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1
1. statement
2. statement
3. not a statement
4. b
5. a.true
b.
c. yes
6. existential
7. universal
8. neither
9. for all, ;; there exists, '
10.holds for all real numbers aand b. Because and
are real, the relationholds when and
. Hence
.
11. Sample:
12. a. squares; x is a rectangle.b. squares; rectanglesc. square; rectangled. x is a square; x is arectangle.
13. a. an even integer; x is prime.b. even integer; primec. even integer; prime
14. ' a real number x such that; real numbers y, .y • x 5 y
x 5 1
60 1 12 5 14775 1(Ï12)2 5Ï12 1
2Ï75 • (Ï75)2 1Ï12)2 5(Ï75 1b 5 Ï12
a 5 Ï75Ï12
Ï75
(a 1 b)2 5 a2 1 2ab 1 b2
(a 1 Ïa 1 1)(Ïa)3 2 1 5 (Ïa 2 1) •
(42 1 4 1 1);p(4) 5 43 2 1 5 (4 2 1) •
15. a. yesb. noc. for 0, 1 and all integersgreater than or equal to 5
16. Sample: All students in mymath class are teenagers.
17. Sample: There exists astudent in my math classwho owns a car.
27. a. The sentence is false.b. The sentence is true.c. No, it is neither true nor false.d. Bertrand Russell (1872–1970) was a mathematicianof many talents and wrotebooks on mathematics,philosophy, logic, sociology,and education.
The paradox is: Let B be theset of all sets that are notmembers of themselves. If Xis any set, then
. Now if X is B, then
.Thus, there is acontradiction.
B [ B ↔ B [/ B
X [ B ↔ X [/ X
Answers for LESSON 1-1 pages 6–13 page 2c
Answers for LESSON 1-2 pages 14–20
1. False
2. False
3. False
4. There exists a person whocannot drive a car.
5. There exists a fractionwhich is not a rationalnumber.
6. ; real numbers x, sin x cos x.
7. p
8. c
9. a. people x; a person y; x loves y.b. ' a person x such that ;people y, x does not love y.
10. ;; not p(x, y)
11. ; functions f, ' realnumbers a and b such that
.f(a 1 b) Þ f(a) 1 f(b)
Þ
12. a. ' a real number x suchthat .b. the negation
13. a. There is a man who isnot mortal.b. the given statement
14. d
15. a. ' n in S such that .b. the negation; 11 is in Sand .
16. a. ; even integers m, m isnot in S.b. the negation; S containsno even integers.
17. a. ' a real number x suchthat ; real numbers y, tan .b. the negation. For ,
tan is undefined so ; real
numbers y, tan .π2 Þ y
π2
x 5π2
x Þ y
11 5 11
n $ 11
2x 1 4 # 0
c
3
18. b
19. The flaw occurs after thefourth line. Since ,one cannot divide bothsides of the equation by
.
20. b
21. e
22. a. True. Every student is onat least one sports team.b. True. No one is in theSpanish club.c. True. Every student is inthe math club.d. False. Each academicclub has at least onemember in the sample.e. False. Raul is not in aforeign language club.
x 2 y
x 2 y 5 0
23. Sample:
24. a. ; real numbers a and b,.
b.25. a. 4
b. -3c. 0
26. a. ; postal charges P $ 88¢, ' n and m,nonnegative integers such that .b. ;
; ;.
c. It is true. You can get 92¢with four 23¢ stamps. Byadding 5¢ stamps to 88, 89,90, 91, and 92 cents, youcan get all charges over 88¢.
2. x is greater than 3 and x isless than or equal to 4.
3. False
4. When p is true and q isfalse, or when p is false andq is true, or when both pand q are true.
5. a.
b. or (p and q)
6. False
7. True
8. True
p ; p
9. See below.
10. orand
and
11. andor
or
12. or
13.14.15. exclusive
16. c
17. a.
b. See below.
7 , x # 11
5 , x # 11
x # 7x . 5
x . 4,(x # 4) ; x # 3x # 4) ; ,(3 , x),(3 , x # 4) ; ,(3 , x
L Þ 12 ; L , 12,(L 5 12) ; L # 12L 5 12) ; ,(L . 12),(L $ 12) ; ,(L . 12
Answers for LESSON 1-3 pages 21–26
p q p xor qT T FT F TF T TF F F
9. p q p or q not (p or q) not p not q (not p) and (not q)T T T F F F FT F T F F T FF T T F T F FF F F T T T T
17. b.(p or q) and
p q p xor q p or q p and q not (p and q) (not (p and q))T T F T T F FT F T T F T TF T T T F T TF F F F F T F
same truth values
same truth values
p q p and q p or (p and q)
T T T TT F F TF T F FF F F F
c
5
18. They are not rectangular, or
are less than inches high
or are less than 5 inches long.
19. a. ; real numbers ,.
b. The statement is true.
20. c
21. sin
22. a. Falseb. True
23. y
x
y = x 21]2
(-2, 2) (2, 2)2
-2 42-4
4
6
-2
Ï32 • Ï2
2 112 • Ï2
2 5Ï6 1 Ï2
4
sin π3 • cos π
4 1 cos π3 • sin π
4 5
(7π12) 5 sin (π
3 1π4) 5
log10 x Þ 0x , 0
312
24. slope , y-intercept ,
x-intercept
25. a. Sample: The waiter givesyou a choice of coffee, tea,or milk. He then comesback to tell you that he hasrun out of all three.Therefore, you can’t havecoffee and you can’t havetea and you can’t havemilk.b. See below.
25. b.p q r q or r p or ,(p or ,p ,q ,r ,q and ,r ,p and
(q or r) (q or r)) (,q and ,r)
T T T T T F F F F F FT T F T T F F F T F FT F T T T F F T F F FT F F F T F F T T T FF T T T T F T F F F FF T F T T F T F T F FF F T T T F T T F F FF F F F F T T T T T T
b. See below.c. The output columns foreach network are identical.
2. (p or (not q)) and r
3. a. In 1999, it was 145 yearsago.b. In 1999, it was 62 yearsago.
p q p OR q NOT(p OR q)
1 1 1 01 0 1 00 1 1 00 0 0 1
4. not(not(p and q) or (not r))
5. ((not p) or q) and not((notq) and r)
6. See below.
7. a. 11¢b. 7¢c. the network in Question 6
Answers for LESSON 1-4 pages 27–34
1. b. outputp q NOT p NOT q ((NOT p) AND (NOT q))1 1 0 0 01 0 0 1 00 1 1 0 00 0 1 1 1
6. The network in Question 6 corresponds to the logicalexpression q or ((not p) and (not r)). Then the input/outputtable for the networks of Questions 5 and 6 is
(NOT (NOT q) NOT output for (NOT p) output forp q r NOT p) NOT AND r ((NOT q) Question NOT AND Question 6
x-intercepts: 2 and 5y-intercept: 10the axis of symmetry:
vertex: (3.5, -2.25)
16. Sample: While working atBell Laboratories (1941–1957), he developed amathematical theory ofcommunication known as“information theory.”
17. Sample: For circuits inseries, consider a string oflights. If one light fails,none of the lights willwork. Each light must workfor the string of lights towork. This is like the ANDgate. For circuits in parallel,consider the lights in ahouse. A light in one roommay work whether or notany other lights in thehouse work. The house iscompletely dark only whenall the lights are off. This islike the OR gate.
8. a. If , then the graphof is not anoblique line.b. True
9. a. If a quadrilateral doesnot have two angles ofequal measure, then thequadrilateral does not havetwo sides of equal length.b. False
10. Converse: If it will raintomorrow, then it will raintoday.Inverse: If it does not raintoday, then it will not raintomorrow.
y 5 mx 1 bm 5 0
x 5 1x 5 2π
p and p q p ⇒ q ,(p ⇒ q) ,q (,q)
T T T F F FT F F T T TF T T F F FF F T F T F
2x2 1 3x3 . 1
x . 111. If two supplementary
angles are congruent, thenthey are right angles.If two supplementaryangles are right angles,then they are congruent.
12.13. a. False
b. False
14. a. yesb. noc. yesd. no
15. If one has been convicted ofa felony, then that person isnot allowed to vote.
16. If one can, then one does.
17. a. If Jon wasn’t at the sceneof the crime, then Jondidn’t commit the crime.b. If one has a true alibi,then one is innocent.
18.
19. If a satellite can stay inorbit, then it is at a heightof at least 200 miles abovethe earth.
20. If an integer is of the form2k for some integer k, thenit is even.
not equivalent
p q p ⇒ q q ⇒ pT T T TT F F TF T T FF F T T
log232 5 5
Answers for LESSON 1-5 pages 35–43
same truth values
c
9
21. If one is elected to thehonor society, then one’sGPA is at least 3.5.
22. a. 1000 3b. -50 THE LOG ISUNDEFINEDc. 0.1 -1
23. See below.
24. a. 1b. 1c. 0
25. a. andb. .c.
26. a. Peggy Sue, Buddy Hollyb. If I Loved You, Carousel(Rodgers and Hammerstein)c. If I Had a Hammer, Peter,Paul and Maryd. Man in the Mirror,Michael Jackson
1. a. If an integer is divisibleby 3, then its square isdivisible by 9. 10 is divisibleby 3.b. is divisible by 9.c. ; integers n, if p(n), thenq(n); p(c), for a particular c; ∴ q(c).d. noe. yes
2. True 3. c
4. Invalid: Sample: Anyindividual under 21 years ofage with a driver’s licenseprovides a counterexample.
5. Valid; Law of Transitivity
6. Valid; Law of Detachment
7. Laws of Indirect Reasoningand Transitivity
8. See below.
102
9. a. p ⇒ q,q∴ ,p
b. See below.
10. a. Mary is not at home.b. Let p and q be thestatements:p: Mary is at home.q: Mary answers the phone.Then the argument has theformIf p, then q.not q.∴ not p.c. Law of IndirectReasoning
11. a. If p, then q.If q, then r.∴ If p, then r.
b. Yes, it follows from the Lawof Transitivity.
Answers for LESSON 1-6 pages 44–52
p q p ⇒ q ,q ((p ⇒ q) and ,q) ,p ((p ⇒ q) and (,q) ⇒ ,pT T T F F F TT F F T F F TF T T F F T TF F T T T T T
9. b.
p q r p ⇒ q q ⇒ r (p ⇒ q) and (q ⇒ r) p ⇒ r ((p ⇒ q) and (q ⇒ r)) ⇒ (p ⇒ r)
T T T T T T T TT T F T F F F TT F T F T F T TT F F F T F F TF T T T T T T TF T F T F F T TF F T T T T T TF F F T T T T T
8.
c
11
12. a. Law of IndirectReasoningb. yes
13. The diagonals of ABCDbisect each other.
14. Sample: If an acrobatic featinvolves a quadruplesomersault, then it is notattempted by the circusacrobats.
15. -3 and -1 are not positivereal numbers, so theuniversal statement doesnot apply.
16. a. yesb. noc. yesd. yese. nof. yes
17. c
18. a
19. yes
20. a. ' a real number y suchthat .b. the statement
b. If an animal is amammal, then it is a whale.Falsec. If an animal is not awhale, then it is not amammal. False
5. a. improper inductionb. invalid
6. a. inverse errorb. invalid
7. a. Law of IndirectReasoningb. valid
8. a. converse errorb. invalid
9. a. improper inductionb. invalid
10. a. q ⇒ pb. ,p ⇒ ,qc. inverse
11. a. p: Peter is not at home.q: The answering machineis on.p ⇒ qq∴ pb. invalid, converse error
Mammals
Whales
12. a. p: q: p ⇒ q,p∴ ,qb. invalid; inverse error
13. a. p(x): x is President of theUnited States.q(x): x is at least 35 yearsold.Let c be Queen Elizabeth.; x, p(x) ⇒ q(x)q(c)
∴ p(c)b. Yes; Noc. invalid; converse error
14. p: the land is covered with ice.q: the land is Antarctica.r: there are researchstations there.s: scientific study is beingconducted.p ⇒ qq ⇒ rr ⇒ ss∴ pInvalid; converse error
x2 5 9x 5 3
Answers for LESSON 1-7 pages 53–60
c
13
15. p(x): x is a real number.
r(x): x is a pure imaginarynumber.Let ; x, p(x) ⇒ q(x); x, r(x) ⇒ ,q(x)r(c)∴ ,p(c)Valid
16. p: You send a minimumorder of $10 to a mail order house.q: You can return asweepstakes couponenclosed with the catalog. r: Your order is one of thefirst 1000 orders.s: You have a chance ofwinning $100.p ⇒ q(q and r) ⇒ ss∴ pinvalid, converse error
c 5 2i
q(x): x2 $ 017. a. See below.
b. inverse error
18. a. yes; yesb. yes; noc. Arguments I and II bothhave the form belowp ⇒ qp ⇒ r∴ q ⇒ rd. invalid
19. a. Let p: Devin is a boy.Let q: Devin plays baseball.Let r: Devin is a pitcher.p ⇒ qq ⇒ r,r∴ ,pb. The argument correctlyuses the Law of IndirectReasoning and the Law ofTransitivity.
20. p ⇒ q, q ⇒ r, hence: p ⇒ r.r ⇒ q, q ⇒ p, hence: r ⇒ p.∴ p ⇔ r
10. About 150 feet from thebatter; it is the mean of
and .
11. converse error
12. improper induction
13. invalid; inverse error
14. a. Sample: If you haveoutstanding school grades,then you will receivefinancial aid.b. Law of Transitivity
15. False. x could be -3.
16. Inverse: If the temperatureinside a refrigerator is notabove 40˚F, then thecooling system is notactivated.Converse: If a refrigerator’scooling system is activated,then the temperatureinside is above 40˚F.Contrapositive: If arefrigerator’s coolingsystem is not activated,then the temperatureinside is not above 40˚F.
x2 5 300x1 5 0
17. a. HELLOb. IF (( ) or ( or
)) THEN PRINT“HELLO” ELSE PRINT“GOODBYE”c. ( and )
18. a. 0 b. 0 c. 0
19. See below.
20. ' real numbers x and y suchthat .
21. ; real numbers x and y,.
22. a. ' a real number value ofu such that cos u 0.b. existential
23. a. -84b.c.
24. Student B is correct,because all steps he usedare valid. Both students Aand C are wrong. Student Amultiplied both sides by
, which is invalid if
. Student C made amistake when he equated
to .0 • t3 2 3t
t 5 1
11 2 t
-4t2 2 11t 2 6-4h2 2 19h 2 21
5
xy Þ 0
xy Þ 2
B . 4A 5 7
B , 4B 5 4A Þ 7
Answers for LESSON 1-8 pages 61–66 page 2
Converse Inversep q q p ,p ,q ,p → ,qT T T F F TT F T F T TF T F T F FF F T T T T
⇒19.
same truth values
c
17
1. universal
2. neither
3. existential
4. a. inverseb. contrapositivec. converse
5. ; countries c, c has notlanded people on Mars.
6. ; intelligence memos m, 'a government official g,such that g reads m.
7. ; composite numbers n, ' apositive integer y, such that
, , and y is afactor of n.
8. If you never practice yourpiano lessons, you will notlearn to play piano.
9. b, c
10. False
11. If one passes a state’s barexam, then one can practicelaw in the state. If one canpractice law in a state, thenone has passed the state’sbar exam.
12. a.b. ( or ) and ( or )
13. If , then log .
14. a. Sue is not wearing a bluesweater.b. Sue is wearing a bluesweater and she doesn’thave brown eyes.
x . 0x . 1
x , 3x 5 3x . -3x 5 -3
-3 # x # 3
y Þ 1y Þ n
15. Some British bobby carries a gun.
16. Some President is notguarded by any SecretService agent.
17. A person wants to travelfrom the U.S. to Europe anddoesn’t travel by plane orby ship.
18. and isobtuse.
19. Excessive bail shall berequired, or excessive finesshall be imposed, or crueland unusual punishmentsshall be inflicted.
20. True
21. p and (not q)
22. True
23. False
24. False; Counterexample:rhombus. The diagonals ofa rhombus bisect eachother, but a rhombus is notalways a rectangle.
25. True
26. a. ' a real number x suchthat .b. the statement
1. The range of a function isthe set of all possible valuesof the dependent variableof the function.Example: The range of thefunction is the setof all real numbers.
2. A real function is a functionwhose domain and rangeare sets of real numbers.Example: is areal function.
3. A real-valued function is afunction whose range is aset of real numbers.Example:
1, if x is a male0,if x is a female
4. a. the correspondencebetween the number ofhours spent driving and thenumber of miles traveledb. the number of hoursspent drivingc. the set of times between0 and 8.5 hoursd. The dependent variableis the number of milestravelede. the set of distancesbetween 0 and 520 miles
10. a. The alphabet can be putinto 1-1 correspondencewith the set of positiveintegers from 1 to 26, which is a subset of the setof integers.b. Any finite set can be putinto 1-1 correspondencewith a subset of the set ofintegers.c. Because a finite set is adiscrete set as explained inpart b.
11. a. 3b. {0, 1, 3, 5, 7, 9, 11, 13, 15}c. Its domain is a finite setwhich is discrete.
12. (-`, `)
13. [-2, 2]
14. (-∞, -3), (3, ∞)
15. (-∞, -2), (-2, 1), (1, ∞)
-2π # x # 2π, x-scale = -2 # y # 2, y-scale = 1
π]2
r 53
Ï3V4π
Answers for LESSON 2-1 pages 78–84
c
21
16. a. $8.25b. The exact length of timecan not be determined, butis at least 1 hour and lessthan 2 hours.c. $11.25d. $1.50e. No. There are elements cin C for which f(c) has morethan one value.f. Yes. It assigns to eachelement of T exactly oneelement of C.g. i. t
ii. ciii. {t: hours}
17. 8a 1 23
0 , t # 24
18.
19. a.
b. The height will be of
its original value.
20.
21. F9(2, -1), I9(1, 0), R9(1, 1),E9(2, 0)
22. Answers may vary. Sample:; ; y 5 z-10x zy 5 z10x zy 5 x2
b. Because the minimumvalue of f is approximately82.9, which exceeds 82.5.
5. a. 58.2 in3
b. 83.1 in2
c. The surface area of a canwith these dimensions isonly slightly greater thanthe minimum.
6. a.
b. r 5.4 cm; h 10.9 cmøø
S(r) 5 2πr2 12000
r
øø
M $ g(z)
(-`, 54G
F7116, )̀(38, 71
16)
-5 # x # 5, x-scale = 1 2 # y # 10, y-scale = 1
7.
Estimated range: [-0.1, 4]
8.
Estimated range: [-1.1, 1.2]
9. a.
b.
c. P(130) 567.7P(140) 565.7P(150) 566.7
d.
(141.4, 565.7)e. 565.7 m
10. radius 2.6 in.height 2.7 in.
11. a. minimum: -3; maximum: 3b. domain: [-5, 6]
range: [-3, 3]c. and d. [-5, -4) and (-1, 4)
x 5 3x 5 0
øø
0 # x # 250, x-scale = 500 # y # 4000, y-scale = 1000
øøø
P 540000
w 1 2w
, 520000
w
-1 # x # 4, x-scale = 1-2 # y # 2, y-scale = 1
0 # x # 2, x-scale = -2 # y # 5, y-scale = 1
1]2
Answers for LESSON 2-2 pages 85–90
c
23
12. (-`, ]
13. a. Yes, because eachelement in R corresponds toexactly one element in P.b. No, because eachelement in P corresponds to more than one elementin R.c. i. 8
ii. not possibleiii. 15
14. a.b.c. (1, `)
15. a. Switch 1 Switch 2 Light
1 1 11 0 00 1 00 0 1
b. p q p ⇔ qT T TT F FF T FF F T
A “1” in the first or secondcolumn means that theindicated switch is up, anda “0” means that it is down.
If 1 corresponds to T and 0to F, it is apparent that thetwo truth tables areequivalent. Hence, thestairway light situation is aphysical representation of p ⇔ q.
If as , , thenas , or -`depending on the functiong. If ,
, and
, for any real
number L, then
does not exist. Similarrelationships occur when
.x → -`
limx→`
f(x)
limx→`
g(x) Þ L
limx→`
g(x) Þ -`
limx→`
g(x) Þ `
f(x) → `x → `g(x) → 0x → `
f(x) → 1Lx → `
g(x) → L Þ 0x → `f(x) → 0x → `
g(x) → `x → `
n $ 5
81-14
x y
76
Answers for LESSON 2-4 pages 97–101 page 2c
14. a. x 100 1000 10,000 100,000 1,000,000
2.70481 2.71692 2.71815 2.71827 2.71828(1 11x)x
27
1. a. 12.08 ftb. very close (differs by 0.06 ft from maximumheight shown in the data)
2. a. 12.11 ftb. very closely (differs by0.03 ft from maximumheight shown in the data)
3. a.
b. (1.50, 7.0)c. (16.50, 4.0)d. T is the time when theball’s motion is altered bystriking the backboard,basket, or floor.
4. a. The horizontal velocity is
b.
c.
d. 1.58 sec
5. No. It takes about 0.995 sec for the ball to travel 12.94 fthorizontally to a point 14 ft from the free throw line. Atthat time it will be 9.5 fthigh and too low to gothrough the hoop.
6. a. b.
-2 # t # 2, t-step = 0.5-10 # x # 5, x-scale = 1-10 # y # 5, y-scale = 1
HI
G
F
t ø
-4.9 msec2
6.5 msec
5.9 msec
13 ftsec
7. a.
b. yesc. Let , then ;so y-intercept is
.Similarly, let , then
or 3; so x-interceptsare , and
.
8. a.
b.
c. They are symmetricabout the line .
9. From , ;
therefore,
. This meansthat the parametricequations are equivalent tothe equation .y 5 2x 1 19
5 5 2x 1 19
y 5 6 • x 1 73 1
t 5x 1 7
3x 5 3t 2 7
y 5 x
-2 # t # 10, t-step = 1-10 # x # 10, x-scale = 1 -2 # y # 10, y-scale = 1
-2 # t # 10, t-step = 1 -2 # x # 10, x-scale = 1-10 # y # 10, y-scale = 1
x(3) 5 3 2 1 5 2x(0) 5 0 2 1 5 -1
t 5 0y(t) 5 0
12 2 3 • 1 5 -2y(1) 5
t 5 1x(t) 5 0
0 # t # 5, t-step = 1-2 # x # 5, x-scale = 1-3 # y # 10, y-scale = 1
Because (cos u, sin u) is theimage of (1, 0) under acounterclockwise rotationof u about the origin and[cos (-u), sin (-u)] can beconsidered as the image of(1, 0) under a clockwiserotation of u about theorigin.cos (-u) cos usin (-u) -sin u
16. a.
b. -125
-1213
55
y
xu-u
(x, y) = (cos u, sin u)
(x, -y) = (cos u, -sin u)
(1, 0)
x 5π2
(-π2, π2)
(-3π4 , 1)(-π
4, -1)(3π4 , -1)
(π4, 1)
-7 # x # 7, x-scale = 1-5 # y # 5, y-scale = 1
17. a. Sample: ;
b.
c. maximum: 23; minimum: -7
18. a.
b. From ,
we have
therefore
. Namely,
c. 2 is half the length ofthe minor axis; 3 is half thelength of the major axis;and (1, -4) is the center ofthe ellipse.
19. d
20. invalid; inverse error
21. 22.23. Cannot be simplified.
24. Answers may vary.
qp(x 1 y)2
(x 2 1)2
4 1(y 1 4)2
9 5 1
sin2 t 1 cos2 t 5 1
(y 1 43 )2
5(x 2 12 )21
Hx(t) 2 12 5 sin t
y(t) 1 43 5 cos t
Hx(t) 5 2 sin t 1 1y(t) 5 3 cos t 2 4
0 # t # 7, t-step = 0.1-2 # x # 4, x-scale = 1-7 # y # 1, y-scale = 1
π2
-7 # y # 230 # x # π
Answers for LESSON 2-6 pages 109–116 page 2c
31
1. b
2. a. the set of all realnumbersb. the set of positive realnumbersc. ;
3. where x is apositive integer
4. Sample: is given.Thus
by AdditionProperty of Inequality.
by property (3)on page 118.
by basic laws
of exponents.
So by substitution.
Since by property(1) on page 118,multiplication by yields
.Since , .
5. $46,966.66
6. a. Falseb. Sample: It was proved inQuestion 4 that isdecreasing on the set (-`, `), if and
, has this form,
where and .b 534a 5 1
(34)x
a . 0
0 , b , 1
abx
abx1abx2 ,a . 0
bx2 , bx1
bx1
bx1 . 0
bx2
bx1 , 1
bx22x1 5bx2
bx1
bx22x1 , 1
x2 2 x1 . 0
x1 , x2
f(x) 5 3x
f(x) 5 0limx→-`
limx→`
f(x) 5 `
7. a.
b. The graphs are reflectionimages of each other overthe y-axis.
8. Domain: set of realnumbersRange: set of positive realnumbersMaxima and minima: noneIncreasing or decreasing:decreasing over its entiredomainEnd behavior:
Model: decaySpecial properties: Values ofthe function are related bythe laws of exponents.
9. a.b. 1990.6 millionc. i. 23.29 million
ii. 102.58 millioniii. 451.89 millionThe formula is a
reasonable accuratepredictor of the populationfor the period from 1790 to1850.
10. a.b. The values increase lessrapidly then the data up toabout 1860, but grow muchmore rapidly after about1890. The values producedby the continuous modelalways exceed thoseproduced by the discretemodel.
11. < 32% of the originalamount
12. a. increasing (-`, 0],decreasing [0, `)
b. relative maximum:
c.
d.
13. a.
b.
c.
d. -1
14. a. the set of real numbersb.
c.
d. no limit
15. 1
π2, 3π
2
0 # y # 1
Ï22
-Ï32
12
0 , y #1
Ï2π
limx→6`
f(x) 5 0
1Ï2π
Pn 5 3.93(1.029655)n 16.17. 12, 16, 20, 24, 28
18. 4, 3.2, 2.56, 2.048, 1.6384
19. Sample:is not necessarily
exponential (if , where and ,
then , which isnot possible for anexponential function).
is not necessarilyexponential (if
, where and.5, then ,
which is not possible for an exponential function).
is not necessarilyexponential
where .
is always exponential,since there is no function
with such
that .(bc)x
5 1
b Þ cf4(x) 5bx
cx
f4(x)
c 51b → f3(x) 5 1 ;x)
(f3(x) 5 bx • cxf3(x)
f2(-x) 5 -f2(x)c 5b 5 2bx 2 cxf2(x) 5
f2(x)
f1(-x) 5 f(x)c 5 .5b 5 2cx
f1(x) 5 bx 1f1(x)
x # -2.29, x $ 0.29
Answers for LESSON 2-7 pages 117–123 page 2c
33
1. the set of all integersgreater than or equal to afixed integer a, if thesequence is infinite, or theset of all integers greaterthan or equal to a but notgreater than another fixedinteger b, where , ifthe sequence is finite
2. , , ,
3. a. neither
b. 2, , , ,
4. a. geometricb. 3, 9, 27, 81, 243
5. a. geometricb. 3, -6, 12, -24, 48
6. a. geometricb. 6, 24, 96, 384, 1536
7. ; integers
8. ; integers
9. a. ; integers,
b.10. a. recursive
b.
c. AA
P4 P2P3 P1B
Hn 5 10(0.8)n21H1 5 10n $ 2
Hn11 5 0.8(Hn)
n $ 1Sn 5 3 2 2(n 2 1)
n $ 1Sn 5 7(12)n21
65
54
43
32
f5 5 18f4 5 11,f3 5 7f2 5 4f1 5 3
b $ a
11. difference equation: ; integers
explicit formula: ; integers
12. a.
b.
13. a. 0, 0, 0, 0, 0; lim 0
b. 1, 0, , 0, ; lim 0
14. ;
15. An ellipse with center (5, -4), minor axis withvertices (3, -4) and (7, -4),and major axis with vertices(5, -1) and (5, -7).
16. a. Trueb. Falsec. Trued. False
17. a. 4b. 2
18. Sample: The resultsdecrease to about 30,000and then level off. Thisincreases faith in themodel, because thepopulation decreases whenthe lake is overstocked.
4. a. 0, 1, 0, -1, 0, 1, 0, -1, 0b.c. or d. 1, 5, 9e. 3, 7, 11
5. a. arithmeticb. decreasing
6. a. neitherb. neither
7. a. neitherb. decreasing
8. relative minimum
9.10.11.
12.
13.
14.
15. 2 log N 3 log M log P21
log 12log 7 5 1.277x 5 log712 5
3-2 52z5 ; z 5
518
t 5 log642 5log 42log 6 ø 2.086
b2 5 9; b 5 3
28 5 x; x 5 256
2x 5 8; x 5 3
5 # n # 71 # n # 33 # n # 5
73
F23, `G23G(-`
16.
17. No, each element in Scorresponds to more thanone element in R.
18. Yes, each element in Rcorresponds to exactly oneelement in S. Yes, it isdiscrete.domain: {x: x an integer,
}range: {18, 56.25, 45}minimum: 18maximum: 56.25
19. a. Yes, it is a function. It isnot discrete since L can takeany real values from 0 to320.b. L, length of skid marksc. speed of the car, sd. {L: }e. {s: }
relative minima at (-3, 1.5),(4, -3), , ,and , c. odd
49. a. {-6, -3, -1, 2, 4}b. in the intervals (-6, -3), (-1, 2), (4, `)c. in the intervals (`, -6), (-3, -1), (2, 4)d. , , e. x in the intervals (-`, -7),(-2.5, -1.5)
50. {y: }
51. a.b. {0, 1, 2, 3, 4, 5}c. Falsed. True
52. Sample:
53. a. decreasing over: (-`, 2.1]increasing over: [2.1, `)
b. relative minimum: c. ,
d. { }e. neither
y: y $ -4
limx→`
f(x) 5 `limx→-`
f(x) 5 `y 5 -4
-5 # x # 5, x-scale = 1-10 # y # 10, y-scale = 1
{x: -6 , x , 6}
0 # y # 4
x 5 -7x 5 -2.5x 5 -1.5
x 5
y 5 -21 # x # 2y 5 2-2 # x # -1
y 5 2-2 # x # -1y 5 -2
-1 # x # 254. a. increasing on the reals;
decreasing nowhereb. none
c. ,
d.e. odd
55. a. i. Domain: the set of realnumbersii. range: iii. increasing over: [0, `)
decreasing over: (-`, 0]iv. no maximum value
minimum 0
v.
vi. models: optics, acoustics(subject to restrictions ondomain)vii. properties: evenb. i. Domain: the set of realnumbersii. range: iii. increasing over: (-`, 0]decreasing over: [0, `)iv. maximum 0no minimum value
v.
vi. model: projectile motion(subject to restrictions ondomain)vii. properties: even, areflection image of
18. a. ' some satellite s suchthat s is not a military spysatellite.b. ; persons p, p was not aleader of a trade union or phas not received the NobelPeace Prize.
x . 0g(f(x)) 5 x
f(g(x)) 5 x
53 1 01 2 0 5 3
3 11n
1 21n2
limn→`
3 11n
1 21n2
5
3n2
n2 1nn2
n2
n2 21n2
5
1n2
1n2
•(u 1 v)n 53n2 1 nn2
2 1
5 10 15 20 25
1
2
3
4
5y
x
(u 1 v)n → 3n → ` 19. a. 22b. 485
20.
21.
22.23. Sample: For ,
; so . As xincreases in the interval {x: }, I(x) increasesfaster than R(x) decreases.As x decreases in theinterval {x: }, R(x)increases faster than I(x)decreases.
2 4 6 8 10
2
4
6
8
10
y
x
y = 1]x y = x
x # 1
x $ 1
I(x) 1 R(x) 5 2I(x) 5 R(x) 5 1
x 5 1
k 5 20
c 594
-3 -2 -1 1 2 3
1
2
3
4
y
x
Answers for LESSON 3-1 pages 148–153 page 2c
41
1. a. b.
c.
d. 7320.5 e. No
2. Sample: ;
;
3. ; domain: all reals
4. , domain:
{x: , 0, 1}
5.
6. ,
,
therefore .
7. a. all realsb.
8. 9.10. 11. x 5 Ï2y ø 8.854
x 5 6Ï7x 5 65,536
1 2 3
1
2
3y
x
f (x)
g(x)
$ 0
f ° g(x) 5 g ° f 5 I
g ° f(x) 5 g(f(x)) 5 (kxk ) 5 x
f(g(x)) 5 k(xk) 5 xf ° g(x) 5
2 4
2
4
6
8
10y
x
f (x)
g(x)
x Þ -1
1
x2 11x2 2 2
6x3 1 1
(cos π2)3 5 0
(cos x)3 5g ° f(x) 5
≈ -.7424cos (x3) 5 cos π3
8
f ° g(x) 5x 5π2
12(-2x 1 3)4
-2398-x4 1 3 12. h is not a 1-1 function.
13.
k is a 1-1 function.
14. :
15. x3 16. 3x
17. a. Yes, 5x is 1-1 for all realnumbers. Therefore itsinverse, log5x is 1-1 over thepositive real numbers.b. 1
5. a. -1 and 0 or 0 and 1b. 2c. -2 and -1 or 2 and 3
6. ; real numbers y0 betweenf(a) and f(b), ' at least onereal number x0 between aand b such that .
7. sin x and e-x are bothcontinuous over the realnumbers. Function additionpreserves continuity so sin is continuous over the real numbers. iscontinuous over the realnumbers and is never equalto zero. Since (nonzero)function division preserves
continuity, is
continuous over the realnumbers.
8. [-1, 0] or [1, 2] or [4, 5]
9. a. [1, 2] b. [1.5, 1.6]
sin x 1 e-x
x21 1
x2 1 1x 1 e-x
f(x0) 5 y0
10. a. 3b. values between [2.347,2.366] are acceptable
11. a. Yesb. No. g is discontinuous at
.
12. a. No
b. is not continuous
on [1,3] so the IntermediateValue Theorem does notapply
13. after minutes
14. a. is tangent to (C at Bso ABC is a right triangle.So
2. Let f be a real function thatis decreasing on its entiredomain. Let x and y be twovalues in the domain of f.Because f is decreasing, if
, then for allx and y in the domain of f.Similarly, if , then
for all x and y inthe domain of f. Thus,whenever , .By the Law of theContrapositive,
, that is, f is a 1-1function.
3. Let f be a real function thatis decreasing on its entiredomain. Assume that onsome portion of its domainf -1 is increasing. Then thereexist some values u and vsuch that implies f -1(u) f -1(v). Since f iseverywhere decreasing, it isdecreasing on [f -1(u), f -1(v)].Therefore we must have f(f -1(u)) f(f -1(v)) or,equivalently, . Thiscontradicts our assumptionthat . Therefore f -1
cannot be increasing on anyu , v
u . v.
,u , v
x 5 yf(x) 5 f(y) ⇒
f(x) Þ f(y)x Þ y
f(x) , f(y)x . y
f(x) . f(y)x , y
y . 12
-4y , -4860 2 4y , 12 portion of its domain.
Hence, f -1 must bedecreasing on its entiredomain.
4. b
5. 1.63
6.7. 0.794
8. 6.31
9. 13 pieces
10. 1,000,000, where x is an integer
11. or
12. 10,000 to 13,000 years old
13. -ln and ln . Thus, by the
Intermediate ValueTheorem, g must have azero between 2 and 3.
14. a. [ , ], [-1, 0], [0, 1]b. any interval of length0.25 that includes theinterval (0.5706, 0.5707)Sample:
5. The function ,where n is a positive integeris a 1-1 correspondencebetween the positiveintegers and the positiveodd integers. Therefore thepositive odd integers havethe same cardinality as thepositive integers, which is :0.
6. The union of the set ofpositive odd integers andthe set of positive evenintegers is the set of allpositive integers. All threeof these sets havecardinality :0. Therefore,
.
7. Suppose a hotel has acountably infinite numberof rooms, all full, and 100new guests wish to checkin. If the guests in Room 1move to Room 101, theguests in Room 2 to Room102, and so on, there willbe 100 vacant rooms for thenew guests. All of the newguests have beenaccommodated withoutdisplacing any of the oldguests. Therefore,
.:0 1 100 5 :0
:0 1 :0 5 :0
f(n) 5 2n 2 1
45, 54, 72, 81
17, 35, 53, 71, 18, 27
8. Sample: 0.211111
9. Label the segments and. Extend a line through B
and D and a line through Aand C. Since and have different lengths,these lines intersect at somepoint P. To establish a 1-1correspondence, pair eachpoint E on with the
intersection of and .
10.11. The distinct rays originating
at the center of thesemicircle and passingthrough each distinct pointof the semicircle eachintersect the real numberline at a distinct point.Therefore, the set of pointsof the semicircle are in a 1-1correspondence with thereal numbers, and, hence,have cardinality c.
(x, y) → (3x, 3y)
P
A E B
C F D
CDPE←→
AB
CDAB
CDAB
Answers for LESSON 3-10 pages 208–212
c
53
12. a. For any two values x1
and x2 in the interval ,
if , then tan tan x2.So tan x is one-to-one
on the interval . For
any real number y, arctan y lies in the
interval and tan x.
So the range of the function tan x with domain
is the entire real line.
Therefore, tan x is a 1-1 correspondence between
and the reals.
b. c
13. Time has cardinality c, and.
14. a.
b. 4c.
15.
16. or
17. a. 2.5b. -7.5c. 316.2
x . 1x , -14
zT 2 M z , 0.001
n . 50
an
n20 4 6 8 103
3.54
4.55
c 1 10,000 5 c
(-π2, π2)
f(x) 5
(-π2, π2)f(x) 5
y 5(-π2, π2)x 5
(-π2, π2)f(x) 5
x1 Þx1 Þ x2
(-π2, π2)18. a.
b.19. ,
20. 10
21. Sample:
22. The Cantor set is the subsetof the real interval [0, 1]consisting of all numbers of
the form , where ei is 0
or 2. Geometrically, it maybe described as follows: oneremoves from [0, 1] its
middle third interval ,
then the middle thirds of
and , and so on.
Among the properties ofthis set is that it is nowheredense in the real line, butdoes have cardinality c.
33. a. The second step, whereboth sides of the equationwere divided by x3, isincorrect. This step is validonly if 0, but 0 is asolution to this equation.b. { -1, 0, 2 }
34. a. No. h(2) h(-2) 16,but .b. No, it is not a reversibleoperation because it is nota 1-1 function.
35. a. ⇒b. ⇔c. ⇔d. ⇔e. ⇔f. ⇔
2 Þ -255
x 5
x 5x Þ
t . -1
x . 2Jx , -13H-8 6-6 2 40-4 -2 x
-7 , x , 5
z .13z , -1
-5 # w # 1
x . 12 , x , 0
x ø
x 565x 5
x 5 H0, π2, 3π2 , 2πJ
z 5
y 5
x 5log2
log3
x3 2 x2 2 6Ïx3
Answers for Chapter Review pages 218–221
c
55
36. True
37. ; . Since
, f and g areinverse functions.
38. log (10x 2 7);
.Since , h and m are inversefunctions.
39. By definition, a decreasingfunction f is such that forall , .Hence, for ,
. Therefore, f is1-1 and has an inverse.
40. a.b.c. No, g is not continuouson the interval [-3, -1]because it is undefined at -2. The Intermediate ValueTheorem does not apply.
41. a. No b. No c. Yes
42. h has a zero between a and b.
43. a.
b. -1 and 0, 0 and 1
-2 -1 1 2
-1
-0.5
0.5
1f (x)
x
g(-1) 5 1g(-3) 5 -1
f(x1) Þ f(x2)x1 Þ x2
f(x1) . (f(x2)x1 , x2
h °m 5 m °h 5 I10log x 5 x10log x1727 5
(m °h)(x) 5x 2 7 1 7 5 x1 7 5(h °m)(x) 5
f °g 5 g ° f 5 I(g ° f )(z) 5 (z3/5)5/3 5 z(f °g)(z) 5 (z5/3)3/5 5 z
c. The ship’s height will nolonger increase once it iscompletely unloaded.
51.52. a.
b.53. possible
54. Not possible. The functionshown is not 1-1 so it doesnot have an inverse.
55. a.
b.
2 4 6 8 10
-8-6-4-2
2468y
x
g
f
f • g
y
x2 4 6 8 10
2
4
6
8
g
f + g
y
x
0.56 , p , 0.60zp 2 0.58 z , 0.02
0 # t # 3.56
5 10 15 20
-0.4
-0.2
0.2
0.4
0.6s(t )
t
(0.5t) 2 0.4s(t) 5 0.05t 1 0.4 c. Because the range of
sin x is [-1, 1], sin x isbounded by the lines
and .Similarly, x sin x is boundedby the lines , .
56.
57.
58.
59. a.
b.
60.
61. -5, -9
62. d
63. Sample:
64. x ø 60.83
.975 # x # 1.025
x2
0.16 1(y 1 1)2
4 5 1
(x 2 2)2
9 1 9(y 1 5)2 5 1
x2
9 1 9y2 5 1
1 2 3 4 5 6
-6
-4
-2
2
4
6y
x
(g + k)(x)
1 2 3 4 5 6
-10-7.5
-5-2.5
2.55
7.510
y
x
(h – k)(x)
(h • f )(x)
x1 2 3 4 5 6 7
-0.4-0.2
0.20.40.60.8
1
y 5 -xy 5 x
y 5 x 2 1y 5 x 1 1
x 1
Answers for Chapter Review pages 218–221 page 3c
57
1. True; there exists aninteger, 12, such that
2. True; there exists aninteger, 1, such that
3. True; there exists aninteger, , such that
4. True; there exists apolynomial, , suchthat
5. (1) If d is a factor of n, thenthere is an integer q suchthat .(2) If there is an integer qsuch that , then dis a factor of n.Statement (1) is used tojustify the substitution of a m for b and b n for c.(2) is used to conclude thata is a factor of c.
6. a. 12 b. 10c. 10 d. 8
7. a. a factor b.c. d. e.f. m is a factor of
8. By substitution, b ca q a r which equalsa(q r) by the DistributiveProperty. Since q and r areintegers, (q r) is aninteger. Then, by thedefinition of factor, a is afactor of (b c).1
1
1•1•
51
n • pq • p(q • p)m • q
n 5 m • q
••
n 5 q • d
n 5 q • d
(n 2 6)(n 2 11).n2 2 17n 1 66 5
n 2 11
4n 5 (nm 1 3n) • 4.2n(2m 1 6) 5 (m 1 3) •
nm 1 3n
17 5 1 • 17.
132 5 12 • 11.
9. Sample: 96
10. a. a(x) is a factor of b(x)because
.a(x) is a factor of c(x)because
.b.
c. The Factor of aPolynomial Sum Theorem
11. a. Sample: , b. ' integers a and b suchthat a is divisible by b and bis divisible by a and .
12. Suppose that a(x), b(x), andc(x) are polynomials suchthat a(x) is a factor of b(x)and b(x) is a factor of c(x).By the definition of factor,there exist polynomials n(x)and m(x) such that
and . By substitution
because polynomialmultiplication is associative.Since polynomials areclosed under multiplication,
is a polynomial;so, by the definition offactor, a(x) is a factor ofc(x).
13. a. degree 6; degree 4. b. The degree of theproduct of two polynomialsis equal to the sum of thedegrees of the individual polynomials. The degree ofthe sum of two polynomialsis less than or equal to themaximum of the degrees ofthe individual polynomials.c. The conjecture does notneed to be modified.d. The degree of apolynomial is determined byits term of greatest degree.The product of a polynomialof degree n and apolynomial of degree m willhave a term of greatestdegree of the form
so theproduct has degree .If the individual polynomialsare both of degree n, theirsum will have a term ofgreatest degree of the form making the sum a degree npolynomial—unless in which case the degree willbe less than n. If theindividual polynomials arenot of the same degree, thelarger degree being n, thenthe term of the greatestdegree of the sum will be
making the sum ofdegree n.an • xn
an 5 -bn
(an 1 bn) • xn
n 1 m(an • bm) • xn1m
14. 4
15. Yes, m! m ( ) ( ) K 4 3 2 1.Because the integers areclosed under multiplication, n m ( ) ( ) K 4 2 1 is an integer.Substituting, we have m! n 3. Thus, by thedefinition of factor, m! isdivisible by 3.
16. For any integer n, the sum
. By the distributivelaw, . Since the integers are closed under addition, is aninteger. Thus, by thedefinition of factor, the sumof any three consecutiveintegers is divisible by 3.
17. a.b.
18.19. a. domain: all real
numbers, range: b.c. neitherd.
20. a.
b. 3
20 40 60 80 100 120
0.51
1.52
2.53y
n
T2,3
{x: x $ 2}{y: y $ 3}
(x 2 3)(x 1 3)(x2 1 9)
32x3 1 180x2 1 97x 2 15(8x 2 1)(4x 1 3)(x 1 5)
n 1 1
3n 1 3 5 (n 1 1) • 33n 1 3n 1 (n 1 1) 1 (n 1 2) 5
•5
••••m 2 2•m 2 1•5
•••••m 2 2•m 2 1•5
Answers for LESSON 4-1 pages 224–230 page 2c
c
59
21. b is not divisible by a and cis not divisible by a.
22. a. ((NOT p) OR q) AND NOT rb. 1
23. 28 5 1 1 2 1 4 1 7 1 14
24. Sample: If two integershave n and m digits,respectively, and if theyhave the same sign, thenthe number of digits intheir sum is either thelarger of m and n or thelarger of and .If the two integers haveopposite signs, then thenumber of digits in the sumis less than or equal to thelarger of m and n. If bothintegers are nonzero, thenthe number of digits intheir product is either
12. a. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12b. any value of the form
(for example, 17,30, 43)c. any value of the form
(for example, 13)d. any value of the form
, (forexample, 27)
13. integer
14. real number
15. integer
16. not a division problem
17. a. 8 pointsb. $2150c. ; the quotient is 8, theremainder is 2150.
18. False.
19. False. .
20. Suppose we have integers aand b such that a is divisibleby b. Then, by the definitionof factor, there is an integerq such that .Squaring both sides gives
. Sinceinteger multiplication isclosed, is an integer.Therefore, by the definition of factor, is divisible by b2.a2
q2
q2 • b2a2 5 (q • b)2 5
q • ba 5
x • (x2 1 x 1 1) 1 1x3 1 x2 1 x 1 1 5
93 5 18 • 5 1 3
22150 5 8 • 2500 1 2150
0 # r , 132 • 13 1 r
q • 13
q • 13 1 4
21.
22.
23.
24.25. a. The vessel that spotted
the boat at an angle of 42°b. < 0.4 mi
26. domain: all reals, range: allreals greater than 0,continuously decreasingover its entire domain,
, .
27. If the citrus crop is notruined, the temperature didnot stay below 28°.
28. a.b.c.d. q 5 6, r 5 1
q 5 -6, r 5 1q 5 -5, r 5 2q 5 5, r 5 2
limx→-`
5 `limx→`
5 0
2Ï23 ø 94.3%
3y2 114
z3 1 z2 1 z 1 1z8 1 z7 1 z6 1 z5 1 z4 1
4x 1 83x5 1 4x4 2 8x3 2 3x2 2
Answers for LESSON 4-2 pages 231–237 page 2c
61
1. 10
2.3. ,
4.
which can be rewritten in the desired form.
5. By the Remainder Theorem,.
6. ,
7.8.9. a. 2355
b. p(13)
10. p(2) 0
11. ;
12.
13. As , the values off(x) become closer and closerto those of the function
.
14. By the Remainder Theorem,since , is afactor.
15.. Both
and are factors of .x4 2 x2 2 2x 2 1
(x2 1 x 1 1)(x2 2 x 2 1)x4 2 x2 2 2x 2 1(x2 2 x 2 1)(x2 1 x 1 1) 5
(x 2 2)p(2) 5 0
q(x) 5 3x3 2 2x2 1 x 212
x → 6 `
h(x) 5 7 1-18
2x 1 1
x Þ -3f(x) 5 (x 1 7)
5
r(x) 5 p(4) 5 623
r(x) 5 p(-2) 5 -73
r(x) 512 x2 1
34 x 2
74
q(x) 552 x2 1
34
r(x) 5 p(-1) 5 -9
(-3x 1 2)(x2 1 1) 1(3x 2 1)3x3 2 x2 1 1 5
r(x) 5 x2 1 x 2 2q(x) 5 x2 2 3
q(x) 5 3x 1 8, r(x) 5 20
16. a.b. ,
17. True, this is a statement ofthe Quotient-RemainderTheorem for .
18.
19.
20.
21. a.
b.
c. The minimum is at,
d. ;
22. Invalid, this is an exampleof the converse error. As acounterexample, observethat 2 10 is divisible by 4but not by 6.
•
lims→`
A(s) 5 `lims→0
A(s) 5 `
7.5 # x # 8.5, x-scale = 0.1397.8 # y # 399.4, y-scale = 0.2
, and -3. At , , -3 but and 4 so .n(2) 5 6 • d(2) 1 1
d(2) 5n(2) 5 25r(2) 5q(2) 5 7x 5 2
r(x) 510x2 2 12x 1 15q(x) 5 x 4 2 5x3 1x 1 2
d(x) 53x 4 1 8x2 2 9x 1 27n(x) 5 x5 2
n(5) 5 329 • d(5) 1 36d(5) 5 55n(5) 5 18,131
Answers for LESSON 4-3 pages 238–243 page 2c
Answers for LESSON 4-4 pages 244–249
1. Step 1: definition of factorStep 2: The RemainderTheoremStep 3: definition of Zero
2. evaluated atis 0. By the Factor
Theorem, we can concludethat is a factor of
.
3. a. Nob. Yes
4. When n is an odd positiveinteger, has a zeroat . By the FactorTheorem, ( ) is a factorof .
5.
6. a.b.
7. a. , b.c. 6d. 0
(x 2 2), (x 1 3)x 5 -3x 5 2
x 5 -1 1 i, x 5 -1 2 ip(-1) 5 0, p(3) 5 0
y 5-32 , y 5
53
cn 1 dnc 1 d
c 5 -dcn 1 dn
t4 2 5t 2 600t 2 5
t 5 5t 4 2 5t 2 600
8. b and e. These are the onlygraphs where a horizontalline can be drawn thatcrosses the graph morethan 4 times.
9.
10. a. {3, 15, 63, 255, 1023}b. Since is a factor of
, is alwaysa factor of . Hence,no value of canbe prime.
11. a. p(x) intersects the linewhenever or
. Since p(x) is athird degree polynomial, thepolynomial is also a third degreepolynomial and thus has, atmost, three zeros. Therefore,p(x) intersects at, atmost, three points.b. A polynomial of degree
will intersect the linein, at most, n points.y 5 x
n $ 2
y 5 x
p(x) 2 xp1(x) 5
p(x) 2 x 5 0p(x) 5 xy 5 x
4n 2 1 . 34n 2 1
4 2 1 5 3xn 2 anx 2 a
x 5 -74, x 5 -15, x 5 1
c
63
12. Suppose we have twopolynomials of degree n,
and , thatintersect at more than npoints. Because polynomialaddition is closed,
is also apolynomial. Assume p(x) is not the zeropolynomial. Because thedegree of the sum of twopolynomials is less than orequal to the larger of thedegrees of the twopolynomials, the degree ofp(x) is # n. By the Numberof Zeros Theorem, p(x) hasat most n zeros. Therefore,
and intersect in, at most, n points—contradicting our initialassumption. Therefore, p(x)must equal 0, and hence,
5. The values in Rn are givenby , where a is anyinteger and .The sum of any value fromRn plus any value from Rmis
.When this sum is divided by3, it has a fixed remainder:n m (or n m 3 if n m 3). Thus, suchsums are always membersof a single set.
6. a: The product is always inset R2.
$1211
3 • (a 1 b) 1 (n 1 m)(3a 1 n) 1 (3b 1 m) 5
0 # n # 23a 1 n
KK5
KK5
KK5
7. Following the notationused in problem 5, theproduct of any value fromRn and any value from Rmis
.When this product isdivided by 3, it has a fixedremainder: nm (or nm 3if ). Thus, suchproducts are alwaysmembers of a single set.
Lesson 1. R1 2. R0 3. R1
4. a. x is congruent to ymodulo 4b. There are four disjointsets: ,for any integer n},for ,1, 2, 3.
5. Monday
6. a. True b. False c. True
7. 2 8. 16 9. 16
10. 9 11. 4
12. a. 624 b. 357
13. Five hours after 9:00, it is2:00.
14. (mod 2)
15. Sample: x(mod 360)
16. 04
x 1 360 ;x ; 0
k 5 0Rk 5 {x: x 5 4 • n 1 k
nm $ 32
9ab 1 3bn 1 3am 1 nm(3a 1 n) • (3b 1 m) 5
Answers for LESSON 4-5 pages 250–257
c
65
17. By the Quotient-RemainderTheorem, for any integer nthere exist unique integersq and r such that
where . If, then, by definition, n
is divisible by 3. If ,either so
orso
. So one of thethree consecutive integers,n, , or isdivisible by 3.
18. Suppose (mod m) and(mod m). The
Congruence Theoremrequires that m be a factorof both and .Therefore, by the definitionof factor, there existintegers and such that
and . By the Subtraction
Property of Equality,
. By theAssociative andCommutative Properties ofSubtraction and theDistributive Property
. Because integersare closed undersubtraction, is aninteger. Therefore, bydefinition, m is a factor of
.(a 2 c) 2 (b 2 d)
(k1 2 k2)
(k1 2 k2)m(a 2 c) 2 (b 2 d) 5
k1m 2 k2m(a 2 b) 5 (c 2 d) 5
k2mc 2 d 5a 2 b 5 k1m
k2k1
c 2 da 2 b
c ; da ; b
n 1 2n 1 1
1 5 q • 3 1 3q • 3 1 r 1n 1 1 5r 5 2
q • 3 1 3q • 3 1 r 1 2 5n 1 2 5r 5 1
r Þ 0r 5 0
0 # r , 3q • 3 1 rn 5
Therefore, by theCongruence Theorem,
.
19. a.
b. 3
20. a. {x: , for allintegers n}b. Their difference is aninteger.
21. , ,
22. a. 8 b.23. a.
b.
c.
24. Assume that a, b, and c areany positive integers suchthat a divides b and adivides . Then by thedefinition of factor, thereexist integers and suchthat and
. By theSubtraction Property ofEquality
. Using theDistributive Property, thiscan be written as
. Since isan integer, by the definitionof factor, a divides c.
congruence classes modulo3, both R1 and R2 are theirown reciprocals, since
and R2. R0 does not have areciprocal, since
, for , 1, and2. For the system ofcongruence classes modulo4, both R1 and R3 are theirown reciprocals, since
and R1. R2 does not have areciprocal, since R0, ,
, and R2. Similarly R0 does nothave a reciprocal.
R2 • R3 5R2 • R2 5 R0R2 • R1 5 R2
R2 • R0 5
R3 • R3 5R1 • R1 5 R1
n 5 0R0 Þ R1R0 • Rn 5
R2 • R2 5R1 • R1 5 R1
Answers for LESSON 4-5 pages 250–257 page 3c
67
1. 6000200300
2. d
3. 2 is not a digit in base 2.
4. 1111112, 778, 3F16
5. 7
6. 50
7. 375
8. 164
9. a. 7b. 10010002
c.10. 173,2558, AD16
11. a. 4 b. 100
12. 1100012
13. 1001012
14. Carry digit is the output ofthe lower AND gate, whichis 0. Since the output of theOR gate is 0 and is one ofthe inputs to the upperAND gate, the sum digit,which is the output of thatgate, is also 0.
15. e
16. It is even if the last digit is 0and odd if that digit is 1.
25. Suppose m is any integer.When m is divided by 4, thepossible remainders are 0,1, 2, 3. By the Quotient-Remainder Theorem, thereexists an integer k such that
, or , or, or .
26. ,
27. a. ab. dc. cd. b
r 5 66q 5 662
m 5 4k 1 3m 5 4k 1 2m 5 4k 1 1m 5 4k
28. a. no (converse error)b. yes (contrapositive)c. no (inverse error)
4. a. There exists a largestpositive integer.b. cannot be aninteger.c. Since n and 1 areintegers and the integersare closed under addition,
must be an integer.This contradicts part b.Therefore, our initialassumption must be false.Therefore, there is nolargest positive integer.
5. a. the product of all theprimesb. and
6. True
7. a. 23 b. No
8.
9.
10.11.12.13. True
14. If and n is even, thenn cannot be a primebecause it has 2 as a factor.
15. Assume there exists anumber n which is thelargest multiple of 5.
n . 2
x(x 1 1)(x2 1 1)
(z 2 Ï17)(z 1 Ï17)
y 98(3y 2 1)(3y 1 1)
(x 211 2 Ï217
3 )3(x 2
11 1 Ï2173 ) •
25 • 3 • 5
p . 1p 5 1
n 1 1
n 1 1
Then, by the definition ofmultiple, , where mis an integer. Because theintegers are closed underaddition, is aninteger. Because theintegers are closed undermultiplication, isan integer which, by thedefinition of multiple, is amultiple of 5. By thedistributive law,
which is an integerlarger than . Thiscontradicts the assumptionthat n is the largestmultiple of 5. Therefore,this assumption is false, sothere is no largest multipleof 5.
16. Assume there exists anumber x which is thesmallest positive realnumber. Since the realnumbers are closed under
multiplication, is a real
number. Furthermore,
because both and x are
positive, is positive.
However, is less than x.
This contradicts theassumption that x is thesmallest positive realnumber. Therefore, thisassumption is false, so thereis no smallest positive realnumber.
20. The prime factorization of1,000,000,000 29 59.Since the FundamentalTheorem of Arithmeticguarantees this factorizationis unique and 11 is prime, 11cannot be a factor.
24. Assume for integers a, b,and p that p is a factor ofboth a and b. Then thereexist integers m and n suchthat and , bydefinition of factor. So
, by thedistributive law. Since
is an integer byclosure properties, p is afactor of .a 2 b
m 2 n
(m 2 n)pa 2 b 5 mp 2 np 5
b 5 npa 5 mp
;5•;5•
;xy ; 12
•5
x 5 2, x 5 5, x 5 -7(x 2 2)(x 2 5)(x 1 7)x2 1 2x 2 35
6(y2 1 1)2(y 1 1)2(y 2 1)2
(x 1 2)(x 2 2)(x 1 3)(x 2 3) 25. a. sine function
b. period second,
amplitude 15 amperesc. sin(120πt)d. 0 amperes
26.27. odd
28. a.
b.
c. inches, inches
29. Assume n is an eveninteger. Then, by thedefinition of even,
for some integer m.Therefore
becauseinteger multiplication isassociative. Therefore, bythe definition of an evennumber, if n is an evennumber then is an evennumber. Consequently, by the Law of theContrapositive, if is notan even number then n isnot even.
30. Answers will vary.
31. Answers will vary.
n2
n2
2m • 2m 5 2(2m2)n2 5 (2m)2 5
2 • mn 5
h ø 5.14s ø 10.29
A(s) 5 s2 12176
s
h 5544s2
x 5 4, x 5 8
c(t) 5 155
5160
Answers for LESSON 4-7 pages 265–272 page 2c
71
1. , 2. ,
3. ,
4. , 5. 66
6. , degree 1
7.8. ,
9. ,
10. ,
11. ,
12. ,
13. ,
14.
,
15. 4 16. 12 17. 10
18. 18 19. R2 20. 2
21.22.23.24.
25.26.27.28.
29. 3, ,
30. , , 31. ,
32. -2, , , 483-15
52-37
13
34-52
114(-3 2 Ï37)1
14(-3 1 Ï37)
66t 1 11)(7t 2 2)(8t 1 1)(-13t2 2
(2x 1 1)(2x2 2 4x 2 9)
(w2 1 z2)(w 1 z)(w 2 z)
2(3v2 1 5)2
(y 1 3)(2y 2 1)(2y 1 1)(y 2 3) •
x(3x 2 5)(x 1 2)
9(t 1 2)(t 1 3)
5(x 2 y)(x 1 y)
r(x) 5 -6732
138 x 2
2916
q(x) 5 x4 112x3 2
54x2 1
r(x) 5 1082q(x) 5 5x2 1 30x 1 181
r(x) 5 24q(x) 5 x3 2 8x2 1 11x 2 15
r(x) 5 x 1 1q(x) 5 2x 2 5
r(x) 5 5q(x) 5 x3 1 3x2
r(x) 5 21x 1 7q(x) 5 7x 2 8
r(x) 5 8q(x) 5 3x2 2 7x 1 2
a 5 -5
5r(x) 5 2x 1 5
r 5 37q 5 -351
r 5 63q 5 83
r 5 2q 5 5r 5 6q 5 5 33. Because and
has no real zeros, 2 is the only realnumber with this property.
34. , , ,
35. True; .
36. False; .
37. True; if a and b are even,then , forsome integers n, m. Then
by the DistributiveProperty. Because theintegers are closed underaddition, is aninteger. Therefore, by thedefinition of factor, is divisible by 4.
38. Let n be an odd integer.Because it is odd,
for someinteger m. Then
by the DistributiveProperty. Because theintegers are closed underboth multiplication andaddition, isan integer. Therefore, by thedefinition of factor, 2 is afactor of . Since anumber is even if it isdivisible by 2, is even.n2 1 n
40. Let a, b, c, and d be anyintegers such that a and care divisible by d and
. Then there existsintegers r and s such that
and bydefinition. Then,
.Since is an integer byclosure properties, b isdivisible by d.
41. This is false. For example,is not divisible by 4.
42. Let a be an odd integer.Then for some integer n,
. The next higherodd integer is
. The number onegreater than the product oftwo consecutive oddintegers is then
by the DistributiveProperty. Because theintegers are closed under
2n 1 1)4n2 1 8n 1 4 5 4(n2 5
1 51 5 (2n 1 1)(2n 1 3) 1a(a 1 2) 1
2n 1 3a 1 2 5
a 5 2n 1 1
33 2 1
(r 1 s)d(r 1 s)c 5 d • r 1 d • s 5
b 5 a 1c 5 d • sa 5 d • r
a 5 b 2 c
both multiplication andaddition, is aninteger. Therefore, by thedefinition of factor, thenumber one greater thanproduct of two consecutiveodd integers is divisible by 4.
43. 11 is a factor of
44. (mod 5)
45. If sin(x) , then
(mod 2π) or
(mod 2π)
46. (mod 11)9 (mod 11)
47. (mod 11)
48. (mod 12)
49. Either r(x) is the zeropolynomial or the degreeof r(x) is less than thedegree of d(x).
56. Assume that there exists aprime p that is a factor of
, but not a factor of n.
57. Assume that n is thesmallest integer. Becausethe integers are closedunder addition, isalso an integer. Since
, n cannot be thesmallest integer. This is acontradiction. Therefore,there is no smallest integer.
58. Assume there are finitelymany prime numbers. Theycan be listed from smallestto largest: , ,
, pm. Multiply these:. . . pm. Consider
the integer . Sinceis larger than the
assumed largest prime pm, itis not prime. By the primeFactor Theorem, musthave a prime factor, p, onthe list of primes. Then p is a factor of both n and . It is also a factor of thedifference .Therefore, p must equal 1.But p is a prime number, sop cannot equal 1. Thiscontradiction proves thatthe original assumption isfalse. Hence, there areinfinitely many primenumbers.
59. 2311
(n 1 1) 2 n 5 1
n 1 1
n 1 1
n 1 1n 1 1
n 5 p1p2p3
Kp2 5 3p1 5 2
n 2 1 , n
n 2 1
n2
60. True
61. six (2, 3, 5, 7, 11, 13)
62. 1000
63. a
64. c
65. Prime
66.67.68. a. 88
b. 47 c.
69. 22 laps, 49.5 m left
70. a. 9 b. 27,500 c.27,500
71. 73 months of 5 days or 5 months of 73 days, 365 months of 1 day, or 1 month of 365 days
72. 5625
73. 143
74. 7
75. 5
76. 10000012
77. 101001012
78. 31
79. 42
80. 1011112, 21 26 47
81. 1111102, 31 31 62
82. d
83. In base 2, a number divisibleby 16 ends in 0000.
4. Suppose r and s are any tworational numbers. Bydefinition, there existintegers a, b, c, and d,where and such
that and . So,
Since by closure propertiesthe product of two integersand the difference betweentwo integers are integers,
and bd are integers.Also, since and
. Therefore, since
is a ratio of integers,
is a rational number bydefinition.r 2 s
ad 2 cbbd
d Þ 0b Þ 0bd Þ 0
ad 2 cb
Addition offractions5
ad 2 cbbd
Multiplicationof fractions5
adbd 2
cbbd
MultiplicationProp. of 15
ab • dd 2
cd • bb
r 2 s 5ab 2
cd
s 5cdr 5
ab
d Þ 0b Þ 0
N 5 0
23 5. The sausage pizza is cut
into N pieces; the area of
each slice is of a pizza.
The pepperoni pizza is cutinto pieces; the area
of each slice is of a
pizza. Since the sausagepizza is cut into morepieces, its slices are smallerthan those of thepepperoni pizza. Thedifference in the areas ofthe sausage and pepperonislices of pizza is:
of a pizza.
6. {N: }
7. {N: }
8. {N: }
9.
{M: and }
10. {N: }
11. {K: }
12. {K: }
13. {a: and }
14. b
15. aa 2 1
a Þ 1a Þ 02a 2 1a2 2 a
K Þ -1K2 1 K 2 1K 1 1
K Þ -1KK 1 1
N Þ -K1N2 1 2NK 1 K2
M Þ 2M Þ 1
M2 2 M 2 1M2 2 3M 1 2
N Þ 0N3 1 KN3
N2
N Þ 0N2 1 KN
N Þ 0N 1 KN
52
N2 2 2N
N 2 2N2 2 2N5
NN2 2 2N 2
(N 2 2N 2 2)1
N1
N 2 2 (NN) 2
1N 2 2 2
1N 5
1N 2 2
N 2 2
1N
Answers for LESSON 5-1 pages 284–288
c
75
16.
17.
18. Suppose r and s are any tworational numbers. Then,there exist integers a, b, c,and d, where and
such that and
.
Because sums and productsof integers are integers,
and 2bd are
integers. Therefore, is
a rational number bydefinition.
r 1 s2
ad 1 bc
Multiplicationof fractions5
ad 1 bc2bd
Definition ofdivision offractions
5ad 1 cb
bd • 12
DistributiveProperty
ad 1 bcbd2
5
Multiplicationof fractions
adbd 1
bcbd
25
MultiplicationProp. of 1
ab • dd 1
cd • bb
25
ab 1
cd
2
r 1 s2 5
s 5cd
r 5abd Þ 0
b Þ 0
2a
2a3b 19. Since the average of two
rational numbers is rational,between any two numbersthere exist a rationalnumber. If r and s are bothrational numbers where
and with
and , then their
average is . If ,
then .
20. The sausage pizza is cutinto pieces, so each
slice is of a pizza. The
pepperoni pizza is still cut into pieces, so each
slice is of a pizza.
Each student who has oneslice from each pizza would
have
of a whole pizza. The twostudents who have 2 slicesof the popular pizza would
and have minimum pointsat (0, 0).b. Odd power functionsinclude negative values andhave no upper or lowerlimits. Even powerfunctions have minimumequal to 0, and include nonegative values.
29. The functions are identical.Relative Maxima (-1.097, -1.486); Relative minimum:none;
. The
functions do not exist when2m, 1, or -1.x 5x 5x 5
g(x) 5 0limx→-`
f(x) 5limx→-`
(7Ï2)2 5 49(2) 5 98
Answers for LESSON 5-2 pages 289–294 page 2
must be a rational number
since it is the ratio of twointegers.
20.
21.
22.
23.
24.
25. a.
b.
26.
ln 3 1.099, ln 2 .693≈x 5
≈x 55 (ex 2 3)(ex 2 2)
e2x 2 5ex 1 6 5 0
400N 2
400N 1 2 5
800N2 1 2N
400N
x2 2 xx2 1 1
361
-(3 1 a)a2
x 2 32x 1 1
H 1 HK2HK2
ba
c
79
1. True
2. d
3.
4. a. The limit of f(x) as xapproaches 0 from the leftis negative infinity.Sample:
b. The limit of f(x) as xapproaches 0 from the leftis positive infinity.Sample:
5. a. T-6, 0
b. -6c. and
6. a. Trueb. Falsec. 0d. i. -.05 .05
ii. -.002 x .002,,, x ,
x 5
h(x) 5 -`limx→-62
h(x) 5 `limx→-61
x 5
-4 -2 2 4-2
2
4
y
x
y
x-4 -2 2 4
-4
-2
2
4
y
x-1 1 2 3 4 5 6
-2-1
1234
7. a. and
b. and
c.
8. a. domain: {x: 0};range: {y: 0}b. decreasing on theintervals (-`, 0) and (0, `)c. and
d. and
e. odd
f.
9. The sun’s brightness seenfrom Earth is 2.25 times thebrightness seen from Mars.
20. a. The total distancetraveled is 2d. The firstaverage speed is 20 mphand the second is x mph.Since the distance traveledeither way, d, is equal, the 2rates divided into d willgive the total time spent.When this value is dividedinto 2d the average speed isobtained.
b.c. 40; the return
trip approaches 40 mph.
f(x) 5limx→`
40xdxd 1 20d
Answers for LESSON 5-5 pages 307–314 page 2
8.an10m1n 1 K 1 a110m11 1 a010m 1 a-110m21 1 a-210m22 1 K 1 a-m100
10m
c
Answers for LESSON 5-6 pages 315–321
1. True 2. False 3. True
4. Assume the negation of theoriginal statement.
5. a.b. and a have a factor of 3.c.d. and b have a factor of 3.e. Both a and b have afactor of 3. However, fromthe beginning of the proof,
is assumed in lowest
terms, thus having nocommon factors. There is aninherent contradiction, sothe negated statement isfalse, and the originalstatement, is irrational,must be true.
Ï3
ab
b2b2 5 3k2a2a2 5 3b2
6. a. Yes, .
b. No, 2 is a rationalnumber.
7. a. irrational; The decimalexpansion neitherterminates nor repeats.b. rational; The decimalexpansion repeats.c. rational; The decimalexpansion terminates.
8. See below.
9.
10.
11. True
3,615,8814950
7,304,81010000
Ï4 521
c
85
12. Assume the negation of theoriginal statement is true.Thus, there is a rationalnumber x and an irrationalnumber y whose quotient
is rational. Then since the
quotient of x and y isrational, both x and y mustbe rational, but thiscontradicts the assumptionthat y is irrational. Hence the assumption is false, and,
must be irrational.
13. a.
b.
c. 1.4854315 and
1.4854315
14.
15. False; counterexample: πand 3 π sum to 3, whichis rational.
16. False; counterexample: Let and . Then,
4, but 4 is a rationalnumber.
17. a. when is zero ora perfect squareb. when andnot a perfect square
b2 2 4ac . 0
b2 2 4ac
a • b 5 Ï2 • Ï8 5 Ï16 5b 5 Ï8a 5 Ï2
2
72 1 26Ï541
ø25 2 5Ï311
ø105 1 Ï3
25 2 5Ï311
5 2 Ï3
xy
xy
18. reciprocal of
19.
20. a. The Quotient-RemainderTheorem states that theremainder r for this divisionis an integer in the range
. Therefore, after54 steps at least oneremainder must repeatsince there are only 54unique remainders.b. The number of stepstaken before repeat of theremainder is equal to thenumber of digits in the repeating number sequencein the decimal expansion.Each time the remainderrepeats, the repeating digitsequence is begun again.c. The Quotient-RemainderTheorem requires theremainder to be an integerr in the range forany long division. Hence,after d steps there must bea zero or a repeated remainder. If there is a zeroremainder, then the decimalexpansion terminates. Ifthere is a repeatedremainder, then the decimalexpansion is repeating.
algebraic if it is a root ofsome polynomial havingrational coefficients;otherwise it is said to betranscendental.
Answers for LESSON 5-6 pages 315–321 page 3c
Answers for LESSON 5-7 pages 322–326
1. tan ; cot ;
sec ; csc
2. a. tan
b. cot
c. sec
d. csc
3. a. b. 2 c. d.
4. Sample: , ,
,
5. a. , n is an
integerb.
6. a. , n an integerx 5 nπ
-3π # x # 3π, x-scale = π -4 # y # 4, y-scale = 2
x 5π2 1 nπ
(3π2 , -1)(3π
4 , Ï2)(π2, 1)(π
4, Ï2)Ï5Ï5
212
u 5hypotenuse
side opposite u
u 5hypotenuse
side adjacent to u
u 5side adjacent to u
side opposite u
u 5side opposite u
side adjacent to u
x 554x 5
53
x 534x 5
43 b.
7. Sample: (0, 0), ,
,
8. a. sin, csc, and tan are oddfunctions. cos and sec areeven functions. cot is neither.b. nonec. sin and cos
d.
sin
cos
tan 1
cot 1
sec 2
csc 2 2Ï33Ï2
Ï22Ï33
Ï33Ï3
Ï3Ï33
12
Ï22
Ï32
Ï32
Ï22
12
π3
π4
π6
(π3, Ï3)(π
4, 1)(π6, Ï3
3 )
-3π # x # 3π, x-scale = π -4 # y # 4, y-scale = 2
c
87
9. d
10. tan ; cot ;
sec ; csc
11. a. The area of the triangle
ABO is (AB)h. So, the area
of the regular n-gon is thesum of the areas of n
congruent triangles,
n (AB)h. .
The altitude h splits into two smaller angles
measuring . tan .
So, tan .Hence,
the area of the n-gon is
.
b. Let the radius, r, be 6and . Then ,and by thePythagorean Theorem.Using the formula frompart a, the area is
tan 72. The area of the
square is 72 and so the formula checks.
12. a. definition of tangentb. sine is an odd functionand cosine an even functionc. defintion of tangentd. transitive property ofequality and definition ofodd function
(AB)2 5 (6Ï2)2 5
π4 5
4(3Ï2)2 •
h 5 3Ï2AB 5 6Ï2n 5 4
nh2tan πnn12 (2h tan πn)h 5
πnAB 5 2h
12 AB
h(πn) 5
πn
/AOB
m/AOB 52πn
12
12
(-π4) 5 -Ï2(-π
4) 5 Ï2
(-π4) 5 -1(-π
4) 5 -1
13.
14. ;
15. a.
b. ;
c. ;
16. a. . So 17 1(mod 2); 1b. , so1000 10(mod 11);
17. No real solution.
18. a. -1 x 4b. All increasing functionsare 1–1 functions, so theinverse of log5x could beused to simplify the equation without changingthe solutions.
5.04The hiker is < 6.32 milesfrom the first and < 5.04miles from the second.
ø
(10)(.4226).8387ø 20. a. iv
b. ic. iid. iiie. vif. v
Answers for LESSON 5-7 pages 322–326 page 3c
Answers for LESSON 5-8 pages 327–333
In-Class Activity1. .2 hour = 12 minutes
2. .25 hour = 15 minutes
3. a. 444.44
b. average speed speedin still air
4. ; ;
444.44 ; average speed
speed in still air
Lesson 1. 42
2. 303 mph
3. a. 20 mphb. 60 mphc. not possibled. not possible
4. a.b. 5 and -5
5. a. Yesb. No: the first equation isundefined when 2, but
2 is a solution of thesecond equation.x 5
x 5
x 5x 54(x 1 5)(x 2 5)
kmhr
,mihr
5400d9d 5
2d900d
20,000
d400
d500
,
mihr
6. No; there are no rational numbers x, , and , but there are irrationalsolutions
; and
7. 16
8. 0
9. 5 or 6
10.11. a. 48 mph
b. Yes, the currentdecreases the length oftravel about 0.05 milesevery 15 minutes.
12. a. < 171 mphb. < 342 miles
13. Your speed returning is not 0 mph.
14. If you go some place at 20mph, you cannot average40 mph for the total trip.
15. Your return rate does notdepend on the distancetraveled.
t 5 -4
x 5x 5
x 5
x 5
h-Ï8, -Ï8 1 2, -Ï8 1 4jhÏ8, Ï8 1 2, Ï8 1 4j
x 1 4x 1 2
c
89
16. a. R 2.73 ohmsb. R2 12 ohms
17. ;
18. 20 minutes
19.
20. tan ; cot
21. sec ; csc 2
22. a. The decimal expansionof x is not terminating andthe decimal expansion of xis not repeating.b. If the decimal expansionof x is not terminating andnot repeating, then x is nota rational number.c. i. rational
8. The area of a rectangle is itslength times its width, ,w.Its perimeter is then
. By setting theseequal, the problem is tofind positive integers , andw such that .,w 5 2, 1 2w
2, 1 2w
47
an 5(n – 2)180°
n
t 5 6
(18, 13)(1
7, 13)(16, 13)(1
5, 13)(14, 14)(1
4, 13)(13, 18)(1
3, 17)(13, 16)(1
3, 15)(13, 14)(1
3, 13)
Dividing both sides by 2,w(which is nonzero) yields
, which
simplifies to . This
is equivalent to theequation in Problem 2,which is also equivalent toProblem 1.
9. Let n be an integer greaterthan 2. Suppose ( ) is afactor of 2n. Then
for some
integer k. Then ,
since . By taking the
reciprocal, . So
, or .
This is equivalent to theequation in Problem 2,which is equivalent toProblem 1.
10. of an hour < 33 minutes
11. 2.744 cm x 2.856 cm
12. a. 3 cmb. 3.75 cm from the mirror
13. ; ,
; csc 5π3 5
-2Ï33sec 5π
3 5 2
cot 5π3 5
-Ï33tan 5π
3 5 -Ï3
,,
611
12 5
1k 1
1n
12 2
1n 5
1k
1k5
n – 22n
n Þ 2
2nn 2 2 5 k
2n 5 (n 2 2)k
n 2 2
12 5
1w 1
1,
,w2,w 5
2, 1 2w2,w
Answers for LESSON 5-9 pages 334–339
c
91
14.
15. a. {x: and }
b. near 5: ;
near -5,
c. ;
d. x-intercept: (-2, 0);
y-intercept:
e.
16. a.
b. , -1, ,
and x Þ -2
25x Þx Þ-17x Þ
x 2 4x 1 2
(7x 1 1)(x 2 4)(5x 2 2)(x 1 2) 5
•(5x 2 2)(x 1 1)(7x 1 1)(x 1 1)
7x2 2 27x 2 45x2 1 8x 2 4 5
•5x2 1 3x 2 27x2 1 8x 1 1
-8 -4
x = 5
x = -5
4 8
-1
1y
x
(0, -225)
limx→`
h(x) 5 0limx→`
h(x) 5 0
limx→-51
h(x) 5 `
limx→ -52
h(x) 5 -`x 5
limx→51
h(x) 5 `
limx→52
h(x) 5 -`x 5
x Þ -5x Þ 5
-10Ï10 1 6013 17.
18. Suppose p and q areintegers such that 4 is afactor of p, and q is even.By definition, there existintegers r and s such that
and . Then. Since
rs is an integer by closureproperties, 8 is a factor ofpq by definition.
19. If two or more differentregular polygons areallowed as faces, there are13 polyhedra for whicheach vertex is surroundedby the same arrangementof polygons. These arecalled semi-regularpolyhedra, and they areconvex.
47. Assume the negation istrue. Thus, the difference ofa rational number p and anirrational number q is arational number r. Then
. So .However, by the closureproperty of rationalnumbers the differencebetween two rationalnumbers is another rationalnumber. Hence, there is acontradiction, and so theassumption is false, whichproves the originalstatement.
48. d
49. The limit of f(x) as xdecreases without bound is1.3.
20. a. sin(α β γ) sin α cos β cos γ cos α sin β cos γ 1cos α cos β sin γ sin α sin β sin γb. Sample: cos(α β γ) cos α cos β cos γsin α sin β cos γ sin α cos β sin γ cos α sin β sin γc. Proof: (Chunk (α β) and apply cosine and sine of sumidentities.) cos[(α β) γ]
cos(α β) cos γ sin(α β) sin γ(cos α cos β sin α sin β) cos γ (sin α cos β cos α sin β) sin γcos α cos β cos γ sin α sin β cos γ sin α cos β sin γ cos αsin β sin γ
11. Left sideSine of a sum identityDouble angle identitiesMultiplicationAdditionPythagorean IdentityMultiplicationAddition
Right side55 3 sin x 2 4 sin3 x5 3 sin x 2 3 sin3 x 2 sin3 x5 3 sin x(1 2 sin2 x) 2 sin3 x5 3 sin x cos2 x 2 sin3 x5 2 sin x cos2 x 1 cos2 x sin x 2 sin3 x5 (2 sin x cos x)cos x 1 (cos2 x 2 sin2 x)sin x5 sin 2x cos x 1 cos 2x sin x
f(x) 17. a. If there exists an x suchthat p(x) is true and q(x) isfalse.b. False
18. a. i. , , , ;
);
; integers
ii. , , , ;
),
; integers
b. );
Yes, the formula works forall noninteger numbers
or 1.c Þ 0
an 51
c 2 1 • (1 2 (1c)n
n $ 1
an 513(1 2 (1
4)n
85256
2164
516
14
n $ 1
an 512(1 2 (1
3)n
4081
1327
49
13
Answers for LESSON 7-1 pages 406–412 page 2c
Answers for LESSON 7-2 pages 413–417
1. a.b. 24
2. a.
b.
3. True, the only difference isthat different letters areused for the indices.
4.
5. a.
b.Sn11 5 S1 1
1n 1 1
S1 512H
ok
j51
1j 1 1
o43
i51i2
1516
2-4 1 2-3 1 2-2 1 2-1
-2 1 2 1 8 1 16 6. a. b.
7. a.
b.
8. a. , ,
,
b.
(oki51
i2) 1 (k 1 1)2
K 1 k2) 1 (k 1 1)2 5(12 1 22 1k2 1 (k 1 1)2 5
ok11
i51i2 5 12 1 22 1 K 1
o4
i51ai 5 30o
3
i51ai 5 14
o2
i51ai 5 5o
1
i51ai 5 1
n 1 12n 2 1 1
n 1 12n 1
n 1 12n 1 1
3715 ø 2.47
on21
j50(12) j
o4
k50(12)k
c
9.
10.
11. It does.
12.
13. Sample:Let . Then
, and
.
14.
15. explicit; -1, 1, -1, 1, -1, 1
16. a. x2 is not defined since
.
b. The definition definesthe first term and defineseach successive term interms of the preceedingterm.
17. ; ,
; integers n $ 1
an 51
n 1 112, 13, 14, 15, 16, 17
x2 510
11000 o
5
i5-5i2
(1 1 1 1 1 1 1)2 5 16
( o4
n51an)2
512 5 4
12 1 12 1 12 1o4
n51(an
2) 5
an 5 1 ; n
0 1 0 216 5 1 2
16
1 1 0 1 0 1(-15 115) 2
16 5
(-14 114) 1(-13 1
13) 1
(-12 112) 1(1
5 216) 5 1 1
(14 2
15) 1(1
3 214) 1(1
2 213) 1
(1 212) 1o
5
k51(1k 2
1k 1 1) 5
o4
j51( j2 1 j 2 4)
(oki51
i(i 2 1)) 1 (k 1 1) k 18. a.3.75, ; integers b. 279° F
19. Left side
Right side; real numbers x and y,
20.
21.
22. Sample: The Arrow Paradoxstates that an arrow nevermoves, because at eachinstant the arrow is in afixed position. Another ofZeno’s paradoxes, known asthe Paradox of Achilles andthe Tortoise, is frequentlysummarized as follows.Achilles, who could run 10yards per second, competedagainst a tortoise which ran1 yard per second. In orderto make the race more fair,the tortoise was given aheadstart of 10 yards.Zeno’s argument, thatAchilles could never passthe tortoise, was based onthe “fact” that wheneverAchilles reached a certainpoint where the tortoisehad been, the tortoisewould have moved aheadof that point.
(g o f )(k) 5(k 1 1)(k 1 2)
2
2k2 2 k 1 1(2k 1 1)(2k 2 1)
cos2 y 2 cos2 x.sin2 x 2 sin2 y 5[55 cos2 y 2 cos2 x5 1 2 cos2 x 2 1 1 cos2 y5 (1 2 cos2 x) 2 (1 2 cos2 y)
1. a.b. Assume that S(k) is truefor a particular butarbitrarily chosen integer
where
.
2. a. and
, so S(1) is true.
b. ;
:
c.
, so istrue.d. S(n) is true ; integers
.
3. a. S(1): 3 is a factor of 3;S(3): 3 is a factor of 33; S(5): 3 is a factor of 135.b. All are true.c. : 3 is a factor of
.
4. a. S(1): ; S(3): ;S(5): b. S(1) is false. S(3) and S(5)are true.c.4 . (k 1 2)2
(k 1 1)2 1S(k 1 1):
29 , 3613 , 165 , 4
(k 1 1)3 1 2(k 1 1)S(k 1 1)
n $ 1
S(k 1 1)(k 1 1)(k 1 2)k(k 1 1) 1 2(k 1 1) 5
ok11
i512i 5 o
k
i512i 1 2(k 1 1) 5
(k 1 2)(k 1 1)
ok11
i512i 5S(k 1 1)
S(k): ok
i512i 5 k(k 1 1)
1(1 1 1) 5 2
o1
i512i 5 2(1) 5 2
S(k): ok
i51(2i 2 1) 5 k2
k $ 1
1 5 12 5. a. ;
b. All are true.
c.
6. a.
b.
c.
d.
e. use inductive assumption
f.
g.
h. The Principle ofMathematical Induction
7. a. , so S(1) is true.b. Assume that S(k) is truefor a particular butarbitarily chosen integer
. S(k):
. Show
is (k 1 1)(k 1 2)(2(k 1 1) 1 1)
6
k2 1 (k 1 1)2 512 1 22 1 K 1S(k 1 1):
k2 5k(k 1 1)(2k 1 1)
6
12 1 22 1 K 1k $ 1
S(1): 12 51(2)(3)
6
(k 1 1)(k 1 2)2
k(k 1 1) 1 2(k 1 1)2
k 1 11 1 2 1 3 1 K 1 k 1
k 1 1 5(k 1 1)(k 1 2)
2
1 1 2 1 3 1 K 1 k 1
k 5k(k 1 1)
2
1 1 2 1 3 1 K 1
1(1 1 1)2 5 1
(k 1 1)(3k 1 2)2
ok11
i51(3i 2 2) 5S(k 1 1):
13 55(14)
2
1 1 4 1 7 1 10 1S(5):
1 1 4 1 7 53(8)
2 ;S(3):
1 51(2)
2S(1):
Answers for LESSON 7-3 pages 418–426
c
true.
c. Since S(1) is true and , then by
mathematical induction,S(n) is true ; integers .
8. a. 1, 4, 13, 40, 121b. 1, 4, 13, 40, 121
c. Let S(n):
; integers .
(1) . This
agrees with the recursiveformula, hence S(1) is true.
(2) Assume S(k): is
true for some integer .Show :
is true.
. 3k11 2 12
3k11 2 3 1 22 5
(3k 212 ) 1 1 53ak 1 1 5 3
ak11 53k11 2 1
2
ak11 5S(k 1 1)k $ 1
ak 53k 21
2
a1 531 2 1
2 5 1
n $ 1
an 5(3n 2 1)
2
n $ 1
S(k) ⇒ S(k 1 1)
(k 1 1)(k 1 2)(2(k 1 1) 1 1)6
(k 1 1)(k 1 2)(2k 1 3)6 5
(k 1 1)(2k2 1 7k 1 6)6 5
(k 1 1)(k(2k 1 1) 1 6(k 1 1))6 5
k(k 1 1)(2k 1 1) 1 6(k 1 1)2
6 5
(k 1 1)2 5
(k 1 1)2 5k(k 1 1)(2k 1 1)
6 1
12 1 22 1 K 1 k2 1 Therefore, for all integers, if S(k) is true, then
is true. Thus, from(1) and (2) above, using thePrinciple of MathematicalInduction, S(n) is true for allintegers . Hence, theexplicit formula describesthe same sequence as therecursive formula.
9. S(n): ;integers . (1) from the recursive definition;
fromthe explicit definition.Hence, S(1) is true. (2) Assume S(k):
for some integer. Show :
is true.
. Therefore,is true if S(k) is
true, and (1) and (2) proveby the Principle ofMathematical Inductionthat the explicit formuladoes describe the sequence.
, so theinductive step is true. Thus,(1) and (2) prove by thePrinciple of MathematicalInduction that S(n) is truefor all integers .
11.
12. a. (-3)2 -3 (-2)2 -2(-1)2 -1 n2 nb. 16
13. a. 1, 2, 3, 4, 5, 6b. Sample: , for allintegers .n $ 1
an 5 n
11K111111
on
i51
in
n $ 1
(k 1 1)(3(k 1 1) 2 1)2
(k 1 1)(3k 1 2)2 5
3k2 1 5k 1 22 5
3k2 2 k 1 6k 1 22 52 5
ok
i51(3i 2 2) 1 3(k 1 1) 2
ok11
i51(3i 2 2) 5
(k 1 1)(3(k 1 1) 2 1)2
ok11
i51(3i 2 2) 5
S(k 1 1)k $ 1
ok
i51(3i 2 2) 5
k(3k 2 1)2
1 • (3 2 1)2 5 1
o1
i51(3i 2 2) 5 1 14. a.–c.
d. i. ii. iii.
15. a. Since is a factor of, and is a
factor of , then by the
Transitive Property ofPolynomial Factors, isa factor of .b.c. Since is a factor of
by part a and is a factor of bypart b, then is a factorof bythe Factor of a PolynomialSum Theorem.
16. True
17. The program contains aninitial condition in line 10and a recurrence relation inline 30. Given an infiniteamount of time andcomputer memory, it wouldprint all the integersgreater than N 1. 2
(x5 2 xy 4) 1 (xy 4 2 y5)x 2 yxy 4 2 y5
x 2 yx5 2 xy 4x 2 y
xy 4 2 y5 5 y 4(x 2 y)x5 2 xy4
x 2 y
x(x4 2 y 4)x5 2 xy4 5
x4 2 y4x4 2 y4x 2 y
5π6
π3
π4
-1 1 3
-1
1
3y
x
y = cos-1 x
y = cos x
y = x
Answers for LESSON 7-3 pages 418–426 page 3c
1., so 16 is a factor.
2.
. It is clear that 2 isa factor of . Since 2is also a factor of ,then 2 will be a factor oftheir sum,
, by the Factor ofa Polynomial Sum Theorem.
3. a. S(1): 2 is a factor of 2;since , S(1) is true.S(13): 2 is a factor of 158;since , S(13) istrue. S(20): 2 is a factor of382; since ,S(20) is true.b. S(k): 2 is a factor of
. : 2 is a factor of
.c.d.
; 2 is afactor of and 2k, so 2 is a factor of
by the Factor of an IntegerSum Theorem.e. 2 is a factor of ; integers .
4. a. Since 5 is a factor of, S(1) is true.
b. S(k): 5 is a factor of.
c. : 5 is a factor of.6k11 2 1
S(k 1 1)6k 2 1
61 2 1 5 5
n $ 1n2 2 n 1 2
(k 1 1) 1 2(k 1 1)2 2
k2 2 k 1 22) 1 2k(k2 2 k 1
k2 1 k 1 2 5k2 1 k 1 2
(k 1 1) 1 2(k 1 1)2 2
S(k 1 1)k2 2 k 1 2
2 • 191 5 382
2 • 79 5 158
2 • 1 5 2
(2(n 1 1))(n2 1 n) 1
n2 1 n2(n 1 1)
2(n 1 1)(n2 1 n) 1n2 1 3n 1 2 5
(n 1 1)2 1 (n 1 1) 5
16 • 753 2 4 • 3 2 1 5 112 5 d.
; 5 is a factorof and of 5, so 5 is afactor of their sum. Thus,
is true. Hence, S(n):5 is a factor of ;
is true by thePrinciple of MathematicalInduction.
5. Since ,substituting and
into the result ofExample 3 yields is a factor of ;
.
6. Since , substituting and into the result of Example 3yields is a factorof ; .
true. (2) Assume S(k): 6 is afactor of is truefor some positive integer k.Show : 6 is a factorof istrue. Now
.Because 6 is a factor of
by the inductiveassumption, a factor of
by the theorem,and a factor of 12, itfollows that 6 is a factor of
. SinceS(1) is true and S(k)
, by the Principle ofMathematical Induction, 6 is a factor of , ;
.
9. Sample: By Example 3 withand ,
is a factor of.
10. a. S(1) is false. S(2) and S(3)are true.b. Sample: All we canconclude is that S(2) andS(3) are true.c. . 8 is a factor of if .Hence, S(n) holds if .n $ 2
n $ 24n12n 2 8n 5 4n(3n 2 2n)
xn 2 yn 5 22n 2 1
x 2 y 5 3y 5 1x 5 22
n $ 1n3 1 11n
S(k 1 1)⇒
(k 1 1)3 1 11(k 1 1)
3k(k 1 1)
k3 1 11k
3k(k 1 1) 1 12(k3 1 11k) 111k 1 11 53k 1 1 1
k3 1 3k2 111(k 1 1) 5(k 1 1)3 1
(k 1 1)3 1 11(k 1 1)S(k 1 1)
k3 1 11k
12 5 2 • 613 1 11 5 11. S(1) is true because
and
. Assume
S(k): is true
for some integer . Show that :
is
true. Now
. Since
S(1) is true and S(k), by the Principle of
Mathematical Induction,S(n) is true for all integers
.n $ 1
S(k 1 1)⇒
[(k 1 1)(k 1 2)]2
4
(k 1 1)2[k2 1 4(k 1 1)]4 5
4(k 1 1)3
4 5
(k 1 1)3 5k2(k 1 1)2
4 1
(k 1 1)3 5 Fk(k 1 1)2 G2
1
ok
i51i3 1o
k11
i51i3 5
ok11
i51i3 5 F(k 1 1)(k 1 2)
2 G2
S(k 1 1)k $ 1
ok
i51i3 5 Fk(k 1 1)
2 G2
F1(1 1 1)2 G2
5 12 5 1
o1
i51i3 5 13 5 1
Answers for LESSON 7-4 pages 427–431 page 2c
c
12. Let S(n): .which
agrees with the recursivedefinition, hence S(1) istrue. Assume S(k):
is true for someinteger k. Show :
is true.
.Since S(1) is true andS(k) , by thePrinciple of MathematicalInduction, S(n) is true for all
. Hence, the explicitformula correctly definesthe sequence.
14. Right side cos (2x)cos2 x sin2 x Formula for cos(2x)(cos2 x sin2 x) 1 Multiplication by 1(cos2 x sin2 x)(cos2 x sin2 x) Pythagorean Identitycos4 x sin4 x MultiplicationLeft side
Therefore, cos4 x sin4 x cos (2x) by the Transitive Property
2. a. basis step: S(1) is thestatement: If ,then .b. S(3): If , then
.c. inductive step: AssumeS(k). So if , bymultiplication, which implies .But so by theTransitive Property .Thus .
3. The sense of inequality isnot changed because .
4. Let S(n) be the statement :if , then . Letcase 1 be that and case 2 be that .For case 1 and 2 we have,
, since ,S(1) is true. Assume S(k).So if , then xk . 0x . 0
0 , xS(1): xn 5 x1 5 x
x $ 10 , x , 1
xn . 0x . 0
x . 0
S(k) ⇒ S(k 1 1)xk11 , 1
x # 1xk 1 1 , xx • xk , x • 1
0 , x , 1
x3 , 10 , x , 1
x1 , 10 , x , 1
a , cb , ca , b Mx
Product ofPowersProperty
Thus, . So, by the Principle ofMathematical Induction,S(n) is true for all integers
.
5. a. basis step: S(1):
b. S(2):
c. inductive step: S(k):;
by multiplication.But
.Thus
6. Let . Then and, so gives
us a contradiction. So wecan conclude that thestatement is false.
7. See below.
3 . 31 1 2(1) 5 331 5 3n 5 1
S(k) ⇒ S(k 1 1)3(1 1 2k) $ 1 1 2(k 1 1)
3 • 3k 5 3k11 $3(1 1 2k)
3 • 3k $3k $ 1 1 2k
9 $ 532 $ 1 1 2(2) ⇒
31 $ 1 1 2(1) ⇒ 3 $ 3
n $ 1
S(k) ⇒ S(k 1 1)
xk11 . 0x • xk . 0 • x
Answers for LESSON 7-5 pages 432–434
c
7. Let S(m) be the statement: , so S(1) is true.
Assume S(k). Then Definition of S(k)M2
Product of Powers andDistributive PropertyDistributive PropertyTransitive Property, since
Thus, . So by the Principle of MathematicalInduction, S(m) is true for all integers .m $ 1
S(k) ⇒ S(k 1 1)k $ 12k11 $ 1 1 (k 1 1)
2k11 $ 1 1 (k 1 1) 1 k
2k11 $ 2 1 2k2 • 2k $ 2(1 1 k)
2k $ k 1 1S(1): 21 5 2 $ 2 5 1 1 1
2m $ 1 1 m
8. Let S(n) be the statement , so S(9) is true.
Assume S(k). Then Definition of S(k)M2
Product of PowersDistributive PropertyTransitive Property since
Thus, . So by the Principle of MathematicalInduction, S(n) is true for all integers .
9. Let S(t) be the statement: S(4): , so and S(3) is true.Assume S(k). Then
Addition Property of InequalityFactorTransitive Property since
Thus . So by the Principle of MathematicalInduction, S(t) is true for all integers .
10. Let S(n) be the statement: 3 is a factor of .S(1): ; since 3 is a factor of 15, S(1) is true.Assume S(k). Then 3 is a factor of . Show : 3 is a factor of is true.
, By the Factor of an IntegerSum Theorem .
Thus, by the Principle of Mathematical Induction S(n) is true forall integers greater than zero.
1. In the Strong Form ofMathematical Induction,the assumption that each ofS(1), S(2), , S(k) are true isused to show that is true. In the original form,only the assumption thatS(k) is true is used to prove
is true.
2. a. Sample: Combine 1 and2, then join 3, then join 4;combine 1 and 2, then 3and 4, then the two blocks.b. 3
3. The 7-piece and 13-pieceblocks needed 6 and 12steps respectively, by theinductive assumption. Thatis 18 steps; joining them isthe 19th step.
4. a. 2, 2, 4, 8, 14b. i. S(n): an is an eveninteger.ii. areeven integers.iii. a1, a2, , ak are all evenintegers. Prove that isan even integer.iv. .ak, , and are alleven integers by theinductive assumption. Sothere exist integers p, q,and r such that
. By theFactor of an Integer SumTheorem, isthen also an even integer.So is an even integer.ak11
ak 1 ak21 1 ak22
ak22 5 2p 1 2q 1 2rak 1 ak21 1
ak22ak21
ak 2 2ak11 5 ak 1 ak21 1
ak11
K
a1 5 2, a2 5 2, a3 5 4
S(k 1 1)
S(k 1 1)K
v. The sum of any evenintegers is even.
5. a. 5, 15, 20, 35b. and aremultiples of 5. Assume a1,a2, , ak are all multiplesof 5. Show that is amultiple of 5. Now
. 5 is a factor of ak
and , so by the Factor ofan Integer Sum Theorem itis factor of their sum, .Hence, by the Strong Formof Mathematical Induction, an is a multiple of 5 ; integers
.
6. and are oddintegers. Assume a1, a2, ,ak are all odd integers.
, whereand
for some integers qand r. Then
sois odd. Hence, by the
Strong Form ofMathematical Induction,every term in the sequenceis an odd integer.
7. a. S(2): ., so S(2) is true.
S(3): . ,so S(3) is true.b. ; integers j such that
is divisible by 3. , which is divisible by 3, soS(1) is true. Assume S(k):
is divisible by3 is true for some integer
. Show :
is divisible by 3 istrue.
3k1
. isdivisible by 3 by theinductive assumption, and
is divisible by3; thus their sum is divisible3(k2 1 3k 1 2)
k3 1 3k2 1 2k3k 1 2)3(k2 12k) 13k2 1(k3 1
6 511k 16k2 12 5 k3 16k 1 3 1 2k 13k2 11
1k3 1 3k2 12(k 1 1) 53(k 1 1)2 1(k 1 1)3 1
2(k 1 1)3(k 1 1)2 1(k 1 1)3 1
S(k 1 1)k $ 1
k3 1 3k2 1 2k
1 1 3 1 2 5 613 1 3 • 12 1 2 • 1
n3 1 3n2 1 2n
Fk12 1 Fk(Fk11 1 Fk) 5Lk11 5 (Fk21 1 Fk22) 1 by 3. Therefore, by
mathematical induction,is divisible by
3 ; integers .
10. S(n): . S(1): a1
4(3)0. a1 4, so S(1) is true.Assume S(k): ak 4(3)k 2 1 istrue. Show S(k 1): ak 1 1
4(3)k is true. ak 1 1 3ak
3(4 3k 2 1) 4 3k. Bymathematical induction,S(n) is true for all .Hence, is anexplicit formula for thesequence.
11. a.
b. tan x
c. See below.
-2π # x # 2π, x-scale = π -10 # y # 10, y-scale = 5
an 5 4(3)n21n $ 1
•5•55
515
55an 5 4(3)n21
n $ 1n3 1 3n2 1 2n
Answers for LESSON 7-7 pages 442–449 page 2
11. c. Left sideDefinition of cosecant,secant, and cotangent
Multiplication
Subtraction of Fractions
Pythagorean Identity
Simplifying fractions
Definition of tangentRight side
Therefore, csc x sec x cot x tan x ; x for which both sidesare defined.
5255 tan x
5sin xcos x
5sin2 x
sin x cos x
51 2 cos2 xsin x cos x
51
sin x cos x 2cos2 x
sin x cos x
51
sin x • 1cos x 2
cos xsin x
5 csc x sec x 2 cot x
c
c
12. a.b. 100 3 100 3 100c. Yesd. Yes
13. , 2
14. a. a1 and a2 must bemultiples of 7.b. a1 and a2 must bemultiples of m.c. a1 and a2 are multiples ofm. Assume a1, a2, K , ak aremultiples of m.
. Because and are multiples of m, theirsum, , is also a multipleof m by the Factor of aPolynomial Sum Theorem.By the Strong Form ofMathematical Induction,every term in the sequenceis a multiple of m.
15. a. Suppose that arecurrence relation defines
in terms of , , ,, and n for each integer
. Then there is exactlyone sequence defined bythis recurrence relation andthe initial condition .x1 5 a
n $ 1x1
Kxn21xnxn11
ak11
akak21ak21
ak 1ak11 5
x 5 -5
P(x) 5 0.15x3 2 1.5x2 b. Suppose there is asecond sequence y forwhich and isdefined by the samerecurrence relation as .Let S(n) be the statement
. (1) Because and , . So S(1)is true. (2) Assume S(1), S(2),
, S(k) are true. Then, , , .
The sequences have thesame recurrence relation;
is defined in terms of , , and k; and is
defined in terms of , ,and k; so .
Thus, S( ) is true. By (1),(2), and the Strong Form ofMathematical Induction,S(n) is true for all .Therefore, the adaptedRecursion Principle isproved.
b. Apply the Filterdownalgorithm to the list. Thesmallest number is now infront. Apply Filterdown tothe sublist which includesall but the first number.Apply Filterdown to thesublists wich aresuccessively one elementsmaller. Continue until thelist is exausted.
9. Sample: 6, 5, 4, 3, 2, 1
10. Quicksort
11. Quicksort
12. Let S(n) be the statement: If, then .
S(1): , which istrue.Assume S(k).
Inductiveassumption
Mx
Product of Powersand SimplificationTransitiveProperty since
for Therefore, S(k) S( ).Hence, by the Principle ofMathematical Induction,S(n) is true for all integers
.n $ 1
k 1 1⇒x $ 1x2 . x
xk11 $ x
xk11 $ x2x • xx • xk $
xk $ x
x1 5 x $ xxn $ xx $ 1
Answers for LESSON 7-8 pages 450–456
6. L = {5, -7, 1.5, -1, 13, 6}
f = 5 Lr = {13, 6} L, = {-7, 1.5, -1}
(L,)r = {1.5, -1}(L,), = Ø f = -7 f = 13 (Lr)r = Ø(Lr), = {6}
((L,)r)r = Ø((L,)r), = {-1} f = 1.5
c
13. ,
14. and are even. Assume, , are all even.
Show that is even.
forsome integers p and r. Since
is an integer byclosure properties, iseven. Therefore, is evenfor all integers by the Strong Form ofMathematical Induction.
15. a.
b. , for all integers
c.
d.
16. Prove S(n):
for all integers . S(1):
is true.
Assume S(k):
is true. Show that kk 1 1
ok
i51
1i(i 1 1) 5
o1
i51
1i(i 1 1) 5
11 1 1
n $ 1
on
i51
1i(i 1 1) 5
nn 1 1
43
(43)(1 2 (1
4)n)k $ 1
Ak 51
4k21
A2 514, A3 5
116
n $ 1an
ak11
cp 1 r
c(2p) 1 2r 5 2(cp 1 r)ak11 5 c • ak 1 ak21 5
ak11
akKa2a1
a2a1
r 5 1q 5 14 : is
true.
. So by mathematical
induction, S(n) is true ;integers .
17. a.
b. ohms
18. The argument is valid. The three premises are: (1) , (2) , (3) .By the Transitive Propertyand statements (1) and (2),
. Then by modustollens and (3), . This isthe conclusion, and so theargument is valid.
19. Sample: e, d, c, b, a; thereare 24 such orderings. Theonly restriction is that amust be the last letter inthe list.
9; sum of powers of 2: 6b. repeated multiplication:9999; sum of powers of 2:26
4. a. pb.
5. E(10) 499,500
6. a. 1 secondb. 1,000 seconds or16.67 minutesc. 1012 seconds or31,710 years
7.
8. S(n): is divisibleby 16. S(1): isdivisible by 16. S(1) is truesince 0 is divisible by 16.Assume S(k): isdivisible by 16 for somepositive integer k. Showthat 16 is a factor of
.
. 16 isa factor of (bythe inductive assumption)and a factor of 16k. Hence,
E(n ) 5 n by the Factor of an IntegerSum Theorem, 16 is also afactor of 16k. Therefore, bymathematical induction,
is divisible by16 for all positive integersn.
9. Let S(n): .For , the formulayields
. This agrees with theinitial condition, so S(1) istrue. Assume S(k):
is true forsome positive integer k.Show that :
istrue.
.This agrees with therecursive formula, so
is true. Hence, bymathematical induction,S(n) is true for all integers
, and so the explicitformula for the sequence is
.
10. Left side
Right side 2(0 1 49 16) 3(0 1 23 4) (1 1 1 11) 2(30) 3(10) 5 85
11. a.
b. o27
n51(34 1 5(n 2 1))
an 5 34 1 5(n 2 1)
5215111121
111111115
26 1 43 5 855 -1 1 4 1 13 1
3n2 1 5n 2 3
n $ 1
S(k 1 1)
ak 1 6k 1 83) 1 6k 1 8 5(3k2 1 5k 25(k 1 1) 2 3 5
ak11 5 3(k 1 1)2 13(k 1 1)2 1 5(k 1 1) 2 3
ak11 5S(k 1 1)
3k2 1 5k 2 3ak 5
3 5 55 • 1 2a1 5 3 • 12 1
n 5 1an 5 3n2 1 5n 2 3
5n 2 4n 2 1
5(5k 2 4k 2 1) 1
Answers for LESSON 7-9 pages 457–462
c
12. a.
b.c.d.
13. a. NOT(p AND (q OR r))b. Sample: , ,
14. a.b.c. -26 1 26i
11 2 5i1 2 3i
r 5 0q 5 1p 5 0
-5 5 10-10
10
y
x
x 5 4y 5 5
h(x) 5 5 123
x 2 4 15. a. 2b. 3c. Each of the n digits ofthe second number aremultiplied by up to n digitsof the first number. Hence,there are at most n2
multiplications.d. 2n column additionse. ; For theproblem, , so theefficiency should be 8. Allfour multiplication stepsshown, must be carried out,as well as the four columnadditions. , so thealgorithm checks.
Hence, by the Principle ofMathematical Induction,S(n) is true ; integers .
30. (1) S(1): 3 is a factor of. ,
so S(1) is true. (2) AssumeS(k): 3 is a factor of
is true for someinteger . Show that
: 3 is a factor ofis true.
Expanding,
(k3 3k2 3k 1)(14k 14)
(k3 14k) (3k2
3k 15)
(k3 14k) 3(k2 k 5).11115
11115
111115
14(k 1 1)(k 1 1)3 1
(k 1 1)3 1 14(k 1 1)S(k 1 1)
k $ 1k3 1 14k
13 1 14(1) 5 1513 1 14(1)
n $ 1
(k 1 1)(k 1 2)(2(k 1 2) 1 7)2
11(k 1 12 )
111(k 1 12 )
2(k 1 1)(3k 1 9)2 5
k(k 1 1)(2k 1 7)2 1
(k 1 1)(3k 1 9) 5
k(k 1 1)(2k 1 7)2 1
3(k 1 1)(k 1 3) 5
ok11
i513i(i 1 2) 5 ( o
k
i513i(i 1 2)) 1 By the inductive
assumption, 3 is a factor ofk3 14k, and 3 is a factor of3(k2 k 5). Therefore, 3 isa factor of their sum by theFactor of an Integer SumTheorem. Hence, bymathematical induction,S(n) is true ; integers .
31. (1) S(2): 3 is a factor of . ,
so S(2) is true. (2) Assumethat S(k): is a factor of
is true for someinteger . Prove
: 3 is a factor ofis true.
2(k 1)3 5(k 1)2k3 6k2 6k 25k 5 (2k3 5k)3(2k2 2k 1)Since 3 is factor of 2k3 5kand 3(2k2 2k 1). 3 is afactor of their sum by theFactor of an Integer SumTheorem. Hence, bymathematical induction,S(n) is true ; integers .n $ 2
Hence, .Therefore, by the Principleof Mathematical Induction,S(n) is true for all integers
.n $ 1
S(k) ⇒ S(k 1 1)
51
k 1 2( 1k 1 1)(k 1 1
k 1 2)5
5 ( 1k 1 1)((k 1 2
k 1 2) 2 ( 1k 1 2))
5 ( 1k 1 1)(1 2
1k 1 2)
5 S(k)(1 21
k 1 2)(12
1k 1 2)(12
1k 1 1)
(1213) K (12
12)
32. Let S(n) be the statement: If , then .(1) S(1): , so S(1) is true, since by the givenstatement. (2) Assume S(k) for some arbitrary integer. Show S(k 1): If , then .
by the inductive assumptionMx
Product of powersGiven SimplificationGiven
Hence, . Therefore, by mathematical inductionS(n) is true for integers .
33. (1) S(2): . So S(2) is true.(2) Assume S(k). Show .
by the inductive assumptionM4
SimplificationDistributive PropertyGiven
So S(k) S(k 1). Therefore by the Principle of MathematicalInduction, S(n) is true for all integers .n $ 2
35. a. 0, 4, 4, 16, 28b. Let S(n): 4 is a factor of
.(1) S(1): 4 is a factor of .
, so S(1) is true. S(2): 4 is factor of .
, so S(2) is true.(2) Assume S(1), S(2), K ,and S(k) are true for someinteger . So 4 is afactor of b1, b2, K , bk.Show S(k 1): 4 is a factorof is true. Since and have 4 as a factor,there exist integers p and qsuch that and
. Substituting intothe recurrence relation,
4q 3(4p) 4q12p 4(q 3p). q 3p isan integer by closureproperties, so 4 is a factorof . Hence, by theStrong Form ofMathematical Induction,S(n) is true for all integers
42. a. 10, 13, 16, 19, 22b. explicit; the nth term ofthe sequence is given as afunction of n in line 20.c. an 5 3n 1 7
S20 5 2(1 2 (12)20)
o20
i51(12)i21
n $ 1
an 5 (12)n21k $ 1
ak11 512aka1 5 1
L = {21, 1, 8, 13, 1, 5}
f = 21 Lr = Ø L, = {1, 8, 13, 1, 5}
(L,)r = {8, 13, 5}(L,), = {1} f = 1
((L,)r)r = {13}((L,)r), = {5} f = 8
k $ 1
Hb1 5 2bk11 5 2bk
Answers for Chapter Review pages 467–469 page 5c
1. a. Leonhard Euler; 18thb. Cardano; 16thc. Wessel; 18th
2. a. 8i b.3. a. 21 b. 4
4. real 5 8, imaginary 5 7
5. 10 1 15i
6. a. True b. No
7. 3 2 4i
8.9. a. 8 1 i b. 8 2 i
c. 2 1 i d. 3 2 2ie. 8 2 i; they are equal.
10.11. a-d.
12. a. 4 1 2i ohms
b.13. a. 5 2 6i
b.
c. parallelogram; Sample:slopes of opposite sides areequal.
imaginary
real0
z + w
z
w
4
-4
-8
4 8-4
65 2
35 i
-12 -9 -6 -3 3 6 9 12
-9-6-3
369
imaginary
real
d (0, -7)
c (-3, 0)
b (-4, 7) a (12, 8)
-35 1
45 i
2 167 i
-
--
Ï20 i 5 2Ï5 i
14. x 5 7, y 5 9
15. a. ohms
b. 5i ohms
16. For two imaginary numbersmi and ni, mi 1 ni 5(m 1 n)i. Since m and n arereal, m 1 n is real, and (m 1 n)i is an imaginarynumber.
17.
5
So 1 1 i is a solution of.
18. 63i, 62i
19.
20. Let . Then 5a 2 bi. By definition ofequality for complexnumbers, ifand only if and b 5
then 2 and .So , andz is a real number.
21. Suppose there is a smallestinteger, s. Then ,and is an integer. Thiscontradicts the assumptionthat s is the smallestinteger. So the assumptionis false, and there is nosmallest integer.
c. domain: {x: 1 2};range: {y: 2 5}d. They are inverses.
# y ## x #
y = f (x)
reflection imageof graph of
y = f (x)
1
1
2
3
4
5
6
2 3 4 5
y
6 x
# y ## x #
# t # ---
25. a.
b. They are complexconjugates.
c.
d. For , the
roots are
and .
So
1
5
. 5ca5
4ac4a2
b2 2 (b2 2 4ac)4a2
-b 2 Ïb2 2 4ac2a
• -b 1 Ïb2 2 4ac2az1 • z2 5
-ba .5-2b2a 5
-b 2 Ïb2 2 4ac2a
-b 1 Ïb2 2 4ac2az1 1 z2 5
-b 2 Ïb2 2 4ac2az2 5
-b 1 Ïb2 2 4ac2az1 5
ax2 1 bx 1 c 5 0
z1 • z2 5 2z1 1 z2 5 -32,
z2 5 -34 2Ï23
4 i
z1 5 -34 1
Ï234 i,
Answers for LESSON 8-1 pages 472–479 page 2c
1. a-d.
2. a. Sample:
b. Sample:
c. Sample:
3. a. By case b of the polarrepresentation theorem in
this lesson,
1
Then by case b again,
for any
integer k is a coordinate
representation of .
b. By case c of the polarrepresentation theorem in
this lesson,
. So by
case b, for
any integer k is acoordinate representation
of .F4, π3G
F-4, 4π3 1 2kπG
πG 5 F-4, 4π3 Gπ
3 1F-4,
F4, π3G 5
F4, π3G
F4, -5π3 1 2kπG
2(-1)πG 5 F4, -5π3 G.F4, π3
F4, π3G 5
F2, -5π6 G
F-2, π6GF2, 19π
6 G
2
ad
c
b
4
4. a., b.
5. Given any point [r, θ].First plot P and Q 5 [1, θ] 5(cos θ, sin θ).
Because [r, θ] is r times asfar from the origin as Q, itsrectangular coordinates are(r cos θ, r sin θ). Thus, therectangular coordinates ofP are given by cos θand sin θ.
6. (0, 4)
7. (1.7, 1.5)
8.9. [ , 56.3°]
10. a. The ship should sail 26.3°East of South.b. .38 hours or
-3 1 4i 20. a. cos (θ 1 φ) 5 cos θ cos φ2 sin θ sin φb. sin (θ 1 φ) 5 sin θ cos φ 1cos θ sin φ
21. a.
b. 6.5 mpg
22. Let the vertices in clockwiseorder be A( 1, 0), B( 5, 5),C( 11, 5), and D( 7, 0). Theslope of 5 0; the slope of 5 0. The slope of
5 ; the slope of
5 . Since
ABCD is composed of twopairs of parallel lines, it is aparallelogram.
23. The north magnetic polelies just north of NorthAmerica and west ofGreenland and is aboutlatitude 76° N andlongitude 101° W onBathurst Island in Canada.The south magnetic pole isin Antarctica and is aboutlatitude 66° S and longitude140° E, just off the coast ofAntarctica due south ofAustralia. The location ofboth magnetic poles varyover time.
-5 2 0-11 2 (-7) 5
54DC
-5 2 0-5 2 (-1) 5
54
ABBCAD
------
400h26 1 h
Answers for LESSON 8-2 pages 480–486 page 2c
1. ;
5
The slope of
the slope of 5
and so .
The slope of 5
the slope of 5
and so .
Therefore, the figure is aparallelogram.
2. ;;;
. The
slope of ;
the slope of 5
5 ; and so .
The slope of
; the slope of
; and so
. Therefore, thefigure is a parallelogram.
3. a.
b. 3Ï2(cos3π4 1 i sin 3π
4 )F3Ï2, 3π
4 GZP z zOW
d 2 0c 2 0 5
dcOW 5
dc
b 1 d 2 ba 1 c 2 a 5
5ZP
WPiOZba
b 1 d 2 da 1 c 2 cWP
5b 2 0a 2 0 5
baOZ
5 (a 1 c, b 1 d) 5 Pz 1 w 5 (a 1 c) 1 (b 1 d)iw 5 c 1 di 5 (c, d) 5 W0 5 0 1 0i 5 (0, 0) 5 Oz 5 a 1 bi 5 (a, b) 5 Z
the diagonals have thesame midpoint, ZOWP is aparallelogram.
15. a.b.
16. a.
b. , and, while , and
Then
, ; and
5 . So the
triangles are similar by theSSS Similarity Theorem.c.
17. a.
b. , and, while , and
. So EFG bythe SSS CongruenceTheorem.
nE′F′G′ùn2Ï2F′G′ 5E′G′ 5 Ï29E′F′ 5 5,FG 5 2Ï2
EF 5 5, EG 5 Ï29G′ 5 3 1 6i
E′ 5 5 1 i, F′ 5 1 1 4i,
Ï5
Ï5B′C′BC 5
Ï202
A′B′AB 5
5Ï5
5 Ï5Ï5
A′C′AC 5
Ï85Ï17
5Ï85.
A′C′ 5B′C′ 5 Ï20A′B′ 5 5,AC 5 Ï17
AB 5 Ï5, BC 5 2C′ 5 8 1 i
B′ 5 4 1 3i,A′ 5 1 1 7i,
-π # x # 2π, x-scale =
-2.5 # y # 2.5, y-scale = 1
π]2
zw z 5 Ï5, -θ ø -63.4°
(7 1 02 , 5 1 0
2 ) 5 (72, 52)PO
(4 1 32 , 6 1 (-1)
2 ) 5 (72, 52)
ZW c.18. a.
5
5 5
and 5
5 .b. ; real numbers c andcomplex numbers
, Proof:
5 5
19. a.
b.20.21. Let , where a and
b are real numbers. Then
1 . Since a isreal, 2a is real. So for allcomplex numbers, the sumof the number and itscomplex conjugate is a realnumber.
22.
23. a. Yesb. domain: the set of realnumbers; range: the set ofintegers
3π8
(a 2 bi) 5 2a(a 1 bi)z 1 z 5z 5 a 2 bi.
z 5 a 1 bi
14 1 8i
x2 1 y2 5 16
4 8
zc z zz z .zc z za 1 bi zÏa2 1 b2zc zÏc2a2 1 c2b2 5
zcz z 5 zca 1 cbi z 5zcz z 5 zc z zz z .a 1 biz 5
3Ï413Ï16 1 253Ï42 1 (-5)2
3 z4 2 5i z 53Ï41,Ï369Ï144 1 225
Ï122 1 (-15)2
z12 2 15i z 5
T1,1
Answers for LESSON 8-3 pages 487–492 page 2c
c
24. center , radius 5 3
25. Geometric SubtractionTheorem: Let and be twocomplex numbers that arenot collinear with theorigin. Then the pointrepresenting is thefourth vertex of aparallelogram withconsecutive vertices
, 0, and .w 5 c 1 dia 1 biz 5
z 2 w
w 5 c 1 diz 5 a 1 bi
5 (-1, 2) 26. Geometric DivisionTheorem: Let z and w becomplex numbers. If
, and ,
then .
That is, dividing a complexnumber z by w applies to za size change of magnitude
15. The chambered nautilus isalso known as the pearlynautilus. It has a smooth,coiled shell 15–25 cm indiameter, consisting of 30to 36 chambers; it lives inthe outermost chamber.
-9 2 2i-20 2 10i
-4 -2 2 4 6 8
-4
-2
2
4z
u
w
u'
v'
v
w'
z '
imaginary
real
z9 512 1 Ï3 1 (Ï3
2 2 1)i(-1 2
3Ï32 )i,
w9 5 -32 1 Ï3 1
v9 5 -32 23Ï3
2 i,
u9 512 1
Ï32 i,
-14 2Ï34 i
512 (cos 4π
3 1 i sin 4π3 )F12, 4π
3 GF2Ï2, π4G
2 4
r = 4 sin u
r = 4 cos u
Answers for LESSON 8-5 pages 500–506 page 3c
c
9. b. cos
cos θ sin θc. cos θ sin θ
⇒
⇒ ⇒ ⇒ ⇒ ⇒
⇒
Hence, the graph is a circle.
10.
2 4 6
1 2
5 ( 1Ï2)2
(x 212)2
1 (y 212)2
y 114 5
12
x2 2 x 114 1 y2 2
x2 2 x 1 y2 2 y 5 0x2 1 y2 5 x 1 y
r2 5 x 1 yx 1 y
r
r 5xr 1
yr
1r 5
15
sin θ • Ï22 )
5 Ï2(cos u • Ï22 1
sin π4)sin θ
15 Ï2(cos θ cos π4
(θ 2π4)r 5 Ï2
12. 625
13. a.
b.
c. 2; 128; 8192; 524,288;33,554,432
14. Sample: Using DeMoivre’sTheorem, for [r, θ] with
, , and 5 r. So and
. Sample: Usingmathematical induction, letS(n): . S(1): , so S(1) is true. Assume S(k): . Then
5So S(k) ⇒ S( ), and S(n)is true for all integers n.
22. Sample:5 REM ENTER A COMPLEX NUMBER, A + BI, AND DESIRED
ROOT, N
10 PRINT “WHAT IS THE REAL COMPONENT, A, OF THECOMPLEX NUMBER”;
20 INPUT A30 PRINT “WHAT IS THE IMAGINARY COMPONENT, B”;40 INPUT B50 PRINT “WHICH ROOT DO YOU WANT”;60 INPUT N70 LET PI 5 3.1415926535980 IF A 5 0 AND B 5 0 THEN PRINT “0 IS THE ONLY
ROOT.”:GOTO 190
85 REM CALCULATE THE ARGUMENT, D, IN RADIANS OF A 1 BI
90 IF A 5 0 AND B 0 THEN LET D PI/2100 IF A 5 0 AND B 0 THEN LET D 5 PI/2110 IF A 0 THEN LET D 5 ATN(B/A)120 IF A 0 THEN LET D 5 ATN(B/A) 1 PI125 REM CALCULATE THE ABSOLUTE VALUE OF A + BI130 LET L = SQR(A * A + B * B)135 REM OUTPUT THE N NTH ROOTS OF A + BI140 PRINT “THE ABSOLUTE VALUE OF EACH ROOT IS”;L^(1/N)150 PRINT “THE ARGUMENTS ARE”160 FOR I 5 0 TO (N 2 1)170 PRINT (D/N) 1 I*(2 * PI/N)180 NEXT I190 END
b. No, the coefficients of pare not all real numbers.
4. False
5.
6. (with multiplicity 2)
7. Sample: 1 15
8. Sample:
9. Sample: y
x
y
x
p(x) 5 x3 2 x2 2 7x
2 2 i, -2
1 2 iÏ32 , 4 2 5i
-25 1 20 1 5 5 020i2 1 525i2 2
p(-5i) 5-1 2 4 15 5p(i) 5 i2 1 4i2 1
-23 2 77i-23 1 77i
4 1 7i 10. Sample:
a. There are four nonrealzeros: either two conjugatepairs (each zero ofmultiplicity 1), or one conjugate pair (each zeroof multiplicity 2).b. There is one pair ofnonreal conjugate zeros,each of multiplicity 1.c. There are no nonrealzeros.d. There is one pair ofnonreal conjugate zeros,each of multiplicity 1.
c. is the reflectionimage of over thereal axis, so 5 .z 1 wz 1 w
z 1 wz 1 w
y
x-2
-4
-2
2
4
6
z + ww
z
z + ww
z
≈ [1.59, 1.65]
≈ [1.59, 3.22]
≈ [1.59, .08]
≈ [1.59, 4.79]
6Ï3 i
c 5
3x 2 28x212x3 2p(x) 5
1 1 3i
14x 2 20x3 2 4x2p(x) 5 19. a. 3
b.20. not ((p and q) or r)
21. a. For , the zeros are, ; for 0, the
zeros are 0, ; for 3,the zeros are , . Asc slides from -5 to 0, its tworeal zeros move closer tothe origin, and its twoimaginary zeros convergeat the origin. As c thenslides from 0 to 3, thepolynomial has 4 real zeros.b. For 4, the zeros are
; for , the zeros
3 are . As c slides from
3 to 4, its positive zeros converge to , and itsnegative zeros converge to - . As c slides from 4 to
, its four complex zerosconverge to two complexzeros.
254
Ï2
Ï2
3 6 i2
c 52546Ï2
c 5
616Ï3c 562
c 56i6Ï5c 5 -5
2t2 2 4t 1 2
Answers for LESSON 8-9 pages 526–533 page 2c
c
22.
For , the zeros areapproximately 2.50, 2.34,and .08 .93i. As cincreases, the real zerosmove slowly toward theorigin and the nonrealzeros move toward eachother until they merge into
6-
c 5 -5
-1 1 real
imaginary
-1
1
p-5p0
p-1p5
p5 p1p-5
p0p5
p5 p-1
p-5
p-5
p-1
approximately .11 (a zeroof multiplicity 2) when c-.05; at that time, the otherzeros are -2.33 and 2.13. Asc increases, the zero at .11splits into two real zeroswhich move apart alongthe real axis, while theother real zeros continuemoving toward the origin.When c 4.69, the twolargest zeros merge into1.53, then split intocomplex conjugates.
6. a. Let be the kth numberobtained (by pressing thecosine key k times). Then isapproaching x for larger andlarger k. But cos ,and cos approaches cos xsince the cosine function iscontinuous. Therefore,
14. a. i. Sample: [6, 110°]ii. Sample: [243, 200°]iii. Sample: [32, 350°]
b.[7776, 550°].
[243, 200°][32, 350°] [(243)(32), 200°1 350°] [7776, 550°]. So .
15. a.
b. θ 5 0°
16.
17. 116.7 feetø1152π2
Ht1 5 1tn11 5 tn 2 4 for n $ 1
2
(zw)5 5 z5 • w55
5 • z5 • w5 5
[65, 5 • 110°] 5(zw)5 5 [6, 110°]5 5
18. If m is odd, then for some integer k.
Since is aninteger, is anodd integer by definition.∴ If m is any odd integer,then is an oddinteger.
19. f(z) is constructed bysquaring the absolute valueof the complex number zand doubling its argumentto obtain . The point isthen translated by adding cto give f(z).
22. a. f(1) < 488.5; that is veryclose to the actual value of 490b. f(305) < 566; that is 17 minutes less than theactual value, or within 3%of the actual time.
c. For the shortest day, Dec. 21, , and forthe longest day, June 21,
. f(355) <483 minutes and f(172) <983 minutes.d. Answers will varydepending on students’latitude.
x 5 172
x 5 355
Answers for LESSON 9-1 pages 552–559 page 2c
Answers for LESSON 9-2 pages 560–567
1. 480 ft/sec
2. a. Samples: 17.5 ft/sec, 22.5 ft/sec, 30 ft/secb.
slope ≈ 40 ft/secc. Sample: 40 ft/sec
3. a.
b.
-3 3
-3
3y
x
y 5 -52x 2 2
2Time (seconds)
Dis
tanc
e (fe
et)
4 6
30
60
90
120
150
d
t
Q1Q 2Q 3
P
c. Sample:
4. The derivative of a realfunction f at a point x is
, provided
this limit exists and is finite.
5. True
6. a. , ,
b. , ;
, ; ,
7. For a discrete function, youcannot find values of x sothat .∆x → 0
this limit does not existbecause it has differentvalues when approachingzero from the right andfrom the left.
12. a. i. during the attack andslope times
ii. during the decay andrelease time
iii. Sample: at the breakpoint and during thesustain time
b. i. sound getting louderii. sound getting softeriii. constant volume
13. a. f9(7:05) ft/min;
f9(7:15) 1 ft/minb. At 7:05, the water level is falling at about 1.3 ft/min; at 7:15, the water level isrising at about 1 ft/min.
14. a. 3b. The average rate ofchange is the slope of thesecant line through and , but f is aline. Hence, the slope of thesecant line is the slope of f.
-2π # x # 2π, x-scale = π-1.5 # y # 1.5, y-scale = 0.5
f9(2) 5 -2f9(1) 5 0f9(0) 5 2f9(-1) 5 4
f9(x) 5 -2x 1 2
-2 # x # 10, x-scale = 2-75 # y # 375, y-scale = 75
(5, 97)
g9(x) 5 8x
f9(t) 5 0
f9(x) 5 -6 14. a.b. Shows tangents at
, 0, and 1.
c. for all x. Thederivative of a constant iszero, so if ,then forany real number c.
15. a. ib. Velocity as a function oftime
16. a. square inchesper inchb. If the value is 2 inches ofwidth, then the area of theborder is increasing at therate of 52 square inches ofborder for an inch of width.
1. 61,800,000 is the averagerate of change in worldpopulation for the years1960 to 1965. It is found bytaking one-fifth of thedifference between the1965 and 1960 populations.
2. per year
3. 5,363,600,000 people
4. 10 mph/sec
5. True
6. False
7. a. increasingb. decreasingc. Neither; it is always -32 ft/sec2
8. a.15 9.8t m/secb.-9.8 m/sec2
9. per minute
or degrees Fahrenheit/min2
degrees Fahrenheitmin
a(t) 5 v9(t) 5 s0(t) 52v(t) 5 s9(t) 5
-440,000people
year
10.
a. < 8.12 degrees/minb. < -14.63 degrees/min2
c. at the beginning; d. at the end;
11. 7,742,000,000
t 5 10t 5 0
graph of f '
graph of f "
graph of f
-2 # x # 10, x-scale = 2-50 # y # 10, y-scale = 20
-2 # x # 10, x-scale = 2-180 # y # 180, y-scale = 80
3 as a factor so S(1) is true.Assume S(k) k3 2k has a factor of 3. Then S(k 1)k3 3k2 3k 1 2k2 (k3 2k) (3k2
3k 3). k3 2k has afactor of 3 and 3k2 3k3 3(k2 k 1) so it has a factor of 3. Thus, S(k) S(k 1) and byMathematical Induction n3 2n has a factor of 3 forall integers n.
15. See below.
16. a. all real numbers except or
b. removable discontinuityat , essentialdiscontinuity at x 5 -2
x 5 1
x 5 1x 5 -2
1
1⇒
11511
111115
1111151
15
515
12
x
y
-2 2 4 6-1
1
f9(6) < 12f9(4) < 1f9(1) < -2
c.17. a. III
b. IVc. IId. I
18. Sample:a. There were more housesbegun (to be built) in thecurrent period than in theprevious period.b. The unemployment levelstarted dropping moreslowly.c. The principle that anincrease in the demand fora finished product willcreate a greater demandfor capital goods.d. Two bodies attract eachother, and that force causesthem to come together atan ever-increasing velocity.
19. a.–b. Answers will vary.
x 5 -2
-5 # x # 3, x-scale = 1-15 # y # 5, y-scale = 5
Answers for LESSON 9-4 pages 576–580 page 2c
15. cos 2x cos2x sin2xcos 2x (1 sin2x) sin2x Pythagorean Identitycos 2x 1 2 sin2x Addition2 sin2x 1 cos 2x
circuits: where E is a voltage source,L is an inductance, R is aresistance, and I is current.
E(t) 5 LI9 1 RI
b 5 e2
Answers for LESSON 9-6 pages 587–591c
Answers for Chapter Review pages 596–599
1. 9
2. a.b. i. 3.31
ii. 3.0301
3. a.b. -4
4.5.6.7.8.9. False
10. increasing
11. . Since for all real x,
for all real numbers.Since the derivative ispositive, the slopes of thetangents to the curve areall positive, and thefunction is increasing for allreal numbers.
12. decreasing
3 $ 0f9(x) 5 3x2 1
x2 $ 0f9(x) 5 3x2 1 3
k9(x) 5 -6x 1 2
g9(x) 5 -6x
f9(x) 5 2
f9(x) 5 -2x 1 1
f9(1) 5 4, f9(-1) 5 -4
-4t 1 5 2 2∆t
3 1 3∆x 1 (∆x)2
13. a. i.ii.iii.
b.
14. a. 1 min/oz; an extraminute of cooking time isrequired for everyadditional ounce ofpotatoes.
b. min/oz; as weight
increases, the rate ofchange of baking timeneeded decreases for everyadditional ounce ofpotatoes.c. Potatoes weighingbetween 10 and 16 ouncesneed less cooking time perounce than potatoesweighing between 4 and 6 ounces.
8. It would be impractical torepresent all 3,276,000possible outcomes
9. 3168
10. a. 10,000b. dw
11. 1716
12. 84
13. a. 555b. < 0.0054
14. a. There are two choicesfor each element: include in the subset, or don’tinclude it. Since there are n elements, by theMultiplication CountingPrinciple there are n factorsof 2, or 2n subsets.b. 15c. a spoonful of cereal (andmilk) only; no fruit
15. S(1) is true, because thenumber of ways the firststep can be done is n1.Assume S(k), the number ofways to do the first k steps,is . Let m bethe number of ways to dothe first k steps and let nrepresent the number ofways to do the ststep. Then by the inductivehypothesis, m n1 • n2 •K • nk and so mn n1 • n2 • K • nk • .So S(k 1), the number ofways to do the (k 1) stepsis n1 • n2 • … • nk • .Thus S(n) is true for all n.
14. For , the number ofpossible permutations is1 1! n! Assume thetheorem is true for n k.That is, there are k!permutations of k differentelements. Consider nk 1. There are k!1
5
555
n 5 1
52n $ 2
10!1! 5
10!1 5
10!0! 5
10!(10 2 0)! 5
10!(10 2 9)! 5P(10, 9) 5
n 5 3
amth
amht
h
t
tm
h
atmh
athm
h
m
mta h
ahmt
ahtm
t
m
mh
t
permutations of the first kelements by the inductivehypothesis. Then there arek 1 places to insert the (k 1)st element into anyof these permutations. Soby the MultiplicationCounting Theorem, thereare k!(k 1) (k 1)!permutations. Hence, bymathematical induction,the Permutation Theorem istrue.
15. a. (n 2)! (n 2)(n 3)(n 4)(n 5) K(2)(1) (n 2)[(n 3)(n 4)(n 5) K (2)(1)](n 2)(n 3)!b. n 7
The sum of the squares forrow n seems to be themiddle element of row 2n.So the sum of squares ofthe elements of the 12throw would be the middleelement of row 24.
11. a. The number of 5-element subsets of S1 plus the number of 4-element subsets of S1 isthe number of 5-elementsubsets of S.b. Let S be a set of elements and let S1 be a setof n of these elements. Thenevery r-element subset of Sis either an r-element subsetof S1 or an (r – 1)-element ofS1 along with the left-overelement not in S1. So
26. a. 56b. 56c. 56d. 56e. The problem of findingthe number of terms in a–dis equivalent to theproblem of finding thenumber of sequences of 4nonnegative integers whichadd to 5.
f. To find that number,
evaluate for
and .
.8 • 7 • 63 • 2 • 1 5 56
(85) 5(5 1 (4 2 1)
5 ) 5
n 5 4r 5 5
(r 1 (n 2 1)r )
Answers for LESSON 10-7 pages 639–644 page 2c
Answers for LESSON 10-8 pages 645–648
1. a. 21b. 10
2. a. m3 3m2n 3m2p3mn2 6mnp 3mp2
n3 3n2p 3np2 p3
b. 8x3 12x2y 12x2
6xy2 12xy 6x y3
3y2 3y 1
3. a. 7560b. 1030
4. 13,860
5. The coefficient of in the expansion of (x1
x2 x3)n is equal to thenumber of choices in thefollowing countingproblem. A set has nelements. You wish tochoose a1 of them. Then,from the remaining n a1,2
11
x1a1x2
a2x3a3
222112
121111
111111
you choose a2. Then, fromthe n a1 a2 thatremains, you choose a3. Thenumber of ways to makethe 3 selections is
8. a. Think of the exponent 6as 6 identical balls. Think ofthe variables x, y, z, and was boxes. Each distributionof all six balls into the fourboxes can be thought of asa term in the expansion of(x y z w)6.b. Suppose that S is the set{x, y, z, w}. The number of6-element collectionsselected from these 4elements of S gives thenumber of terms in theexpansion of (x y zw)6.c. 84
9. 495
10. < 0.19
11. 4,838,400
12. 4,096,000,000
13. ; u < 5.44y < -4.47
111
111
14. S(1): a1 c, so a1 is divisibleby c.S(2): a2 c, so a2 is divisibleby c.Assume S(k): ak is divisibleby c, for all .Then
for someintegers p and q
, where is an integer.So is divisible by c.So S(k) S( ), and,therefore, every term isdivisible by 3.
15. a. The degree of d(x) is oneless than the degree of p(x).b. p(x) (3x 2) d(x)x 2, so some samples are:d(x) p(x)
15. The tester prefers A to C,and prefers C to P. But in adirect comparison of A andP, the tester prefers P.
16. a.
b. < .396c. < .474
17. a. 18 b. 17 c.18. Sample:
7x 1 15p(x) 5 x3 2 x2 2
V 5 E 1 1
ChocolateChip
PeanutButter
ChocolateChip
Vanillawafer
redjar
greenjar
5]]12 5
]]24
7]]24
3]]16
5]]16
7]]12
3]8
5]8
1]2
1]2
v4e2
e1
e6
e5
v1 v2
e4
e3
v3
19. quotient: ; remainder:
14,043
20. a.
b.n
21. a. i. 1ii. 4iii. 6iv. 4v. 1
b. i. 1ii. 3iii. 3iv. 1
c. Pascal’s triangle;binomial coefficientsd. 1, 2, 1 is row 3 of Pascal’striangle.
Fcos (nu)sin (nu)
-sin (nu)cos (nu)G
5Fcos usin u
-sin ucos uG
Fcos 2usin 2u
-sin 2ucos 2u
G287x 2 2006
6x3 2 42x2 1
Answers for LESSON 11-2 pages 666–672 page 2c
1. a. ; ;;
b. 8
2. The statement does notinclude the case whenedges are loops which arecounted twice.
3. a. 2 b. v1 c. 1d. v2 e. 1 f. v1
g. contributes 1 to thedegree of v3 and 1 to thedegree of v1.h. contributes 2 to thedegree of v3.
4. 253
5.
6. a. 28b.
c. It is equivalent to 8 people shaking handswith each other.
7. 42
8. 8
deg(v4) 5 3deg(v3) 5 0deg(v2) 5 3deg(v1) 5 2 9. Assume that each of the
nine people could shakehands with exactly threeothers. Represent eachperson as the vertex of agraph, and draw an edgejoining each pair of peoplewho shake hands. To saythat a person shakes handswith three other people isequivalent to saying thatthe degree of the vertexrepresenting that person is 3. The graph would thenhave an odd number ofvertices of odd degree. Thiscontradicts Corollary 2 ofthe Total Degree of a GraphTheorem. Thus, the givensituation is impossible.
10. The total degree of anygraph equals twice thenumber of edges in thegraph.
11. counterexample:
12. Impossible; it can’t have anodd number of oddvertices.
15. Impossible; one of thedegree 3 vertices goes toeach of the other vertices.But the other degree 3vertex cannot connect toitself (the graph is simple),to the first degree 3 vertex(no parallel edges), or tothe other two vertices (theyalready have one edge).
16. a.b.
c.
17.
18. a.
b. It is the same problembut with one more person.
19. “Twice the number ofedges” must be an evennumber.
(n 1 1)n2
n(n 2 3)2
n(n 2 1)2
n(n 2 1)n 2 1
v1 v3
v4v2
20. The set of even verticescontributes an evennumber to the totaldegrees of the graph. Sincethat total degree is even,the set of odd vertices mustalso contribute an evennumber to the total degree.An odd number of oddvertices would contributean odd number, so thenumber of odd verticesmust be even.
21. See below.
22. 10 hoursA
B
C
D
E
F
G
H L
I
J
K
2
3
36
3
28
9
4
5
3
4 110
3
1
2
Answers for LESSON 11-3 pages 673–678 page 2
21. F MiamiMilwaukee
Minn./St. Paul
Miami012
Milwaukee107
Minn./St. Paul290G
c
c
23. , , ,
24. a. , b. , c.
25.
26. a. If G is a graph with medges and n vertices and
, then G is not
a simple graph.
m .n(n 2 1)
2
F 718
1231G
-1 , x , 2x . 3-3 , x , 1
1 , x , 3x , -3
-iÏ32
iÏ32-Ï5
5x 5Ï55 b. Proof of statement:
Given a simple graph Gwith n vertices and medges, let maximumnumber of edges possible(that is, ). We knowfrom the lesson that
8. c. 3d. No, you can use pairs ofe1, e2, and e3 as many timesas you wish.
9. Yes, if the graph is notconnected, there is no waya circuit could contain everyvertex.
10. Not necessarily; there is aconnected graph (whichmust have an Euler circuit)and a nonconnected graph(which cannot).
11. Yes; think of replacing eachbridge by two bridges. Sucha walk exists by thesufficient condition for anEuler Circuit Theorem, sinceevery vertex will have aneven degree.
12. No
13. Yes, if the walk repeats anedge, then there is a circuit.Remove edges from thegraph until there is nocircuit. Then connect v to w.
v1 v2
v5
v3
v1 v2
v4
v3
v4 v5
14. Sample:
15.
16. a.
b. < 2.48%c. < 19.76%d. < 80.24%
17. Leonhard Euler, eighteenth
18. 120
19. < -.204
20. (x, y) → ( , )
21. n must be an odd number.For the complete graph ofan n-gon, the degree ofeach vertex is . Thatdegree is even if n is odd.
n 2 1
3y 1 4x 1 5
0.005
0.0049
0.02
0.98
0.98
0.02
Hascancer
Test Positive
Test Negative
Test Positive
Test Negative
Doesn't have
cancer
0.995
0.0001
0.0199
0.9751
F0111
1011
1101
1110G
v1 v2
v3v5
v4
Answers for LESSON 11-4 pages 679–686 page 2c
1. False
2. 2, 0
3. a. 0b. 0c. 1
4. False
5. False
6. e2 e3; e3 e2; e3 e3; e2 e2; e5 e5
7. e4e1e4, e4e2e5, e4e3e5, e5e2e4,e5e3e4
8. e2e3e3, e3e2e2, e3e3e2
9. 5 walks: e1e2, e1e5, e4e2, e4e5,e6e3
10. ;
11. If the main diagonal is allzeros, there are no loops,and if all other entries arezero or one, there are noparallel edges, so the graphis simple.
12. 105
13. 4
F 39
11
94442
114219
GA3 5
F231
3136
16
10GA2 5
14. a. If the adjacency matrixfor a graph is symmetric,then its graph is notdirected.b. Sample:
21. Left network , (p and (qor , r)), p or , (qor , r), p or (, qand r)Rightnetwork
Therefore, the left networkis equivalent to the rightnetwork.
22. a. ,
b. There are an infinitenumber of solutions. For all
real x, .y 549x 1
23
y 51711x 5
1411
;
;
;
;z Þ 1z2 1 1
z 2 123. a. ,
,
,
, , .The pattern is .b. The paths betweenvertices are circular.c. v1
v3
v2
An 5 An(mod3)A6 5 A3A5 5 A2A4 5 A1
F001
100
010GA4 5
F100
010
001GA3 5
F010
001
100GA2 5
F001
100
010GA 5
Answers for LESSON 11-5 pages 687–692 page 2c
1. a. 15%b. 75%
2. a. 36%b. < 22.9%
3. Yes
4. a. the probability that itwill be cloudy 10 days aftera cloudy dayb. No matter what theweather is today, theprobabilities for theweather in 10 days areabout the same.
5. a. 0.7b. 0.6
6. ;
7.
8.
In 20 generations, of the
seeds will produce pale
flowers, and brilliant
flowers, no matter whatseeds you start with.
9. a.
b. MBC SBS.9.2
.1
.8GMBCSBS F
.9 .8
.1
.2MBC SBS
47
37
F3737
4747GT 20 5
47 5 b
4070 5
2870 5
1270 1
710 (4
7) 5
410 (3
7) 1.4(37) 1 .7(4
7) 5
a 537b 5
47
c. MBC: 67%; SBS: 33%
10. a. 15%b. 27.4%c. 32.1% tall, 32.7%medium, 35.3% short
11. Let , and
. If A and B are
stochastic, then each rowsums to 1 and each entry isnonnegative.
.
Row 1 sums to a1b1
a2b2 a1b2 a2b4 a1b1
a1b2 a2b3 a2b4
a1(b1 b2) a2(b3 b4)a1 a2 1, since b1 b2
b3 b4 a1 a2 1. Row 2 sums to a3b1
a4b3 a3b2 a4b4 a3b1
a3b2 a4b3 a4b4
a3(b1 b2) a4(b3 b4)a3 a4 1, since b1 b2
b3 b4 a3 a4 1. Eachentry of AB is nonnegativesince it is the sum of twoterms, each of which is the product of twononnegative numbers.Hence, the product of two2 2 stochastic matrices isstochastic.
d. Over the long term, theproportion of occurrencesstabilize to a and b.
13. See below.
14. 24
15. 10
16. a. Yes, because all verticesare even, and it isconnected.b. 4
17. No, a graph cannot have anodd number of oddvertices.
18. a. v(1) 18 ft/sec
b. 2.125 sec
c. 1.5625 sect 55032 5
t 56832 5
5
b 559a 5
49
F.4444.4444
.5556
.5556GT 16 ø
F.4444.4444
.5556
.5556GT 8 ø
F.4445.4444
.5555
.5556GT4 5
F.45.44
.55
.56GT 2 5
d. The time in part c ismidway between the timesin parts a and b.e. Never, it is always -32 ft/sec2.
19. no solution
20. ;
21. Let n, n 1, n 1, and n 3 represent the fourconsecutive integers. By theQuotient-RemainderTheorem, n 4q r whereq is an integer and r 0, 1,2, or 3. If r 0, then n isdivisible by 4. If r 1, n 3 (4q 1) 34(q 1) is divisible by 4. Ifr 2, n 2 (4q 2)2 4(q 1) is divisible by4. If r 3, n 1 (4q3) 1 4(q 1) is divisibleby 4. So, exactly one ofevery four consecutiveintegers is divisible by 4.
22. A good source is MarkovChains: Theory andApplications by DeanIsaacson and RichardMadsen.
1511515
1511515
151151
55
515
111
limn→`
f(n) 513lim
n→-`f(n) 5
13
Answers for LESSON 11-6 pages 693–699 page 2c
13. T stabilizes to since if [a b] [a b]
.6a .3b a implies -.4a .3b 0 implies -.4a .3b 0
.4a 7b b a b 1 .4a .4b .4.7b .4
and a 537b 5
47
5515151515151
5F.6.3 .4.7GF3
737
4747G
1.2. a.
b.
c.3. a.
b.
c.4. a.
V 2 E 1 F 5 6 2 9 1 5 5 2
V 2 E 1 F 5 6 212 1 8 5 2
V 2 E 1 F 5 9 2 16 1 9 5 2 b.
c.5.6.7. a. The vertex of degree 1
and its adjacent edge wereremoved.b. Both V and E werereduced by 1, so stayed the same. Since F didnot change, didnot change.
8. a. An edge was removed.b. Both E and F werereduced by 1, so wasnot changed. Since V didnot change, didnot change.
9. True, by the contrapositiveof the second theorem ofthis lesson: Let G be a graphwith at least one edge. If Ghas no vertex of degree 1,then G has a cycle.
b. Since the graph in part ahas no crossings and 6 edges, holds true. Remove a vertexof degree 1 and its adjacentedge does not change thevalue of . This wasdone in part a, and F didnot change, so holds for the originalgraph.
12. a.b. It is not connected andcontains crossings.
13. a. 9 b. 5
14. a. 6 vertices, 9 edgesb. impossible c. 5d. In this graph, a facecannot have 1 edge, sincethis would mean a lineconnects a house or utilityto itself. A face cannot have2 edges, since this wouldmean a house and a utilityhave two lines connectingthem. Finally, a face cannothave 3 edges, since therestrictions prevent twohouses or two utilities frombeing connected to eachother. So a face must haveat least 4 edges.
V 2 E 1 F 5 6 2 6 1 8 5 8
V 2 E 1 F 5 2
V 2 E
V 2 E 1 F 5 2
e. Since each edge bordersexactly two faces, if we sumfor every face the numberof edges bordering it, weget 2E. Since there are atleast 4 edges borderingeach face, there must be at
most faces. Hence, .
Multiplying, , or .
f. By part a, . If thereare no crossings, by part c,
. This contradicts parte, since 2(5) 9 does nothold true. Therefore, it isimpossible to connect threehouses and three utilitieswithout lines crossing.
15. a.
b. R U
c. Urban 71%; Rural 29%
d. .95a .02b a-5a 2b 05a 5b 5
7b 5
;
16. 5
a 527 ø .286b 5
57 ø .714
5515151
øø
F.95.02
.05
.98GRU
.95 .98
.02
.05Rural Urban
#F 5 5
E 5 92F # E
4F # 2E
F #2E4
2E4
Answers for LESSON 11-7 pages 700–706 page 2c
c
17. a. No, there are twovertices with odd degrees.b. Sample:
18. No; for example, let v1, v2,v3 have degree 3, and v4
have degree 1. Then {v1, v2},{v1, v3}, and {v1, v4} are thethree edges from v1. {v2, v1}and {v2, v3} are 2 edges fromv2. Now there must beanother edge from v2. Butthat edge cannot connect v2
to v4, since v4 must remainwith degree 1. It cannotconnect to v1 or v3, since asimple graph cannotcontain parallel edges. Andit cannot connect to itself,since a simple graph doesnot have any loops. So,there is no such graph.
19. < 20%
School
20. To prove S(n):
.
S(1): ,
and , so S(1) is
true. Assume S(k):
.
Then
.
So, S(k) ⇒ S(k 1), and bymathematical induction S(n)is true for all integers n.
12. Impossible; a graph cannothave an odd number of oddvertices.
13. Sample:
14. Sample:
15. Impossible; a graph cannothave an odd number of oddvertices.
16. No, the sum of the entriesmust be the total degree ofthe graph, which must beeven.
17. The graph has an Eulercircuit by the sufficientcondition for an EulerCircuit Theorem, since it isconnected and every vertexis of even degree.
18. No, the graph is notconnected.
19. No, because v2 and v3 haveodd degree.
20. It cannot be determinedsince the graph may not beconnected.
21. a.
b. < 61.4%
22. a.
b. ≈ 0.48%
23. a.
b. 33 hours
A6
B39
D5
16C11 E
4G1228
F5
H5
I5
33
Broken into
Not broken into
.96
.04
.02
.98
.0001
.9999
Alarm.000096
No alarm.000004
Alarm.019998
No alarm.979902
Pac
Lux
Rudder .035
Rudder .022
.03
.97
.04
.96
.05
.95
.07
.69
.31
.93
Gauge .02
None .64
Gauge .01
None .28
Answers for Chapter Review pages 713–716 page 2c
c
24. No; this situation may berepresented as a graph with25 vertices, each with 5 edges. This is not possiblesince a graph cannot havean odd number of oddvertices.
25. No, a graph cannot have anodd number of oddvertices.
26. Yes; below is a samplegraph:
27. a. Vertices F and G haveodd degrees, so there is notan Euler circuit.b. the edge between F andG
28. a. Yes, sample:
b. No, two of the verticeshave odd degree, so noEuler circuit is possible.
a3
a2
a1
a6
a5
a4
29. a.
b. B NB
c. T8 <
They bowl on about 56% ofthe Tuesdays.d. ≈ 56%
30. 38% Democrat, 36%Republican, 26%Independent
31. v1 v2 v3 v4
32. The diagonal has zeros, andall other entries are eitherzeros or ones.
b. (55 cos(-22.5°), 55 sin(-22.5°)) < (50.8, -21 .0)
5. (cos 218°, sin 218°) <(-0.788, -0.616)
6. a. (5, -6)b.
4-2
-2
-4
-6
2 6
(5, -6)
y
x
4-2
-2
-4
-6
2
(3, -6)
y
x
150
N
E
S
W
180
10˚
N
E
S
W
7. a. length , direction< 308.7°b.
8. units
9.
10. [ , tan-1 ] <[24.8, 130°]
11. The standard position arrowfor the vector from (-1, 2) to (4, -1) has endpoint (4 (-1), -1 2) (5, -3);the standard position arrowfor the vector from (3, -2) to (8, -5) has endpoint (8 3,-5 (-2)) (5, -3). So thevectors are the same.
17. a. -0.267 words per sec; thesubjects forget 8 words per30-second period when thewait time changes from 0sec to 30 sec.b. -.13 words per sec; thesubjects forget 6 words per30-second period when thewait time changes from 30sec to 60 sec.
18. The initial conditions donot hold, because S(1), S(3),. . . are not true.
19. b
20. and
21. Since and, when ,is determined.
When , ,is determined.
Create a number line with Pat 0 and Q at 1. Then eachvalue of t will determinethe corresponding point onthe number line.
y0 1 v2)Q 5 (x0 1 v1t 5 1
P 5 (x0, y0)t 5 0y 5 y0 1 v2t
x 5 x0 1 v1t
AD←→
AC←→
Answers for LESSON 12-3 pages 733–738 page 2c
In-Class Activity
1. a. ;
b.2. a. < .361 b. 68.8°
3. cos u 0; u 90°
4. Sample: The product of theslopes of perpendicularlines is -1, and the ratio ofsecond to first componentsis the slope of the linecontaining the vector. Since
, the lines
containing the vectors areperpendicular.
Lesson 1. 70 2. (-350, -420)
3. 73 4. 150°
5.
6. The meaning of the dotproduct of two vectorsbeing negative is the sameas the meaning of cos ubeing negative, where u isthe angle between the twovectors and 0 u π. Forthese values of u, cos u , 0
only when .
Therefore the dot product
π2 , u # π
##
u 5 cos-1 38 ø 68°
63.5 • 7
-12 5 -1
55
Ï74
zs→
z 5 Ï85r→ 5 Ï13
of two vectors is negativeonly if the angle betweenthem is obtuse or if theyhave opposite directions.
is parallel to (2, -1, 3),since it is a scalar multipleof it. (2, -1, 3) isperpendicular to the
plane N, and hence isperpendicular to N.
6. Sample: ;;
7.
8. Since and lie in M,
3 gives the vectorperpendicular to M. Thus, Q
is on M (so that is on M)
if and only if is
perpendicular to 3
.P1P→
3
P1P→
2
P1Q→
P1Q→
P1P→
3P1P→
2
P1P→
3P1P→
2
2x 2 y 1 3z 5 -2
z 5 -1 1 3ty 5 3 2 tx 5 2 1 2t
w→
w→
5w→
5
22
2
4
(0, -4, 0)
(2, 0, 0)4
y
x
z
(0, 0, )4]3
2x 1 y 2 2z 5 -1z 5 3 2 3ty 5 2 2 2tx 5 1 1 3t
z Þ -8t 5 4x 5 10
z 5 -2 1 4ty 5 5x 5 1 2 3t
t(-3, 0, 4)5z 1 2y 2 5x 2 1 9. a.
b.10. M1, is perpendicular to
(a1, b1, c1), and M2 isperpendicular to (a2, b2, c2),so < is perpendicular toboth (a1, b1, c1) and
(a2, b2, c2). Since is alsoperpendicular to both (a1, b1, c1) and (a2, b2, c2),
is parallel to <.
11. a. (3, -2, 1) and (1, 2, -1),the vectors perpendicularto the planes M1, and M2
are not parallel.
b. Sample:
c. Sample: ; ;
12.13.14.
(u1, u2, u3) (v1 w1,v2 w2, v3 w3)
(u1v1 u1w1) (u2v2
u2w2) (u3v3 u3w3)(u1v1 u2v2 u3v3)
(u1w1 u2w2 u3w3)
15. e
16. a. W I
b. < 7.4%
F.96.5
.04.5GW
I
u→
• v→
1 u→
• w→
5111115
111115
111 • 5
u→
• (v→
1 w→
)
p 5 6Ï5
5n 1 10d 1 25q 5 1000
z 534 1 8t
y 5 1 1 4tx 574
(74, 1, 34)
w→
w→
7x 2 2y 1 z 5 6
7x 2 2y 1 z 5 6
Answers for LESSON 12-7 pages 760–766
c
17. a.
b.
18.
19. a. y
x2-1
2
-14
x 567
(n 1 1)(2n 1 1)6n2
on
j51(1n)2
• (1n) b. a cone
c. cubic units
20. a. a plane parallel to and 1unit above the xy-planeb. As a increases, vectorsperpendicular to the planerotate from the z-axistoward the x-axis, so theplane tilts more steeply.
derived. These points arethe same as those on theintersection line shown onpage 768 of the StudentEdition. Since two pointsdetermine a line, this is thesame line.
2. a.
b. Sample:
3. a. Sample: b. See below.
3x 2 2y 1 4z 5 2
Hx 5 t
y 512 2
12t
z 512 2
12t
z
11
1
x
y
y – z = 0
x + y + z = 1
(79, 2, 50
9 )z 5509
(319 , 0, 89)
z 589
y 59z 2 8
21x 5-12z 1 83
21 4. the line
5. the plane
6. z
z
z
x
y
M1 = M2 = M3
x
yM1 = M2
M3
x
yM1 M2
M3
-x 1 3y 2 2z 5 -6
Hx 549 1 11t
21y 5 t
z 57 1 8t
21
Answers for LESSON 12-8 pages 767–772
3. b. subtracting plane Nsubtracting plane M
Therefore, the system has no solution.0 5 1
3x 2 2y 1 4z 5 13x 2 2y 1 4z 5 2
c
7. a. the first and secondequations, because thevector (1, 1, 3) isperpendicular to bothb. (2, -1, 1) is perpendicularto the third plane, so thatplane is not parallel to thefirst two, and must intersectboth of them.
8. a. Subtracting the secondequation from the first, weget . Subtractingthe third equation fromtwice the second equation,we get , or
. Substituting forx, , whichsimplifies to . Hence,there is no solution to thissystem.b. The first two equationsintersect at the line
, , .
9. the point
10. the line
11. the line
Hx 5 ty 5 7z 5 5 1 t
Hu 5 ty 5 9 2 tw 5 -27 1 8t
(12, 12, 0)
z 5 ty 55 2 t
5x 510 2 3t
5
-8 5 1-(3y 1 8) 1 3y 5 1
x 5 3y 1 8x 2 3y 5 8
-x 1 3y 5 1
12. ,
13. mi, mi, mi
14. a plane parallel to thex-axis containing (0, 0, 10)and (0, 10, 0)
15.
16. a.b. The planes are parallel.
17. is perpendicular to
both and .
if and only
if is also perpendicular to
. Thus , , and ,being all perpendicular tothe same vector at the samepoint, are coplanar.
52. (-8, -5, -2); This vector canbe pictured by an arrowstarting at the endpoint of
and ending at the
endpoint of , providing
and have the sameinitial points; or putting thevector in standard position,it is the diagonal of thefigure having vertices (-3, -4, 1), (-5, - 1, -3), (0, 0, 0), and (-8, -5, -2).
25. An Archimedean screwconsists of a spiral passagewithin an inclined cylinder.It is used for raising waterto a certain height. This isachieved by rotating thespiral in the cylinder.
26. Using wood with areasonably consistentdensity, Archimedes couldhave weighed a rectangularpiece of wood andmeasured its area. Then hecould have cut out aparabolic region andweighed it. The weights ofthe parabolic region andthe rectangular piecewould be proportional totheir areas.
2. Many different functionshave the same derivative.
3. 1 4. antiderivative
5. ln x 6. c
7. ln 20 ln 5 < 1.386
8. a. 0.5b. 0.5043; the result agreesvery well; the relative erroris only about 0.8%.
9. Fundamental Theorem ofArithmetic: Suppose that nis an integer and that . Then either n is a primenumber or n has a primefactorization which isunique except for the orderof the factors. FundamentalTheorem of Algebra: Everypolynomial of degree with real or complexcoefficients has at least onecomplex zero.
n $ 1
n . 1
2
10. a.
b.
11. 112 ft3
12. -47.5
13. 13
14. (1 sin 2x) dx
15. Sample: e1e2e3e4e5e7e6
16. No Euler circuit exists.
17. Answers may vary.
1E0
5π4
E4
0
y
x
(4, 4)
2 6
2
4
6
4
Answers for LESSON 13-7 pages 826–831
1. 3600
2. 2025
3. a.
b. 0.500
4. 32
5. 4
6.
7. a.
b. < 5.59
8. 72
9. 221.
10. 21
11. 240
12. 2x dx
13. 1 dx
14. False
15. y
x
f (x )
I II
a c b
E3
1
E10
4
6
1 3 5
1
3
5y
x
y = 25 – x2
π4
π600 sin (π
3 1 i π600)o
100
i51
Area I is f(x) dx, Area II is
f (x) dx, and the union of
the regions represented by
Areas I and II is f(x) dx.
Since Area I Area IIequals the area of theunion of the regionsrepresented by Areas I and