wy k +j---dscheib/teaching/summer2011/mac1147_files/… · (4) the given pattern continues. write...
TRANSCRIPT
MAC 1147 Exam #6b
Name: ----'-~--'--_ jWY K__+----
ID# Sv~tw\lf dD II--~~~~-~~-------
HONOR CODE: On my honor, I have neither given nor received any aid on this examination.
Signature: _______________ _ _
Instructions: Do all scratch work on the test itself. Make sure your final answers are clearly labelled. Be sure to simplify all answers whenever possible. SHOW ALL WORK ON THIS EXAM IN ORDER TO RECEIVE FULL CREDIT!!!
No. Score 1 /24 2 /6 3 / 18 4 /4 5 /12 6 .._·0 7
/16 /8
8 /12 Bonus / 10
I Total I /100 I
(1) Solve each triangle. (4 points each)
(i) b = 3, C = 4, A = 95° ·1..
(a) no triangle (b) al = 4.2 , BI = 49.9°,C1 = 35.1° & a2 = 5.2, B2 = 35.1°, C2 = 49.9°
(c) a = 4.2, B = 49.9°, C = 35.1° = 5.2, B = 35.1°, C = 49.9°@a
b-l-~ (;.1. ~ (.7. - u,..c. M BfA?" ...,.~.,.. +(,'t._ ~k (,o~A
\) 3~= *5.)'1- +~'L- .1C~.I)('1) CtJSB 130.,0 t C. -:. tfsOD
Q'J.: ,~.r~"1. ~ J.(~'('t\ ~ 'i'S q:; 21.o~ ~t'o- It(·<' C(b e, C~tt~ ·~·
a1- -:.~ H\O -l\{ (- [) .C\".}) q:: 4~.o~ - l{t~ tD~~
al- -:. 1.'!;' "" J,. oil -~'1 ,O\{ :=: - ~f . (, Cir.I ~
~'L=-'L'1jOU c.o~ \3 = .l6lib
(.t -;. ~ -:: '5 .'2 B-= ()S-' (.~\<i)~35· lO
• • ,.-;. "\0 ~ e'\.i-: C
(ii) a = 4.6 , b = 7,B = 80°
0.-:.5
0.. ~'\.\# (a) no triangle
\, ::..1
~"to° _ ~\ ... Pt_ "). .... tt.~
'01 0 A~ 4.(.;,'11)' =".{'C'ltS) -:: [,Ifl1- .
ft ~15t'n-\ (.(,,~:}\ = 40.~o
(b) Al = 40.3°,C1 = 59.7° , CI = 6.1 & A2 = 75.4°, C2 = 24.6° , C2 = 4.6
(d) Al =40.3°,C1 = 59.7°,CI =4.6& A2 = 75.4°, C2 = 24.6°, C2 = 6. 1
A=-\{o.-;O Of' A:: \'60 -l{o .~o = r~'L'=lD
! ~ 'to :1,0 -t '60' it-::\to· \~/":)O + 00" t C. :: liDO
I'JO ,1,,' t t ~ (~" ~lt1.1° +t -= (tOi)•C~~:=.) ~
V\ot ~'i\lti -foo l:Ij --ft, ht 'I~ -tl. "'<if
(iii) a = 15, b = 10, B = 20°
'6,,, - - to
(a) no triangle
<5)'1'\ duo _ fI,,,A- -'0 \'5
. A - f5 &"\" ;)..{)c -:::- 15( .1lt1 -; .5""1 10
(iv) b = 6, c = 7, B = 70°
® no triangle (b) C = 36° , A = 74° , a = 17& C = 75°, A = 35°, a = 13
c= S\i\~ \ (\ . ~O\
~~\- ~ ~\n.tt ~\"- o~ ~u"i l 'X. ;S -- l ~ '( f- \ .
(v) A = 105°, B = 25°, b = 7°
(a) no triangle (b) C1 = 50°,al = 16 ,c1 = 12.69& C2 = 50°, a2 = 12.69, C2 = 16
® C = 50°,a = 16 , c = 12.69 (d) C = 50°,a =12.69,b=16
IO~ +2~o ~c. ::: I~oo
1'3(f tt. ':: If00
C z.t;l)0
~.... e, -;:. ~C \, c
c= 1 Si,,9Jo ~i,,2SO
(vi) a = 8, b = 6, C = 4
(a) no triangle ~~A = 104.5°, B = 46.5°, C = 29°
(c) A=46.5° , B=29°, C=104.5° (d) A=104.5° , B=46.5°,C = 29°& A = 46.5°, B = 29° , C = 104.5°
\'1 2.. 1- IO\{,~ +~~.So +c.::: '(t/~t? -:: II +C - J.G.c.. C£ls B 0.2. =b'l..tC,'t - 2~ CtbA
151 0 t<.. ~ tiDO
(g~';:. ~~ -rlf~ - .l(g)(~) CDS B ~.,..= Cot t"'1.- J(~)('1)co<;A c. = .;z,o 30~ (,'{-tHo - Co\{ ~ &
C9"t ~ 3~~ I" -~ tDSA
3~ ~ gO - ~\{CO~ ~ (g"{ -=- S~ - llt CDS A
-~'i ~ -~~~ B l2. -:: - ttt tos A
Co'b A-= - •1,5"
- -
(2) Evaluate the expression. (2 points each)
(i) 7!
(a) 28 (b) 40,320 (c) 7 @ 5,040 (e) None of the above
( .. ) 10' II ST
(a) 2 ~90 (c) 10 (d) ~ (e) None of the above
10 1 . -- ..,
(iii) (~) (a) 1680 (b) 35 ~70 (d) 140 (e) None of the
above
1. \ 0~! g! f 1"'ts: ~- . --:. '1 ' 1-S" ="1-0 - --( ~)~ \{I."t~ ~.~(. /y1.(~ .,\~) I,
I l
(3) (i) Write out the first four t erms of the sequence. (4 points)
an = 6n + 5
@ 11, 17, 23 ,29 (b) 5, ll, 16,21 (c) 6, ll , 16,21
(d) 5,11,17,23 (e) None of the above
~~ ~(,(lhr-::: tlt?::.t"i
U, ::.. "" ('Sl' ~ ::: If l-S"" ::- '2)
(ii) Find the common difference of the previous sequence. (3 points)
(a) 5 (b) -5 /(;;06 (d) -6 (e) None of thel7 above
(iii) Write out the first four terms of the sequence. (4 points)
3n
a =n 2n +1
D( ) 3 9 27 81 3 .2 II 81 1 3 9 27(b) (c)V;;!:}) ;), 8' 16' 32 2'4'8'16 2' 4> 8' 16
(d) :3 9 27 81 (e) None of the above 2'2'2' 2 ;; -;'\ z,":}----.. J \~I 4\= l ~tl - --- =- 2'4 II..
L-_~.:E--_ . -0'1 - ~ Z.. L
tl -")1. 8 C1 -:: - - 31"t 2"' .... •
3 ~ .B:. -1L ~ I l J \~' ~"2.
(iv) Find the common ratio of t he previous sequence . (3 pOints)
@~ (b) ~ (c) 3 (d) 2 (e) None of the
n ., above
blf\1. Q _ _ \4 - ---..... " 3 ~ '_ 1.-p. l-\ - T f. · ~ ~ l--~--- ~ - Lat 4 y \ 11
ho 1--! - .1a~ - J-.. ~\Q
l.
(v) Write out the first four terms of the sequence. (4 points)
(a) 1,4,2,5 (b) 2,0,1,2 (c) 2,4, -1,5
(d) 4,7,11,16 @ None althe above
Z r'1 , -=t,t\
(4) The given pattern continues. Write down the formula for the general nth term of the sequence suggested by the pattern. (4 points)
(5) (i) Write out the sum. (4 points)
5 1
L 2k+1 k=O
(b) 1 + .! + .! + .. . + ...l. (c) .! + .! + 1 + .. . + ...l.2 4 32 2 4 8 32
@ None of the above
'5 \ ~ , + .... .} -I o~\ + J ;t~1 14 t-IL. ;J. ~...+\ ;;2 ,.... , :;z ~"'" J
:l';- tlJ.. ~
f-"VO , 1 ~ .-L +- --'-- + j -\- +- -l~ JIiI. J~- -;).' ;;.~
, -L--LtJ..{ ~+- t, -t~ -;;;2 + c, It
....- 2- '1
(ii) Express the sum using summation notation. (4 points)
. '{..-~
1
1
3(~\ C\\{-:..- ~ = -~
(iii) Express the sum in part (ii) in summation notation but wi th an index starting 4 higher than it started in (ii). (4 points)
tvo~. O~ ~Mrs Fs~i~~" ~)-()\ ()rI\~W -h> ~art- (ii)
(6) Find the sum of the series. (4 points each)
(i) 12
L (k2 + 2k - 4) k=l
(a) 680 (b) 650
t ~:: 10l-(1~"1)~ ' n ·h)
~~I
(ii)
k=6
~(5-Yt~ \1 -\'5\'1.
"1 J'( l'i,7.l \ \05'1
1(1\/ olc;,\-::: ll, oSll .l(~J.~\ -== ,\'51)
--
-\d. -\ ~ ~~ ~J,. (ii i) ~fV\0
10 + 12 + 14 + 16 + 18 + . . . + 68
(a) 1140 @}) 1170 (c) 1320 (d) 1425 (e) None of the above
().;i-lt,,,,,*,<- ~utK\lR w...l{., a '" 10 t. d~ :l. (041Zf(H . . . +~~ :. 3g (IOf{o~)
:: \~(~ilQ,,:: a4- (V\-l) d
:: II:}O
~::~ =9V\~ 30 \ \ \
(iv) ...:'" ..."'" ~"" fYY\t 2+-+-+ - +···
2 8 32
(a) ~ (b) ~ (c) ~ (d) ~ @ None of the above
"V" \\,. a~ ;). &c:, -;: ~
C~l~ ..
(1 ~~ ~'1 4-l
(7) (i) Find the 48 th t erm of the arithmetic sequence with initial t erm a = 1 and common difference d = 12. (4 points)
(aJ 589 (bJ 577 @565 (d) 553 (e) None of the above
:: \ + y~ lr2.)
-::.. \ 4- S10 \{
~ SW$"
(ii) Find the 19t h t erm of the geometric sequence with initial term 2d48 and common ratio - 2. (4 points)
(a) -128 @128 (c) -64 (d) 64 (e) None of the above
(1\- 1
(-2)
t ( ~(o~, t~l{) ~O~~
~ (?-\ :: l'l-~ -~,
(8) (i) Expand the expression using the Binomial Theorem. (4 points)
(3x + 2) 5 \e:-"'~ I ,~I'\ ':-1
(a) 243x5 + 810x4 + 1080.1:3 + 1080x2 + 810.1: + 32 I 1- 1~f\"''2.. (b) 243x5 + 162x4 + 108x3 + 72x2 + 48.1: + 32 I ~ ; \ e-~\'::: 3
{(c))243x5 + 810x4 + 1080x3 + 720x2 + 240x + 32 I '1 ~ ~ \ ~V\-= ~ '(d) 3x5 + 810x4 + 1080x3 + 720x2 + 240x + 32
\ ., 10 \0 ~ \ 4:- h-=~ (e) None of the above
(3x~a)5::: t- (3)(Yi _;;..0 + 5 -(»)<)\:Z\ (O· (3)t.)3 . ,,7. 4- fO ' (3)(Sl-' J~~ '5 'B?'-)' .2"i -+ (-(3Jtt-ZS"
:: \ ' (2~)f} . ltS(~\)C'f) ' 2 -\- \O ' ('2-:})t3)'~+fO(qx~) ' 8 + - ~(3)c) ' 1~ t-{ · l' 32
(ii) Use the Binomial Theorem to find the coefficient of x 2 in the expansion of (4x + 3)6 (4 points)
(a) 23 , 328 (b) 34,560 (c) 11,664 @ 19,440 (e) None of the above
To r t I wf NUt -!k 4V'" ( 'IS C~)~ I ~ CP~ thU t<.t's (~)wi-l'k
_ ~ ::."s :: I!:"
\
(iii) Use the Binomial Theorem to find the 5th term in the expansion of (4.T+5y)5. (4 points)
(a) 15,625y5 (b) 10, OOOX 2y3 @12, 500xy'
(cl) 2500xy4 (e) None of the above
s·· ; :: 5
~. ,
'5 . ( It xl' .("51J~
'5'("\"'\ (c.~S'7~)
Bonus. Use series to represent the repeating decimal 0.67 as a fraction in lowest terms. (10 points)
~ - .2..-+ --4100 /000
l (\ -' J- J- + , .' \ 100 -\ 10 ((10 J
~~ (00-10(\ - +0,)(00