Solution6 - Sow“ o Hmwwé 6 I fAzPoBIIéLQ UdManar-g 4 2 I O"l ‘2 pzxw = p 2 Home we m sewed 2(3‘ i'2‘ ‘ W P2X21}=

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Unformatted text preview: Sow“ o Hmwwé 6 I. fAzPoBIIéLQ UdManar-g 4/ 2/ I) O) "l/ ‘2._ pzxw} = p{ 2 Home we m sewed} 2 (3‘) __ i ('2‘) ‘ W ' P2X21}= Pi/b/eac/cbwdwllmanjzbwmfl’I/W} : (20(7) __ i ('29) ~ 7' ' fl X=t} = PiléledCWIW‘abmwmw/W} 2 GMT) fl __3_2_ ('2‘ ‘ 7! P%X=o} = 73gwa Ind!» Mesa/wind} = Li). .2 _L. (I?) 7! Pixr’l} : P{J“£3&AQM¢M/W¢%& bWa/rew/W} : mm _ i (9‘) ‘ w P2X *2} = P%2%&9Wm>dwd} 2 £2. ... 28* (I: 9: 4' MPSSI‘W Ua’pmofx M1/2/325‘,5/6.Z‘Aus PM”? P2X=3}= p} m} :pzxm =0 Mweovw : _ 558’ Pix 2} (0’ = 75:87) 1 -5457’ 5‘ P2X 3} (0' = 32;, sz=4}=§43561 " 5 10' ~ 3‘71?) P2X=5}=5432§5l =_£_, 10‘ 2511 Pike} — ELL’ 2; lol 252 l2. (a). MQWMWPDSSPWuman waé‘j Main/W has“? 5 W Mitten 15 ‘50 754.2. r. 50% swag gym to 532(fl/b26.) a" b) c, A =1’2} mam Madgé /6 outcome/.5. A/o/am {Xalf} :— i(2.,2/ 2, 1)} , 2x543=flziw»vh {X= 3} ={(?-,I, 1,1), (ha/2%)}, {X2 -3} 2 {(ua,» I), <2», 2/2)}, {X22} = {(I,l,s,2)}, 1X:.2}:{(|,L2/1)}) {st=ivmfl[email protected][email protected] 2 {(1H, 2,2)/ U,2,l,z)) (2,1,2/|))(2/z/1,x)) (um, I) ) (1,2)2/0) (2/552), {2/23/29} T‘Am’gme PEXW} = P2X: ~4} = 75;) Plx=s}=~r={x=—33 =- 7*; =ng PIXH} =P1X=~zf< = #32 P1X=fi =~% = é~ U0) l¥ bow W L»? #42. )W W 0* fipc‘hfievs , \ooHa boxes; Corral/“40, $0 X30) 1’? ‘01»fo W in? aja-deflt WWW 4 £0806 M‘ “but/75 Wren? So Xzb. Thu, dd. W93w$>cfl1 vovbu. :250. HW Plxzok 2|, f4. Pike} = = .5 ) pZXM} 3‘+2-:}2+3 2 = f) PU”}= “'5?” = —.‘5, WM} = if, = 315, P1X=43= :3 - if (“9- PZX=13 = P?X£:}— p2){<z} = F<I>—/ff;tv F(l~#) = % um, w—ww—g’ = #- P2x=2} = Fm wm wt) 71-300 ~ ._1. ~ 12— Ewe—m = ~[z’4—fl = g. PiY=3}= Ravage—71;) : "/53: 7’5 =I~ 7';- I d” PEER)“ 3*} =1>ZX<-i~}— P2X F{-§~—i)~ m‘) Meta [%+#<%~#—Dj— 25: - .L _L .L 2.2L .2 *2*&”8 1?. MIDDS$596C UM 8£ X m o) I, 2/ 3) ¢¢°>= WM} = How/m F(o—-,,L)= 4:, 73—31» 490): Xu} 2 Fm— flag/:(z—fi): 33:_é=7% 9 Wm M):— ?<59=PiX'-‘3} = PUP/:32: F(3‘£J=%—-Zi = 72, we: WNW} = HW’xfifi)’: F($‘-717)=1~ lo a 1J5. 2l(b)' =3 4.0 + 35.P[ X: 3331-25 y=z;-} +57>P2X=5o} - ° .32. 33:. - maxi? + 33x1”, + 25x “*8 +§Dxfl .. 58/ _ * 7:84 * 37.28. em = 4o :92 Y m} + 35-sz M} +2st fur} “WV-w} = 403(‘42’1' 331% +25x$ +Soxi 2.] Aim/{224% m away 8W4 44.94;,(4/6419 cashmer MMomut 3C. MX hflw PMQ‘Ie'O‘iE W Mfiky VW x/ x-"A; M P8X”? = 1—4)» PiX=x—A} =¢~ luv) T and = x m a} + w-m PZX=Z~A} = 1&4)? + (an—A)./p = x~ x419. In mar % em = A = A40, we We have M1» = ’4/“3, TAM-219M 12A!” '4/10 2 AC?” 1/10). 28. Au: X 94 m W 0% cicada/ave L‘éW am 54%. Th. Fossiélz Influx: 512 X (we 0,/, 2) 3/ and /6 P2 =0} = (5) a :—~/.op = --—---—-—~ 2 fl 1 '5 P? (I; 4: 7 (a9 5ij =/(o pin-uo} + («In») PM“ 4”} .. o i .. .w H _. g _ 4, -IIX7 / x7 —. 76 ~~Zé1~ooé7 0” 50?} ~<II°)‘°PZX=H° +C-Ioo>‘P{X=-w «=3 -_-: i 5 5‘4 (21X? 7“ -§- 2 '9‘?- T4m1ew ,_ woo: EEX‘B- (EN) = — ( #4le“? W 4W— 5: EI W —1 , raw; Efx‘J = 4. (Q) g[(2+ X)? = Z (2+1)2/P(x) 2 “flaw/Fm 1,111)» 2,111)» = 4 2 pam- L, :xfix)+ 5: x’vycx; X W” aw)» x pm» ...
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This homework help was uploaded on 04/13/2008 for the course M 362k taught by Professor Berg during the Spring '08 term at University of Texas at Austin.

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Solution6 - Sow“ o Hmwwé 6 I fAzPoBIIéLQ UdManar-g 4 2 I O"l ‘2 pzxw = p 2 Home we m sewed 2(3‘ i'2‘ ‘ W P2X21}=

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