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• 1. Chapter 3: Set Theory and LogicL e s s o n 3.1: T y p e s of S e t s a n d S e t Notation,page 1541. a) e.g., Y e s , those explanations make sense.b) T h e universal set is set C.C = {produce}O = {orange produce} = {oranges, carrots}Y = {yellow produce} = {bananas, pineapple, corn}G = {green produce} = apples, pears, peas, beans}B = {brown produce} = {potatoes, pears}c) e.g., S c F because all fruits you can eat withoutpeeling are also fruits. S c C because all fruits you caneat without peeling are also produce.d) Sets S and V are disjoint sets, as are F and V.e) Yes. e. g., C = { F and V}, F = V.f) n{V) = n{C)-n{F)n{/) = 1 0 - 5n{V) = Sg) oranges, pineapple, bananas, peas, corn, carrots,beans, potatoes2. a)b) Sets E a n d S are disjoint sets, as are sets F a n d S.c) i) True, e.g.. Multiples of 8 are also multiples of 4 .ii) False, e.g.. Not all multiples of 4 are multiples of 8.iii) True, e.g., All multiples of 4 are multiples of 4.iv) False, e.g., F = {all numbers from 1 to 4 0 that are notmultiples of 4}v) True, e.g.. T h e universal set includes natural numbersfrom 1 to 4 0 .3. a).1. A A A A A Al l i l S i i t l i l i l l S # f ^ ^A A A A A AA A i A . A iA . A - . A - A / A , -1 . .;-;A>;AIOAIA;,ia-..A / •) lo j ob) Subsets of set B: C c B a n d S c Bc) Subsets of set R: H c R anti D a Rd) Y e s , the sets S a n d C are disjoint, e.g., A cardcannot be both a spade and a club.e) Y e s , the events in sets H and D are mutuallyexclusive, e.g., Y o u cannot draw a card that is aheart and a diamond at the s a m e time.f) Y e s , that statement is correct, e.g., Becausethese sets are disjoint, they contain no c o m m o nelements. Therefore, w h e n the numbers ofelements in each set are added, no element will becounted twice.niS or D ) = n{S) + n ( D )n{S or D ) = 13 + 13n{S or D ) = 265. a) e.g., C = {all clothes}, S = { s u m m e r clothes},W = {winter clothes}, H = { s u m m e r headgear}b) e.g., In set C, but not in set S or set W, becausethey would be w o r n year round.c) N o , set S is not equal to set W. Set S includesthe jacket, but W does not.d) Sets S and Ware disjoint sets. Sets H a n d Ware disjoint sets.e) e.g., C = {clothes},H = {headgear} = {cap, sunglasses, toque},6 = {clothing for body} = {shirt, shorts, coat,jacket}, F = {footwear} = {sandals, insulated boots}6. n{X) = n{U) - n(X)n{X)= 100 0 0 0 - 1 2niX) = 99 9887. Not possible; e.g., there m a y be s o m e elementsthat are in both X and Y.walleye northern pike 8. n(L/) = n(X) + n(X)n{U) = 34 + 4 2lake trout Arctic charArctic grdylinq lake whitefishb) e.g., N a 7 m e a n s that aft the fish found in Nunavutare also found in the Northwest Territories. Tct N m e a n sthat not all the fish found in the Northwest Territories arefound in Nunavut.niU) = 769. a) S = {A, E, F, H, I, K, L, M, N, T, V, W , X, Y, Z}C = {C, 0 , S}b) False, e.g., B is not in S or C.F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-1
• 2. 10. Let U represent ttie universal set. Let L represent ttieset of land transportation. Let represent ttie set ofwater transportation. Let A represent ttie set of airtransportation.lu-.t .ill K>,il!(>cu!s .".ai-kiny i.ikir,!;•.kiinii riiivinijpO.Ml lUi.lt,11. a)1 2 .5 4 S lO 9 3 / hb) Sets A and B are disjoint sets.c) i) False, e.g., 1 is not in B.ii) False, e.g., - 1 is not in A.iii) False, e.g., 0 is in A but not in B.iv) True, e.g., n{A) = 10, n{B) = 10.v) True, e.g., No integer from - 2 0 to - 1 5 is in U.12. a) S = {4, 6, 9, 10, 14, 15, 2 1 , 22, 25, 26, 33, 34, 35,38, 39, 46, 49}W= { 1 , 2, 3, 5, 7, 8, 1 1 , 12, 13, 16, 17, 18, 19, 20, 23,24, 27, 28, 29, 30, 3 1 , 36, 37, 40, 4 1 , 42, 43, 44, 45, 47,48, 50}b) e.g., £ = {even semiprime numbers}E = {4, 6, 10, 14, 22, 26, 34, 38, 46}c) n{W) =n{U)-n{S)n{W) = 5 0 - 1 7n{W) = 3 3d) No, it is not possible to determine n{A). e.g., Ttiere isan infinite number of prime numbers, so there is aninfinite number of semiprime numbers.13. e.g.. Let U represent the universal set. Let Erepresent the set of entertainment items. Let T representtechnology items.equipment televisionluniputv.-14. Agree; e.g., A (z B means that set A s a part ofset B, and it could be that set A and set B areequal. f Ac: B, then set A will have the s a m enumber or fewer elements than set B. Withnumbers, x < y m e a n s that x is less than or equalto y. Or, or if y4 c B, then n{A) < n{B). The numberof elements in a subset must be equal to or lessthan the number of elements in the set.15. a) S = {x I - 1 0 0 0 < X < 1000, x e I }r = { f | f = 2 5 x , - 4 0 < x < 4 0 , X G 1}F = { f | f = 5 0 x , - 2 0 < x < 2 0 , x e 1}Fez Td Sb)16. a)U = { H H H , HHT, H T H , T H H , HTT, THT, T T H ,TTT}b) £ = { H T H , HTT, T T H , TTT}c) n{U) = 8, n(£) = 4d) Yes, e.g., because each element of £ is also anelement of U, and there are some elements of Uthat are not elements of £ .•^••••^••IT THH IMIe) For example, £ is the set of elements of Uwhere the second coin turns up heads.n(£0 = n((y)-n(£)n ( £ 0 = 8 - 4n{E^ = 4£ = { H H H , HHT, T H H , THT} and n(£) = 4f) Yes. e.g., A coin cannot show both heads andtails at the s a m e time.17. a)b) e.g., N is the set of all non-natural numbers. Wis the set of all non-whole numbers. 1 is the set ofnon-integer numbers. Q is the set of numbers thatcannot be described as a ratio of two integers. Qis the set of numbers that can be described as aratio of two integers.3-2 C h a p t e r 3: S e t T h e o r y a n d L o g i c
• 3. Set C o m p l e m e n tN W = {x 1 X € f?, X g N}.1 / = { x 1 X e R , X « 1}Q QQ Qc) Sets N and Q are disjoint sets. Sets lA^and Q aredisjoint sets. Sets / and Q are disjoint sets. Sets Q andQ are disjoint sets.d) Yes. e.g., Q is ttie set of numbers that cannot bedescribed as a ratio of two integers, which is the set ofirrational numbers.e) W, I, Q, Rf) No. e.g., The area of a region in a V e n n diagram is notrelated to the number of elements in the set.18. a) S = { 1 , 4, 9, 16, 25, 36, 49, 64, 8 1 , 100, 1 2 1 , 144,169, 196, 225, 256, 289}n ( S ) = 17E = {4, 16, 36, 64, 100, 144, 196, 256}n{E) = 8b) n ( S ) = 17, n(£) = 8n(0) = n{S)-n{E)n ( 0 ) = 1 7 - 8n ( 0 ) = 9c) n{U) = 300, n(S) = 17n(S) = niU) - n{S)n(S) = 3 0 0 - 17n(S) = 28319. a) e.g., /A c S if al! elements of A are also in B. Forexample, all weekdays are also days of the week, sow e e k d a y s is a subset of days of the week.b) e.g., A consists of all the elements in the universal setbut not in A. For example, all days of the week that arenot w e e k d a y s are w e e k e n d days. So weekend days isthe complement of weekdays.20. e.g., Disagree; since both the subsets are empty,they both contain the same elements and are thereforethe s a m e subset.L e s s o n 3.2: E x p l o r i n g R e l a t i o n s h i p s b e t w e e nS e t s , p a g e 1 6 01 a )uIS n 10 14b) i) n{A) = 5ii) n{A but not S) = n{A) - n{A and B)n(A but n o t e ) = 5 - 2n{A but not S) = 3iii) n{B) = 7iv) n{B but not A) = n{B) - n{A a n d B)n{B but not A) = 7-2n{B but not yA) = 5v) n{A and 6 ) = 2vi) n{A or S) = n(A but not B) + n{A and 6 )+ n{B but not yA)n(>AorB) = 3 + 2 + 5n{A or B) = 10vii) n{A) = 5, therefore n{A) = 52. a) 8 students are in both the drama club and theband.b) 11 students are in the drama club only.6 students are in the band only.c) Drama: 1 1 + 8 = 19Band: 8 + 6 = 14d) Drama club or band: 1 1 + 8 + 6 = 2 5e) 38 students in grade 12 - 25 in drama club orband = 13 students in neither drama club nor band3. a) hockey or soccer: 45 - 16 = 29hockey and soccer: 20 + 14 = 34overlap: 34 - 29 = 55 students like hockey and soccer.b) only hockey: 20 - 5 = 15only soccer: 1 4 - 5 = 915 + 9 = 2424 students like only hockey or only soccer.c)4. a) ski or snowboard: 55 - 9 = 46ski and snowboard: 25 + 32 = 57Overlap: 57 - 46 = 1111 guests plan to ski and snowboard.b) only ski: 2 5 - 11 = 1414 guests will only ski.c) only s n o w b o a r d : 32 - 11 = 2 121 guests will only s n o w b o a r d .5. a) n{U) - n{U but not A or B): 25 - 4 = 21n{A) + n{B): 13 + 10 = 23n{A and 6 ) : 2 3 - 2 1 = 2n{A only): 1 3 - 2 = 11n ( B only): 1 0 - 2 = 8b)2 8F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-3
• 4. L e s s o n 3 . 3 : I n t e r s e c t i o n a n d U n i o n o f T w o S e t s ,p a g e 1 7 21. a) ^ = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4, 6, 8, 10}B = {0, 1,2, 3, 4, 5, 6, 7, 8, 9 , 1 0 }AuB = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10}b) n{AuB)= 16c) AnB = {0, 2 , 4 , 6, 8, 10}d) n{A n B) = 62. a) Let A represent ttie universal set. Let N representttie set of tundra animals. Let S represent the set ofsouthern animals.N = {arctic fox, caribou, ermine, grizzly bear, muskox,polar bear}S = {bald eagle, Canadian lynx, grey wolf, grizzly bear,long-eared owl, wolverine}Nu S = {Arcticfox, caribou, ermine, muskox, polar bear,grizzly bear, bald eagle, Canadian lynx, grey wolf, long-eared owl, wolverine}Tr, S = {grizzly bear}b)Arctii, fox bald eagletcuibou Canadian lynxermint; grizzly bear grey wolfrTin^kox lonc]-eared owlr^otar bear wolverine3. a) / u C = { - 1 0 , - 8 , - 6 , - 4 , - 2 , 0, 2, 4, 6, 8, 10, 12,14, 16}niA u C) = 14A nC = {2,4, 6, 8, 10}n{A nC) = 5b)2 412 144. a) 7 u C = {half-ton trucks, quarter-ton trucks, vans,SUVs, crossovers, 4-door sedans, 2-door coupes, sportscars, hybrids}b) n ( r u C ) = 9c) T n C = {crossovers}5. a) Let U represent the universal set. Let Frepresent the set of African animals. Let Srepresent the set of Asian animalslionqiraffehippoi..Vni;ielephanttKjerb) F = {lion, camel, giraffe, hippo, elephant}S = {elephant, tiger, takin, camel}F u S = {lion, giraffe, hippo, camel, elephant, tiger,takin}F n S = {camel, elephant}6. a)0 C!6 12 3 I15b)AuB = { - 1 2 , - 9 , - 6 , -4, - 3 , - 2 , 0, 2, 3, 4, 6,9, 10, 12, 15}n{AuB)= 16AnB = {-6, 0 , 6 , 12}n{A n B) = 47. Let U represent the universal set. Let Hrepresent the set of people w h o liked SherlockHolmes. Let P represent the set of people w h oliked Hercule Poirot.n{H uP) = n{U) - n{(H u P))n{H u P) = 25 - 4n{H u P) = 21n{H nP) = n{H) + n(P) - n{H u P)n ( H n P ) = 16 + 11 - 2 1n{H n P) = 66 people like both detectives.n{H only) = n{H) - n{H u P)n ( H only) = 1 6 - 6n ( H o n l y ) = 1010 people liked Sherlock Holmes only.n(P only) = n(P) - n{H u P)n(P only) = 1 1 - 6n{P only) = 55 people liked Hercule Poirot only.3-4 C h a p t e r 3: S e t T h e o r y a n d L o g i c
• 5. 8. Let U represent ttie universal set. Let V represent ttieset of people wtio liked vanilla ice c r e a m . Let Crepresent ttie set of people wtio liked ctiocolate icecream.n(C u VO = n{U) - n{(C u V))n ( C u VO = 80 - 9n ( C u V0 = 71n{C only) = n ( C u V) - n{ V only) - n ( C n VOn ( C o n l y ) = 71 - 1 1 - 2 0n{C only) = 4040 people like chocolate ice cream only.9. Let U represent the universal set. Let K represent theset of people w h o like to ski. Let W represent the set ofpeople w h o like to swim.niKuW) = n{U) - n{(K u W))n{KuW) = 26-5n{Ku 140 = 21n{KnW) = n{K) + n{W) -n{KuW)n{Kn t V ) = 1 9 + 1 4 - 2 1n ( K n MO =1212 people like to ski and swim.10. e.g., She could draw a V e n n diagram showing theset of multiples of 2 and the set of multiples of 3. Theintersection of the sets would be the multiples of 6.11. a) U = {all customers surveyed}C = {customers ordering coffee}D = {customers ordering donuts}N = {customers ordering neither coffee nor doughnuts}b) For the following Venn diagram:The rectangular area labelled U represents the universalset.The shaded area labelled D represents the set of peoplew h o ordered doughnuts.The shaded area labelled C represents the set of peoplew h o ordered coffee.The shaded area labelled D n C represents the set ofpeople w h o ordered coffee and doughnuts.The unshaded area labelled N represents those peopledid not order coffee or doughnuts.customers ordering botti coffee and a d o u g h n u tI — ^ _customers ordering neitherc) Determine n{D n C) using the information available.niU) = 100, n(D) = 45, n(C) = 65, n{(D u C)) = 10n{D u C) = n{U) - n{(D u C))n{DuC)= 1 0 0 - 1 0n{D u C) = 90Therefore,/7(D nC) = n{D) + n(C) - n ( D u C)n(D n C) = 4 5 + 65 - 90n(D n C) = 20There were 20 people w h o ordered coffee and adoughnut.12. Let U represent the universal set. Let Trepresent the set of seniors w h o watch television.Let R represent the set of seniors w h o listen to theradio.n{R only) = n ( L / ) - n ( T )n ( R o n l y ) = 1 0 0 - 6 7n{R only) = 3333 seniors prefer to listen to the radio only.13. Let U represent the universal set. Let Crepresent the set of people w h o attended theCalgary S t a m p e d e . Let P represent the set ofpeople w h o attended the P N E .n(C uP) = n{U)- n{(C u P))n ( C u P ) = 5 6 - 1 4n ( C o P) = 42n{C n P) = n(C) + n(P) - n{C u P)n ( C n P) = 30 + 22 - 42n{CnP) = 1010 people had been to both the Calgary S t a m p e d eand the P N E .14. Of the 54 people, 31 o w n their home, so54 - 31 = 23 people rent their home. Of that 23, 9rent their house, so 23 - 9 = 14 rent theircondominium. O f t h e 30 people w h o live in acondominium, 14 rent, so 30 - 14 = 16 must o w nthe condominium in which they live.15. Let U represent the universal set. Let Rrepresent the set of people w h o like reality shows.Let C represent the set of people w h o like contestshows.n{C u R) = niU) - n{(C u P))n ( C u P ) = 3 2 - 4n{C u P) = 28n{C nR) = n{C u P ) - n{C only) - n{R only)n ( C n P) = 2 8 - 1 3 - 9n{C n P ) = 66 people like both type of shows.16. No. e.g.. The three numbers do not add up to48. There is an overlap between sets B and C, butB<xC.The s u m of the three values in the problem is 59.59 - 48 = 1111 students must drive a car and take a bus.31 - 1 1 = 2 020 students drive a car but do not take a bus.1 6 - 11 = 55 students take a bus but do not drive a car.There are a total of 15 + 12 = 27 students w h o donot take a bus.F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-5
• 6. 17. a) Sets A and B are disjoint sets.b) Sets /A and C intersect.c) Yes; B and C; e.g., C intersecting A and /A and Sbeing disjoint says nottiing about the intersection, if any,of 8 and C.18. e.g.. The union of two sets is more like the addition oftwo numbers because all t h e elements of each set arecounted together, instead of those present in both sets.19. a) e.g., indoor, outdoor, racesb) e.g., U = {all sports}/ = {indoor sports} = {badminton, basketball, curling,figure skating, gymnastics, hockey, indoor soccer, speedskating, table tennis, volleyball, wrestling, Arctic Sports,Dene G a m e s }O = {outdoor sports} = {alpine skiing, cross-countryskiing, freestyle skiing, s n o w s h o e biathlon, ski biathlon,dog mushing, s n o w b o a r d i n g , snowshoeing. DeneG a m e s }R = {races} = {speed skating, alpine skiing, cross-countryskiing, biathlon, dog m u s h i n g , snowboarding,snowshoeing}c)b^clrnii i k n i bdskotlj-allcurling hockuyfuiure ^k:Jtln()c)ynuui!>li(.hiiKk>or sui.i,(.r.vr-sliii!ijArc!i< Sp.iftiDens?Gamesjlpin(> skiing(ros-tountry skiinytrcHstyk skiinydoq rTUishinysnowlsoijrdiiiysnowshoiifiyMiowshoe bidtdlonki ri.jirl.•!:•,•d) Yes. e.g.. My classmate sorted the g a m e s asindividual, partner and t e a m games.H i s t o r y C o n n e c t i o n , p a g e 1 7 5A . e.g.. The "barber paradox" can be stated as follows:Suppose there is one male barber in a small town, andthat every m a n in the town keeps himself clean-shaven.S o m e do so by shaving thennselves and the others go tothe barber. So, the barber stiaves all the m e n w h o do notshave themselves. Does the barber shave himself? Thequestion leads to a paradox: If he does not shavehimself, then he must abide by the rule and shavehimself. If he does shave himself, then according to therule he will not shave himself.B. e.g.. O n e remarkable paradox that arises fromCantors work on set theory is the Banach-Tarskitheorem, which states that a solid, three-dimensional ballcan be split into a finite number of non-overlappingpieces, which can then be p u t back together in a differentw a y to yield two identical copies of the original ball of thesame size.M i d - C h a p t e r R e v i e w , p a g e 1 7 81. a) V c N, M d N, F c N, F cz Mb) e.g., N = {all foods}, V = {fruits and vegetables},M = {meats}, F = {fish}c) No. e.g.. Pasta is not part of M or V.d) Sets V and M are disjoint, Sets V and F aredisjoint.2. a)) i-s M> .8 -lUb) Sets F and S are disjoint sets.c) i) False, e.g., 6 is in E but not F.ii) True, e.g.. All elements of S are in E.iii) False, e.g., 9 is not a multiple of 15.iv) True, e.g., F = {15, 30}.v) True, e.g., A set is a subset of itself3. e.g., S = {summer sport equipment} = {baseball,soccer ball, football, tennis ball, baseball glove,volleyball net}W = {winter sport equipment} = {hockey puck,skates, skis}B = {sports balls} = {baseball, soccer ball, football,tennis ball, hockey puck}E = {sports equipment worn on body} = {baseballglove, skates, skis}baseball!;-o:h.:-l!baseball glovevolleybdll netVl^ ki-y (ii.i kA.tCSskis4. a) beverage or soup: 40 - 5 = 35beverage a n d soup: 34 + 18 = 52overlap: 5 2 - 3 5 = 1717 students bought a beverage and soup,b) only beverage: 34 - 17 = 17only soup: 1 8 - 1 7 = 118 students bought only a beverage or only soup.3-6 C h a p t e r 3: S e t T h e o r y a n d L o g i c
• 7. c)5. a) sunglasses or hat: 20 - 5 = 15sunglasses and hat: 13 + 6 = 19overlap: 1 9 - 1 5 = 44 students are wearing sunglasses and a hat.b) only sunglasses: 1 3 - 4 = 99 students are wearing sunglasses but not a hat.c) only hat: 6 - 4 = 22 students are wearing a hat but not sunglasses.6. a) e.g., Tanya did not put any elements in theintersection of A and B.n{A u B) = n(L/) - n{{A u By)n{AuB)= 4 0 - 8n{A u 6 ) = 32n{A nB) = n{A)+ n(B) - n{A u 6 )n{AnB)= 1 6 + 1 9 - 3 2n{A n 6 ) = 3n{AB) = n{A)- n{A n B)n{AB)= 1 6 - 3n{AB)= 13n ( B l 4 ) = n(B) - n{A n B)n{BA) = 1 9 - 3n{BA)= 16b)! S7. Let ty represent the universal set. Let D represent theset of students w h o have a dog. Let C represent the setof students w h o have a cat.n{C u D) = niU) - n{(C u D))n(C u D) = 20 - 4n ( C u D ) = 16n(C n D) = n(C) + n(D) - n{C u D)n ( C n D) = 8 + 8 - 16n(C n D) = 0No students have a cat and a dog.L e s s o n 3.4: A p p l i c a t i o n s o f S e t T h e o r y ,p a g e 1911. n(P) = p + 16, n(Q) = g + 2 1 , n(R) = r + 18e.g., p Can be any number. Suppose p = 14. T h e nn(P) = 30.n(Q) = 30, so q = 30 - 21 or 9n(P) = 30, so r 301 - 1 8 or 122. a) n ( ( F u M ) ^ ) = 9 + 1 5 + 8n ( ( F u M ) ^ ) = 32b) n{{A u F) A/f) = 9 + 11 + 7n{{A uF)M) = 27c) n((F u /) u (F u M))= (9 + 11 + 7 + 9) + (15 + 8 + 4)= 36 + 27= 63d) n{AFM) = 73. e.g., Staff could look at how many David Smithswere on that bus route or they could look at thebooks in the bag and see how many David Smithsare taking courses that use those books.4. P = {population surveyed}n(P) = 641L = {people wearing corrective lenses}L = {people not wearing corrective lenses}n ( L ) = 167G = {people wearing glasses}C = {people wearing contact lenses}n{L) = n{P)-n{nn(L) = 641 - 1 6 7n ( L ) = 474n(G u C) = n{L)n(G u C) = n{G) + n{C) - n(G n C)474 = 442 + 83 - n{G n C)51 = n(G n C)51 people might make use of a package deal. This51is = 10.759...% or about 10.8% of all574potential customers.5. e.g., "Canadian Rockies," "skiaccommodations," "weather forecast," "Whistler."By combining two or more of these terms, Jacquescan search for the intersection of w e b pagesrelated to these terms. For example, "skiaccommodations" and "Canadian Rockies" is morelikely to give him useful information for his trip thaneither of those terms on its o w n .F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-7
• 8. 6. Using the principle of inclusion and exclusion for threesets:32 + 35 + 3 8 - ( 9 + x ) - ( 1 1 + x ) - ( 1 3 + x) + x = 5 81 0 5 - 9 - X - 1 1 - X - 1 3 - X + X = 5 87 2 - 2 x = 5 8- 2 x = 5 8 - 7 2- 2 x = - 1 4X = 77 teens are training for the u p c o m i n g triathlon.7. • s a m e numbers, s a m e s h a p e s , different shadings• s a m e numbers, different s h a p e , different shading• s a m e numbers, different s h a p e , different shadingo •8. a) e.g., T h e dealer might use exterior colour, interiorcolour, or year.b) e.g.. The dealer might prioritize the search accordingto options Travis wants or b y distance from where Travislives.9. e.g., John a s s u m e d that 9 0 people ate at only onerestaurant for each of the 3 restaurants. He did notcalculate the correct n u m b e r of people eating at only oneof each of the 3 restaurants.I defined these sets.C = {students w h o like only Chicken and MoreF = {students w h o like only Fast Pizza}G = {students w h o like only Gigantic Burger}I listed the values I knew a n d entered t h e m in a V e n ndiagram.n ( C n P B ) = 37; n { C n e P ) = 19; n ( P n S C ) = 11n{CnBnP)= 13Chicken Fast Pizzaand MoreGigantic BurgerI used these figures and diagram to determine theu n k n o w n values.n(C B P) = 90 - n(C n P e ) - n{C n B P)-n{CnBnP)= 9 0 - 3 7 - 1 9 - 1 3= 21n(B P C) = 90-n{CnBP)-n{PnB C)-n{CnBnP)= 9 0 - 1 9 - 1 1 - 13= 47n ( P B C ) = 90-n{PnBC)-n{CnPB)-n{CnBnP)= 9 0 - 1 1 - 3 7 - 13= 29Fast PizzaGigantic Burgern ( P ) = 21 + 29 + 4 7 + 37 + 19 + 11 + 13n(R)= 177177 students like at least one of these restaurants.2 4 0 - 177 = 63So, 63 students do not like any of the restaurants.10. a) e.g., He can search for colleges and(Calgary or Edmonton).b) He should use "and" to connect the words.c) He should use "or" to search for one or the othercity.d) e.g., colleges and (Calgary or Edmonton) and"athletics programs" -universitye) e.g., about 1500f).iii,.ac>F..c!niontonAthleif. Irocjiain,)3-8 C h a p t e r 3: S e t T h e o r y a n d L o g i c
• 11. c) Conditional statement: If a number is a repeatingdecimal, then it can be expressed as a fraction. Thestatement is true.Converse: If a number can bo expressed as a fraction,then it is a repeating decimal.The converse is false. Counterexample; The decimalnumber 0,3 can be expressed as the fraction ^10but 0.3is not a repeating d e c i m a lThe statement is not biconditional9. a | Conditional statement; If AB and CD areparallel, then the alternate angles are e q u a lConverse; If the alternate angles are e q u a l then andCD are parallelb| Proof of conditional statement:I drew two lines crossed by a transversal and numberedthe angles as shown.///- /^ i -l_ _AlUrn.ii.! rfi-fjiot.eQy.ll.Ith(-iven.Alf-rnate angles.Z 2 and ,^1 ciEcsupplementary.niOy form a straightline._ 4 A^vl esupplementary.They form a straightline.Z I = Z 3 Supplements of equalangles are also equal.AB II CD Corresponding anglesare e q u a lTherefore, the conditional statement is true.Proof of converse:1 used Inti i-iwn di-it/.-iinAB CO Gien.Z I = , lines are parallel,corresponding anglesare equal.Z 2 and Z I aresupplementary.They form a straightline.Z 4 and Z 3 aresupplementary.They form a straightline.Z I = Z 3 Supplements of equalangles are also equal. 1 and . 3 (Hialternate angles.Therefore, the converse is true.10, a) Converse; If your pet is a dog, then it barks.The statement and its converse are true, so thestatement is biconditionalb) Converse; If your pet w a g s its tail, then it is adog.The converse is false. A cat w a g s its t a i l Thestatement is not biconditional11. a) True.x + yXb) True,P - QP • Q + QP= z= z-y= z-y I= r+q12. e.g., If a number appears in the same row,column, or large square as the shaded square,then it is not in the shaded square. The numbers 1,4, 5, 6, and 8 must go in column 4, If I were to put1, 4, 5, or 8 in this square, then I could not put 6 inany other square in column 4, I conclude that 6 isthe only number that can go in this square. As aresult, 5 can only go in the square above. 4 canonly go in the square below. 8 can only go in thetop square, and 1 must go in the remaining square.The numbers in the column should be. from top tobottom: 8. 3, 9, 5, 6, 4. 1. 7, 2.13. a) i) If a figure is a square, then it has four nghtangles.li) If a figure has four nght angles, then it is asquare.ill) The statement is true. The converse is false.The figure could be a rectangle.iv) The statement is not biconditionalb) i) If a triangle is a nght tnangle, then a + b = cii) If, for a triangle, + = c^, then it is a righttnangle.iii) The statement is true. The converse is true,iv) A triangle is a right tnangle if and only if+b =c) i) If a quadnlateral is a trapezoid, then it has twoparallel sides.ii) If a quadrilateral has two parallel sides, then it isa trapezoid.iii) The statement is true. The converse is false, Aregular hexagon has two sides that are paralleliv) The statement is not biconditionalc) The onginal statement is true, because both thestatement and the converse are true, so the statement isbiconditionalF o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-11
• 12. 14. Use the finance application on a calculator. Note:Mortgages are c o m p o u n d e d semi-annually in C a n a d a .a) i) The number of payments is 25 • 12 or 300.T h e interest rate is 6.5%.T h e present value is \$250 000.The payment amount is unknown.The future value is \$0.The payment frequency is 12.T h e c o m p o u n d i n g frequency is 2.They should pay \$1674.559... or \$1674.56 per month.ii) The number of payments is 25 • 24 or 600.The interest rate is 6.5%.The present value is \$250 000.The payment amount is unknown.T h e future value is \$0.The payment frequency is 24.The compounding frequency is 2.They should pay \$836.163... or \$836.16 bi-monthly.b) 2 payments/week • 52 weeks/year = 104The number of payments is unknown.The interest rate is 6.5%.The present value is \$250 000.The payment amount is \$836.16 - 4 = \$209.04.The future value is \$0.The payment frequency is 104.The c o m p o u n d i n g frequency is 2.They will make 2164.088... or 2164 mortgage payments.If Michelle and Marc m a k e one payment each month for300 months, they will pay\$1674.56 300 = \$502 368 in total.If they pay two payments each month for 300 months,they will pay\$836.16 600 = \$501 696 in total.If they make 2164 payments of \$209.04, they will pay2164 • \$209.04 = \$452 362.56 in total.They will save\$502 368 - \$452 362.56 = \$50 005.44 by paying morefrequently, so they should do that if they can.15. e.g., a) M: If it is December, then it is winter.U: If a number is even, then it is divisible by 2.b) Let W represent winter, and D represent December.Let £ represent even numbers, and D represent beingdivisible by 2.c) e.g.. If the sets are the same (i.e., there is one area inthe V e n n diagram), then the converse is true. If there aretwo or more areas in the V e n n diagram, then theconverse is false.16. a) e.g.. If the first letter is a consonant, thesecond letter is a vowel.b) E is the most c o m m o n letter used in the Englishlanguage. X is very frequent in the puzzle.SubstituteJ = A and X = E.EK S Q Q S C A X H B M V T D T YA E ED K J D C S N S A U C X X AA A EQ J T D J Y T C L V XE E EP S P X C D N X B S H XA EY D J H D T C L D S T P Z H S W XED K X Q S H V A .A E A- J C C X B H J C GThe last two words are someones name. W h a tname starts with A, has the same two letters, andthen ends with E? Anne. Substitute C = N.N EK S Q Q S C A X H B M V T D T YA N N E ED K J D C S N S A U C X X AA A N EQ J T D J Y T C L V XE N E EP S P X C D N X B S H XA N EY D J H D T C L D S T P Z H S W XED K X Q S H V A .A N N E A N- J C C X B H J C GW h a t could N E E _ be? Need. There are threevowels left: I, O, and U. Two-letter words usuallyhave a I or an O. If T w a s a vowel, it wouldprobably be an I rather than an O. Substitute A = Dand try T = I.N D E I IK S Q Q S C A X H B M V T D T YA N D N E E DD K J D C S N S A U C X X AA I A I N EQ J T D J Y T C L V XE N E EP S P X C D N X B S H XA I N 1 EY D J H D T C L D S T P Z H S W XE DD K X Q S H V A .A N N E A N- J C C X B H J C G3-12 C h a p t e r 3: S e t T h e o r y a n d L o g i c
• 13. i r-«i fi • I - < u s c d . Try " i f and•, . > . J r . * . i i | l - ; ) .sM-t { SI T I SV s . , I , x 1! ;: f.. y I D T Yi , ^ . E DI. • 1 ^ : !. • n i; f ... /.••J i I* J Y I i i V >.. F- • . : • ;-J X ( !i ^ : E/ .11 : i i : : L •• I /: H s w Xr) i- > ••/ Af . s i . U • ; r .• :.•,( .-l^v ni^/S- • Af St i - . V,",., I ^ < ; i , - .= i b - . | . t . ; t > - S = O andK = H. • • • • ^. • I sV :J l l (, A X i i b M ^/ 1 i YD K ,1 Li ( .> fl A U c. >, X h0 J I D f f; L y /•p f. p X c l l rj X b P. H XV D J H n r f ! ri L. r p / n vv xK X O h •/ AJ r c /. ^ , . .H O ^ is probably "how," so ^ A I T is "wait." which makessense . J I -vT uses the s a m e letter twice. All I can thinkof is "moment." STA^TIN^^ probably ends in "ing." TryQ = W, P = M <"rn t .H O W W O N D E I T I SK S O O S sX A X H B M V 1 D "1 YD K J U C S N S A U G X X AO J 1 D J Y T C L V XP S P X C D N X B S H XS T A T I N G T O I y O EY D J H D T C L D S T P Z H S W XD K X Q S H V A .A N N E A N- J C C X B H J C G- > I M f I X r -Kihr;..- XX(. M A TING is: ifiMHj H . H v V o N i i i _ : " l i i o De "wonderful."X-iljs]i»i;l X i P P I ! . M = U,H O W W O N D E R F U L I T I SK X .J (J ; (, A X H s M v i y T Yh K .1 r t. X, /t pi c X X Ai j I I I ) .1 f f s . L V /P X P X t; P N X H f. H >• • : • R O E P 1 P 1, i L P> G I P Z H S W XP K X i } I , H V AJ { i ; X p, H J (;F R A N ^ is "Frank." ^ E F O R E is "before." SoNO O D could be "nobody " W h a t letters are left?C, J, P, Q, V, X. Z, IM^RC^ L could be "improve,""iutiolilut,; (J K.N p. IJ V X - P and VV -H O W W O N D E R F U L I T I SK X (J u X X A X H B M ^ I D T YP P ; f: X. -. IJ i^, / 1^ < X < AX» J r PI I V 1 C i V XP X X C iJ X B S H X i I h i l I P L Is X T P Z H S W XIJ K X u G ! l V AKJ C (; X B H J C G17. a) Use the finance application on a calculator.Ihe number of payments is unknown.T h e interest rate is 4 % .The present value is \$265 233.48.The payment amount is \$1400 + \$250 or \$1650,T h e future value is \$0,T h e payment frequency is 12.T h e compounding frequency is 12.It will take them 230.631 or 231 months to pay offthe mortgage.b) At \$1400/month; \$1400 • 300 = \$420 000At \$1650/month: \$1650 ^ 231 = \$381 150\$420 000 - \$ 351 150 = \$38 850They will save \$38 850 over the life of themortgage by paying \$1650 per month instead of\$1400.F o u n d a t i o n s o f M a t h e m a t i c s 12 S o l y t i o n s M a n u a i 3-13
• 14. A p p l y i n g P r o b l e m - S o l v i n g S t r a t e g i e s , p a g e 2 0 7C . e.g., In this solution, squares are numbered from 1 to9, from the top left to the bottom nght, as on the numehcpad of a telephone. First, I examined the colouredsquares. I know from the first clue that either square 5 or6 must be blue. T h e second clue tells me that eithersquare 6 or 9 is blue. The fifth clue tells me that eithersquare 7 or 8 is blue. So I know that either 7 or 8 mustbe blue, and one or two of 5, 6, and 9 must be blue.Also, either square 4 or 5 is red, either square 6 or 9 isred, and square 2 or 5 is yellow.Since squares 6 and 9 are red or blue, they cannot beyellow.I know that there is a heart in either square 1 or 2, andtwo hearts in either 2 and 4, or 3 and 5. The yellow heartis in either 2 or 3. All of the hearts are in the first tworows.I decided to begin to place the colours and symbols,knowing that I might need to move them around. I puthearts in squares 1, 3, and 5.Since the ttiree hearts must be three different colours, Ithink that square 5 will be a blue heart, which makessquare 1 a red heart, because the blue colours appear tobe in the second and third rows.redheartyellowheartblueheartred orbluered orblueI know that either square 4 or 5 must be red. Since thereis a blue heart in square 5, square 4 must be red, whichmakes square 7 blue, according to the fifth clue.The sixth clue indicates that either square 1 or 4 is astar. Since I have a heart in square 1, this means that astar must be in square 4; so, it is a red star.I now have three red squares and three blue squares, sothe remaining two squares, square 2 and 7, must beyellow.redheartyellowyellowheartredstarblueheartred orblueblue yellowred orblueThe fourth clue tells me that either 3 or 6 is apentagon. Since I already have a heart in square3, square 6 must contain the pentagon. This cluealso tells me that a star must be in square 8, andsince I know square 8 is yellow, it is a yellow star.T w o yellow shapes have been placed. The yellowpentagon belongs in square 2.redheartyellowpentagonyellowheartredstarblueheartred orbluepentagonblueyellowstarred orblueI used the sixth clue to determine whether square9 is red or blue. The star is in square 4, so square9 must be red. T h e only missing red shape is thepentagon, so it is a red pentagon, and square 6 isthe blue pentagon. Finally, the blue star belongs insquare 7.red yellow yellowheart pentagon heartred blue bluestar heart p e n t a g o nblue yellow redstar star pentagonI double-checked my clues. My puzzle is correct.Solution:KV yellow ? , . _ J^ yp||(..)vv / /A ,hi blue / { blue 1 JX y e l l o v ^ y A / /3-14 C h a p t e r 3: S e t T h e o r y a n d L o g i c
• 15. D. The solution is; Inverse; If a quadrilateral is not a square, then itsdiagonals are not perpendicular,Contrapositive; If the diagonals of a quadrilateralare not perpendicular, then it is not a square,cl) Converse: If 2n is an even number, then n is anatural number.Inverse; If n is not a natural number, then 2ri is notan even number.Contrapositive; If 2n is not an even number, then nis not a natural number.2. a) Converse: If an animal is a giraffe, then it hasa long neck.Contrapositive; If an animal is not a giraffe, then itdoes not have a long neck.b) No. e.g., Ostriches and llamas have long necks,so the contrapositive is not true.3. a) Converse; If a polygon is a pentagon, then ithas five sides.Inverse; If a polygon does not have five sides, thenit is not a pentagon.b) Since pentagons are the only shapes with5 sides, both of these statements are true. Theyare logically equivalent.4. a) I do not agree with Jeb. If / = 25, thenX = 5 or X = - 5 .b) Converse; If x = 5. then x^ = 25, This statementit true,c) Inverse; if / ^ 25, then x ^ 5, This statement istrue,d) Contrapos!t!ve:1f x t 5, then / # 25. Thisstatement is not true, because x could equal 5, and/ would still equal 25.F. To make the puzzle easier, I could give more clueswhere both the colour and shape are given. Or, I couldgive the shapes in a diagonal, or a group of colours orshapes that would show one square in every row andcolumnTo make the puzzle harder. I could not give any cluesWith both the colour and the shape, or I could make thepieces smaller or without angles so there are morepossibilities for their location in the 3 by 3 grid.L e s s o n 3.6: T h e I n v e r s e a n d t h e C o n t r a p o s i t i v eo f C o n d i t i o n a l S t a t e m e n t s , p a g e 2 1 41. a) Converse: If you are looking in a dictionary, thenyou will find success before work.Inverse; If you do not find success before work, then youare not looking in a dictionary.Contrapositive: If you are not looking in a dictionary, thenyou will not find success before work.b) Converse: If you can drive, then you are over 16.Inverse: If you are not over 16, then you cannot drive,Contrapositive: If you cannot dnve, then you are not over16.c) Converse: If the diagonals of a quadnlateral areperpendicular, then it is a square.5. a) I) I ne statement is true.ii) Converse; If you are in Northwest Terntones,then you are in Hay River,The converse is false. You could be in another cityor town in Northwest Territories, for example,Yellowknife.iii) Inverse: If you are not in Hay River, then youare not in the Northwest Terntones.The inverse is false. You could be in NormanWells, Northwest Territories for example.iv) Contrapositive: If you are not in the NorthwestTerntones, then you are not in Hay River.The contrapositive is true.b) i) The statement is true. A puppy is either maleor female.ii) Converse: If a puppy is not female, then it ismale.The converse is true.iii) Inverse: If a puppy is not male, then it is female.The inverse is true,iv) Contrapositive: If a puppy is female, then it is notmale.The contrapositive is true.c) i) The statement is true.ii) Converse: If the Edmonton Eskimos are number 1in the west, then they w o n every g a m e this season.F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-15
• 16. T h e converse is false. T o be number 1, they must winmore g a m e s than the other western teams, but they donot have to win t h e m all.iii) Inverse: If the Edmonton Eskimos did not win everyg a m e this season, then they are not number 1 in theThe inverse is false. T h e y m a y have w o n more g a m e sthan the other western t e a m s and would still be number1.iv) Contrapositive: If the Edmonton Eskimos are notnumber 1 in the west, then they did not win every g a m ethis season.The contrapositive is true.d) i) The statement is false. The integer could be 0. Zerois neither negative nor positive.ii) Converse: If an integer is positive, then it is notnegative.The converse is true.iii) Inverse: If an integer is negative, then it is notpositive.The inverse is true.iv) Contrapositive: If an integer is not positive, then it isnegative.The contrapositive is false. The integer could be 0.6.Conditional StatementInvers. C o n v e r s e _ _C o n t r a p o s i t i v ea) 1 b)T1F 1 1T j | Tc) d)JY r fF_ ^ fF 17 a) If the - l a t o m o n ! is true, the contrapositive -s ahr.hill- If th.; slateriHjnt is false, Ihr- contrapositive i-^- ak-(^b) If the inverse is true, the converse is also true. If theinverse is false, the converse is also false.The pairs of statements are logically equivalent.8. a) No, I cannot draw a conclusion about theconditional statement and its converse. There is norelationship between the two statements.b) No, I cannot draw a conclusion about the inverse andthe contrapositive. There is no relationship between thetwo statements.9. a) Converse: If a polygon is a quadnlateral, then it is asquare.Inverse: I f a polygon is not a square, then the polygon isnot a quadnlateral.Contrapositive: If a polygon is not a quadnlateral. then itis not a square.b) The conditional statement is true. Every square is aquadnlateral by definition.The converse is false. A counterexample is aparallelogram, which is not a square, but is aquadnlateral.The inverse is false. A counterexample is a rectangle,which is a quadnlateral, but is not a square. The polygoncould be a rectangle, which is not a square, but is aquadnlateral.The contrapositive is true. If a polygon is not aquadrilateral, then it cannot be a square.10. a) Converse: If a line has a y-intercept of 2,then the equation of this line is y = 5x + 2.Inverse: If the equation of a line is not y = 5x + 2,then its y-intercept is not 2.Contrapositive: If a line does not have a y-interceptof 2, then the equation of this line is not y = 5x + 2b) The onginal statement is true, because they-intercept of that line is 2.The converse is not true. If a line has a y-interceptof 2. its equation could be y = 2. or infinitely otherequations.The inverse is also not true. A line could not havethat equation, and still have a y-intercept of 2. Forexample, the equation y = x + 2 has a y-interceptof 2.The contrapositive is true, because if a line doesnot have a y-intercept of 2, it cannot have thatequation.11. e.g.. If a conditional statement, its inverse, itsconverse and its contrapositive are all true, I knowthe conditional statement is biconditional.12. a) i) e.g , The statement is true. Pins can burstballoons.ii) Converse: If a pin can burst the M o o n , then theMoon is a balloon.The converse is false, e g , The M o o n could be asoap bubble, for example.iii) Inverse: If the Moon is not a balloon, then a pincannot burst the Moon.The inverse is false, e.g.. Again, the M o o n couldbe a soap bubble.iv) Contrapositive: If a pin cannot burst the Moon,then the Moon is not a balloon, e g.. T h econtrapositive is true.b) i) The statement is true. The negative of anegative number is a positive number,ii) Converse: If x is a positive number, then x is anegative number.The converse is true. The negative of a positivenumber is a negative number.iii) Inverse: If x is not a negative number, then -xis a not positive number. The inverse is true.iv) Contrapositive: If - x is not a positive number,then X IS not a negative number. Thecontrapositive is true,c) i) The statement is true.ii) Converse: If a number is positive, then it is aperfect square. The converse is false. 3 is apositive number, but it is not a perfect square.iii) Inverse: If a number is not a perfect square,then it is not positive. The inverse is false. 3 is nota perfect square, but it is positive.iv) Contrapositive: If a number is not positive, thenit is not a perfect square.The contrapositive is true. Negative numberscannot be perfect squares.3-16 C h a p t e r 3: S e t T h e o r y a n d Logic
• 17. d) i) The statement is true.ii) Converse: If a number can be expressed as a fraction,then it can be expressed as a terminating decimal.The converse is false. For example, - , written as adecimal, is 0.333... This is a repeating decimal.iii) Inverse: If a number cannot be expressed as aterminating decimal then it cannot be expressed as afraction.The inverse is false. For example, 0.333... is a repeatingdecimal. It is also — .3iv) Contrapositive: I f a number cannot be expressed as afraction, then it cannot be expressed as a terminatingdecimal. The contrapositive is true.e) i) This statement is trueii) Converse: If a graph is a parabola, then the equationof this parabola is f{x) = 5x^ + 10x + 3.This statement is false, because there are manyparabolas that do not have that equation, such as f(x) =iii) Inverse: If the equation of a function is notf(x) = 5x^ + 10x + 3, then its graph is not a parabola.This statement is false. For example, a function can havethe equation f{x) = x^, yet it is a parabola.iv) Contrapositive: If a graph is not a parabola, then theequation of this parabola is not f(x) = 5x^ + lOx + 3.This statement is true, because only a parabola canhave that equation.f) i) This statement is false. For example, - 1 is aninteger, but not a whole number.ii) Converse: If a number is a whole number, than it is aninteger.This statement is true,iii) Inverse: If a number is not an integer, than it is not awhole number.This statement is true.iv) Contrapositive: If a number is not a whole number,than it is not an integer.This statement is false, for example - 1 is not a wholenumber, but it is an integer.13. a) e.g.. The contrapositive a s s u m e s as its hypothesisthat the original conclusion is false, which m e a n s that theonginal hypothesis must also not be true. If the onginalhypothesis is not true, then the conditional statementmust be false.b) e.g., The converse of a conditional statement isformed by stating the conclusion before the hypothesis.The inverse is formed by negating the hypothesis andconclusion of a conditional statement. Since negatingboth parts of the statement is the same as reversingthem, the converse and inverse are logically equivalent.The inverse of a statement is the contrapositive of thestatements converse.14. e.g., a) Conditional statement: If you are tall, thenyou like chocolate.Contrapositive statement: If you do not like chocolate,then you are not tall.Counterexample: I a m tall and do not likechocolate. Both the conditional statement and thecontrapositive are false.b) Conditional statement: If a traffic light is green, itis not red. Contrapositive: If a traffic light is red, it isnot green. Both the conditional statement and thecontrapositive are true.15. e.g., a) Conditional statement: If it is Saturday,then it is the w e e k e n d .Inverse: If it is not Saturday, then it is not thew e e k e n d . The inverse is false. Counterexample: itcould be Sunday and be the w e e k e n d .Converse: If it is the w e e k e n d , then it is Saturday.The converse is false. Counterexample: it could bethe w e e k e n d and be Sunday.b) Conditional statement: If a polygon has sixsides, then it is a hexagon.Inverse: If a polygon does not have six sides, thenit is not a hexagon. The inverse is true bydefinition.Converse: If a polygon is a hexagon, then it has sixsides. The converse is true by definition.C h a p t e r S e l f - T e s t , p a g e 2 1 71. Let L/ represent the universal set of writers. LetP represent the set of poets. Let N represent theset of novelists, and let F represent the set offiction whters.Subset fiction writer, F = {Armand Ruffo, RichardVan C a m p }p i s i l ^ ^ M f l ^ l l i i i MA j i n . i n d Rijfi;:f;< b.-iid V.ui ,inip2 ^} 14 1 1- 17 •. 1^ -• 21 22. • : 24 i )Set C is inside set A, therefore C czA.b)A = { 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 1 1 , 12}B = { 1 , 2 , 3 , 4 , 5, 6, 7, 8}A u e = { 1 , 2 , 3, 4, 5, 6, 7 , 8 , 9, 10, 1 1 , 12}F o u n d a t i o n s of Mathematics 12 S o l u t i o n s Manual 3-17