course Mth 158 ??R????Y?f??????assignment #021021. `query 21
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12:06:40 **** query 2.5.8 / 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.
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RESPONSE --> y= k/sqrt x 4 = k/sqrt9 4 = k/3 12=k y=12/sqrtx
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12:06:52 ** The inverse proportionality to the square root gives us y = k / sqrt(x). y = 4 when x = 9 gives us 4 = k / sqrt(9) or 4 = k / 3 so that k = 4 * 3 = 12. The equation is therefore y = 12 / sqrt(x). **
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RESPONSE --> ok
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12:08:36 query 2.5.12 / 2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1 and x=2 and y=3.
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RESPONSE --> z= (x^3 + y^2)k 1 = (8 + 9)k 1 = 17k k = 1/17 z = 1/17(x^3 + y^2)
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12:08:47 ** The proportionality is z = k (x^3 + y^2). If x = 2, y = 3 and z = 1 we have 1 = k ( 2^3 + 3^2) or 17 k = 1 so that k = 1/17. The proportionality is therefore z = (x^3 + y^2) / 17. **
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RESPONSE --> ok
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12:10:35 query 2.5.20 / 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)
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RESPONSE --> T = k(sqrt l) T = (2pi/sqrt 32)(sqrtl)
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12:10:55 ** The equation is T = k sqrt(L), with k = 2 pi / sqrt(32). So we have T = 2 pi / sqrt(32) * sqrt(L). **
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RESPONSE --> ok
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12:11:25 **** What equation relates period and length? ****
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RESPONSE --> p = k(L)
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12:13:32 query 2.5.42 / 2.7.42 (was 2.7.34 (was 2.6.30). Resistance dir with lgth inversely with sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with resistance 1.44 ohms and diameter 3 mm? Give the details of your solution.
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RESPONSE --> r = k(L/d^2) 1.24 = k(432/4^2) .046 = k 1.44 = .046(L/3^2) 12.96 = .046L 281.74= L
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