Business Calc Homework w answers_Part_59

0825e 0 04 25 332965 billion barrels 39 22 0 936 t 24

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Unformatted text preview: ,000 2x. 12 kx 2 kx dx 9 81 N cm 0 k(1), so k (b) For total distance: W 16 (10r 3 16 0 1244.07 in3 sec 45 81 4 W in3 sec r 3) dr 0 14 r 4 2 and F(x) 9 0 2rr 3 r2)(2 r) dr 5r 2 2(110) 18N F(x) dx (a) W Cross section area 0 16 2(112) d 10,000. 12 kd 2 0 1 (10,000)(0.5)2 2 1 (10,000)(1)2 2 5000 For second half of distance: Inches per second in. sec 2(119) 1250 inch-pounds r, Length r 2) 2(115) 2(115) k(3), so k d 0 (b) Volume per second 8(10 2(120) 0 2 83,776 24. (a) Width 2(115) 2(9) (b) W (2r 0 kx; 6 (a) F(9) 0 2 2(110) 9 2 r)(2 r) dr f (xn) 121] Area 0 ∑ 2 f (xi) 1156.5 2rr Population density 10,000(2 798.97 thousand 0.5 i1 2(120) 0 2 miles. 2 r: Area 2 (d) 5.0x n1 f (x0) 24 t cos 12 93.6 kilowatt-hours r, Length (c) Population 28.8 a 2n 18 8 [120 2(10) 0 24 10.5 2.3 2 x 2 28. Treat 6 P.M. as 18 o’clock: 19.5 meters 10 5.0) dx (b) The answer in (a) corresponds to the area of midpoint rectangles. The curve now gives a better approximation since part of each rectangle is above the curve and...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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