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M122SecF02Test3key

# M122SecF02Test3key - l‘viat‘nematics 122 Section F02...

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Unformatted text preview: l‘viat‘nematics 122, Section F02 Test 3 December 28, 2007 Last Name, First Name: 7 Student: Number: lnstructor: M i Schurch You have 50 minutes to complete this test, The only calculator allowed is the Sharp ISL—.510 R. (But no calculator needed!) Do not simplify your EmHVfﬁfI‘S. In each case to receive any credit at all you must show enough work that it is possible to determine Where your answers came from. J. State vheztther each of the following is line or l’alse. DUPOH'IAN'FI‘: Each correct response is worth one mark, each incorrect response is worth —?; of a mark and if a question is left blank it is worth zero. lt is possible to score less than zero on this question. Let f : Z'* —> R and Ar] : Z+ a) R. (a) ,Jjﬂggm li' E Elk/(71)) then 9(71) 6 (St/(Til) (d) ___EJ5'LW 32007 E l(lIlOd «Mo ((3) MELALW list a, b e: Z”. ll lcrrﬁai 3,4 ‘lc their gcilhrj) 7.4 l. 1‘ Caruijtlwx. 2i [3] Let (1.7[370 e Z Prove that if (rib and oil’brl c), then ale, guppm at”; amok at! 00%)» Time warm Obi la+c:cc~m mum who é2_ 7L9» bu, “WWW :7 0'”*C :aiv‘“ =7 L; arm' GL-KL :j C :0t(m"\—) JIMLH VH“V\ =7 ale. 3. Let a.., b, C E Z and n E Z“? Prove that if a E b(mod n); then ac E bc(mod n). Suppma CLE‘O (“ax”) TL“. “I (“40> :7 RICCR‘E’) C E g 2 I‘LIchcla z QC‘IOL C E '06.. (VAN/gm) : 0L Al, (21) Use L110. Eudidean algoriUnu L0 dctenuine g<"<1(z:192,154). 2 7‘ ~ 50 ‘ r s w 5 + q 50 z 7* f1 Lt 14% ﬁciO/tcll) SW): 3* (h) [2] Determine integers :r: and y such that 492:1; 154:4, :: gem/492154). 1 2 so - ?- Li 2 BO «105% 530) {who} _ 4cm) : BQDWLI— 346% ~ 7R4“) 1 (390ml) r (413)030 5. Use the deﬁnition to prove LhaL 1072f“ ~ir 5"" ~‘r 1} is Om,“ 125C, (in?!) a: 67 Mon," +5“ H! = lan“‘+5“+tf ,\7n3’ \$ 10%“ +r{‘+4“nh Vnzs é Iann [Er/“35.. 6, 1L3} Use the deﬁnition to prove that 2n2 ~+ 3n is not C(71) \ A ‘ t ,u + badger/x Hm IS ‘ ‘Leh me‘l and; m C: '2 MK [4‘ (3 Z: I '3“ka \AAVQI 5 m.n ) \fﬂ 2 [RbSyll/Vt‘ vale—z I‘s hV+ Madam SN”? [OL'H‘ 3 . i m i ‘7“ ’l/‘ CLN \0L«S'\“\\L\ C13 nim€_ I VnZl‘ 77 1n ( ( i'. The Fibonacci numbers may be deﬁned recursively by , 1‘0: \KJ [255 {Low- 64 {\3 1., F; I J and for integers n 2 Fri—1 fin—2- r\ (I) 1]] UP'tPrminE? Fa, @23 b) [4] Use mathematical induction to prove F“ > (ié‘ﬁ')'“"" % 7 case you do not; have your calculator; (‘3)5 : 7.593751 ‘ {5m mus, 1615(4) FU:§ 7 L: 69?) FT: l3 WiMom’ {£9 for ad] r: I?» 6. marij (Just in gar 30M 14 2 _/ 'SVHSk \mx) ‘1 \ "r FK_‘ C Hm Mum'in oﬂmbv» 0‘ FW> -\ Vi-l ‘ . 705‘ 4%.) U33 iL I H) E, W 8. [2] Find the general solution for the. recurrence relation an : Saw] — 2cr.n,g., n 2 2. a“ * 3am Mini: (7 (b) Solve the recurrence relatdcn a,1 : CLO ,_ an ‘ 1(er Swnger‘: (REP): (‘VL C wer Cc-vaanA’)‘ 01830:} \l/vk‘C’ \ 1! pr, ilk 6‘- V\ K W WNAA a 89m: SUIWLCO'N a,» vbaw-I + Jami :5 Saw] — 21:17]“? + 5, n 2 27 a0 : 0,0,1 : 3 /-\-;L“ H5 «Cr-m axiom. 803 is a. j N h H4 (“Whig/(Mons W.V.> WL- “400‘” “rib-1(va =5 M4M 'r3lr +7):ka “LN/r :5 c V5-5 mph—3n. Zj)aw’;%im’-SV\‘<K n In .A“Mu.n A v "m .M A ‘ -“Lﬁvh- ‘ A1 :1. [1L] Dettuulm: the mmle UlL buugmph « 1 00 & m5+<1>r+ <1) 2“ I . 7/. ,[Lg‘ 10. [3] Let T r (V, E) be a tree with 7 vermies of ('iegree ‘2 or more. If the sum of the degrees of the 7 vertices with degree ‘2 or more is ‘23. dem-rminc the number of vertjces of degree 21 in T. W[ 2 A aux [A [1» H: Raj; WV‘LMS A ...
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M122SecF02Test3key - l‘viat‘nematics 122 Section F02...

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