CCF08182010_00014 - Solutions for Exam l— Morning Seclion...

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Unformatted text preview: Solutions for Exam l— Morning Seclion 69 Solutions for Exam l—Morning Section fl _ (l i NO — 2 0.01 = (%)m t In (0.01) = (4-3 days)1n§ 1n0.01 t: (4.3 days) ( ln0.5 ) = 28.6 days (29 days) Answer is C. 2- (1X2) - (40(3) : —10 Answer is A. 3. Solve by factoring. 2_ _ lim ac 9 = lim (x+3)(r 3) :r—>3 Jr— 1—)?) x—3 =1irr§(ac+3) Answer is C . 4. The magnitude of vector V is v — (1)2 + (2)2 + (1)2 = x/é l The a:-directi0n cosine is cosqfi —E—i 1 0=cos_1<—> 6 :65.9° (66°) Answer is C. 0° 1 °° —1 — 5 fight" :___1 2 :1: x2 00 2 =1/2 Answer is C. 6. The magnitudes of the two vectors are V1 = x/(1)2 +(2)2+ (1) = x/6 V2: (1)2+(3)2+(77)2=¢E _ W 2 o 43“” < WW) > 90 Answer is C. 1 281—8 0 0 1 7. A=/ exdxzex 0 m 1.718 (1.7) Answer is B. 8. The first and second derivatives are y’=m3—3x+2 y/l:3m2_3 For a critical point, y’ = 0. By inspection (based on the four answer choices), y’ : 0 at x = 1 and :L‘ = *2. "(1) = (3)(1)2 - 3 = 0 y"<—2> = <3><—2>2 — 3 = 9 W) = (i) <—2>4 —<1.5)<—2>2 + <2><—2> + 5 = —1 Answer is A. Answer is C. 10. (y2+y+i)+(x2—2x+1)=5+%+1 = 25/4 (y + %)2 + (a: — 1)2 = 25/4 This is a circle centered at (1, —1/2), so ymax is at m = 1. Answer is C. — Professional Publications, Inc. ...
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