MAC 1147 Exam #6 Review
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- Rudolph Bruce
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1 MAC 1147 Exam #6 Review Instructions: Exam #6 will consist of 8 questions plus a bonus problem. Some questions will have multiple parts and others will not. Some questions will be multiple choice and some will be free response. For the free response questions, be sure to show as much work as possible in order to demonstrate that you know what you are doing. The point value for each question is listed after each question. Notice that all four parts of question 10 are multiple choice but you will get partial credit based on your work. So it is possible to get the correct answer and not receive full credit and it is possible to get the wrong answer and still receive some credit. The bonus problem will be worth 10 extra credit points. Attempting the bonus problem can only help you; it can't hurt you. A scientific calculator will be necessary on this test but you may not use a graphing calculators or calculators on any device (cell phone, ipod, etc.) which can be used for any other purpose. The exam will be similar to this review, although the numbers and functions may be different so the steps and details (and hence the answers) may work out different. But the ideas and concepts will be the same. Additionally, you will be allowed to use the Trigonometric Identities and Unit Circle which I have posted on my website, along with the formulas on the last page of this review. The formulas provided on the review will also be provided on the test but the Trigonometric Identities and Unit Circle will not. So if you wish to use either or both, you must bring them with you. The Trigonometric Identities and the Unit Circle may not be shared nor can they be used if they have any writing on them.
2 (1) Solve each triangle. (4 points each) Q71.l'S" (i) B = 10, b = 9.5, a = (a) no triangle 51"\ \(1 -= '''' A..'5 J.l.; /()o -\}OO +C ';:.o \{bo tc- -:;.(w> C= I,\Oo,. \':\,c;(j b -:;, (a) no triangle Q A = 87.4, B = a'l. '=: 6'1. -t- Ct - a\ac. Cd> A Iq'1. If.Dt +u -J. (I) (1\) CO'.» A. 3(, \-:: -\ \2 \-?,t;' (.(:b A _ I\' - '3C}2 toa o. Olts =(\So A (b) A = 57.3,B = 87.4,C = 35.3 & A = 87.4,B = 57.3,C = , C = 35.3 (d) A ,B = 35.3,C = 57.3 ;( -:: 31 H1.\- """to\. O,S\{ = B 0 'B-= n\(o,9t) =5T.--S 'ti,l.{ t.1 -r C:/lio Il{::t 1"L -=-I C -=?l5.3
3 C ( a) no triangle ri Q c = 60,b = 6.13,c = Pr -;,u L:..:.. -- \,---\0.\'\ c./' fii"a- - - \0 (b) C = 60, b = 5.65, c = 6.13& C = 60, b = 6.13, c = (d) C = 60, b = 5.65, c = '(f.. -::roo -tc:: ll(jt POo f C. :: (()o CCoOo ::: t'\c... (iv) a = 7, b = 9, B = 49 b A'l,C; Q {,\<i >..,c... ot S'\f\ = <t;\f'a_ (a) no triangle (b) Al = 76, C I = 55, CI = 7.60& A2 = 36,C 2 = 95,C2 = 11.9 A = 36, C = 95, C = 11.9 (d) A= 76, C = 55, C = 7.60 It"=-.'o (Y A== m>0 -?1S.'o =- I.lo l x;.o t<{.{- ( :::- (in Q <rt.'o +t two q 1 c -:.. q I D +-\,.. '{,b _ :(.1-$5)' _ sa' C'{iWl. _ &\Y\ C. -(W\. A- -= " - - C \\{. \0 -+ '1o +C (8tt l1),11i +C (D < \t. -t.o leij -Iu -h+ -to ",01-.,\, / -rl.
4 (v) B = 112,b = 3,a = triangle (b) A=56,C=12,c=27& A = 18, C = 50, c = 20 (d) A=18,C=50,c=20 5\ \lao 1;>'iY\ A - 4- '5\ t\ II tt -:' '-\-s:,,," \\o "b :: J C ;Ql ::: ). ({o "!, (+ -=- \f\-\ (1. Y'J 6-\- \t\u \i\ J. "vt_\ )( \'t> -I!:.. X I (a) no triangle (c) c = 9, A = 52.8,B = 57.2 (b) c = 9, A = 52.8, B = 57.2 & c = 8.2, A = 43.5, B = c = 8.2, A = 43.5,B = 66.5 c'--= c.'- +l-'z. - 2 o.\'cot 4'3-l1r +- &-\::ro o -= IroO c"l -=- 'Z. t8"'l - J.()C!)Ca =too '1- f. 1..-\1.-C -;2(,)(. j) (l)')a It,.\.(SO +() ::: l&t> 0 c"1-= - 'J\ (34J) 30 :: \{. Co1.2-Y-13l.2 tj'i»a - qs. li ':. - l3l J.t05 A o. 1 7k -= <»5 A AU>.-\ (o.t) """ql 13 =--(p(o.
5 (2) Evaluate the expression. (2 points each) (i) 9! 362,880 (b) 18 (c) 72 (d) 13 (e) None of the (") 7! 11 (a) 72 (c) (d) 2 (e). -:=. -' 1- \ '- ' A-1 - j.g t None of the (iii) G) (a) 1 (b) 5 (c) 0 {fy1o (e) S\ \ l5 -\( r -- None of the ( ) J. \ (s-;;y. :2 '. I. \- ( -7 'l. :: S Z,O
6 (3) (i) Write out the first four terms of the sequence. (4 points) an = 9-6n (a) 3, -1,-7,-13 (b) -6, -12, -18, -24 (c) 3,9,15,21 3, -3, -9, -15 (e) None of the qi q-- (t') :=- l q"3 -:: 4' - G,b) :: i- (7-4 -z. q- ('l,) 1'-\ ':" -J l1''1 4- U'1) -= - Zl( -::. -(,) 3 / - 3, -'1, - (5 (ii) Find the common difference of the previous sequence. (3 points) (a) Ci( : G'!, - 'l. -:= - Ii - (- )) -, F3 -:' (c) 9 (d) -9 (e) None of the - G :: - tj - (-) - ts" -H ::.- Ct, (iii) Write out the first four terms of the sequence. (4 points) (a) t,5,25,625 (b) 5,lO,15,20 (c) -5,5,10,15 (d) -5,125, -625, N one of the,}- l - I Q3:::tl) 5 :: -5 -= - 1'2 - c;, (ZS') - '3'ZS I rzs (iv) Find the common ratio of the previous sequence. (3 points) (a) t (b) -5 (c) (e) None of the V
7 (v) Write out the first four terms of the sequence. (4 points) an -l al = -8 Un =- n+l (a) (b) l:j.\-8 (c) - - _2 ) ) 3) 3 ) ) 3) 3 ) 3) 3) 15 (d) _1 ) ) ) 3 (e) None of the Ct.-, f1, - Q tl "'5 3 G,_, G"Z.. 3 -L l- - X- 'l ::: "?, >( rt '1 I 2- I L ') 5 fs - '2 -, - - I "!>
8 (4) The given pattern continues Write down th of the sequence suggested by' the tt (4 e?rmula for the general nth term pa ern. pomts) '1' 4' -9' 16' - 25' 36"" :, 3 +-ZCO) Z( (-I) 3 - -, - - l, Q: , ---,1... )J. 2 (Z.-l) es 3fl. 1.. :;,.\ "ll'\ :::. - '1.. \ = - 2'1- Zl. "3+ 2 (oz (3- ) 5'TL _ 3... Z tz ::.- ' ,;- '1 3 -=- - 3' :t.n.. 3t ' \ q G1 :: - '1"l. 'l. 3 2-l) \- 2(' ttl 2l. { - - t _ '3to ',( Fl (.'1.
9 (5) (i) Write out the sum. (4 points) k=l (a) (b) Q t (k-('l.:: ( I.f-S -I- (z.t)' (3-I' t."::1 (d) (e) None of the (ii) Express the sum using summation notation. (4 points) -\-\ ( : y..-, Q, 7, i =- - '"... :)... Q1, ::. ) n-:;:..i c- C.,.. 3 a :: t ::() tc _ \' 4,\ ::' - J(I..\) (iii) Express the sum in part (ii) in summation notation but with an index starting 2 lower than it started in (ii). (4 points) b";' S fi) 5!i\"t I D.M -hl (i';). J Ot"\
10 (6) Find the sum of the series. (4 points each) (i) 15 L (k 3-4k + 3) k=l (a) 7743 (b) 8505 (c) 7785 (d) None of the,,' '" :; r(l:: r UNj' = IwL-:.I'\,oo I '-{ r: "- "':- rls"('\ (llo ') ::. \.t v l 'L') t "3 -::., (\t)) -;-\.{$' --:..\ t (\?... k-\- =- {\)O.-l(tD+\{S" =13fDS' (ii) 1113 (b) 1155 (c) 1197 (d) 371 (e) None of the f C3l1.) -;: f. (3l) - L (,&-) I5'S -l{l :-'i ".:;. l '\w I [?th)('3h1 ::,:t)16.] " )(k\'i'l? tio(ioto '(Otl'>1 "\ :? r{( II 1"-3fu.n 'tis" lis!) \
11 +'-.r r -4' (iii) (b) 184,041 (c) 182,329 (d) 183,612 (e) None of the ar'd'h...wc.. 5 W a=-l &ct H' 3}5l1+' +fs} (I+-S)) - ( ) -=- l+- C = ft Z- 2 ::: "t;j (iv) A 9 27 (a) - (b) (d) - (e) None of the S.('.Q w\\-t, Q ;).. Ix c = - _Co (0 -.l _ (;2. _ -- - I-H) - Gd)
12 (7) (i) Find the 50 th term of the arithmetic sequence with initial term a = 6 and common difference d = -9. (4 points) (a) 285 (b) -444 (8)-435 (d) -429 (e) None of the o a.j- 1\ (V'\-l) 01 Q Co./- (so 1) (--4J S1> ::: Co +- \.((- C,) :::: - lt\ :::. --tf3s' (ii) Find the l3 th term of the geometric sequence with initial term 177,147 and common ratio. (4 points) 8) 1 (b) l (c) 27 (d) 59,049 (e) None of the -
13 (8) (i) Expand the expression using the Binomial Theorem. (4 points) (4x-2)4 (a) 256x x x x + 16 / -z.. I 4::-":; L (b) -256x x 3-384x x '1'\ 1 (c) 4x x x x x + 2 'i G, If Ie:- 1\ I.f»Q. 4x 5-40x x 3-80x x - 2 None of the )c -2') =I-0.S (-Z)o t- lj. ("IX»(--1.)' + Co (). (-2)' -I- l(. ('/)' (-d + /- (t r-1-) I _g.l<")., -t "f ((.,) (-l)..-<0 (I(gk:)() + 'f (h)(-f) + (-, -Ow,) _ 2S1o K'f _ t;/2x'3 T 3S<.( 'l. - \2 x f { (ii) Use the Binomial Theorem to find th. (2x + 5)5. (4 points),, e coefficient of x in the expansion of (a) 15,625 (b) 2500 (c) 1250 D(d) 6250 (e) None of the YJt- os (;). S'l -'--! (t
14 (iii) Use the Binomial Theorem to find the 8 th term in the expansion of (4x-2y)9. (4 points) (a) 36,864x 2 y7-73, 728x 2 y7 (e) None of the j Th -6 -\<rm CO'\ 10.\1> h. y) 1- (- vk.y cot.f.&; ;5 ()?Jo-(y Y:Y (-ill 1. (11.-<')(- \2:r) ) 1 ( -))\ \ q.. "X" \( -:;: 3lo ":f'. '1 \. -t.. I
15 Bonus. Use series to represent the repeating decimal as a fraction in lowest terms. (10 points)... ) - 1'1 rooo 0.. '1_') 1000 ( \000 (
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