-0.1Py = -36
Py = 360 Nox N(3)"=73
(b) (360) + Dy = 400 (b) Since this is a 30°-60°-90° triangle,
Dn = 40 1x SF =4
360 mL 10%, 40 mL 20% SF= 4
4. Downstream: (B + W)Tp = Dp (a) x = ¥3(4) = 4V3
Upstream: (B — W)Ty = Dy (b)
y = 24) =8
(a’) 12B + 12W = 168
(b’) 9B — 9W = 54 m> + mp + p”
m — p)m pr
3(a’) 36B + 36W = 504 10.
4(b’) 36B — 36W = 216 m — mp
72B eg hZ4u)
hie
B = 10 mph
(a’) 12(10) + 12W = 168
mp? — p>
mp” — p
12W = 48 0
W = 4mph
11. Sets (a) and (c) are functions. Set (b) is not a
5. Selling price = purchase price + markup function, because —2 has two images.
2400 = 400 + My
$2000 = My 12. y =x 4
% of cost = om x 100% = 500% By adeei rand
-+y=(x-
2) -2
% of Sp = om? x 100% = 83.33% y=-(x-2° +2
2400
From this we see:
10° ea 512
6. “ORO000SI ete (a) Opens downward
10° — 100,000,000
(b) Axis of symmetry is x = 2
7. N = 0.01432|32|32 ...
LOON *=" 1A32> 32832°32"... (c) y-coordinate of vertex is 2
N = "0014-32 32532 .:.
99N = 1.418
_ 1418
99,000
1
8. Atriangle a Bee
H = (413)? - (23)
H = /48
—-12 = /36 = 6cm
1
12/3 cm?
ATriangle Fi 5643 cm)(6 cm) =
296 Algebra 2, Third Edition
Problem Set 104
13. -|x| + 2 > -2; D = {Integers} Substitute these values of x into (a) and solve for y.
=x)
> —4
In) <4
@ yi +B) 2-414 iB
a <4 ant x > 4
@y=4t-22)
na 2.418
-4-3-2-10 12 3 4
[+ a1 + via
ana
14. N= + (31-2)
1 V13
8
1
ee
(4
8
)
ff-Ba-ve)
——
8 DSS 8 17.
(a)x—y = 2z=-14
1 (2) (b) 24 + y= z= 2
= — + —| —
8 TAVRS
(Cc)x+y-z=-4
= rapa “A (a). .% = Y= 22 =
56 56 28
(Dayaee ee
1 2 abe — 3z = -12
iS, Oe—xX
—- 5) —y=-5
(d) x-z=-4
(b) 0.005x — 0.04y = —0.755 (a) x -y-—2z =-14
fa oe oF = 75 ()-x+y- z= -4
—3z
= -18
(b’) 5x — 40y = -755
z= 6
fg We 5 re 6y= (d) x — (6) = -4
—1(b’) -5x + 40y = 755 Lie
34y = 680
(ayy
gan Ore 16
y-=420
P= <=
(a’) 5x — 6(20) = -75 (2,
4, 6)
Se = 45
ea 8 18. (a) x — 3y S$ -6
(9, 20) 7 Son
3
16. (a) 4x -y=2 (b) x 2 -2
The next step is to graph each of these lines.
(b) xy =
3
y= =
x
(a’) ax - (2) =a2
Xx
Ai? 35a 2x
Ae I 8 0
Solve this equation by using the quadratic formula.
ee Dt T2
2) = 4(4)(-3)
2(4) The region we wish to find is on or above the solid
slanted line and on or to the right of the solid vertical
line. This region is shaded in the figure.
297
Algebra 2, Third Edition
Problem Set 104
19. O<x+2
4 4; D = {Reals}
aD ee Sa)
3-2-1041 2 3
iH]
GH=Z0icos20e u 18.79
5D (aatye ye e RG ty ae
: —al2 ary) D = 20sin 20° = 6.84
ae y
= x24 - 4y-Sal2 21 21R 221,210
5 hele eeee 2 40.00R — 28.05U
ehCa 1 = al
_ 28.05
05
ae SY, |aea tan @ =
k ~ 40.00
00
BUGS al) Sie kOe @ = —35.04°
PAL eps pe pale Epis be
= (40)? + (-28.05)? = 48.85
». > 4°93 (22)
SqI/IS _ 72/5QI/5 _ 27/15
48.85 /-35.04°
A it eT ee
Se es ak ee 28.
a6 18) +7 Be elt. % y C b
at-9 10 10 bcemy + bemx = bxy + acxy
bemy + bcemx — axcy = bxy
Ba 2+3V2 _ 2+3/2 4+3,2
" 4-18 4-3/2 44 3/2 bxy
+12v2 +18 43 _ gg
_ 8+ 6V216-2818 bmy + bmx - axy
29. -4x7
= x —5
25. -i + /-4V4 - 3-9 + 2i4 Age + 5 0)
=-1+4-914+2=2
- 6i
W 1 5
+e
= f=
26. ogee eae ee 7
= —_—
2/2 = An AO:
(@+dret)-5es
4 64 4 64
os 5
( al 81
= ee x+=—-] = —
8 64
_ 1510 fl4VJ10 80/10 _ 6110 gomsetye in
10 10 io 10 8 8
Pa ee oat
R Pilg
a BO
4
4000 mL 1 cm? 1 in. 1 in.
30. ——— x —— x x
s 1 mL 2.54 cm 2.54 cm
1 in. . 1 ft e Late . 1 ft . 60s
2.54 cm 12 in. rine 12 1 min
> 30 cos 45° = 21.21
Il
4000(60) hie
B = 30sin 45° ~ 21.21 ~ 2. 54)(2.54)(2.54)(12)(12)(42) min
298 Algebra 2, Third Edition
Problem Set 105
PRACTICE SET 105
a. —5x - 12 + 2x7 = 0
RrIy + RywTw = 64; Rr = 10;
ax* = EADS 6
Rw
= 6; Ty + Ty = 8
2--12 = -24
10(8 — Ty) + 6Ty = 64
FACTORS OF —24 SUM OF FACTORS ATy = 16
(-24)(1) —_ 93 Tw = 4
(-12)(2) <a —10 Tr =8-(4 =4
Dw = RwTw = 6(4) = 24 miles
(-8)(3) aa -5
Dy = R7Tr = 10(4) = 40 miles
2 — &y + 3x — 120
2x(x — 4) + 3x — 4) = 0 0.2(140) + (Dy) = 0.44(140 + Dy)
(2x + 3\(x — 4) = 0 28 + Dy = 61.6 + 0.44Dy
0.56Dy = 33.6
2G 4.0) = oe ()
Dy = 60 mL
a i je Sa!
2 Po MAY
M, (Fy)
b ~TIx — 6 = -3x?
5(5)°
GoS
is) |
~a | nN I oO Ae SAO
je‘) iH] 6000 potatoes
FACTORS OF —18 SUM OF FACTORS Selling price = purchase price + markup
Sp = 2400 + 800
(-18)(1) Rae oa -17
Sp = $3200
(-9)(2) seers =i
800
cfg ae ep eed iH =)
% ofSp= =~ x 100% = 25%
3x — 3) #2 — 3) = 0 800
% of cost = x 100% = 33.33%
(3x + 2)(x - 3) = 0 2400
ay 2 = 0 50 =e ty (g + h)%) = g(x) + h@)
=x-¢1 42-5
2
=-— x = 3
43 =x7+x-4
(g + AQ) = (2)? + 2) 4 512
hg(x) h(x) - g(x)
PROBLEM SET 105
(x - 5x" + 1)
1. Carbon: Boule a= 36 XP oh ae eS
Hydrogen: 3% T= 3 Coles 5 tae Ge = 5; D = {Integers}
Chlorine: Si 4 Sipe I N = 0,0001234|234|234 ...
Total: 36 + 3 + 175 = 214 1000N = 0.1234 234 234 ...
N = 0.0001 234 234 ...
Ww 1050 999N = 0.1233
214 C,H,CI, at 1233
C3H3Cls = 1284 grams 9,990,000
299
Algebra 2, Third Edition
Problem Set 105
9. N = 0.01651|651|651 ...
14.
1000N = 16.51651 651 651 ...
Nes 2001651 26515 651. oa. oe
999N = 16.5 TN Cg ee!
_ 165
9990
Be Sc
4 S\4 20 20
m? — mp + p D 1
10. m+p m + Pp
15. (a) =
ae ed = 6
m + mp
(b) 0.07x + 0.14y = 6.58
-m’p = mp” (a’) 8x — Ty = -168
mp” + Pp (b’) Ix + 14y = 658
mp” + p 2(a’) 16x - 14y = -336
0
(b’) 7x + 14y = 658
x“ -— xy + y? 23x = 322
11. x+y e + y? x= 14
+ xy (a’) 8(14) — Ty = -168
-x*y —Ty = —280
-x’y = xy*
y = 40
xy? ay:
xy* + y
(14, 40)
0
16. (a) 5x -y=2
12. y=x? + 6x +8 (b) xy = 6
y = (x" +.6x 409) + 89
yea a8) 1
ae x
From this we see: Substitute (b) into (a) and get:
(a) Opens upward
WN8 | II Zz
os
=e
—
(b) Axis of symmetry is x = -3 aNOA
ae|& Nr
5x?
— 6 = 2x
(c) y-coordinate of vertex is —1
5x2 — 2x -6 =0
Solve this equation by using the quadratic formula.
_ 2)#27
= 46)C-6)
2(5)
stlgp at
5 5
Substitute these values of x into (a)
DiaSi
@) y= 5) = a re ei
y's vag Ol
a) yy = 5}E
(a) — - —]-2=-
5 ) 1 - ¥31
1 31
(3+ os -1+ aan
x <1 and x > -l
1 31
ee
2-160 2 G-“Ba- ai)
300 Algebra 2, Third Edition
Problem Set 105
17. (a) x7 + y2 = 8
(ob) x= y = 2
y=x-2
Substitute (b) into (a) and get:
(a’) xo + (4 2)*4= 8
e+ x? — 4x 4428
a 2x = 0 20.
Solve this equation by using the quadratic formula. ES i Is
y= Rages al (eye
DHE VC) SE =! rig
2(1) (b) x 2 -3
Substitute these values of x into (b) and solve for y.
(b) y= (1+ V3)-2=-1+V3
Gj = C= ¥3) — 2= -1 = 3
(1 2i3F-1 + 43) and (1 — /3%-1—-,3)
18. (a) 2x —-y —- 2z =2
(Be) see y= 2 = 7
(c) 2x -y-z=0
(a) 2x -y-2z=2
(Dyer oy = Seay
3x —-3z=9
(d) Bi 2 The region we wish to find is on or below the solid
faye =p = 2 slanted line and on or to the right of the solid vertical
line. This region is shaded in the figure.
-l(c)
-2x + y+
21. 3x3 — 5x2 — 2x 0
N | | i)
x(3x7 5x?) 0
(d) x - (2) iH
Now we multiply the coefficient of be by the
5a GO
me
constant term.
(b) (1) + y - (-2) ll
3(-2) = -6
y f
4
Find the factors of —-6 whose sum is —5.
(1, 4, —2)
1(-6) = -6
1: (a) 3x + Sy = 4 1 + (6) = —-5
(b) 10x — 15y = —50
Replace —Sx with —6x + x and get:
A x(3x" — 6x + x — 2) =.0
res 3 MCKIE ee) or lx ee) =a0
x(3x + 1)\x = 2) = 0
a
eS —_
Ww =4 Complete the solution by using the zero factor
aaa
theorem.
= a x+ arn |[ts =& re
4 iZels
iI| L ae) 3x+ | Il So * | i) Il So
| & ll | | O° (eS) te se =
0, ey
Algebra 2, Third Edition
301
Problem Set 105
22. 6x? — 10x — 4 0 Find the factors of 12 whose sum is 8.
DOK = 5x 2) 0 6(2) = 12
Now we multiply the coefficient of sie by the 6+2=8
constant term.
Replace 8x with 6x + 2x and get:
3(-2)- =r 56
3x(3x? +. 6x + 2x + 4) = 0
Find the factors of —6 whose sum is —S5.
3x(3x(x + 2) + 2(x + 2)) = 0
1(-6) = -6
3x(3x + 2)(x + 2) = 0
1 + (6) = -5
Complete the solution by using the zero factor
Replace —5x with —6x + x and get: theorem.
Bi = Oneeer ete 0 3x = 0 3x +2 =0 x aD
3x(x — 2) + 1(@ —- 2) 0
(3x + 1) - 2) 0 Xx =t0 x a35-£Z3 x = 2
Complete the solution by using the zero factor
2
theorem. ar ae
3x + 1 ll oS s
x= = 25. Op ap oO
Now we multiply the coefficient of p? by the
= constant term.
2(-5) = -10
23. 3x7 + 8x +4 =0 Find the factors of -10 whose sum is —3.
Now we multiply the coefficient of x by the 2(-5) = -10
constant term. 2 + (-5) = -3
3(4)
= 12
Replace —3p with 2p — Sp and get:
Find the factors of 12 whose sum is 8.
Op + op — Sy Se 0
6(2)
= 12
2p(p + 1) - S5(p + 1) = 0
6+2= 8
(2p - S)(p + 1) = 0
Replace 8x with 6x + 2x and get:
Complete the solution by using the zero factor
3x7 + 6x + 2x +4 theorem.
3x(x + 2) + 2(x + 2) ll
2p -— 5 p+i1=0
(3x4 2)(X4 2) =>
SS)
Dp |
Complete the solution by using the zero factor
theorem.
bye oe A) Lately = ()
a
2
2
i= == x = -2
3 26. Ax? 4.418x)428
oF, 2(2x* + 9x + 4)
Now we multiply the coefficient of x? by the
constant term.
24. 9x3 + 24x? + 12x 0
2(4) = 8
3x(3x? + 8x + 4) 0
Find the factors of 8 whose sum is 9.
Now we multiply the coefficient of x? by the
constant term. 8(1) = 8
3(4) = 12 8 +1 9
302 Algebra 2, Third Edition
Problem Set 106
Replace 9x with 8x + x and get: 2(a) 6x + 4y = 12
2x Ox ex 4 = 0 ; —3(d) -6x + 9y = 3
2x(x + 4) + lw +4) =0 = 15
13y
(2x + 1) + 4) =0 _b15
13
Complete the solution by using the zero factor
theorem. are (=) ¥
2x +12=0 x +4 =,0 13
3x = 6- el
a pots Lea 13
a -4 Bye = .
<
_ 16
Sol nad allied y? cy alee ie = 3B
27.
y2/3 4/6 i yb/3 ,al6 HF
2a +8 b (b) 2}—|-z=9
thas: Y_ +8, 25/3
_ ylla/6 13
eer Ta] we ae J
¥4:" 32
2-3 _ 24+3i -2i+2 seus
i+ 2i7+ 317 -2i-2 -21+2 pee
eer Seb Ss 13
7 =A 94 " egtevg
29. go ti oF
345 PROBLEM SET 106
sass 5 4 4 3./5
1. Selling
_ 20 + 12V5 + 10/5 + 30 ee price =
: purchase P price + marku
P
- 25 — 45 715 = Pp + 0.3Pp
_ -25 - 11/5 11S =" L3P ps
10 $550 = Pp
30. Area of circles = 2(2(1)?) + 2(2)° : 5
= 2n + 4m = 6xunits” La
1200
Area of circles = Area of triangle ,
(6;
is
ae
Au) be
250
4x units = H RpTp = 2005 RcTc¢ = 250;
Rp = 6R¢; Tp =Tc- 1
PRACTICE SET 106 6Rc(Tc - 1) = 1200
Fyteoy = 62.66) 6(250) — 6Rc = 1200
2x-z=9 (b) 6Rc = 300
ota Te Rc = 50 mph; Rp = 300 mph;
i capa. = 4hr
To = Shr; Tp
—(c) —- 3y+z=-10
(d) 2x - 3y = -]
Algebra 2, Third Edition 303
Problem Set 106
3. Downstream: (B + W)Tp = Dp (a) (d) Bri yer
Upstream: (B - W)Ty = Dy (b) (ec) a+y=1
(a’) (B + 2)Tp = 56 4x =o
GS Dp
(b’) (B = 2)2Tp = 80
2(a’) 2BTp + 4Tp = 112 (e) 2) +y=1
(b’) 2BTp — 4Tp = 80 y=-l
“BT,
= 192 (c) 2):4.El ea 2
(c) BTp = 48
ha=-3
(a’) (48) + 2Tp = 56
Tp = 4 (2, -1, -3)
(c) B(4) = 48
bH
B = 12 mph 7. Atriangle = “a
P\V, = PoV> H = Jay — (4ny = V49n? — 1672
700(500) = P,(1000)
= (332
Py, = 350 torr
82(/337) 22
(a) 3x - 3y = 9 Atriangle = Tin cee a 4/332 cm
(b) 4x +z=5
(c) 4y + 2z = -10 8. (a) Since this is a 45°-45°-90° triangle,
—2(b) —-8x — 2z = -10 J2 x SF = 3
(c) 4y + 2z = -10
ape oN2,
—8x + 4y = -20 2
(d) —2x +y=-5
nh
3/2
S|
3/2
3(d) -6x + 3y = -15
2) 2
GQ) 3x 3y = 9
—3x = -6 ie Ss 3v2 = 3V2
x= 2
i. 2 2
(d) -22) +y=-5 (b) Since this is a 30°-60°-90° triangle,
y=-l 1x SF=7
(c) 4-1) + 2z = -10 SF =7
LS
p = V3(7) = 73
(2, -1, -3)
q = 2(7) = 14
(a) 2x —-2y-—-z=9
(b) 3x4 3y = 7 = 6 9. 1000N = 0.7013 013 013 ...
N = 0.0007 013 013 ...
(Clix oy +2-=22
999N = 0.7006
(a) 2x -2y-z= 9
_ __7006
(ey WRsas Nn icPie
—a 9,990,000
(d) 3x - y ="i
10. 100N = 410.26 26 26 26 ...
(b) 3x + 3yneezrSa86
N= 4.10 26 26 26 ...
(C) ek Wy sa
99N = 406.16
4x + 4y = 4
ny = 40,616
(e) fea
aad Stas | 9900
304
Algebra 2, Third Edition
Problem Set 106
11. -~y =x? +2x+ 3
14. (a) =x - LS =)
— pet Ge 29 psd es 4
y= & tel + 2 (0) 0.012454 °0.07y ="2:26
vis.
Gal) <2 (a) 12 5y = 00
From this we see: (b’) 12x + 70y = 2200
(a) Opens downward
=1(a’) 101 8 bye=e 180
(b) Axis of symmetry is x = —1 (b’) 12x + 70y = 2200
75y = 2100
(c) y-coordinate of vertex is —2
Wider}
(a’) 12x — 5(28) = 100
ie = WY
(20, 28)
15. (a) 5x -
(b) xy MT
Se
ef
-|x| - 4 < 0; D = {Integers} Substitute (b) into (a) and get:
0 9-(9
12.
it a4 3
|x| 2 —4
5x?
—4 3x
All integers satisfy this inequality because any
integer (even zero) has an absolute value greater Sep 3 40
than —4.
Solve this equation by using the quadratic formula.
~(-3) + (-3) - 4(5)C4)
2(5)
13. | w |— xe— 3 +t, —
89
ey,
SS 10 10
aut aR\e*)
Ne Substitute these values of x into (a) and solve for y.
ll +
|—
a
caer
@y-f5+)-s 3 , 489
wlS
=wo 10; 2F10 2 2
a|S
|
. 4 3° 80)
(A) 9b 10ies aad
10
|e Sle
2
a eae»)
89 3 BD.)
10 10° 2 y:
3
te
89 3 89
VE 9 3 2
Algebra 2, Third Edition 305
Problem Set 106
16. (a) x7 + y’=7 (b) y> 2
(b) 2x -y=2 The next step is to graph each of these lines.
y= 2-22
Substitute (b) into (a) and get:
(a’) Ox Sd = 7
x + AN? = By $4 =F
5x? — 8k -3 = 0
Solve this equation by using the quadratic formula.
ee TOS ECB = 43)
2(5)
© 4a NBL
i en The region we wish to find is on or below the solid
line and above the dashed line. This region is shaded
Substitute these values of x into (b) and solve for y. in the figure.
eye ye el Ree: mie 1
5 5 5 5 19 3 - 2i Dae ery i
: 3 : a =
An BAl31 Leesan SE ig oler 2 2 Ia Oe2
(b) y= 272-2]
5 5
-2=-2-
a)
55 1 7
20. Xe Xe
E Bt 2 IE)a —
oS 5
(¥ -Fr+a| paar ee
4 _ N31 2 _ 2V31 336k a6 | 36
oe
17. oa(ayo)
ie - 5
= = 27
5
5
(a+ x--
:
12
(peat =
85
2
+—
y= et a9) 6 6
x= 1 + 85
(b) 2x — 5y = 26 ME rg
Substitute (a) into (b) and get: 41. A
a
(b’) 2x - 5(33 ~ 9]=026 yy
ig
a = =f a 45 = 26 B 10
ie = 19
g:
& Il a \o Il A = 10cos 70° = 3.42
e
Sle
GTNSe
f B = 10sin 70° = 9.40
OP FOS Purtet
12
D
(3, 4) Pee
G
18. (a) 3x - 4y = 8
Q i] 12 cos 15° = 11.59
lA | be |
y i) S Il IZsimlo. = 3.41
306
Algebra 2, Third Edition
Problem Set 106
11.59R + 3.11U ; 5
8.17R -— 6.29U 2z~ + 10z + 3z + 15 =
2z(z + 5) + 3(z + 5) =
tan@ = = 6.29
—= 9 = -37.59° (22 + 3\¢ +5) = =
SSS
¥(8.17) + (6.29) = 10.31 2z 4-3 = 0) dinghy I
F = o
10.31 /-37.59° ie 5
22. Yes, the set designates a function, because no two
pairs have the same first element and different “>? -5
second elements.
3 2 =
23. Since i is not a member of the domain of O(x), it is 28. 6p" + 33p" + 45p = 0
not a member of the domain of ‘¥@(x). Therefore, 3p(2p* +-lip*+ 15) = 0
the answer is eithe
© or
r { }. 3p(2p? + 6p + 5p + 15) = 0
m? + mp + p” 3p(2p(p + 3) + 5(p + 3)) = 0
2p la ee 8 3p(2p + 5\(p + 3) = 0
2 2
3p = 0 ap +is= 0 p+3=0
4 a
mp ie 3
Seg oe
aes p=0 aes
pear
ip — "Pp i
0 0, -=,-3
2
3x2 + 7x +2=0 29. 3n°%= 13p — 10 = 0
25
ae eee ke SO Bpaaui (=) 1080
map spy
app — 2) + 2-5) = 0
Axx +2) + Ix 4-2) =0
(3x 41) + 2) = 0 (Gp s22)\p = 3), = 0
3p +2=0 Dantd: 320
ax + th =O yt 2=0
2
Pa p=5
x= -5 x =-2
2
ae (ies
2
30. 90° Jia + 15,5,0
26 aye +e — oe
Aye i Ae Ox = Dm. 2a2 - 6a - 5a + 15=0
2a(a — 3) — 5(a —- 3) = O
3x(x + 1) -2@ + 1) =0
(2a — 5a - 3) = 0
eae) Cae ae) ean)
Ria ZS ‘jee Sees)
3yP 2 =10) eeriy— O
5
Ori Qa -=83
res 36 = ll
hie Cink
ao 2
307
Algebra 2, Third Edition
Problem Set 107
PRACTICE SET 107 Substitute (a) into (b) and get:
T qr (b’), 10U +.(15 = U)s=0G@i=—U)atrgU + 9
ree 9U + 15 = -9U + 159
U = units digit isu =-144
107 + U = original number ee 8
10U + T = reversed number i é 4
Tali a7 Cus ysaiee 3
igi = 78
Ula aman lOne ret = 27 el eae
107 -— T) + T = 107 + 7 - T) - 27 4,.. T-= tens digit
70.— 100 T = 10T +°7 — P27 U = units digit
-10T + T - 107 + T = -70 + 7 - 27 10T + U = original number
-18T = -90 10U + T = reversed number
T=5 Qe S13
(5) + U =7 Caer
Usa2 (b) 10U + T= 10T + U-9
Thus the original number was 52. Substitute (a) into (b) and get:
(b’) 10U + (13 - U) = 1013 - U) + U- 9
9U + 13 = =—9U + 121
PROBLEM SET 107 seh =a o8
1. (a) 0.3Py + 0.6Dy = 144 U=6
(b) Py + Dy = 400 (a) T = 13 —-(6) =7
Substitute Dy = 400 — Py into (a) and get: Original number = 76
(a’)
a’) 0.3Py 0.3Py + 0.6(400
( — Py)w) = 144 5 Spe Py Mime
-0.3Py = -—96
1400 = Pp + 0.7(1400)
Py = 320
$420 = Pp
b) (320) + Dy = 400
a My = 0.7(1400) = $980
Dn = 80
320 mL 30%,80mL 60% Oe ane
(b) 3y 4 2z = 12
2. Carbon: rex Dee 12
(c) x - 3z = -11
Hydrogen: 3x 16543
Bromine: 1 x 80 = 80 Pete ae
Total: 12 +3 + 80 = 95 Se ei a Ore
36 (d) =) + 67 = 16
Bromine: 95 x 100% = 84.21% (b) 3y +27 ="12
3(d) -3y + 18z = 48
3. fF -= tens digit 20z = 60
U = units digit z= 3
107 + U = original number (d) -y + 6(3) = 16
10U + T = reversed number y=2
(a) T+ U = 15 (@) 2x8 (2) = 6
f= 153 y= 2D
(b) 10U + T= 10T+U+9 (-2,
2, 3)
308 Algebra 2, Third Edition
Problem Set 107
a. Yayenteeyitag I Nw x =) 2°=4>D = {Reals}
(b) x — Sy + z = -2 Be trae |
(c) =v + y= z= 2
(b) x = 5yi+ cS? i ale?
(¢c)-x+ y-z==-2
Pee ae SS: 12. N=2+ 462-2)-2+ (4)
3 4\ 3
viel
4 ch 11
(a) Sx- y-z= 2 oS ee
(b) x-S5y+z=-2
(d) 6x — 6y =a 13. (Q)E2 Xe ay ial)
(d) 6x — 6(1)
=0 2a)
FSI
(b) -6x - 40y = —432
(che (alg= =2
Substitute (a) into (b) and get:
Y= 2
(i, 1, 2) (b’) =6x — 40Qx - 15) = —432
86x =" 1032
8. 1LOQON = 1:213 213 213. 4
Xo elZ
N = 0.001 213 213 ...
999N = 1.212 (ajey = 2(12)2—"15r=—9
4» = 1212 (12, 9)
999,000
x? = 29) 5s y? 14. (ay 4y=-x7= 2
ee hs y)x3 + y (BD) cy s="5
5 ie x*y
Bhs
Y
ax?y - xy’
see a Substitute (b) into (a) and get:
xy” sie y
0
10. wien Zt. 4y?
—5 <
eS eo Oe en Ay" + Oye 5 SIay
AS)
y=(- av, Solve this equation by using the quadratic formula.
y= See 17"
From this we see:
(a) Opens downward
“he 2(4)
(b) Axis of symmetry is x = 1 yet
(c) y-coordinate of vertex is 2
Substitute these values of y into (a) and solve for x.
@ x-fi ) 224 ai
4 4
@ x= 44-4) 2-1 — i
[1a ai,4 +
1 2.
2h)an
[-1- 211. 2h)
Algebra 2, Third Edition 309
Problem Set 107
15. (a) x? +y?=4 The next step is to graph each of these lines.
My
(b) x — 2y = 1
reeLy 1
Substitute (b) into (a) and get:
(a’) (y+ 1) Hy =84 x
4y>+4y+1+y%=
5y* + 4y -3 = 0
Solve this equation by using the quadratic formula.
ie -(4) + ¥(4)?_- 4(5)(-3) EL 19 The region we wish to find is on or above
the solid
2(5) 5) 5 line and below the dashed line. This region is shaded
in the figure.
Substitute these values of y into (b) and solve for x.
232 -i+2 = -3i+2 3-4i
18.
Cine et eOe
5ts 5 34 4i FM3 al
cits
Brie Dfves: 12) toi eS yee) SE EF
ate Sees 9 + 16 25 25
x= 4-2-8
ip _ 19. [x5 y? Pay? = 3535231434.
3 4
423/12, 17712
1 2h 19 ee 19
[3+ le eal Jana 20. vVs— 48 =8-Vs
19 - — 48 = 64 — 16Vs-+
Vs-+s
12 ©2192
Sums TF,asi
Bue w5IO) 55 16V=s112
Vs =7
s = 49
16> (at 5x-— By =32
Check: ¥49 — 48 = 8 - 49
(b) 2x — 2y = 16 as
=a
Zi; y
Substitute (b) into (a) and get:
(a’) 5x - 3(x - 8) = 32 Hh
IN
2x = 8 Be
He
(b) y = (4) -8
“ae
= -4
an
he]
(4, —4)
17. :(a) 3x - 8y > —x
ig ere H=
=
(8) a + (4% pas= 4/5
i tan@ = 2
IWRS a
4
6 = 63.43°
<
(b)x sy Since @ is a second-quadrant angle,
ares 3 6 = -—(63.43) + 180 = 116.57°
4/5 /116.57°
310
Algebra 2, Third Edition
Practice Set 108
22. Write the equation of the given a in slope- 27. 4x3 + 2x? — 30x = 0
intercept form.
5x - 3y = 4 Ox (2x78 x —415)p=10
2x(2x° 4+ 6x = Sxenals y= 0
2x(2x(x + 3) — S(x + 3)) = 0
Since the slopes of perpendicular lines are negative 2x(2x -— 5\(x + 3) = 0
reciprocals of each other, 2x=0 w-5=0 x +3 Il i=)
a
ee
x= 0 sommes Yaa 35
p
3
¥ Sar rd eB
3
3 = -=(-2)
56 +b
28. Rye 0
| Dee
cae A(2x* + 3x + 1) = 0
3
xs 2x tie + 1 = 0
2h BD) 4 tee Die 0
= 5
Oxantlr cect 0
1 1
23. hi ——
mr mp
Pp
pe m? px -1
800 liters | 1000mL 1 cm? 29. 1O000N = 74,213.213 213 213 ...
N= 714.213°213 213 5.3
min 1 liter eras B
999N = 74,139
1 in. ‘ 1 in. tetas
_ 74,139
254cm 2.54cm 2.54cm
999
#1 Te ahh fF Toit ogy 1 tna
12m 12 in. | Assia. 60 s
30. (hits gx) = (eo + 2)+ (x3 4.2)
800(1000) ihe
(2.54)(2.54)(2.54)(12)(12)(12)(60) ry
(h+ g(x) =x 4x44
(h + g)(-3) = (-3)° + (3) + 4
25. Area of rectangle = Area of right triangle (h + g)(-3) = -26
10(#7)
10(5) ,
MG fim. YY
PRACTICE SET 108
FG = (oy + (10)? = 10V2m
64p%a? — x3y!2
= (4p’a*)> — (xy*)?
26. Area of circle = Area of square
n(2V2)* = s? This expression is a difference of 2 cubes and can be
factored as
An? = s*
(4p7a? - x y4)(16p4a® + 4pa>xy4 +x 2yey
27cm = s
Algebra 2, Third Edition 311
Problem Set 108
Variation method:
PROBLEM SET 108
D
Sp = Pp + My Vie pees
B
6400 = Pp + 0.4(6400)
k eee = 500
$3840 = Pp
(1)
T = tens digit 2
ee 500(10)
U = units digit 10
107 + U = original number B = 5000 blues
10U + T = reversed number
Downstream: (B + W)Tp = Dp (a)
Qn +20 =16
T=6-U Upstream: (B —- W)Ty = Dy (b)
(b) 10U
+ T = 107 + U - 18 (a’) (Bue3)T = 92
Substitute (a) into (b) and get: (b’) (B - 3)T = 68
ib’) 100 + (6.="U)= 106—
U) + U— 18 (a) Bist 37 =F 92
9U + 6 = -9U + 42 (b')S87 337 = 268
18U>=36 2BT = 160
Ui=2 (c) BT = 80
(a) T=6-
(2) =4 (a’) (80) + 3T = 92
Original number = 42 Tf = 4:he
(c) B(4) = 80
T = tens digit
B = 20 mph
U = units digit
107 + U = original number Tp = Ty = 4hr
10U + T = reversed number
27a®p!? ae ie a (3a’p*)3 a (yy?
(a) T+ U-=1%5
= (3a*p* + y)(9a4p® — 3a*p4y + y?)
T=15-U
8x12 76 S my? = (2x42?) a (my*)
(b) 10U-+ Tf = 107 + U =223,
= (2x42? — my>)(4x824 + 2mx4y2z? + m?y®)
Substitute (a) into (b) and get:
(b’) 100e+*(iS—1U) = 1015 Sa — 97 my? _ 2 = (my?) — (2?)
9U + 15 = -9U + 123
= (my? - 27)(m4y® +. my3z2 4 z)
18U = 108
We=e6 v1 SOS
(a) T = 15 = (©) = 9 bs = 105
Original number = 96 5 = 125
nasil 25"
Equal ratio method:
ies é:
y _ (G)*B,
10. LON, =, 41223430355
Y, (Gye,
N= 4.123°35
100 = (1)"-B, ON? = 37.11
10. (10)°5 _ 3711
By = 5000 blues 900
312 Algebra 2, Third Edition
Problem Set 108
1 y = x? 2x key Solve this equation by using the quadratic formula.
y=(?-2x+1)-1-1 SoG
ERGY =4OED 9 4, ¥21
one ya? 2(5) mG
From this we see: Substitute these values of x into (b) and solve for y.
(a) Opens upward (oye {-2 4 = 4057 2 ig 2/21
(b) Axis of symmetry is x = 1 ; P
(c) y-coordinate of vertex is —2 (b) y= 2 ae Vai +2= eins
5 5 5 5
ste! + V2 2 + 221 d
5 Rh 5
<4 sN21" 23geznl
ese 5
15. () 4x4 2y = 8
(b) Sune 32, =.-9
(c) —3x + y= i
(b) aye = Be =
(C)\=3x += <I
12. -2z = -8
z=4
(c) -3x + (4) = 1
XS" 2.06 x <.—2 36 = ak
+9 ++ + (a) 4(1) + 2y = 8
ae WG 1 2S ee
1
(1,
2, 4)
135 (a) 5 —--y=2
= 16. (a)x-y-3z=-2
(b) —0.008x — 0.2y = —1.68 (b) 3x +y + z= 12
(a’) 8x — 5y = 40 (c) 2x -y+z=5
(b’) -8x — 200y = -1680 (a). xx = y = 3z = -2
-205y = —1640 (b) 3x + y+ z= 12
y.= $8 (d) 4x -— 2z = 10
= 40
(a’) 8x — 5(8) (b) 3x +y + z= 12
8x = 80 (ce2Qxieny #2) z= 5
(e) 5x +2z=17
x = 10
(d) 4x -— 2z = 10
(10, 8)
(e) 5x +27 = 17
iG ey = 3 Ox = 27
re 3
(b) y- 2x =2
y=2x+2 (d) 43) — 22 =410
z= 1
Substitute (b) into (a) and get:
(yan rt 2) = 8 (a) 3) - y - 301) = 2
x? + 4x7 + 8x +45 y=2
5x2 Tove = (3, 2, 1)
313
Algebra 2, Third Edition
Problem Set 108
17. (a) -x - 3y > -9 Ha i(4 404? = 24/53
1
SoeeK-+3 14
ue tan 0
4
(b) y < 2x @ v= —74.05°
The next step is to graph each of these lines. 2/53 /-74.05°
y
AO ft 1 mi 60 s 60 min
23. x x
S 5280 ft 1 min 1 hr
_ 40(60)(60) mi
~ (5280) ~—shr
x
24. (x,y) = C3,7)
(x2, y2) = (5, 3)
Dh (-3eSy eee J80
= 4/5
25. There is never an exact solution to these problems.
The region we wish to find is on or below the solid
One possible solution is given here.
line and below the dashed line. This region is shaded
in the figure. W = mE + b
18. see per Pel2 Use the graph to find the slope.
[i =P oP Se 3 i ee ds
-j-2-24+ 4i Dae 10
1+4 ~ &§ W = 15E
+ b
4+3V¥5 2-5 Use the point (40, 100) for E and W.
19.
2+75 2-45 100 = 15(40) + b
as
erSe Ene) —500 = b
4-5
= 7-2/5 W = 15E - 500
1933 = (32) 631/18 _ 31/33 1/18 _ 37/18 -
20. ab 2 ab(a> + 1)
ee ab* a +a + ab?
21. Site ae360
SO na 3
Ge sel
pe Are b(a> +1)
at +a+t+b
_ ISIS | 1OVTSR “308/15: «= a5 7
See Lesson 71.
22.
28. 2x* + 9x +9 =
Dn? Gl tea Len! =
2x(x + 3) + 3@ + 3) =
(2x + 3)%
+ 3) = 4S)
ey
©
24,
+ 3= 0 xe
Ss = 0)
ae ed
314 Algebra 2, Third Edition
Problem Set 109
29. 2b(3b7 + 10b + 3):=/0 ; 2. T = tens digit
2b(3b* + 9b + b + 3) = 0 U = units digit
2b(3b(b + 3) + 1(b + 3)) = 0 107 + U = original number
2b(3b + 1b + 3) = 0 10 U + T == reversed number
(WY a at 9 ia
2b = 0 36 + 1=0 b+3=0 T=11-U
b=0 oe. oa 3 (b) 100 + T= 107 + U - 27
3 Substitute (a) into (b) and get:
‘ Les (b’) 10U+\(11 "= V))= 10011’ = U) +. Ui - 27
a3? 9U + 11 = -9U + 83
18U = 72
30. (h + p)(x) = x7 + x? SA
(h + p)(-3) = (-3)? + 3) = -18 (a) T= We (Qyeee?
Original number = 74
PRACTICE SET 109 st Dy
a. mi/2 + qi2
/-—_—
63
+}
1,25 Zip D,
petinlgl2
m/2ql/2 +a 60
m + 2m'2q)? 4 a R,T, = 63; RcTc¢ = 60;
T, = Te +11; Ro-= 2R
B a¥2a,513 oe lai c ig
2R,(T; — 11) = 60
: Zoae SpPB
—— 2(63) — 22R, = 60
peta
zl2pl/3 4 52/3 -22R, ib =ie -66
sion tip) pec
ae 2712 ,%4 + a3
ha.
1h
R, = 3mph; Rc = 6mph; 7, = 21 hr;
Tc = 10hr
PROBLEM SET 109 iba
4, eS are
1. T = tens digit Tala
P.
U = units digit S007 Ee 2s
jue 400 1200
107 + U = original number P, = 2400 torr
10T + U=87T7+U
lk a ; 5. 0.46(600) — 1(Dy) = 0.4(600 — Dy)
(b) 6U=T+5 276 — Dy = 240 - 0.4Dy
T = 605 = 36
0.6Dy
; Dy = 60 mL
Substitute (b) into (a) and get:
(a’) 10(6U — 5) + U = 8(6U — 5) + 8U 6. Apply the power rule and get:
27x3/4 3/2,3
61U — 50 = 56U — 40 af
5U = 10 7. xl4 4 yl
y=2 x4 4 yl
xl 4 yl/4y
14
(b) T = 62) -5 =7 ly1/4 4 yl/2
ei/
Original number = 72 Fr teedaan ee
Sy bee
315
Algebra 2, Third Edition
Problem
Set 109
8. xl4 x‘ yy" (b’) 30x — Sy = 255
xl4 + >"
—l(a’) -6x + Sy = -15
xl? > xt
xl, A y 2
(b’) 30x — Sy = 255
1M, -14 + yi?
24x = 240
yl2 4 2
x = 10
9. xy — 27m = (ay -— GmpP (a’) 6(10) — Sy = 15
= (xy? — 3m)(x2y* + Smry* + 9m) y=9
(10, 9)
10. 649° + pes = (477 + (p*zy
= (4x)? + p*) Ue? - 4p*ysy7z + pS?) (a) x — 2y = 5
11. 100N = 102.342 42 42 ...
(b) xy = 3
N= 1.023 42 42 ... 3
x=-
99N = 101.319 y
y — 101,319 Substitute (b) into (a) and get:
~ 99,000
(a’) (2]— 2y 5
12 spax* aes 1
3 ~'2y
= Sy
~y = (x* - 44 +.4)4+1-4
2y7 + Sy -3 =0
-y=(«- 2) -3
2 °
Solve this equation by using the quadratic formula.
y = -—@ - 2 + 3
~(5) + ¥(5P_- 4(2)(-3)
From this we see: 2(2)
(a) Opens downward
= _5 ed = Ls =
(b) Axis of symmetry is x = 2 4 4 2
(c) y-coordinate of vertex is 3 Substitute these values of y into (a) and solve for x.
(a) x t)+s=6
(a) x = 2(-3) + 5 = -1
16.
(b) 4x +y+z=4
(c) 3x +y-—z=9
(a) x+ yu=z= 7
(b) 4x + ya=
(d) 5x + 2y =.1]
(b) 4x + y $zi=- 4
(c) 3x + ys3z= 9
(e) 7x + 2y = 13
2 1
14. (a) =x - —y=1
> 2
—l(d) -5x - 2y = -ll
(ec) Txt 2y S13
(b) 0.3x — 0.05y = 2.55
2 aD
(a’) 6x — 5y = 15 x=1
316 Algebra 2, Third Edition
Problem Set 109
(d) 5(1) + 2y ll —_
20.
SND 5/9 =D
= 35 5/2 2 M5 EB
(a) (1) + (3) -z
(a
Zz Woof
_ 15V2 +6 + 20+ 4/2 _ 26 + 192
oe)
50 - 4 m 46
(1, 3, -3)
21. Y27V3 = (33)'3313 = 3(33) = 343
17. (a) 2x + 3y = 15
(ob) x — 22 = -~3 22. [By fx2y = S2yl2~yll2 _ Ty
(Cy Sy = 2.=.0
xa/3 ifs) 2/2
z= 3y - 6 23. My — ,—5a/3 - 2/3,,3b/2
x24 y-b m2 v4
Substitute (c) into (b) and get:
(b’) x — 2(3y - 6) = -3
cee 43) = bboy me NONE MIIG ne
x — 6y = -15
V3 2 3 2
2(a) 4x + 6y = 30
pals =, Be _ RV6 _ 296
(b’) |x = 6y = -15 6 6 3
es = 15
x = ka* 4 kee
25.
(a) 2(3) + 3y = 15 ifs so3 ror
ka are ka eea
y=3 k*a—1
(c) z = 373) —6 = 3 _ kalk*a - 1)
a= kaa
(5:3, 3)
18: (a) 2x — 3Sy2"i5 26.
B 15
VS, hea Go
ele a : 20
A
(b)ys-
The next step is to graph each of these lines. A = 15\cos 47? = 10.23
B = 15sin 47° = 10.97
—10.23R + 10.97U
20.00R + 0.00U
9.77R + 10.97U
10.97
tan@ = ——
9.77
@ = 48.31°
F = (9.77 + (10.97% ~ 14.69
14.69 /48.31°
The region we wish to find is on or below the solid
lines. This region is shaded in the figure.
1000 liters _ 1000 mL . 1 min
27.
19.
2? -J-9+2 _ -3i 3+2i min 1 liter 60 s
3 - J-2 2 3-2) 342i od 1000(1000) mL
Z (60) s
Algebra 2, Third Edition 317
Problem Set 110
28. 6p* — 3p - 30 = 0 b. x? — 3x > 4; D = {Integers}
3(2p* - p - 10) = 0 x? —-3x -420
2p* + 4p - Sp - 10 = 0 eae Lye
2p(p + 2) - S(p + 2) = 0 (Pos)(Pos) = 0
(2p - 5\(p + 2) =0 a= 42>) Ocand «tote 0
2p —"5é=.0 p+ 2= 0 x = 4 and x 2 -l
5 (NEG)(NEG) > 0
os san WPA Ueander eo
5 x < 4 and xs -l
yon Thus, the solution is x = 4 or x < -1.
29. Ax? — 14x? — 8x = 0 =| (0). {| Peakietin
Ix QxUt tix =A) = 0
Ix(2x*'—8t + x - 4) = 0
PROBLEM SET 110
2x(2x(er='4) + 1@= 4)) = 0
1, 7 = tens digit
2x(2x + 1x — 4) = 0
U = units digit
erS OF. gx gt tes. 0 et, 107 + U = original number
x=0 x= pal x=4 10U + T = reversed number
2
(a) T+ U=9
0, 5.4 T=9-U
(b) 10U + T = 10T + U — 27
30. 3x2 - 7x -6 = 0 Substitute (a) into (b) and get:
3x? — Ox + 2x -6 = 0 (b’) 10U + © — UV) = 109 —.Y) +0 27
340 = 3) + 26 = 3)s= 0 9U +9 =-9U
+ 63
(3x + 2) - 3) =0 18U = 54
3x + y = 0 x - 3 = O U = 3
2 (a) T = 9: — (3) = 6
Nee x= 3
3 Original number = 63
5, 3 2. T = tens digit
U = units digit
107 + U = original number
PRACTICE SET 110 10U + T = reversed number
a. (x + 1)(x - 2) > 0; D = {Reals} (aj T+ U=7
(Pos)(Pos) > 0 SIM a's
4 ts 0 and ee oo (b) 10U.4+-7
= 10 PeteUae4ss
SI and pee e) Substitute (a) into (b) and get:
GO: and.
x —.2 <0 9U + 7.= -9U +115
x < -1 and 5ak 18U = 108
aa Ul= 6
‘Thus, thesolutionis x« > 2 ot 7 <-—1.
(a) T= 7-6)
=1
Sed Ohh (2s 3 Original number = 16
318
Algebra 2, Third Edition
Problem Set 110
3. Potassium (K): 1 x 39 = 39 (NEG)(NEG) = 0
Chromium (Cr): 2 x 52i= 104 x- 5s 0 and x+1.<s 0
Oxygen: 7% 16 j=. 172 ese x Ss —1
Total: 39° +104 + 112\— 255 Thus, the solution is x => 5 or x < -1.
112 336 =| tO ele
Ae SG
255 2 KCr, 0,
KCr,0, = 765 grams Apply the power rule and get 4x%y426,
x2 _ yli4
4. Downstream: (B + W)TD = Dp (a) xll2 v4
Upstream: (B — W)Ty = Dy (b) 7 yi
oe J ye
(a’) (B + W)8 = 120
x — 2xW2y14 yl
(b’) (B — W)9 = 63
10. x2 4 yl/2
9(a’) 72B + 72W = 1080
x2 _ yl
8(b’) 728 - 72W = 504
sae glee
144B = 1584 _ xli2yli2
B = 11 mph
x —-y=x-y
(a’) 8(11) + 8W = 120
W = 4mph 11. 8x” — y®p* = (2x)? = (y*p)?
= (2x3 - y*p)(4x® + 2x3y2p + y4p?)
5. Sp = Pp + My
27x!4,? + p°m'> = (3x4y3)3 + (p?m°)?
12.
Sp = 1400 + 700 = (3x4y3 & p’m*)(9x*y6 a 3x4y3p2m5 + p'm°)
Sp = $2100
13. 1O00N = 13.62. 362°362)...
x 100% = 33.33% N= 0.01 362.302
brs es ane 999N = 13.61
700 _ 1361
% of cost = = x 100% = 50%
99,900
6. (x + 2)a% — 4) > 0; D = {Reals} 14. Yee Kee Ae be
(Pos)(Pos) > 0 y = (x*- 4n 4+ 4043-4
x+2>0 andx-4>0 ye(Qe- 2-1
x > —2 and xa 4 From this we see:
(a) Opens upward
(NEG)(NEG) > O
(b) Axis of symmetry is x = 2
x+2<0 andx-4<0
(c) y-coordinate of vertex is —1
Kaan goreal
Thus, the solution is x > 4 or x < -2.
3-2-1012 345
7 5° . ay = 5 2 OS {Integers}
(x — 5a + 1) 2 0
(Pos)(Pos) 2 0
X eo 2 Uren x + 1 IV =)
x = 5 and x 2-1
Algebra 2, Third Edition 319
Problem Set 110
15. 3 > xt 2 orx-+ 5S] 82 Ds=—"(Reals} Substitute these values of x into (b) and solve for y.
i>s0 or 5 wy=(f+Ht)-1-
‘ai AG)ae canker
7
O Wutees SPa4
pi i ge iu
as lae ae ek 2 S92
16. (a) te2aan
Ea 1ey
SY
(3 + 5 + an
9 Z 2 2
1_ Mnll 1 _ vai
(b) 0.03x + 0.07y = 1.12
(a’) 10x - 7y = 70 jee 2
(b’) + 7y = 112
3x
13x oe. 19. (a)x-z=3
x =14
(b) x + 2y = 5
(a’) 10(14) - 7y = 70
(cc) y+z=0
Veo)
(a) + (c): x ty =3 @)
(14, 10)
—2(d) -2x -— 2y = -6
17. (a)xt+y+z=1 (ec) x+2y= 5
x aed
(b) 4x - 2y -z = 6
(c) 3x -y+z=-l
(b) (1) + 2y
(a) 4e(b): 5x =)y =A —.@)
y NO
(b) + (c): 7x -— 3y = 5 (©)
(a) “(=z iH es
—3(d) -15x + 3y = -21
z= -2
(C)y ax = 3) Set 1S
(1, 2, -2)
—8x = -16
x= 2
20. (a)
2x 2°6
(d) 52) --y=7 wa
wea3 (b) x+y<3
(a) 2) + G+ 2 = 1 y<—x+3
z=-4
The next step is to graph each of these lines.
3, -4)
(2,
18. (a) x7 + y? =6
(b) x =y = 1
Yow!
Substitute (b) into (a) and get:
(a’) x? + (7 -2)e =! 6
x? 4x7 2x +1 = 6
04" —. Vie Saat)
Solve this equation by using the quadratic formula.
The region we wish to find is on and to the right of
ye DEP = 4S) , Mil
the solid line and below the dashed line. This region
22) 2 2 is shaded in the figure.
320 Algebra 2, Third Edition
Practice Set 111
21. 2i2+ %. 2
set Py2i at oO 28. 3s* + lly + 10 = 0
Jeo + 835i + Gao i 37 3a 3s* + 65 + 5s + 10 = 0
6 + 61 1 1, 3s(s + 2) + 5(s + 2) = 0
= =-—
-—- —!
= t= 2 (3s + 5\(s + 2) =0
35 + 5) = 0 Sac e—e()
Bo rd1 a5
bn Cae eon yea s=-—2
3
. 34+3V5 +4V5 +20 -23- 7,5 5
1-5 - 4 ae
Ae
7 we = gq’ — 5b/2,.2
- b/2 29. 8x3 + 6x2 — 2x = 0
x2 ge!2
Dx(Ax?t+23x\— 1) = 0
24. 18/2 = (23)'4gU8 = 93/4y18 _ 27/8 Qx(Ax® +o40p eevied en)
De Ax 471) =a 1) = 0
25. eee aa le 2x(4x — 1)\(x + 1) = 0
aa. fo Dee = Ase — jl =O eh
il = ©
_ 2V6 _ 15V6 , 36V6 _ 23/6
ae () ae |
6 6 6 6 4
1
26. Jk Oo = vk — 46 Dae
k - 12Vk + 36 =k - 48
12Vk
= 84 30. Since —2 is not a member of the domain of g(x), it is
not a member of the domain of (h + g)(x).
Jk =7 Therefore, the answer is either © or { }.
k = 49
Check: ¥49 — 6 = V49 — 48 PRACTICE SET 111
N+D+Q= 35 (a)
5N + 10D + 250 = 500 (b)
27.
IN = O (c)
(a) N + D + (2N) = 35
(a’) 3N + D = 35
(b) 5N + 10D + 25(2N) = 500
(b’) SSN + 10D = 500
-10(a’) -30N — 10D = -350
(b’) 55N + 10D = 500
25N = 150
H = (15% + (4) = V241 N =6
15 (c) Q = 2N = 12
an
tan@d = —4
(a) (6) + D + (12) = 35
@ = -75.07° (Dy
= 7/
\ 241 /-75.07° There were 6 nickels, 17 dimes, and 12 quarters.
Algebra 2, Third Edition 321
Problem Set 111
Consecutive integers: N, N + 1, N + 2
PROBLEM SET 111 4.
MN + 2) = 5(N + 1) + 35
1. (a) Ny + Np + No = 28
N* + 2N = 5N + 40
(b) 5Ny + 10Np + 25Ng = 250
N2 - 3N - 40 = 0
(c) Ny = 5No
Solve this equation by factoring.
Substitute (c) into (a) and (b) and get: (N + 5)(N - 8) = 0
(a’) Np + 6Ng = 28 N = -5,8
(b’) 10Np + 50Ng = 250 The desired integers are —5, —4, —3 and 8, 9, 10.
-10(a’) -10Np - 60Ng = —280
Equal ratio method:
(b’) 10Np + 50Ng = 250
D)
-10Ng = -30 Lge=; oeBa 2)
No = 3 quarters S, T,(A,)
(c) Ny = 5(3) = 15 nickels
1000 _ 5(2)*
eile 6) c!
(a) (15) + Np + (3) = 28 S, = 400 students
Np = 10 dimes
Variation method:
(a) Ng + Ng + Ny = 10 S = kTA2
(b) Ng + 4NG + SNy = 39 1000 = k(5)(2)*
50 =k
(c) Ny = Ng + 2
Substitute (c) into (a) and (b) and get:
S = (50)(8)(1)*
S = 400 students
(a’) Ng + 2NG = 8
(b’) Ng + ING = 29 (x + 4)(x — 2) > 0; D = {Integers}
—l(a’) -Ng - 2Ng = -8 (Pos)(Pos) > 0
(b’) Ng + 9NG = 29 x+4>0 andx-2>0
ING = 21 x > —4 and xXe>A2
Ng = 3 greens (NEG)(NEG) > 0
(c) Ny = (3) + 2 = S yellows x+4<0 andx-2<0
x < -—4 and Kime
(a) Ng + (3) + (5) = 10
Ng = 2 blues Thus, the solution is x > 2 or x < —4.
T = tens digit =6 5-4-3)
20 On See oe
U = units digit
107 + U = original number x*-5x+6>0;D= {Integers }
(x — 3)\(x - 2) > 0
(a) 107 + U = 4(7+ U)
(Pos)(Pos) > 0
(b). U =F7e+4t
x-3>0andx-2>0
Substitute (b) into (a) and get:
x > 3 and n> 1)
(a’) 107 + (T + 1) = 47 + 4(7 + 1)
(NEG)(NEG) > 0
117 + 1 = 87 + 4
x. = 3°<)0"and
%¥ = 272.0
ot oaes
x-< 3.and bse aey
pat
Thus, the solutionis x > 3 or x < 2.
(b) U=(1)+1=2
Original number = 12 OLmilee2s
(Sie 4s
322 Algebra 2, Third Edition
Problem Set 111
(x2 x ey
13. -|x| — 3 = -7; D = {Reals}
x1/2 = phe
ex) 4
x2 = ape
|x| < 4
x + ry
xa la as yl? x <4 and x 2 -4
x + ya amy ahs Ps yl?
5-4-3-2-1012 3 4 5
(xi4 . np ia
xi2 _ yn 16. 25 x+5 <4; D = {Integers}
x2 _ 12
—1sx<-l
x— x12
ne ee
-8 -7 -6 -5 -4 -3 -2 +1
x — 2x12, 12, 1
10. Apply the power rule and get xywl, 17. (a) =x - =y = -10
11. x3 — my® = @) — (my?)3
(b) 0.003x + 0.2y = 1.97
= (x -— my?)(x? + mxy? + my‘)
(a )e3x = 2y Ss)
12. 8x6y3 ve 27m>p'? = (2x?) +9 (3mp*)
(b’) 3x + 200y = 1970
= (2x2y — 3mp*)(4x4y? + 6x?ymp* + 9m?p®)
=I (a) 35.4) Dy = - 50
a3. 1OON =-19215- 1I3MS5—-3 (b’) 3x + 200y = 1970
N te L02 15813 dt 202y = 2020
99N = 101.11 hoa 8)
n = 10,111
9900 (a’) 3x — 210) = -50
x = -10
14. —y ee a et
(-10, 10)
Je +4 44 14
weG@+ oy = 3 18. (a) x + 2y = 10
y= - + ay + 3
(b) x — 3z = -16
From this we see:
(c) y +°2z = 16
(a) Opens downward
(b) Axis of symmetry is x = —2 2(b) + 3(c): 2x + 3y = 16 (d)
(c) y-coordinate of vertex is 3 —2(a) + (d): -y = -4
y=4
(a) x + 2(4)= 10
x 2
(c) (4) + 2z = 16
eeroO
(2, 4, 6)
323
Algebra 2, Third Edition
Problem Set 111
19. (a) x27 +y =4 y _ _-5x/2,,1 + 3x/2
24. 3x .—2x ud J
(b) x-y=1
yex-1 [xy tye= x2 yli2, 1/2 Bs xy
25.
Substitute (b) into (a) and get:
(ee ee 1) 4 op wide ag
V2 37
po? Oa 4
26.
LT. Mee ‘iad ee"
Oye x 3.0 _ 2vi4 _ 2114 , 84V1 4
_ 65/14
Solve this equation by using the quadratic formula. = 14 14 14 14
© (2) 4 {-2)? = 403) _ 1, V7 27. (a)
-y < 3
22) De ai)
y>-3
Substitute these values of x into (b) and solve for y.
(D344 nS
1 i
(b) y= : oF v7 =ilbs == v7 y S$ -3x
+3
2 2 D; 2
UNeS
ima 2 18 AF
ime 2
The next step is to graph each of these lines.
1 ane ~ i + v7 and
2. 2 2 2
ei icp
Pita Wee 2 2
20. ox" = 3x. —ihe 0
ER Sips
2
iat
2,
(2 =| 40 9
x* —- =x +— — + —
2 16 16 16
The region we wish to find is on and below the solid
( sy 49 line and above the dashed line. This region is shaded
x--—]| =—
4 16 in the figure.
ade
2,Ol 28.
4 4
20
x= 3 Se of
4 4
5
x= >? = | A
A = 20.cos 45° = 14.14
40 in. 2.54 cm 1m 60 s 60 min
21. x - se x x 20 sin 45° = 14.14
Ss 1 in. 100 cm 1 min 1 hr
_ 40(2.54)(60)(60) m —6.00R + 0.00U
(100) hr 14.14R + 14.14U
8.14R + 14.14U
a3 we 98 SD
De re eg et =-—-—-—--l
14.14
4 — 3i i] Tt te tan @
8.14
Payeee Bebe ar NR VE a ak SMoprMORI) @ = 60.07°
23. ——— - A een
gia VS AE ee
se ee. o-73 F = (8.14)? + (14.14)? = 16.32
_ 6+ 4V3
3
16.32/60.07°
324 Algebra 2, Third Edition
Problem Set 112
29. Dxhed nits 10 PROBLEM SET 112
2x% ApS cololeI
BX 2) oe Sees 12g = ae
Se 1. (a) Ny + Np + Ng = 20
(2x -— 5)\(x + 2) =0
(b) SNy + 10Np + 25Ng = 325
2x-5=0 Seti24=0
(c) No = 2Np
5
LS = x= 2
2 Substitute (c) into (a) and (b) and get:
>,en (a’) Ny + 3Np = 20
(b’) 5Ny + 60Np = 325
30. 2x 5 7x?" 15x = 0
x(2x? — Ix — 15) = 0 -5(a’) -5Ny — ISN, = -100
x(2x7 — 10x +.3x.— 15)= 0 (b’) SNy + 60N, = 325
mM2xtx —5) 4+ 30 = 5)y= 0 45N,, = 225
x(2x + 3)\(x-— 5) =0 Np = 5 dimes
x=.0 2x 4+ 3=0 Veo =.0 (c) Ng = 2(5) = 10 quarters
3
- =f 5 x= 5 (a) Ny + (5) + (10) = 20
3 Nw = 5nickels
0, re
2. 1 = tens digit
U = units digit
PRACTICE SET 112
107 + U = original number
a. (x + 4)@ - 1) < 0; D = {Reals}
(NEG)(Pos) < 0 10U + T = reversed number
Ger
4 <a) rand x. — I>" 0 (a) a Ue 7,
x < —4 and » eae | i
hg
Pos)(NEG) < 0
atl (b) 10U + T= 10T + U-9
x+4>0 andx-1<0
x > —4 and all Substitute (a) into (b) and get:
There are no real numbers that satisfy the first (b’) 10U + (7 —-U) = 107 -— U)+ U =9
conjunction, so the solution must be -4 < x < 1. 9U +7 =-9U + 61
Oo
————<
a5 24S Oe 0 1 2 = 54
18U
ie)
b. x” — 3x — 10 < 0; D =s{Integérs}
x -— S)\a + 2) < 0 (@) Tse7 S03)
(NEG)(Pos) < 0 Original number = 43
> << GQeandee
20> 0
iy< ) -and ri> <2 as Sp = Pp + My
(Pos)(NEG) < 0 1800 = Pp + 0.2(1800)
%— 5. > 0 and reo GO 0 — 360
> 5: ane x < 2 cee ities
There are no integers that satisfy the second Pp = $1440
conjunction, so the solution must be 298 Pals Be ey
Eg er
0.1 2 3 4 596
Algebra 2, Third Edition 325
Problem Set 112
4. Downstream: (B + W)Tp = Dp (a) There are no integers that satisfy the second
Upstream: (B - W)Ty = Dy (b) j i
conjunction, i
so the solution must tbe be -1 < x < 6.
Tp = 2Ty {9.08 of 22) ore 6
(a’) (B + 3)2Ty = 230
(a’) 2BT,, + 6T, = 230 (Pos)(Pos) > 0
2(b’) 2BT,, - 6Ty = 170 x+2>0 andx-3>0
4BT, = 400 x > —2 and x > 3
(c) BT, = 100 (NEG)(NEG) > 0
(a’) 2(100) + 6Ty = 230 T+ 2<0 3 <0
aidx
6Ty = 30 x < —2 and bar ee,
Ty = Shr; Tp = 10br Thus, the solution is x > 3 or x < -2.
(c) BS) = 100
B = 20 mph =O 2m) (Oda se ous
Dr 9g, x2 4 yl?
5.
|_________+| 2 _ \-1/4
1800 eee
pee xl/2y1/2
-
| H | 2 xl/2y-1/4 ~yl/4
1200 Pe : y/2y 1/2 4 xV/2y-1/4 be yl4
RrTp = 1800; RyTy = 1200; ee
10. Pr =k airy — &
Ry = Rr + 200; Tr = 3Ty
= (p*x” — k)(p4x4 + kp*x? + k*)
(Ry — 200)3Ty = 1800
3(1200) — 6007 = 1800 11. 100N = 401.43 43 43 ...
—600T} = —1800 = 401 43:43...
Ty = 3hr 99N = 397.42
T; = 9hr; Ry = 400 mph; Rp = 200 mph Nie aan
6. ( + 3)\~ = 4) < 0; D = {Reals}
(NEG)(Pos) < 0 2 y= x? Oy 43
x+3<0 andx-4>0 Pee er o>, Wee at
x < —3 and Ae Ve tel ithe 2
(Pos)(NEG) < 0 From this we see:
ee)> Ol angie
64 n<an() (a) Opens upward
x > -3 and x<4 (b) Axis of symmetry is x = —1
There are no real numbers that satisfy the first
(c) y-coordinate of vertex is 2
conjunction, so the solution must be -3 < x < 4.
4-3 =2:=1 0-41 2 3-4 5
7. (x - 6)a + 1) < 0; D = {Integers}
(NEG)(Pos) < 0
x = 6 <O;and
x + I S*0
x < 6 and x SS!
(Pos)(NEG) < 0
xe 6e-0 and x + 1 <0
x > 6 and x<-l
326 Algebra 2, Third Edition
Problem Set 112
13. |x} +3 <0; D= {Reals} 20. (a) @i=r2y = 10
~|x| <,-3 (b) 3x27 = 11
|x|>3 (c) 2y — 3z = -9
ce LOE xX S 3 (a)o* (C): x ="3z = Xd)
—3(d) + (b): 8z = 8
4-3-2-1012 3 4
z= 1
14. x-2S0o0rx+4>
8; D = {Integers} (d) a (1) = 4
< OL ge! x= 4
(a) (4) — 2y = 10
y=-3
LS Sets (a) and (b) are functions. Set (c) is not a (4, -3, 1)
function, because —4 has two images.
21. (a) 2n=s =" 7
16. ihe {s-22)=2 (2-2) (b) xy = 4
3 8 8 3\ 8 8
eaex
GL
5220.038
peg 2A lee 12 Substitute (b) into (a) and get:
47. 5x7 +x+4=0
(at) 2x - (=) =,
ai
DSO? = 464)
Ix? — 4 = Ix
2(5)
On TES 4S
1 79 ,
x = —-— + —i
10 10 Solve this equation by using the quadratic formula.
-(-7) + y(-7)? — 4@)C-4)
18. Write the equation of the given line in slope-
2(2)
intercept form.
x+3y-42=0
io ee ee
AS A
es
3 3
Substitute these values of x into (a) and solve for y.
(ay y= 24)-7=1
Since the slopes of perpendicular lines are negative
reciprocals of each other, (A) ye 2-5] — 7] =-8
m, =3
y=3x+b (4, 1) and (- -8]
“y= 3(5) + b
—22 =D 22. (a)xt+yt+z=1
(b) 2x -y+tz=-5
y = dr, -,22
(c) 3x +y+z=5
19. (a) 2x + y = 28 (a)
+ (b): 3x + 2z = -4 (d)
y = 28 = 925
(b)
+ (c): 5x+
(b) 70x — 2y = 536 =1(d)
+ (ce): 2%.
=4
Substitute (a) into (b) and get: x=
(b’) 70x — 2(28 — 2x) = 536 (d) 3(2) + 2z Il | TS
14x = 592 re KS} <3)
5 aI. (a) (2) + y + (-5) = 1
I] a
(a) y = 28 — 2(8) = 12
y
(8, 12) (2, 4, -5)
327
Algebra 2, Third Edition
Problem Set 113
23.
377-29 — 34+2i -3+3i PRACTICE SET 113
f=3ai=3 =f=9 -3 i -=3.4,3i x = In 0.0052 = -5.26
= es9+91-6i-6 5,1
949 6 6 (ites 1
/2-1..2 x = @BBe* = Ine* = 1n(51.4) = 3.94
24. a* idl Shop
ye a
3 Ixy [x5 y? = /3y1/3,5/2, = 525/6)4/3
Inx = —4.16
25. x = @Dinx = el = & +6 = 0.0156
= 1,56 x 10°?
26. Nee +
7
— -—-
356 = WA, ME ig
V7 V2
2vi4 | 7V14 _ 8414 _ _ 7514 log x = -—4.16
14 14 tA ee pa at x = @BBlogx = 10'°8* = 10-416
= 6.92 x 10%
27.
3V2 -2 72 +3
TAD S36 Owe 3 (0.000612)(576) _ In (0.000612) In (576)
_ 424+ 9/2 - 14/2 -6 36 - 52 0.0512 x 1074 ~— In. (0.0512 x 1074)
Deo 89 e740 6.36
= e7 1 -40,6.36,+35.21 e417
28. 3x° + 5x2 + 2x = 0 = e521
x(3x?+ 5x + 2) =0 uv 6.92 x 1014
x(3x7 + 3x + 2x + 2) =0
x(3x(x + 1) + 2% + 1))
=0
x(3x
+ 2)(x + 1) =0 PROBLEM SET 113
aa) ox? = 0 x +1 ll S
(a) Ny + Np + Ng = 19
Ke, ee x =-l
(b) S5Ny + 10Np + 25Ng = 200
ates 3
(c) Ny = 2Np
Substitute (c) into (a) and (b) and get:
29. xe 23x ES (a’) 3Np + No = 19
DeteAieHe= 2-= Ng = 19 - 3Np
2x(x — 2) + lw - 2) =
(2x + 1)\x
- 2) = (=)
ey
Se
a
(b’) 20Np + 25No = 200
see io=a0 x-2=0 Substitute (a’) into (b’) and get:
i= = x=2
(b”) 20Np + 25(19 — 3Np) = 200
-55Np = -275
Np = 5 dimes
(a’) No = 19 — 3(5) = 4 quarters
30. 3x7 + 8x + 4 =
3x? + 6x + 2x +4 =
(a) Nut (Sd) = 19
SAT)
OK 2 = Ny = 10 nickels
(3x+ 2)(x+ 2) = =>
SS]
5k =U ee ll =)
Ry = By (ey?
R, Z B,(M,)?
2
x= -= x —2
10 _ 4(20)?
peerrey
By = 120 blues
328 Algebra 2, Third Edition
Problem Set 113
3. (a) 0.7Py + 0.6Dy = 126 (Pos)(NEG) < 0
(b) Py te Dyn = 200 Late? sn) wanda
x 0 sae ()
x > —2 and es 3
Substitute Dy = 200 — Py into (a) and get:
(a’) 0.7Py + 0.6(200 - Py) = 126 There are no real numbers that satisfy the first
conjunction, so the solution must be -2 < x < 3.
0.1Py = 6
Py = 60 Sees il Hop ail ek eh 7!
(b) (60) + Dy = 200
10. (x = 3)@ + 2) 2 0; D= {Integers}
Dy = 140
(Pos)(Pos) = 0
60 mL 70%, 140 mL 60%
x-320andx+220
Se 2 3) Buin 3 2)
(NEG)(NEG) = 0
Dp x-3<0andx+2<0
RcTc = RpTp; Tc = 8; Tp = ee eS 3S) Brace Se Soe2
Ry = Rr — 20 Thus the solution is x > 3 or x < -2.
R(8) = (Rc — 20)12
3 =2 <j) © 7-28 a
8Rc = 12Rc — 240
-~4Ro = -240 11. (x — 4) + 2) < 0; D = {Integers}
Rc = 60 mph (NEG)(Pos) < 0
Rp = 40 mph; Dc = Dp = 480 miles x-4<0 andx+2
x < 4 and x
gutit wud tetan Vt (Pos)(NEG) < O
TY 7, x-420andx+2<0
1000 _ 2000 x 2 4 and x <2
ro) T;
There are no integers that satisfy the second
T>, = 3400K conjunction, so the solution must be —2 < x < 4.
6. (a) -4.68 = 24a 0) Agee! Sei A 7S
(b) 4.14
12. |x| — 1 > 0; D =' {Integers}
7. (a) e263 >
2 or |
(b) 136.77
32-1012 3
8. AE x SF = AB
6 x SF = 24 13; | Seow
/\ ale De Reals}
SF =4 Dee?
AG = ADSCSF Se ee,
ee
SOO e213
AC = 4(4) = 16 meters
x3 2
ry erie
14.
9. @o+2e
— 3) < 0; D {Reals} 2/3 “ ue
(NEG)(Pos) < 0 x + gy?o)
Gei Oeeand: 3 EX aaaey
x + eye 4 7/3 2/3 +y
x < —2 and Vv
Algebra 2, Third Edition 329
Problem Set 113
15.
ah
8p°k
ee
xm
ee RO Up Y
OT eo em) ; ae 20. (a) wx x es) = 30
= (2p7k> — xm)(4p4k!? + 2p7kPxm? + x?m*) 5 3
(b) —0.18x -— 0.02y = -3.78
16. 100N = 0.316 16°16 ...
N = 0.003 16 16 ... (a’) 18x + 10y = 450
99N = 0.313 (b’) -18x - 2y = -378
nv = 313 Bye 2
99,000 vieeg
"7 ie eer i (a’) 18x + 10(9) = 450
ree = 360
18x
yey Ax + 4) +, le ey
yee = Dees
(20, 9)
y = -(x — 2)*2 + 3
2 oy) a
From this we see: 21. (a)x° + y =4
(a) Opens downward (b) 3x —- y = 2
(b) Axis of symmetry is x = 2 eae. o
(c) y-coordinate of vertex is 3 Substitute (b) into (a) and get:
(a’) a? Ox en) ed
x? + 9x7 - 12x #4254
10x? - 12x = 0
2x(5x — 6) = 0
see (0) .
5
Substitute these values of x into (b) and solve for y.
(b) y = 3(0) - 2
y=-2
y= 3(2)-2
8.
oh ae
N=2+2(42 -2) 8
,
hee ree
Bera vriP
SENe
Vereere:
: 6 8
11 (0, —2) and | —, —
Brot 9 es
33° 33 33 22. (a) x -— 4y = -15
19. 59 a = P= G (6) 334-20
eel 7 (c) 2y-z=5
bE eb PE ie
3 (b) + (c): 3x + 2y = 25 (d)
(P-sr+t)-H+2 2(d) + (a): 7x = 35
Zz. 5 436 36-36 i
Se i) Oia ee ae485
(« =| = (d) 3(5) +
d) 3(5) 2y =
+ 2y
1 _ 085 sd
fe Sa (b) 3(5) + z = 20
~ hemi ss za
SG +6 (5;,5,:5)
330 Algebra 2, Third Edition
Problem Set 114
23. (a) x - 2y-z=-8 29. Sitee
tome 2
(b) 3x14 y — 22 = 5 | ax? a) Sheds22 10
3x(x - 1) + 24 - 1) = 0
(c)x+y+z=9
(3x + 2) - 1) =0
(a) + (c): 2x-y= 1 @d)
(b) + 2(c): 5x + y = 13 (e) 3x°+ 2:= 0 x= 1'="0
7% Ses 2
» ied x= 1
Ses 3
+ y = 13
(e) 52) B2y
9
y=3
(c) (2) + 3) +z= 30. 3x3 + Ix? + 2x = 0
Fe a Oxo
+! Tx 2) 0
(2, 3, 4) x(3x” + 6x +x +2) =0
x(3x(x
+ 2) + 1(x + 2))
=0
24. thi" — jes a gt x(3x + 1)\(x + 2) =0
Tena ee a
x =0 aha eee) x+2= 0
oe eo. 22A4 nae eee
Ao SB ad ed 3
44442 — 10/2 = 20
ps LE SA Saat Siena
1
0, -—,-2
8 — 16
_ -16-6¥2
_ 8+ 32
-8 4
PRACTICE SET 114
26. ofxy3 3!xy? = for y2/3 = rey Region A (This region includes all points on or
above the line that lie outside the circle.)
jaf Seen eae23, 3 - 123
PROBLEM SET 114
_ BP aIe) 3693) aongn/3
c 3 3 3 3 1. Consecutive even integers: N, N + 2, N + 4
NW + 4) = 8(N + 2) + 16
2s. Ie + x 3x = 0
s N2 + 4N = 8N + 32
KQx0,P x - 3) = 0
N? — 4N - 32 = 0
Min = Ont oe = 3) = 0
(N + 4)(N — 8) = 0
x(2x(x
Ixlx - 1) + 3(v.=—
— 1) el)1)) =) take
x(2x + 3\(x —- 1) = 0
The desired integers are —4, —2, 0 and 8, 10, 12.
=. 2/46 =O x-1=0
3 2. 0.2(240) + 1(Dy) = 0.52(240 + Dy)
xX — He =
2 48 + Dy = 124.8 + 0.52Dy
0.48Dy = 76.8
yee Dy = 160 mL
2
Algebra 2, Third Edition 331
Problem Set 114
Region A (This region includes all points on or
above the parabola and all points below the line.)
Pe
Region B (This region includes all points on or
inside the circle and all points on or above the line.)
2000
Rrlr = 4800; RsTs = 2000; (a) —4.77
Tr = Ts + 1; Rp = 2Rg
(b) 4.47
2R(Ts + 1) = 4800
10. (a) 9185
2(2000) + 2Rs = 4800
2Rs = 800 (b) 1541.70
Rs = 400 mph
11. Agsgquare = s? = 9cm?
Rr = 800 mph; 7; = Shr; 7; = 6hr
cj Shon
Downstream: (B + W)Tp = Dp (a)
AB = 3cm + 1/72 — 32 cm = (3 + 2V10)cm
Upstream: (B —- W)Ty = Dy (b)
12. 715° 10° = 10 *
Ty = 3Tp
log 7.15 x 10-8 = -7.15
(a’) BTp + 4Tp = 34
(b’) 3BTp — 12Tp = 54 13. Sp = Pp + My
3(a’) + (b’): 6BTp = 156 140 = Pp + 0.2(140)
BTp = 26 (c) $112 = Pp
(a’) (26) + 4Tp = 34 14. (x + 4)@ - 1) 2 0; D = {Integers}
Tp S22
(Pos)(Pos) = 0
(c) B(2) = 26 x+420 andx-120
B = 13 mph
x 2 -4 and Seg |
T = tens digit (NEG)(NEG) = 0
U = units digit x+4<0 andx-1<0
10T + U = original number x < -4 and S11
10U + T = reversed number Thus, the solutionis x > 1 or x < —4,
(aya Uae 8
= -4 -3-2-1 0 1 2
F=8-U
(b) 10U + T = 107 + U + 54 15; (x — 4)~ + 1) < 0; D = {Integers}
Substitute (a) into (b) and get: (NEG)(Pos) < 0
(b’) 10U +°(8:= VU) = 106 — Dt Ui 54 x-— 45 0 and x + 1.20
9U + 8 = -9U + 134 x < 4 and x 2-1
1SUa— 126
(Pos)(NEG) < 0
Ug,
x-420andx+1<0
(a) T=8-(M)
=1 x 2 4 and x <-l
Original number = 17 There are no integers that satisfy the second
conjunction, so the solution must be -1 < x < 4.
Region A (This region includes all points on or
above both the line and the parabola.)
=27=1 0. 108 Gas
332 Algebra 2, Third Edition
Problem Set 114
16. -|x| - 3 > -7; D = {Integers}
1 1
|x| > -—4
2. (a) 2x + zy = 10
|x| < 4 (b) 0.03x — 0.03y = —0.36
wispAvand x > =4 (a’) 35x + 3y = 150
+ (Bmax 2 ayBeis
= BEAN.
656
tee
21H 1 2 3.4 root 38x 4
17. x=3
6 <x -4 < 8; D = {Integers}
10<x< 12 (b’) 3(3) - 3y = -36
Weald
8 9 10 11 12 (3, 15)
9
18. 100N = 0.1056 56 56 ... 22. (4) 3% =9 z= 8
N= U00010 56 56° =.
99N = 0.1046 (byes peed
w = getbto» (c) 2y + 32 =2
990,000
(b) + (c): 2x + 3z ==—2) (@)
19. y= x = 4 47 3(a)
+ (d): llx
= 22
aia — A+ 4) 47-4 eae
y=(- 27° 4+3 (d) 2(2) + 3z = -2
From this we see: Zo ee
(a) Opens upward (c) 2y + 3(-2) = 2
(b) Axis of symmetry is x = 2 yiaa
(c) y-coordinate of vertex is 3 NN,
y 23s. (avery. =. 6
bia
225005 ee oe Oa
a } 1
ae
n (a’) xX +(-=] =
v
ocaemie ae OX
x= 6xeeel =
1 Py ot 1 Solve this equation by using the quadratic formula.
. N= 45+ (65-45) ee pele: 6) + (6)? a- 4-1
(=0) (1)¢ ) 3410
9 (2 18 2(1)
=—+— /— -—
= BNA 5 Substitute these values of x into (a) and solve for y.
bette (a) y= 6-(3+ V10) = 3 - V10
Ze 3
(a) y= 6-(3-
10) = 3+ 10
_ 144 , 21 _ 165
= 39 732 32. (3 + /10,3 — /10) and (3 - /10,3 + 10)
Algebra 2, Third Edition 333
Practice Set 115
24. (CONers
SE Re) 29. 5x7) toTx?
+ 2x 0
(b) 2x —-y + 2z7 = 9 x(5x2
+ 7x + 2) =0
(cjiar
ty + 2 Set x(5x2 + 5x + 2x + 2) = 0
—l(a) + (b): x +z=6 (Td)
x[5x(x+ 1) + 2@ + 1)] = 0
(b+
(c)2 x4 3z7,= 10° (©)
x(5x
+ 2)(x + 1)
=0
-(d)
-x - z=-6
(e). x 3z.='10 a) Swe do Os == (0) Xe =0)
eg
5 ia _2 x=-l
z=2 5
(d) x -+9@)
=-6
g)
x= 4 0, - % -1
@)4)= ys)
yrsus 30. x2 _ 12
zl _ 12
(4, 3, 2)
ye lie
25. There is never an exact solution to these problems. — xl 4 yl
One possible solution is given here.
x 2 ayll2y tes,
O = ml + b
Use the graph to find the slope.
225
m= a es)
10 PRACTICE SET 115
O= 22.5! +b
27 = 10**>
Use the point (85, 400) for J and O.
= 19*+5
400 = 22.5(85) + b
-1513 = b x +5
O = 22.51 — 1513 a wm~ TT x
Bie len keise
26. ApeuPe™
ee) ee oP ee fee
= (980) ¢(0.07)
1+2i+i-21
= 980 0-63
= 2i4 5
Bb KI
> |R 980(1.88)
AS
Ro
27. Hb SANZ fap
oN Weal Ag = $1842.40
A, = Age”
_ S¥10 , 16V10 _ 120/10 (1600) = (400) eX)
eae (3) 20 20 4 ll
®
3k
99/10 a ee
139 Il WwW>
28. RW e a
0.46 =k
25 —- 10Vx + x
= E1ONia Ato = 400 €9-46(10)
Vx Ajo = 400 e*6
16= <x
Aig ~ 400(99.48)
Check: J16 - 15 = 5 —- 16
1= 1
Ajo = 39,792 rabbits
334 Algebra 2, Third Edition
Problem Set 115
PROBLEM SET 115 -5(a’) + (b’): 35No 35
No 1 quarter
1. Potassium: Ly X59 = "39
Manganese (Mn): 1 SS0e 35 (a’) Ny + 4(1) = 14
Nw = 10 nickels
Oxygen: 4x 16 = 64
Total: 39 + 55 + 64 = 158 (a) (10) + Np + (1) 14
Np 3 dimes
Oxygen _ 64
790 158 i T = tens digit
Oxygen = 320 grams Us—sunits digit
107 + U = original number
(a) 107 + U = 4(T+ U)
(b) 4U= T + 14
Rplz + RsTs = 280; Rp = 20; i= 4U
- 14
Rs
= 45; Tg + Ts = 9 Substitute (b) into (a) and get:
20Tp + 45(9 — Tg) = 280 (a’) 10(4U — 14) + U = 4(4U - 14) + 4U
—25Tp = -125 41U — 140 20U — 56
Tz = 5
21U 84
U=4
Ts = 9 - (5) = 4
(b) T= 4(4) - 14 =2
Dp = Rplz = 20(5) = 100 miles
Original number = 24
Ds = RsTs = 45(4) = 180 miles
6. -4.88
a) D
—— 7) 4.53
1200
10:93 = 10% +3
a P
480
heh ll Pt ae Ww
-1.47 = x
RyTy co 1200; Rplz = 480;
9,
102412 ~ 2582.26
Ty = Tg + 1; Ry = 2Rp
2R,(Tp =e 1) = 200) rr (aaa eon) ek o7 7:89 + 6.18 + 39.63
(e777 9?)
2(480) + 2Rz = 1200 ;
2Rp = 240 = e979? = 2.94 x 101%
Rz = 120 mph 1.
Agee Pe”
Ry = 240 mph; Tz = 4hr; Ty = 5 hr Ag =) 1002?"
> Ne) iH] $224.79
4. (a) Ny + Np + No = 14
12. A, = Ane”
(b) 5Ny + 10Np + 25Ng = 105
60,000 = 16,000e*)
(c) Np = 3NQ 3.75 =
Substitute (c) into (a) and (b) and get: 1.32 =
(a’) Ny + 4Ng = 14 0.44
(b’) 5Ny + 55Ng = 105 Aio = 16,000e°*40 ~ 1,303,214 inhabitants
335
Algebra 2, Third Edition
Problem Set 115
2D pF 20. ml5 — gp3y° = (am>)> — (2py’)?
£3; jaf = (=| ae (=| = SA units = (xm> - 2py?)(x?m!° + 2xmpy” + 4p’y*)
yy 2 2
2 S y
2 ED 100N = 104.747 47 47 ...
Anerc = H -( 5 )= *2 units? 21.
N= 1.047 47 47 ...
99N = 103.7
14. Region C (This region includes all points on or
inside the circle and on or below the parabola.)
_ 1037
990
iS: (x + 5)\(x — 2) > 0; D = {Reals}
22. y= i oe od
(Pos)(Pos) > 0
y= (x7 +2r4+1)+1-1
x+5>0 andx-2>0
ve omand x SD
yea 441)
From this we see:
(NEG)(NEG) > 0
aon<a ONeanGmey.
2, <0) (a) Opens upward
ean Kee (b) Axis of symmetry is x = —1
Thus, the solution is x > 2 or x < —5. (c) y-coordinate of vertex is 0
—=<—_o + OT
=6 594 olen te OMe 82S
16. G+ oKe +. 2) <= 02 D= {Reals}
(NEG)(Pos) < 0
se o> See(pinta! Se es A Sal)
x <= —) and ~S —)
(Pos)(NEG) < 0
cate)
>) 0) cande«
bt 2. < 0
x > —5 and x < —2
There are no real numbers that satisfy the first
conjunction, so the solution must be —-5 < x < -2.
23.
LON 3
36 S554 231-0 1
zie, ue naa
17; 4; D = {Integers} 3 10\ 6 6
=>
68
—4+—
1 =
23
—
went 4
=o 2) OR 123 24
18. 6<x-— 5 < 10; D = {Integers}
(b) -—0.2x + 0.05y = -1.8
11 <q lS
teh
(a!) 20x-— 27y= -84
11 12 13 14 15 (b’) -20x + Sy = -180
Gu a aly ~22y = ~264
19.
y = I2
x4 ye
x14 ve (b’) -20x + 5(12) = -180
1/2 xll4 —1/4 20x = -240
= a + ye pe
xl caf 2x 1/4,-14 & ka (12, 12)
336 Algebra 2, Third Edition
Problem Set 116
25. (a) 4x —z = 12 30. Bx" iat 2x CU8RE“0
(b) x — 3y = -10 : 3x* + 6x — 4x —-8 = 0
(c) 3y+z=8 3x(x + 2) — 4x + 2) = 0
(a) + (c): 4x + 3y = 20 (d) (3x — 4) + 2) = 0
(b) + (d): 5x = 10 3x —-4'= 0 be lin DD (G)
2S) ae 4 eh
(d) 4(2) + 3y = 20 3
My =4 4
—, 2
(c) 3(4) +z = 8
= -4
(2, 4, -4)
PRACTICE SET 116
a. |7|6|5]=7-6- 210
=5
26. (a)xt+y-—-z=4
(b) 2x +y + z= 10 b. There are 36 possible outcomes in the sample space
(c) 3x + y+z= 14 when rolling two dice. Six of these outcomes add to
Jeet hus:
(a) + (b): 3x + 2y = 14 (d)
le OL Egor
(a) + (c): 4x + 2y = 18 (e) 36 6
-l(d) + (e&): x = 4 There are also only 6 outcomes with a total greater
(d) 3(4) + 2y = 14 a total
than nine, so the probability of rollin g greater
than nine is equal to
y Fh
6 1
(b) 2(4) + (1) + z = 10 Oar
z= 1
Thus, the probability for the given sequence of
(4, 1, 1) events is
27. -—3 — 2i — (-2) + (1) — 3i + 2i - 6 DO ae ee
; 6 6 36
= -8 - 3i
4/2 +1 1432
23.0 ES
——— = +: pr PROBLEM SET 116
Fp LY ORO Pn Reeto aaa 1 fey i,) 1. Multiples of6: 6N, 6(N + 1), 6(N + 2)
, 1- 18 17 6(6N + 6N + 12) = 10(6N + 6) — 84
29. 7 72N + 72 = 60N — 24
P 12N = -96
Ln ee wb
(# bs uy. a x = 72 if i The desired integers are —48, —42, and —36.
3 36 36 = 36
( a 73 Ven,
er et)6 oto 2 —=—
36 T, T,
shone
B
hae6 400
800
_ 200
‘T,
os :4 2 T, = 400K
Algebra 2, Third Edition 337
Problem Set 116
3. Downstream: (B + W)Tp = Dp (a) (CRC SS = 9-740
+6.36 +35.21
7. (e3521)
Upstream: (B — W)Ty = Dy (b) =. MAT a egy C1018
(a’) (B + W)5 = 65
(b’) (B - W)8 = 56 8. (a) pH = log H’
pH = -log 0.00204
8(a’) + 5(b’): 80B = 800 oH ~ 2.69
B = 10 mph :
(b) pH = -logH
(a’) (50) + SW = 65 ieee at
We 3 mph 1073-2 = Ht
4. Equal ratio method: Ht = 6.31 x 1074 more
2 liter
TZ _ RA)
JI...
T, (A) Oe ANatre”
1000 _ 2(1)* Ay = 1400e°!1 = $3023.67
Tilsen = sel2) 2 4
T> = 2000 tomatoes 10. A, = Age
=. 1,200,000 = 400,000e*°?
Variation method: 3 Oe
bees 1.10 ~ 3k
1000 =- k(2)(1) 2 a =k
037
=nk
co Ag = 400,000e°77) = 7,719,188 paramecia
T = 500(1)(2)*
T = 2000 tomatoes il. [9/8{7[6|5| = 15,120
5. T = tens digit 12. [9]9]9]
10] 10] 10 |10| = 7,290,000
U = units digit
13. ‘mZBAC =. 180-935 = 35'5=- H10-
10T + U = original number
mZ3 = 180 — 110 = 70°
(ay 10r + OU S27 + Uy
+7
mZ4 =
Il 180 — 35 = 145°
(b) U = 3T + 3
mZ3 + mZ4 = 70 + 145 = 215°
Substitute (b) into (a) and get:
(a’) 107 + (37 + 3) = 27 + 27 + 3) +7 14. Region A (This region includes all points on or
137 3.87 4.13 above the line and on or above the parabola.)
Sf = 10 15. x2-462x 2-3 > 07D"= {Integers}
tog (x + 3% - 120
(b) U= 32) +3=9 (Pos)(Pos) > 0
Original number = 29 xe 3270. ane 5 — Ie. 6)
x = -3 and x>} 1
eed? (NEG)(NEG) > 0
(b) 5.86 x 10° x +35 Oand xe Ie)
x < -3 and eel
(c) 10!-92 zy 10*+4
Thus, the solutionis x > 1 or x < -3.
192=x%+ 4
—2.08 Il Ps 423
2-1 0 1 2
338 Algebra 2, Third Edition
Problem Set 116
160 ro eS eo: De {Integers} 20e Ge taee4)4
(x + 3)@ - 1) < 0 x? = 4
(NEG)(Pos) < 0 cls
Yn ao
**32< 0 and x ali>,0
23/4 3/2
= Beery
x < -—3 and x al 1/2,,3/4
(Pos)(NEG) < 0
SaaS 0° hd & ae < 6 21. 27m’p? — x!*y> = (3m'p)? - (xy)?
x> Bem ee = (3m3p — x4y)(9mp? + 3m3px4y + xy?)
There are no integers that satisfy) the first 1 at 1
conjunction, so the solution must be -3 < x < 1. ba NS | Oo
4 2 4
—<—+— 909}
8 1 O 1 -7+{4-3]
4 94 4
17. -|x| + 2 > -1; D = {Integers} eral - 26
= |x| Sas 4 36
|x| < 3 i pre =
36 = 36 36
cece Seana cago
3 1
SF -F D0 4 2 3 23. er eyes
18. 100N = 104.76 76 76 ... BO Taeae) 204 eel
Ne =a 1:04 70" 16. (a) 21a Sy = 210
NAP 2 (b’) 21x + 2y = -273
ee e.a72 Tay oS
9900
y = 21
19. -y = x7 + 4x + 6 (a’) 21x. 5(21) = 210
y= +Ar +4) + 6 =A m1 ¢°£1315
yet Oe + 2 Beplo
y= -(x + 2)? -2 (15, 21)
From this we see:
24. (a)x -2z=7
(a) Opens downward
(b) Axis of symmetry is x = —2 (b) y + 22 = ~-9
(c) -x + 2y = -7
(c) y-coordinate of vertex is —2
(a) + (b): x + y = —2 (d)
(d) + (c): 3y = -9
ei
(d) x + (-3) = -2
vet Leh
(b) (-3) + 2z = -9
z=-3
di, -3, -3)
339
Algebra 2, Third Edition
Problem Set 117
25. (a) 2x —y + 2z = -9 PRACTICE SET 117
(b) 2x + 2y + z= -15 {xe Z|x+3>5}
(c)x —-2y+z=0 5dat Sts 3)
(b) + (c): 3x + 22 =-15 d) x > 2:>x 1s al integer.
2(a) + (b): 6x + 5z = 33 €) ee a
—2(d) + (e): z = -3
ee Nee PROBLEM SET 117
x= =3
1. (a) Ny + Np + No = 16
Omer) = 0 (b) 5Ny + 10Np + 25Ng = 150
= -3
7 (c) Np = 3Ng
-3, -3, -3
aaa Substitute (c) into (a) and (b) and get:
20
tS Sie 3. eas (a’) Ny + 4Ng = 16
1-4/4 1-4i 1+ 4i (b’) SNy + 55Ng = 150
Beet eS ree oe 14 Ny + 11Ng
= 30
le 17> 17 -1(a’) + (b’): 7Ng = 14
a Bie Guut V6 No = 2 quarters
2 ae es eae (a’) Ny + 4(2) = 16
636 =o lee eNO Ny = 8 nickels
e=6 2 (a) (8) + Np + (2) = 16
F Np = 6 dimes
Ronee es [pitaxi
Fore 2
IP) 2 Se = Pp = My
5000 = 4000 + My
29. jp—-45 =9-
, VP $1000 = My
p — 45 = 81
— 184 petrip
-126 = -18,/p % of Pp = soon x 100% = 25%
Weed
49 = p % of Sp =~ x 100% = 20%
Check: 49 - 45 = 9 — /49
2=2 gE:
a
4 5 200
30. 3x" = X= 24 = 0
x(3x° Bx S2)e='0 Ou
Max = 3x 4+ 20 2)5=40 650
x(3x(x — 1) + 2@ - 1)) = 0 RgTg = 200; RyTy = 650;
Hox +o 2)(x% = 10 Ty = 2Tg; Ry = Re +225
Gems Bre
2 =) 0 eRe!) 2Tg(Rg
+ 25) = 650
) 2(200) + SOTg = 650
am) ae x=1
3 S0Te 3-250
2 Tc = 5hr
0, -—,
3 1 Ty = 10hr; Rg = 40 mph; Ry = 65 mph
340 Algebra 2, Third Edition
Problem Set 117
(a) 0.3Py + 0.6Dy = 234 13. Region E (This region includes all points on or
(b) Py + Dyn = 600 outside the circle and on or below the parabola.)
Substitute Dy = 600 — Py into (a) and get:
14. V = Apase X height
(a’) 0.3Py + 0.6(600 — Py) = 234
-0.3Py = -126 270 Eee
E (x(2 m)?) + 52 m)(2 m)|x 10m
Py = 420 (3m + 2)(10) m3
(b) (420) + Dy = 600 _ __114.25(100)(100)(100) 4,3
Dyn = 180
(2.54)(2.54)(2.54)(12)(12)(12)
420 mL 30%, 180 mL 60% = 4034.70 ft?
My = 0.8(4320) = $3456 15. 6x SF=9
SE ode5)
Sp = Pp + My
4320 = Pp + 3456
$864 = Pp
BMSE =
|S]4]= 6720
s[7[S peg
2
4 26 13 Ree
36 «636 162 3
x = 3(-5.50) = -16.50 aerfeor =+ (Hr
e
3 3
4e* = 24
e = 6 $2 F sarc
=n 0
=—21:79
82 3 sierr
3 2 2
10. (a) pH = —log H* 2AV2 9.802.
10°PH = Ht 6 3
1079-34 = ia 4V2 >
3
Ht ~ 4.57 x 10° —
iter 16.
x +1 > 12D {Integers}
(b) pH = -log H* xipn0
1p? = Ht
1970-00263 _ yt (yg) Pes Zt
Ht ~ 0.994 mole 17. x? + 4x + 3 > 0; D = {Reals}
liter
(ae a) + 12,0
11. Aer (Pos)(Pos) 2 0
Bene 12609-0801)
x + one OUtand yet Toe 0
Ay, = $3037.73 ae 2 2) Bhavel x 2 -l
12. A; = Age“ (NEG)(NEG) 2 0
1700 = 85e*!2) x + 3.5 0) Jand x + 1°s 0
20 = elk xs =o. and x <-l
3.00 = 12k Thus the solution is x 2
Os) =
Ay3q = 85e%25(130) ~ 1.11 x 10!° rabbits =A Gua Ok=
341
Algebra 2, Third Edition
Problem Set 117
18. x2 + 4x +3 <0; D {Reals} From this we see:
(a) Opens downward
(x + 3)\(x + 1) < 0
(b) Axis of symmetry is x = 2
(NEG)(Pos) < 0
(c) y-coordinate of vertex is 2
+ one )Meand=+
I 0
<a —Somand 5 > col
Ba
(POS)(NEG) < 0 ae
ee
x+3>0 andx+1<0 ne
|
Ve
x > —3 and eo Ke I=
There are no real numbers that satisfy the first
conjunction, so the solution must be -3 < x < —-l.
BE ok
22. Mev,
xhl2 cay lll
19. (ajeyeour
+. 3
ave
(b) y<—= By ey
Pa UE SLI yd!
The first step is to graph each of these lines.
23. m> — 8° = (m)> — (2p°k°)?
= (m — 2p7k?)(m? + 2mp?k? + 4p4k®)
24. w= 24412 - 2)
) VANES) 5
3 (2)
+—|—]=
2a
—+—=
16 43
—
5 9\.5 45° 45 45
25.
(b) —0.15x + 0.2y = -0.6
The region we wish to find is below the negatively
(a’) 15x - l6y = 96
sloped dashed line and above the positively sloped
dashed line. This region is shaded in the figure.
(b’) -15x + 20y = -60
4y = 36
ye= 9
20. 100N = 204.25 25 25 ...
N= 02.04.25 .625 08 (a’) 15x — 16(9) = 96
99N = 202.21
x = 16
nae 20,221
(16,9)
9900
26. (a) Sx + y = 2
PAB Sa
= ae (b) 2x -z=0
Syne (aan
a AY ED) ed (c) 2y + z= -4
oe) Gia DS? (b) + (Cc) 2k + Dyes 4a)
<<
Ht -(x — 2)? +2 —2(a) + (d): -—4x oo
Xx slWel
342 Algebra 2, Third Edition
Problem Set 118
(d) 2(2) + 2y = -4 ie) tO —e) gew & Il —fo)gaw = ON
y= -4 log; x” = log; 16
(c) 2-4) +z = -4 x? = 16
; 2% x= +16
(2, -4, 4) tli
eee St
i+2 Pay PROBLEM SET 118
St
the Sao 24,8, 9
Ee?ier we 1. = xT = 6399
ae 24 w 4 T = 11,376
233. ————
4-J2 4+~72 S = 11,376 — 6399 = 4977 squatted
— 8¥2+4-16-4/2 — -6
+ 2V2
———“-2 °° = 2. Carbon: 2x12 =24
Hydrogen: 4°x l= 4
29.
3/5
we
413 _ _5/3,3/4, 1/4 _ ,29/12, 1/4
xy Oxygen: 1 x 16 = 16
30. ae —dix — 6 =.0 Total: 16 = 44
24444
Ox* 12 x —"6= 0 Ae earl
2x(x — 6) + l(x - 6) = 0 44 396
(2x + 1)x - 6) = 0 H = 36 grams
ee Gert 3. (a) Ny + No = 18
1
dar x=6 (b) 5Ny + 25Ng = 270
1 —5(a) + (b): 20No = 180
>? 6 No = 9 quarters
(a) Ny + (9) = 18
PRACTICE SET 118 Ny = 9 nickels
a. logs (x + 8) + logs4 = logs 80
4. Downstream: (B + W)Tp = Dp (a)
logs (x + 8)(4) = logs 80
(B — W)Ty = Dy (b)
Upstream:
4x + 32 = 80
Tp = 2Ty
Ax = 46
+= (a’) (B + 5)2Ty = 160
10 (b’) (B - 5)Ty = 40
b. log7 (x + 7) — log7 (x — 2) = log7
ed (a’) + 2(b’): 4BTy = 240
log ——, = |lo27 10
‘athe BTy = 60 (c)
a _
T,,U = 4hr;r; Tp4p = 8hr
x +.7=/10x.=.20
D1 = 99x (c) B(4) = 60
3 x B = 15mph
343
Algebra 2, Third Edition
Problem Set 118
5. T = tens digit 14. Since ABCD is a rhombus,
U = units digit mZABC = mZADC
107 + U = original number mZABD = mZCBD = mZBDA = mZBDC
(a) 107 + U = 87 + *UY?) + :'1 2(mZABD) + 34 = 180
(ob) 3T = U + 11 2(mZABD) = 146
U = 3T - 11 mZABD Y=
Substitute (b) into (a) and get: mZBDA = mZABD = 73°
(a’) 107 + (7 — 11) I= 8T + 8GT- 11) + 1
15. log; (x + 5) + log73 = log; 60
13T — 11 = 327— 87
16 = 19T log; (x + 5)(3) = log7 60
oe 3x + 15 = 60
3x = 45
(ob) U = 34) - ll=1
x = 15
Original number = 41
16. 4 logex = log, 64
. Ap= Pe" :
- loge x* = log, 64
Ajo = 15,000e°°U = $38,785.64 5
ae Oo
7. [11]10]9/8]7]6] = 332,640 x = V64 = 2V2 = 2.83
; 3 21 - 17. JOON = .01463,6363 >.
ee kg eR ey =. 0.00165° 63
36 36 = 144 oO
99N = 0.162
9. Inx = 0.0144 n = 162
eA Tol 99,000
-1.45 18. --y = x?
10. (e Tee) sp B45 419.94 — 10.47 oe oe ae
(210-47) ~y = (x? + 6x + 9) + 10-9
= e& ~ 3941.18 pei 3y 41
y=-(@
+ 3)? —1
1. A, = Age
From this we see:
20 = 4e®
ce ok (a) Opens downward
1.61 = 5k (b) Axis of symmetry is x = -3
0322) —) k (c) y-coordinate of vertex is —1
Aggy = 49327040) ~ 1,569,542
12. pH = —logH*
pH = —log 3.14 x 10°
pH = 2. mn
13. H* = 10?°H
H™ = 107>-42
Ht = 9,08 x 19-6 Mole
liter
344
Algebra 2, Third Edition
Problem Set 118
19. |x| — 3 > -7; D = {Integers} (Pos)(NEG) < 0
|x| > -4 x-3>0Oandx+1<0
The solution consists of all integers because any x > 3 and x<=l
integer (even zero) has an absolute val t
thies‘ ) a os a There are no integers that satisfy the second
conjunction, so the solution must be -1 < x < 3.
3-2-1012 3 SS eK Se ee nee
=F 0. 71-28
20. (a)x
+ y 2-3
23. x°'+ 6x ¥°10 = 0
y2—x-3
(weet iGy +H 9)P= 10-49
(b) x — 2y < -4
Cae te = al
x See at
Te >: 4
The next step is to graph each of these lines. Sige ee 182
¥i 4. 14 = yl4
pl Se
xl/2 _ 14, 1/4
al ney oe we
x12 al Days a ye
25. xy? — 27p°m? = (x2y)? — (3pm)?
= (xy - 3p?m>)(x4y2 + 3x’yp?m> + 9p4m®)
26. N= 4 BS
A 5
The region we wish to find is on or above the solid
line and above the dashed line. This region is shaded _ 21 *(61 42
in the figure. 5 8\10 ~=10
21. x* —2x—3 IV 20; D = {Integers}
4
ae
yes
F 80
(x —- 3\(x + 1)20
339, 4/574393
(Pos)(Pos) > 0 aie ake a450
eo
Onan aoe 1)
x = 3 and 5 21. (ay 9
(b) 2x +y-—z= 12
(NEG)(NEG) = 0
x-320 ands + 1250 (c) 2x-y+z=0
x <3 and x<-l (a) + (b): 5x - 2z = 21 (d)
Thus, the solution is x = 3 or x < -l. (b) + (c): 4x = 12
ers OR) 2°34". xe
(d) 5(3) — 2z = 21
22: x2 — 2x — 3 < 0; D = {Integers} ibs
(x — 3)@+ 1) <10
(b) 23) + y - (3) = 12
(NEG)(POS) < 0 yaa
xo 3%" ORand x“ fe>"0
x <"3 “and x >-l ' (3, 3, —3)
345
Algebra 2, Third Edition
Problem Set 119
28. 2 2 2 3. Equal ratio method:
VY,
i+] i- 1 -l1-1 EAE+ nih
E,2
=
Py (B)°W,
29, 127.3 = 39/531/10 = 37/10 s2yulos eyed
P, — (10)°.20
30. D5 ee aor iy
Py = 1000 pinks
NaOH22 Variation method:
= Wea Swe ee
kW
IB = RB
> Hoe)
(10)
PRACTICE SET 119 SO F=VI6
a. {xe R| |x - 3] <5}
= ae = 1000 pinks
oe
ande sn Ss)
i) ane Se << te
(a) Ny + Np + No = 24
(b) SNy + 10Np + 25NgQ = 425
29-2 =1 OV 1 Loa Gud Su GETS 9
(c) Ny = Np
b. {xe Z| |x + 3| < 2} Substitute (c) into (a) and (b) and get:
x+3 22 andx+3<2 (a’) 2Np + No = 24
x 2-—5 and xs -l No = 24 — 2Np
(b’) 15Np + 25No = 425
=6 -5 -4 -3 -2'-1. 0
Substitute (a’) into (b’) and get:
(b”) 15Np + 25(24 — 2Np) = 425
PROBLEM SET 119 -35Np = -175
Np = 5 dimes
1 ain uy (c) Ny = Np = S5nickels
LY 7, (a) (5) + (5) + No = 24
00 a No = 14 quarters
700 2100
V> = 1500 mL T = tens digit
U = units digit
2 Do
107 + U = original number
250 10U + T = reversed number
Cad eee
D,
———eee T=5-U
400
(oy 10U +7 2 70r 2 U = 77
RoTg = 250; R7Tr = 400;
Substitute (a) into (b) and get:
Tr = 2Tg; Rg = Rr + 10
(b’) 10U + (5 — U) 10(5 - U) + U - 27
2Tg(Rg — 10) = 400
9U + 5 = -9U + 23
2(250) — 20Tg = 400
ist? ="18
—20Tg = -100
Ural
Tg = 5hr (a) The Sos psed
Tr = 10hr; Rg = 50 mph; R; = 40 mph Original number = 41
346 Algebra 2, Third Edition
Problem Set 119
6. A, = Ape” 14. (a) -5.45
640 = 40eK°)
heen (b)ney ct eed GI xelOnr
2.77 = 3k
0.923 =k 15.
(e778 Ree)
Wig
)(0597)= Ae, a8 >
4.6.07 td442.29
Ax9 == 40e 0.923(20) _
= 4,160,409,959 amoebae = 940.58 ~ 499 x 1917
7. |x + 1] < 3; D = {Integers}
16. mZCBD = 180 — 130 = 50°
x+12-3 andx+1<3
x > -4 and x<2 mZCDB = 180 — 90 — 50 = 40°
mZx = 180 — 40 = 140°
5-4-3-2-10 123
8. |x - 3| > 5; D = {Reals} 17. Region B (This region includes all points on or
above the line and on or below the parabola.)
YS > ow F = 3 <5
2 >*S or xer<2 18. pH = -log H*
aio
ie oe Jere See Se 6.7 28.09 pig 08 2 322 10
pH = 11.02
9. In(x + 3) + In4 = In40
In (x + 3)(4) = In 40 19. H* = 10-?H
4x te 12 = 40 Ht = 1072-23
4x = 28 mole
ay x 10> —
H* ~ 5.89
liter
10. 3 log)5x = logs 27
ae 5, 5000L , 1000 mL lcm? ¢ an,
£15 : =P ES * ‘nin 1L ImL 2.54cm
ae / : E
ae 1 in. 1 in. 1 min
x — XK x
2.54 cm De AeCin 60s
1 *( cee .) 5000(1000) in.?
11. N = 6—+—)12
Woccdt-~ 20 10; ~ (2,54)(2.54)(2.54)(60) s
61 4/( 243 122
ER10 Ag5NAL boar20 21 :
61 484
= — + —
10 100
610
100
,~—:100
484 _547+
50
————
12. xl/2 ae
ie yin
Ve Fee 2
_ xyla Dy
x — 2x22 yy
13. xy? a 64p!?m? ay (xy?) = (4p4m?)P
= (xy? — 4p4m3)(x2y® + 4xy3p4m? + 16p8m®)
347
Algebra 2, Third Edition
Problem Set 119
22. Aghaded = Asector AOE ~ ATriangle POE From this we see:
(a) Opens downward
Asector AOE = 2G
360 cm)*)= “
7 em?
(b) Axis of symmetry is x = —1
ADOE is equilateral because
(c) y-coordinate of vertex is 2
ZDOE = ZOED = ZEDO = 60°. It follows that
PE = 3, since OA is the bisector of ZDOE and,
therefore, the bisector of DE.
Since triangles POE and POD are 30°-60°-90°
triangles,
2x SF 318
oF =32
BETS
PO = 3(5) = 33 cm HREEE EHH
2 2
A Triangle
rangle POE == 5|5)
(3 52) em) re
22 en 26.
See Lesson 71.
9/3 ®)
= —cm
8
4B 27. N= 324452 -32)
3a 0 9/3 EWEN Chae
Ashaded = (22= 23) cm?
10, (lege)
3? Sse eS
23. (a) 2x -y-z=8
=—_—+
10. 36
(b) 3x +y-z=9 TS
(c) 6x -y+z=0
_ 250 ,31_ 281
(a)
+ (b): 5x — 2z = 17 (d) in Sic 1G) 75
(b)
+ (c): 9x= 9
28.
27-2 — -2i-2 342i
x=1
Sena) oT 349i
(a5)
922 "=
_ -6i+4-6-4i 2 10,
z=-6
9+4 de Shes
(b) 3+
y= 6) I
b
y I =) 29. gixe = ghl2,bl2
qo!2 b/2
(1, 0, -6)
24. 100N = 1.3 13 13 ... 30. 3x3 — 3x2 = 6x = 0
N = 00 13 13: 4. 3x(x2
- x - 2) = 0
99N = 1.3
3x(x? -2x+x-2)=0
apne
990 3x(x(x
— 2) + I(x
- 2)) = 0
3x(x + 1)(x
— 2) = 0
25. 62 PEO
(574) ee x-2 I So
Ge +e =e —1 w= 2
wv 0g
il 1
348 Algebra 2, Third Edition
Problem Set 120
PRACTICE SET 120 | 3. Now + 15 Years
10 Years Man: My Mn + 15
Now 5 Years Ago From Now on Sy Sy + 15
Ben: B B= 5 B+ 10 (a) My = 18Sy
Kris: K K—-5 K + 10 (b) My + 15 = 3(Sy + 15)
2 Substitute (a) into (b) and get:
B-5 = =(K - 5) )
3 3 (b’) (18Sy) + 15 = 3Sy + 45
B+ 10 = (K +10) hire
() let
6 Sn = 2
3(a) 3B - 15 = 2K — 10 (a) My = 18(2) = 36
(a’) 3B — 2K = 5
4. Since the triangle is equilateral, each side must
6(b) 6B + 60 = 5K + 50 equal 8 meters.
(b’) 6B — 5K = -10
~2(a’) -6B + 4K = -10
(b’) 6B — 5K = -10 8m 8m
Jena
K = 20
(a’) 3B — 2(20) = 5 au
3B - 40 = 5 H = (8m)? - (4m)? = 4/3m
B= 15
A= +(8m)(4V3 m) = 16/3 m?
PROBLEM SET 120 5; Now —6 Years —20 Years
1. 0.76(300) — (Dy) = 0.2(300 — Dy) Lucie: Ly Iy-6 = Ly - 20
228 - Dy = 60 — 0.2Dy Myma: My My-6 — My - 20
168 = 0.8Dy (a) Ly — 20 = 2(My — 20) + 2
210 liters = Dy 3
(b) My — 6 = qhn 5)
2. T = tens digit .
U = units digit (a’) Ly — 2My = -18
107 + U = original number (b’) 3Ly - 4My = -6
10U + T = reversed number —2(a’) -2Ly + 4My = 36
(a) T+ U=11 (b’) 3Ly - 4My = -6
T=11-U Ly = 30
(b) 10U + T = 3(10T + U) + 5 (a’) (30) - 2My = -18
My = 24
Substitute (a) into (b) and get: N
+ (11 — U) = 30(11 - U) + 3U + 5
(b’)10U Ly = 24
= 30; My
LU Co 6. |x - 3| < 2; D = {Integers}
36U = 324
x-3>-2andx-3<2
U=9
x > ID and x= 5
Abani TO &
Original number = 29 ipe 3.455
349
Algebra 2, Third Edition
Problem Set 120
|x + 3| < 2; D = {Reals} 18. (-3 + /2,4 - V2)
eo 3 2-2 -and x) +°3 2
and
A, = Pe™
Ag = 2000e°8) = $3232.15
26 |25|24|10|9]8]= 11,232,000
e 15.91
10. = e 15-91 — 12.83 — 9728.74
e!2.83 19, yea"
Ir 3
= 3.30 x 107-8 =y = (77> 24 4.1) Hye. al
y= (x — 1)* +2
(ey = 10.665 ~ 428 x 104
11.
y=-(@- 1)? -2
12. 4x = 3 In 0.0037 From this we see:
3 In 0.0037 (a) Opens downward
x = —— = -4,20
4
(b) Axis of symmetry is x = 1
13. Sin x= =4013 (c) y-coordinate of vertex is —2
4.13
lige SS ===
3
x = 0.25
14. pHi = los Ht
pH = -log 0.062
pH = 1.21
15. Ht = 10°PH
H* = 1973-13
Ht ~ 7.41 x 19-4 mole 20. 5x- + 2x — I S
liter
& a | * ll
AN fey ine Win © 2on, umn
16. — xX x x x
min 1 ft 1 ft EKG 60 s
_ 40(12)(12)(12) in.? Rl-
oe
= (60) s NO
& + I
nS Ma|R
er
17. 1000N = 21.63 163 163 ...
N = 0.02 163 163 ...
999N = 21.61
2161
S I
99,900
350
Algebra 2, Third Edition
Problem Set 120
3 3.76R + 1.37U
21. (a) os + y = 13 < 3.86R + 4.60U
7.62R + 5.97U
(b) 0.2x — 0.02y = 1.12
F = (7.62)? + (5.97)? ~ 9.68
(a’) 3x+2y = 26
— 2y = 112
(b’) 20x OEE Es
23x = 138 7.62
x=6 0 = 38.08°
(a’) 3(6) + 2y = 26 9.68 /38.08°
y=4
24. 5x? + 9x2 — 2x = 0
(6, 4)
x 5x" + 9x = 2) =O
22. a 5xoP + 10x! x) — 2) 0
x(5x(x + 2) — 1 + 2)) = 0
x(5x — 1)\(x + 2) = 0
x=0 sx -1=0 x +2,
=0
GS f x = 2
3
geet;
3
H = \(6)?+ (4)? = 2v13
(©) 4) V13 25. log, (x — 3) — logy 7.= log, 31
6
tan@ = —
4 log, Sa -, pe losaat
@ = -56.31° oy
x - =
2/13 /-56.31°
yy =.220
Se | 7/8
C 26. In(x + 5) + In5 = In65
A In (5)(x + 5) = n65
= 4cos 20° = 3.76 5x 25? 2°65
B = 4sin 20° = 1.37 5x = 40
be the
27, (32 _ y32y2
3/2 _ 32
2 _ pr
Fee
_ 3/2312 4 53
x2 — 2x32y32 4 3
28. m°p!? <# 8y%z!° = (mp*)? oo (2yz°)
C = 6cos 50° v 3.86
= (mp4 — 2yz*)(m4p® + 2m*p4yz> + 4y?z av)
Il 6 sin 50° u 4.60
Algebra 2, Third Edition 351
Problem Set 121
PROBLEM SET 121
29. Obit Be = 6
2 1. 0.4(80) + (Dy) = 0.52(80 + Dy)
AB = (12)? + (9)? = 15 32 + Dy = 41.6 + 0.52Dy
0.48Dy = 9.6
Since ABPO and ABCA are similar triangles,
Dy = 20 liters
6 x SF = 15
Zee
L 1,
10 20
|
Sv x | Ne}
600T, ~= 1200K
7,
Downstream: (B + W)Tp = Dp (a)
Upstream: (B —- W)Ty = Dy (b)
30. mZB = (180) = 90°
(a’) 6B + 6W = 84
mZBAD = 180 — 90 — 54 = 36° (b’) 7B — 7W = 42
Ts /BAO =5--(G6). = 18° 7(a’) + 6(b’): 84B 840
2 2 B 10 mph
x= 5mZBAO = 18° (a’) 6(10) + 6W = 84
W = 4mph
Now +10 YEARS
PRACTICE SET 121 Garfunkel: Gy Gy + 10
—2 Spot: Sn Sy + 10
<x 2D Reals}
x -—3
(a) Gn = 2SN
The value x = 3 cannot be a solution, because
(b) 4(Gy + 10) = 3(Sy + 10) + 15
division by zero is not allowed. Multiply both sides
by @ 4a)? Substitute (a) into (b) and get:
32 < —2(x -— 3)(x —- 3) (b’) 4(2Sy) + 40 = 3S, + 45
(Co
x —3 5Sy = 5
x7 - 6x +9 <-2x+6 Sy=1
x” 4x + 3.< 0 (a) Gy = 201) = 2
(x — 3x - 1) < 0
Now + 10 YEARS
(Pos)(NEG) < 0
Rover Boy: Ry Ry + 10
*,.— 3 > O.and x = 1.<.0
Yolanda: Yn Yy + 10
x > 3 and o< |
(a) Ry = Yy + 5
(NEG)(Pos) < 0
(b) 4(Ry + 10) = 2(¥y + 10) + 50
x — 3 <"0' and x = I >20
Substitute (a) into (b) and get:
ce Sean x1
(b’) 4(Yy + 5) + 40 = 2¥y + 70
No real numbers satisfy the first inequality, so the 2Y¥y' = 10
solution is given by 1 < x < 3.
Yy = 5
0 1 2pacye4 (a) Ry = (5) + 5 = 10
352 Algebra 2, Third Edition
Problem Set 121
6. |9|8|7]6] = 3024 14. x = 61n 0.003 ~ -34.85
Pe ue ae 13 15. 6Inx
= -2.78
26, 16° «(16 2.78
Kiley ae
S A; = PeM
Az = 1,000,000 e137) = $9,484,322.53 ‘wig
11.62
9 A, t = Ape”
0 oo 16. e ir. L ¢ 11624811 =e
19.73 = 3.70 <x 10
8
140 = 40e e
Baus
& . E 17. ZPOR + ZPRQ = 180 — 80
25 = ZPOR + ZPRQ = 100°
0.25 =k : :
1
Ayo = 40692500 ~ 487 5(ZPOR + ZPRQ) = +(100°) = 50°
10. |x + 2| < 3; D = {Reals} mZQSR = 180 — 50 = 130°
Xe 2 SS Scand tested 3 td
x>-S5and x<1 seen obra
pH = -log 0.053
—=}+—_o—__+_—_+——_o—
6-5 -4-3-2-1 0 1 2 pH = 1.28
11. Multiply both sides by (m + 3)" 107 teen eH
(m + 3)(m — 3) < 20m + 3) Ht = 1977.24 x10
m2 — 9 < 2m? + 12m + 18 vie
m-
2 + 12m + 27 => 0 Ht = 1.00 liter
(m + 9)\(m + 3) = 0
m = —3 cannot be a solution, because division by Al, JP = 2 62, = 1p)
zero is not defined. x = 180 — 62 = 118
(Pos)(Pos) 2 0 eee BO pe 18) 084
m+920 andm+3>0 ee
mBD = 124
— 2(34) = 56
m = —9 and m> —3
1
(NEG)(NEG) = 0 mz A= ay) = oe
m+9<0O andm+3<0
Le 2D alt 7 45 mZBDA = 180 — 62:— 28 = 90°
Thus, the solution is m > —3 or m < -9. 1
21. Atriangle = rays
10-9 -8 -7 -6 -5 -4 -3 -2
Since AOAB is equilateral,
12. In(x + 2) + In6 = In 36 b = 8cm
In (6)(x + 2)2) = In 36
= In
h= \82 - 42 = 4/3cm
6x + 12 = 36
a 1
6x = 24 Atriangle = 5 8 cm)(4/3 cm) = 16V3 em?
xeaad
2
a ai fog 27 Atexagon = 6(Atriangle) = 6(16V3 cm”)
: . 0811 %aes
= 11 |. 96/3 cm 2
0811*% = 10811
5 22. Region B (This region includes all points above the
a ;' line and below the parabola.)
Algebra 2, Third Edition 353
Problem Set 121
23. ay = xe Sag a7 25. (a) 3x ~y +z =1
ay a@-—-4y-4+°4)¢
7 =A (yy
Heh orl
a = O32) 48 (c) x —-2y-z=-2
y=-—(x- 27-3
(a) + (b): 4x — 2y i N
From this we see:
(a) + (c): 4x - 3y Il | —
(a) Opens downward
(b) Axis of symmetry is x = 2 (d) + (Ife): y = 3
(d) 4x -— 2(3) = 2
(c) y-coordinate of vertex is —3
x = 2
(ob) (2) = 37 = 2 I —
z —2
(2, 3, -2)
26.
l
~x + 1 ee
@) ri 3
(b) 0.4x -— 0.2y = -0.2
(a’) 3x + 2y = 30
(b’) “4x — 2y
1 il InN]oe\)|nN
Ny
x=4
= 4sin 15° = 1.04 (a’) 3(4) + 2y = 30
= 4cos 15° = 3.86 yond
(4, 9)
27. (a) Sx + y 7
6 D (b) 2x -z = -1
(c)y+z=5
(b) + (c): 2x + y
ees
C —l(d) + (a): 3x I ee)
soa |
D= 6sm67° = 5.52
C =.6c0s 67°, = 2.34 (a) SQ) + y = 7
-3.86R — 1.04U
2.34R + 5.52U (c) (2) + re S
-1.52R + 4.48U
F = \(-1.52)? + (4.48)? = 4.73
(1, 2, 3)
fanion= hohe
1.52 28.
4-5 5 9
@ ui= -71.26 + 180 = 108.74° (S42 aoe
4.73 /108.74° Bl Soh Bes Se a
$d
354 Algebra 2, Third Edition
Problem Set 122
3
29. 3° — 4x" 2 - Tx rY
= 0 ; PROBLEM SET 122
x(3x* - 4x — 7) = 0
x(3x? + 3x —- 7x —- 7) =0 1, (@) 3Ny + 1OND = 255
x(3a(x + 1) = 7 + 1)) = 0 (b) Ny = Np + 25
x(3x — 7)(x.+.1) = 0
( I ) Substitute (b) into (a) and get:
eS 8x = = 0 tes ()
4 * (a’) 5(Np + 25) + 10Np = 455
same: x =-1 ISNp = 330 |
; 7 ; Np = 22 dimes
at (b) Ny = (22) + 25 = 47 nickels
30. (a) 1.74 x 104
2. Equal ratio method:
b)
(b) 4.844.84 xx 10 1073 M, A.(Y,) 2
. a
M, A,(V,)
PRACTICE SET 122 400 prey
a. MAP 1600 a
The points that are members of both M and P a2
A> = 128 apes
Variation method:
M = kA(Vy*
400 = k(2)(2)*
b MAX a ae
: 2
The points that are members of both M and X 1600 = (50)( a\(3]
A = 128 apes
3. T = tens digit
U = units digit
¢. Ar M 10T + U = original number
The points that are members of either Xor M (a) 10P eer + Be 8
(b) 4U = T + 30
T = 4U — 30
Substitute (b) into (a) and get:
(a’) 10(4U — 30) + U = 2(4U —- 30) + 2U + 8
d.°X AP 41U — 300 = 10U — 52
The points that are members of either X or P 31U = 248
U=8
(b) T = 4(8) — 30 = 2
Original number = 28
Algebra 2, Third Edition bs)
Problem Set 122
Now + 10 YEARS (NEG)(Pos) < 0
Yehudi: Yn Yy + 10 x-2<0andx-1
Mohab: Mn My + 10 x <= 2eand ih IV ©-
(a) Yy = My + 4 (Pos)(NEG) < 0
(b) 2(¥y + 10) = My + 10 + 24 ee Onandaxt—slmsa0
ye 2) and 5 Se il
Substitute (a) into (b) and get:
(b’) 2(My + 4) + 20 = My + 34 There are no real numbers that satisfy the second
conjunction, so the solution must be 1 < x < 2.
2My + 28 = My + 34
Mn = 6
OF 1 23 4
(a) Yy = (6) + 4 = 10
11. (m — 3)(m + 3) < 1(m
— 3)
Now + 10 YEARS m —9 < m* —- 6m +9
Petunia: Pn Py 10 -18
< -6m
Daisy: Dy Dy + 10 3>m
(a) Py = 2Dy m = 3 cannot be a solution, because division by
(b) 2(Py + 10) = Dy + 10 + 25 zero is not defined. Thus, the solution is m < 3.
Substitute (a) into (b) and get:
(b’) 2(2Dy) + 20 = Dy + 35
3Dy = 15 12: lx — 2] < I3.Du=_fReals}
Dy = 5
x —2>-l1 andx-—-2<1
(a) Py = 2(5) = 10 > and w< 3
A, = Age“ OP Ries
2a a4
1,500,000 = 1000 ek)
1500 = e 13. |x + 3| < 4; D = {Integers}
2.44 =k x +3 2-4 andx+3<4
Ag = 1000e**48) =~ 3,00 x 10! euglenae en and x <a
7T6Ts
[4]=840 -8 -7 -6 -5 -4 -3 -2 -1 0 12
102-7!
729 14. cts = 102-71 + 7.15 = 794 1016
(a) log MN log M + log N 15 (ig? =)? = 10!-57 = 37.15
M
log M log N 16. (10!4-%) (19-19-52) = 10!4-66 - 19.52
= 10-486 ~ 1.38 x 19°75
(c) log MN=N logM
17. pH= -log H*
10. (x - 2)°(2)< —2(x — 2)
pH= log 3.26 x 10-°
257 Rx 4h —2x + 4
pH = 8.49
x7 8xy-+ 2 lA 0
(t= 2) =P) lA 0 18. pH= -log H*
x = 2 cannot be a solution, because division by pH= -log 7.04 x 10>
zero is not defined. pH = 4.15
356 Algebra 2, Third Edition
Problem Set 122
19, H = -lo Ht
| s P8 -34+i
ie3-
techV- 242i
aan ESAS) BOP BORE Tophrey
Se
ae Tas] Ate
OL eG
;
20. ne=e 10-PH
Ht
H* = 1074-02
28. 22-
4v 5 : 1+-72
a
A)
5
16
op:ai cs
5 mole
H* t = 9.55 x 10 re
4y2 +8 -—5 - 5v2
PE Eh
op et
21. H* = 10-PH
Ht = 1078-23
4 6 N2
29. Vxy> a3 y = x /4 5/4 1/2),1/6 S x/4yIT
Ht ~ 5.89 x 1079 mole
liter
30. (ayer OZ
22. ae —pH
The points that are members of both P and Z
Ht
at = 10710-13
Ht = 7.41 x 19-11 Mol e
liter
23. Ashaded = ABig Circle — 2Asmall Circle
m(12 yd)> — 2(x(6 yd)’)
= 72 yd* ~ 226.08 yd? (bd) POS
226.08 yd? = 226.08(3)(3)(12)(12) in.” The points that are members of both P and S$
1000 in. _ 2.54 cm Im 1 km
% pay ema a
S 1 in. 100 cm 1000 m
60 s 60 min
1 min 1 hr
_ 1000(2.54)(60)(60) km
(100)(1000) hr
(E)= 27 GFP
25. 100N = 0.168 68 68 ... The points that are members of either
Z or P or
N = 0.001 68 68 ... both
99N = 0.167
eer
99,000
26. 5x7 +x+4=0
trie Bas -:
(srg
(2 ++ |< 100a+100
x 5 100) The points that are members of f either
eitherZZ or
or S
S or
fs both
1 79
x+—| = -—
10 100
XY tes = =
1 179
10 10
oer),
= -—— t ——i
10 10
Algebra 2, Third Edition es
Problem Set 123
PRACTICE SET 123
RyTy = 800; RoTo = 650;
Ty = 2Tg; Ry = Ro - 50
The locus is a pair of parallel lines that are 6 feet (Ro — 50)2Tp = 800
apart. (Each is 3 feet from the given line.) 2(650) — 1007) = 800
-100Tg = -500
To = Shr
Ty = 10hr; Ro = 130 mph; Ry = 80 mph
Dy
a Dw
RyTy = RyTw; Ry = 16:
Ry = 10; Ty + Ty = 13
16(Ty) = 10(13 — Ty)
The locus is a pair of concentric circles, one with a
radius of 5 cm, the other with a radius of 11 cm.
26T = 130
(Both circles are also concentric with the given Ty = 5
circle.) Distance = RyTy = 16(5) = 80 miles
10. T = tens digit
U = units digit
PROBLEM SET 123
107 + U = original number
See Lesson 123. (a) 107 + U = 3(7 + U) + :13
(6) V=T +
See Lesson 123.
Substitute (b) into (a) and get:
(a’) 107
+ (7 4% 1) = 3F + 30 41) + 13
See Lesson 123.
117 + 1 = 67 + 16
iY be
See Lesson 123.
T=3
(b) U=4
See Lesson 123.
Original number = 34
See Lesson 123. 11. Now +5 YEARS
Man: Mn My + 5
Consecutive odd integers: N, N + 2, N + 4 Son: Sn Sn + 5
NN + 4) = 10(N + 2) — 13 (a) My = 6Sy
N* + 4N = 10N + 7 (b) My + 5 = 3(Sy + 5) + 2
N? - 6N-7=0 Substitute (a) into (b) and get:
(b’) (6Sy) + 5 3S + 17
(N + 1)(N - 7) = 0
3 Sw 12
N = -1,7
Sw 4
The desired integers are —1, 1, 3 and 7, 9, 11. (a) My = 24
358 Algebra 2, Third Edition
Problem Set 123
12. P ev R=, 7,13} 21. Region C (This region includes all points on or
below the line and inside the circle.)
13. AV B= {1,3, 8 753,10}
22. logy (x + 2) + logz3 = log, 15
14. PU K = {2,3,5,7, 8, 9, 10, 11, 13, 15} log, 3(x + 2) = logy 15
Tee te Lh)
2 Se 23) 3x
+ 6 = 15
Oe ee) eee | bye ea)
x7 —-x<0 x=3
xx-1)<0
23. In@ = 9) + In2 = 1n45
x = 1 cannot be a solution, because division by In 2(x -— 9) = In 45
zero is not defined.
2x — 18 = 45
(NEG)(Pos) < 0
2K = 63
kes Ocandex >) |
63
(Pos)(NEG) < 0 ee 5
ee cand 9. < a1
24 A/S S yn
There are no real numbers that satisfy the first 4/5 ys
conjunction, so the solution must be 0 < x < 1.
yes = 44/55
0-9 4/5. 4/5, .8/5
T'psa te 2 ealeie Bd ete ae
x85 _ 2,45 y4/5 8/5
16. (m — 1)(m + 1) < 1(@m — 1°
Me TS” 2 PM 25. am?! - py = at = yy oe
pre ne = (am? — p*y'?)(a’m'® + am?p?y"? + py)
12m 26. 1000N = 1.2352 352 352 ...
m = 1 cannot be a solution, because division by Ne= 0001293525852 P.:
zero is not defined. Thus, the solutionis 1 > m. 999N = 1.234
2 eee aes 1234
reap OTT 999,000
Lig. |x == 3| = 33 Dv= {Reals} Die y= x2 aie Be Yl
x 2 -—6 and x = 0 er eo) ae
yr yey 4 5
From this we see:
18. pH = -log H* (a) Opens downward
pH = -log 1.42 x 1071!
(b) Axis of symmetry is x = 2
pH ~ 10.85
(c) y-coordinate of vertex is 3
19. Ht = 10H
Ht = 1072-27
Ht ~ 1.07 x 19-4 mole
liter
20. A, = Age“
480 = 240eM1
= =)N \o U =
Aig = 240e9-°6%18) ~ $31 bacteria
Algebra 2, Third Edition 359
Problem Set 124
WS mo3a? ses Ser Die. 0 4. ADCB = ADAB_ SSS
_ ~(5) + 46 — 4302) ZCBD = ZABD CPCTC
ae 2 BD 1 CA :
-—(5) +1
vee A eels. au Two equal adjacent angles whose sum is 180° are
right angles.
29. 3-61 + 2i- 26+ 22=3-4i
5. AEAB = AEDC SAS
30. (a) Since this is a 30°-60°-90° triangle, BE = CE CPCTC
— oe o 6. See Lesson 123.
SF =7
m= i) = 4 7. See Lesson 123.
n ='/3(7) = 73
8. See Lesson 123.
(b) Since this is a 45°-45°-90° triangle,
1 x SF =3 1Gm
9. (m =:3)¢n +1320 +23)
SF =3 m= —9 > m= — 6m + 9
18 > -6m
aA > Fe
c = ¥2(3) = 3V2
3<m
d = 1(3)
=3
m = 3 cannot be a solution, because division by
zero is not defined. Thus, the solution is 3 < m.
PRACTICE SET 124
Pe PBL as
a. The conditions of congruence are SSS, AAAS,
SAS, and HL. 10. Minor seine B)
2
b. Since ZE = ZH, AEDH is isoceles. SS ee ee
Dee Wak ses
ED = HD Definition of isoceles triangle . ar
2x(x - 1) < 0
AEDF =Rt AHDG SAS
x = 1 cannot be a solution, because division by
D
zero is not defined.
(NEG)(Pos) < 0
x-s 0 and x.>"1
fe iF G H (Pos)(NEG) < 0
x 20 and x < 1
There are no integers that satisfy the first
PROBLEM SET 124 conjunction, so the solution must be 0 < x < 1.
1. AWAX = AZBY SAS
LWT CPCTC “r) 90 6h 2
2. AABD = ACBD SAS 11. |x + 2| < 3; D = {Reals}
AD = DC CPCTC Yoh 2. —3 and pie? 263
Y >) —omand ial
3. ABAG = ADEF SAS
Snail SONA: A0n SE ne eee eee eeeexseeen eee
BG = DF CPCTC -6 -5 -4-3-2-1 01 2
360 Algebra 2, Third Edition
Practice Set 125
12. 38 = 10**3 20. Sy Sx? 2 +14
19!:58 Pe 10* +3
ayes (Xe
+ 1) 4 og
1.58.= 2+ 3
~y=(-
1)? +3
-1.42 =x
i i] ie
Ll) a
1S. In (x + 2) + In3 = 1n39
From this we see:
In 3(x + 2) = In39
(a) Opens downward
3x + 6 = 39
x= 11 (b) Axis of symmetry is x = 1
(c) y-coordinate of vertex is —3
14. 4 loggx = loge 48
logg x* = logs 48
x* = 48
x= 148
15. (a) log MN = logM + logN
(b) log = = logM - logN
(c) log MN=N logM
16. Now +12 YEARS
Man: Mn My + 12
Son: Sn Sy + 12 PRACTICE SET 125
(a) My = 11Sy
f ireie:@ 1. Given
(b) My + 12 = 3(Sy + 12)
2. ZOAP =,90° 2. Given
Substitute (a) into (b) and get:
(b’) (11Sy) + 12 = 3Sy + 36 3an2OBP) =u90° 3. Given
8Sy = 24 4. OA = OB 4. All radii of circle O
Sy = 3 are congruent.
(a) My = 33 R
5. OP-=OP 5. Reflexive
17. 6. AAOP = ABOP 6. HL (4, 5)
7. APHESBP 7, GPCTC
18.
19. 3
8 ‘
3 (= = |
=-—+-|—_—--—
8 7\40 40
3 bE
8 140
los
280
, 146
280
_ 280
251
Algebra 2, Third Edition 361
Problem Set 125
b. 1. Circle O 1. Given PROBLEM SET 125
2. EG = FG 2. Given 1. 1. Circle Q 1. Given
3. Line segment EF 3. Given 2. PR 1 ST 2. Given
4. OG = OG 4. Reflexive 3. OS = OT 3. Radii of same circle
5. OE = OF 5. All radii of circle O 4. OR = OR 4. Reflexive property
arene vergctt 5. AQRS = AQRT 5. HL (3,4)
6. AOEG = AOFG 6. SSS (2, 4, 5) 6. RS = RT 6. CPCTC
7. ZOGE = ZOGF 7. CPCTC 7. ZPRS’= ZPRT 7. EPCTC
8. ZOGE = ZOGF 8. Two adjacent equal 8. PR = PR 8. Reflexive property
= 90" angles whose sum is 7%
— 10, Z8°= ZT 10. CPCTC
9. OH 1 EF 9. QED
2. ZBED = ZBDE Given
ZAEB = ZCDB_ Given
ZAEB + ZBED = ZCDB + ZBDE
ye ae ZAED = ZCDE
sae re eae 3. AABE = ACBE AAAS
CE = AE CPCTC
e« IOrko 1. Given
4. AOAM = AOBM _ SSS
25 PR IST 2. Given
as. 3 ZAMO = ZBMO_ CPCTC
3. S0 = TQ 3. All radii of circleQ ae ee
are congruent. OM 1 AB
4. RO = RO 4. Refiecve Two equal adjacent angles whose sum is 180° are
right angles.
5. ASRQ = ATRQ S-Ni ;.4) 2. oe te
> 5. Begin by drawing OA, OB, and OP.
6. SR = T. 6. CPETC
AOAP = AOBP HL
7. ZSRQ = ZTRQ Tt CPCTC ee ee
Ratios AP = BP CPCTC
8. RP = RP 8. Reflexive
6. See L :
9. ASRP = ATRP 9. SAS (6,7, 8) Semin £3
10. ZS = ZT t0Y CPQTC 7. See Lesson 123.
S S228 8. See Lesson 123.
eo 4 9. See Lesson 123.
T Aq = 1150e°8C) ~ $2773
362 Algebra 2, Third Edition
Problem Set 126
11. A; = Age 18. = 24
1600 = 1000e*)
56 \76,
1.6 = e
19. (eee )GSk2 ym o-1.56 + 6.82 + 27.89
0.157 =k (e279) =
= e275 = 6.18 x 101!
Ajo = 100021570) ~ 4791 synaptic reactions
12.
20. 1O00N = 31.54 154 154 ...
Sx = 3 In 0.0035
N= 0.03 154 154...
x=-_—_—_—_
3 In 0.0035 999N = 31.51
5 _ 3151
x = —3.39 99,900
13. Inx =
Ss
———
PROBLEM SET 126
3
ABCE = ADCE SAS
x =~ 0.18
BE = ED CPCTC
14. |x — 5| < 3; D = {Integers}
Draw radii OX and OW.
y—-)
> — 5 and x — 5 < 3
AOXY = AOWY SSS
Fae ee STvo | Xs
ZALL = ZW CPCTC
253 4-5: Geer 8 ZAC eee WY AA — = AAA
5. x? — 6x + 9 > 0; D = {Reals} OY 1 Wx
(x — 3)@ — 3)> 0 Two equal adjacent angles whose sum is 180° are
right angles.
(Pos)(Pos) > 0
sc SS Draw radii AP, BP, CP, and DP.
(NEG)(NEG) > 0 AAXP = ABXP = ADYP = ACYP HL
Sere 1)
AX = BX = CY = DY CPCTC
Thus, the solution is 3 < x < 3.
AAPB = ACPD SSS
1 25s eS AB = CD CPCTC
16. Now — 6 YEARS —20 YEARS ABED = ABFD SAS
Melvina: Mn My —- 6 My — 20 ABAE = ABCF AAAS
Beula: By By - 6 By — 20
AAFD = AABD_ SSS
(a) My — 20 = 2(By — 20) + 4
Z3 = 24 CRCTC
3
(b) By - 6 = = (My — 6) ADEF = ADCE SAS
(a’) My — 2By = -16
CPETS
(b’) —3My + SBy = 12 See Lesson 123.
3(a’) —— (b’): Bn = 30
See Lesson 123.
By = 36
(a’) My = 56 See Lesson 123.
17, =40
Ppp 22221221212] See Lesson 123.
Algebra 2, Third Edition 363
Problem Set 126
10. |9|8[7[6[5] = 15,120 18. ‘x? - 5x +4 0; D = {Reals}
= Ae = 1) IA 0
i: = 6561 (NEG)(Pos) < 0
x =< 4 andx 2 1
12. (a) log MN = logM + logN (Pos)(NEG) < 0
ye anda S |
M
(b) log —— = logM - logN There are no real numbers that satisfy the second
N eee
conjunction, so the solution must be 1 < x < 4.
(c) log MY = NlogM
Orglinc®s3 Ave
13. 2 = nee Sea 19. T = tens digit
. U = units digit
107 + U = original number
14. Inx = = 10U + T = reversed number
(a) T+ U= 14
x = 0.29
T=14-U
15. pH = -loeH” (b) 10U + T = 2010T + UV) — 23
pH = —log 0.0053 Substitute (a) into (b) and get:
pH ~ 2.28 (b’) 10U + (14 - U)
= 20014 — U) + 2U = 23
16. Ht = 10 °H 9U + 14 = -18U + 257
Ht = 10-619 21 = 243
U9
le
Ht = 6.46 x 10-7 “°° k
liter Ue ee
Original number = 59
17. @ =] Dp 32).s.
2g: =>) es ie te
- y=x" —- 6x + 3
Pao a = 2p? = Spas ys Goa "6rs 9) Se
p” - 8 + 1220 vate ay a6
(p= 2)(p = 6) 2 0 From this we see:
p = 2 cannot be a solution, because division by (a) Opens upward
zero is not defined. (b) Axis of symmetry is x ="3
(Pos)(Pos) = 0 (c) y-coordinate of vertex is —6
p-2>
0 and p - 6 IV io)
os
p > 2 and p26
(NEG)(NEG) = 0
p=2.<
0 andp-6<sra oS
p < 2 and ps6
Thus, the solution is p = 6 or p < 2.
anh
hort
oat
hears
N
Ui yey ey ay 7
364
Algebra 2, Third Edition
Problem Set 127
PRACTICE SET 127 ZBYZ = ZC Corresponding angles are equal.
ZC = ZYAC Definition of isosceles triangle
a. PF
= PF Reflexive
ZYAC = ZAYZ Alternate interior angles are
APFG = APFH HL equal.
4G = ZH CPETG ZBYZ =I ZAYZ Reflexive property
P YZ bisects ZAYB Definition of angle bisector
EZisa perpendicular bisector of XY.
Therefore, EZ XY
AABE = AACE HL
BE’2CE CPCTC
Definition of bisector
AE bisects BC Definition of bisector
Reflexive
See Lesson 123.
All right angles are congruent.
See Lesson 123.
SAS
See Lesson 123.
CPCIC
cs & Reflexive
See Lesson 123.
ZABD = ZABC All right angles are congruent. 10. Af="Pe™
Ag = 1460¢%-°) ~ $3282
AD = AC CPCTC
1B be A; = Age
AADC is isoceles Definition of isoceles triangle.
560 = 40ek@)
A 0.88 =k
Ag = 40e°°8) ~ 110,070 incidents
12. pH = -log Ht
pH = —log 0.081
pH = 1,09
13. Ht = 10 PH
Ht = 190711
6 mole
PROBLEM SET 127 Ht = 7.76 x 107 —-
liter
1; AAPB 2 ACFE Given
14. 91 = 10*+2
AB
= CB CPCIC 10!-96 = 10*%* 2
[9Gr= x 2
2." KABAPLSTACAP HL -0.04 Xx
AB
= CB CPGTE e720.49
= eo 20-49 — 13.16
15. els. 16
Thus, A is equidistant from B and C.
ie 22.43 X05
Algebra 2, Third Edition 365
Practice Set 128
16. (a) |x + 1| < 3; D = {Reals} 5 \13f,4 5)
18. =—-+—3—-—
* 5 33 7
-2435-2)
Ways as. 28
atic
=-+e__
7 140
(b) (p + 3)(p
- 3) = Ap + 3) 100, 213 _ 313
140-140 140
pr’ O'S 2p? -utap > 218
pill p07 =O 19. [6[sTa] = 840
(p + 9)(p + 3) S$0
I
p = -3 cannot be a solution, because division 20. Atriangle = moe
by zero is not defined.
2
(NEG)(Pos) < 0 es (5)? ~ (= = 5V3 cm
\ 2 2
p < -9 and p > -3
I 53
553cm = 253 2
Atriangle = sh cm
(Pos)(NEG) < 0
p = -9 and p < -3 Aghaded = Acircle — Atriangle
There are no real numbers that satisfy the first m(5cm)? — 25/3 cm
2
conjunction, so the solution must be
Wey
-9 <p < -3.
[25x= 24) cm?
=10-0 6-7 <6 —5 4 <9 =2
PRACTICE SET 128
We Sy Se
yey + ay + Bed = 1
L n (vata
prereset
se 5
From this we see:
(a) Opens downward
(b) Axis of symmetry is x = —l
(c) y-coordinate of vertex is 5
366 Algebra 2, Third Edition
Problem Set 128
PROBLEM SET 128 8. ABXP = AAXP = ACYP = ADYP HL
1. ACBD = AABD SAS ABPA = ACPD SAS
ZADB = ZCDB CPCTC al = CD CPCTC
, §- B= 6
2 1 ABTS BC 1. Given
a BD ale CA 2. Given
3. ZCBD = ZABD 3. Perpendicular angles
are equal (2). D
4. BD = BD 4. Reflexive property 9, AAOP = ABOP HL
5. ACBD = AABD __ 5. SAS (1,3, 4) AP = BP CPCTC
6. ZADB = ZCDB_ 6. CPCTC A
5
3. AXDA = AYCB SAS are P
ZAXD = ZBYC_CPCTC a
Therefore, by definition, AOXY is isosceles.
=
B
4. 1. ABCD is arectangle. 1. Given 10. AADB = ABCA SSS
2 XD CY 2. Given ZADB = ZBCA_ CPCTC
3. ZXDA = ZYCB 3. From1 A B
4. AD = BC 4. From 1
5. AXDA = AYCB 5. SAS; (2, 3,4)
6. AOXYis isosceles. 6. From 5 D C
5. AKNL = AMNL SAS li. = Ue RS
2x +. 6 = xo 8
6. ASOT = APOT SAS x= 2
ST = PT CPCTC Props = 2(8) + (2x + 6) + ( + 8)
Ppors = 16 + 3x + 14
Therefore, by definition, ASTP is isosceles.
Prors = ope cS bf
7. 1. NPRSisaparallelogram. 1. Given rors 3/22) 6 20 = 36
750 = OF 2. Given 12. AC = DB
. poe xt? = 0.2%
3, ZSOT?= ZPOT . Given a]? ==) Re
4°70. TO 4. Reflexive 15 =x
property DB = 0.2(15) = 3
5. SAS (2, 3,4
YR
. =
Se APOT aa 13. x + 60 = 90
6. ST = PT 6. CPCTC x = 30
7. ASTP is isosceles. 7. From6 Porrc = 430 + 1) = 124
367
Algebra 2, Third Edition
Practice Set 129
14. V = ar-h 23. 4x SF = 10
250m. cm? = mr?(1000 cm) SF = :
250iem" »
1000 cm (2) =x +7
y
0.5cm =r
21
> —— (ex
15. Aus Ur wD
An? cm? = mr? (3) 45
2/x em =r ae Dilan a2
16. Since this is a 45°-45°-90° triangle, 24. 4x SF = 6
(2 x SF = 10 VF = ;
SF = 5/2
A = 1(5V2) = 5V2 Mx>=7
B = 1(5V2) = 5V2 14
M= =
17. Since this is a 30°-60°-90° triangle,
Pas 3 9
Nes 3x >
—-= —5
il 3x a
i a 5
SHie=m)
= 2(5) = 10 25. ABC = 52 (2m(6 cm))
D = V3(5) = 5V3 ABC = 107cm
1
18. A = —bh
2 PRACTICE SET 129
= \(4n)? — (2m) = 2V3zin. STEM LEAF
a. 4 899) 7
A = 1(4min.)\(2V3xin.) = 4V3x2 in?
2 5 2)
6 3, 8; 2,.8, 0 do 3393.0
40 + 80
19. xX = ) = 60 r 0, 2, 4, 4, 0, ie 4, Vs eit > 4, Bs 9, 7,4
8 ©; 67'6,.75 Oy Ds ky te OOy Sree
20. = 40 al 9 3h OL Oe)
Range: 99 -— 47 = 52
2A. 1k =a) b. 2,.304¥505-8. 16,15
20
eee ise ee ed he ae)
8
Mean = 6.5
22.. 60) 4,6) t= 15 7
Gy Aaa Medians 2° as
2
6y=)43
. 1B Mode = 5
’ 2 15 - e
Ra=ng 2 = 13
368 Algebra 2, Third Edition
Problem Set 129
Cc. Mesq st ee at I ea) ee,
6 ‘
DISTANCE SQUARED DISTANCE << Il
FROM MEAN FROM MEAN
Dh by = ole Ss 9
1 he e/a tl 1
ve CS eS al i
14: 14-8 =6 36
8: 8-8 =0 0
Be ay Shi eas) Dy
56
Standard deviation = Ie(56) = 3.06
PROBLEM SET 129
Ag Loe
2 A es cm)? = ou cm?
360 9
Since this is a 45°-45°-90° triangle,
J2 x SF = V6
SF = V3
= I i(v3) = V3
NS
iH 1(/3) = V3
8 = 2x 10.
AEX,
2x SF = 7
p = (4+ 1) + (4 + 2) + B® —- 6] = 17cm
5. NO+ 20 _ ¢
4
4x = 5(3)
x=
15
—
4
l 1 ue
11. Atrapezoid = Pie 7 5 ONS) = 9 units
a 10 0
a
Acie = ar = 9 units”
4 SF = 11 r=
1 :
units
ripe
ead
12. See Lesson 123.
eens ake
4 13. See Lesson 123.
36
x= — 14. See Lesson 123.
i|
Algebra 2, Third Edition 369
Problem Set 129
15. AABX = ACBX SAS 22. 1Sealoe 4
1o!8s = 10**4
Aree CX CPCTC
188 =x +4
Bx & bisector AC —2.12 =x
Points (B and X) equidistant from the ends of a line
segment (AC) lie on the perpendicular bisector of ea -2h.27- 13.34
wae
the segment.
es e776 = 4.64 X 107!”
B
24. x + 3 2 2; D = {Reals}
x2-l
2-101 2
A Xx C y=xt—- 2-2
y = (x? eee hd) x24.
16. LSA
= 2zrh
y= (x- 1-3
6n2 cm? = 2n(3 cm)h From this we see:
mem =h (a) Opens upward
(b) Axis of symmetry is x = 1.
a7. mr? = 25n7 cm
r2 = 25a cm? (c) y-coordinate of vertex is —3.
ip = 5x cm
d = 2r = 10/xcm
18. PQ
= RS
Dyecke
Bis (oye ==
Gy=s4x%
3
— = xX
2D
PeaeePORS == 20 (2 5 4 el ty hilelela
Ppors
= 28
19. 2x + 40 = 90
ye OS
PaBcp = 4(25 = i) = 40
20. A, = Fe"
> a |= 1400e9°°) = $2195.64
21. A; = Ape?
400 = 50e3*
0.69 = k
Ajy = 50e°°2) ~ 197,209 bugs
370 Algebra 2, Third Edition
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