math103hw6solns

math103hw6solns - ’17[J’zixéuaqi 17v 0>5“ “29...

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Unformatted text preview: ’17 \[J’zixéuaqi 17v 0 >5“ “29 76/3 Fhrsf weHLQ SL4»! “ML VCWLV’. L61” V'eV~ M "j W, 1; 3?6 \/J’ Hum 1720. EA” Jr‘vd’aflMLW VéQ/QLJ M 5m ‘7. was ML; Hwy} CWLY’ New 5mm, (QM/.NV' (L‘mQ/J’):n, 0?wi 0/4.) )kanM (QJ'Y): I4/ war-no we, jaw” OLLmQ/L 09M g/LyL Li LymWL VC (VJ-3L \qul'35 V 'Q/Ly’h 2 VCR felRV‘ LLL “PM! NM £9“ +4“ W W‘Lriaflj? M va'zhivev iLevL. W “iv: 7 X ' CVL LXJ‘) ‘iVLx 4'): 'Pv&')"L+L]RJ-flz+l§‘£¥# BuLEZ 2L :0 $9 ml“: 11%)] § WWO“ 554* “L 7 “Yvéfllén “>2“ Gte-«wlrfiwflb/ 14A,. “H. A. ved‘or M 1W .1 1 :3 owe/0%Qr9ecf if {Ms/op some” :' 10‘ fi-o"= 2+3k+4=fi+3k. The WOVectors enclosearightangleiffilfi':6+3k=0, that is, if .3: = —2. 12. ”6+1D'H2 = (13+1m - {5+tfi) (by hint) =“fiF+HflF+2W4m[mmkfimMHfihmwfl E Hi?“2 + ”fill” + 2HfiTHltfill {by CauchyHSCth} = (llfill + “one, so that “5+ fill? 5 [Iran + Illa")? Taking square roots of both sides, we find that ”1? + 13“ 5 “if“ + ||tfi||t as claimed. 14. The horizontal oompouents of 131 and 1'32 are —|[F‘.1||siufi and |EF2|| sing. reapeotively {the horizontal component of F3 is zero}. Since the system is at rest, the horizontal compononts must add up to I], so that — || F1 || sin 13+ t ~_ 4 . fl _- |ng|| sino = 0 or ||F1i| 5111.6 = ||F2|| sum: or || || _ 3.343. To find %, note that m 2 Emma: and E = Emmi? so tlmt g-% = “m“ =' tan 9—“2 - “—“i-E =- "f 1' fig Since a and {3 are two distinct acute aligies, it follows that :iln ,5 cos c! "Fa ll cost: ' .g—é 5,5 figfi, so that. Leonardo was mistaken. 2 -§-‘ = 0.20 + 0315 + ll5c : 76, o lilinhm‘ze quantit at" - at“ = 2 3 {swarm} tells us that (55. 3})? 5311531]ng o + b + c2. The Cauchy-Sc] 1 ’ i « - o2 + .52 .f. C2 3, m? [[2, or T62 5 (a? War-z meliuélhty ___ ' + 62 + 03(1),? + I132 2 033‘ Quantity :12 + 62 + c2 is minimal when a2 + b” + 2 -— $20.5 J {’1' is the case who 1’ E 0.2% - c H m- This 13 a, :.-. : 0.3%; for some ' ' I _ pofittl‘b'e constant t: It ' ' I c 0‘ .. . lS molar-ad th 02a+0.3b+05c: CI 5;” at ($%+ww%+052 . . it ~.: : The Student mus { J 76, 90 that k = 200. Thus the first exam, 60 hours for 34. rref['A) = [ [.‘r 1 2 A basis of ker(A) is 51 = 10—1—2 3 DHMH 2 1 _ 1 W1! — $5 6* firm-e72 £1 _ 1 ‘ ifwzii — Ire—tapas M — m ...
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