#$&* course mth 158 7/20 5pm 021. `* 21
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Given Solution: * * The inverse proportionality to the square root gives us y = k / sqrt(x). y = 4 when x = 9 gives us 4 = k / sqrt(9) or 4 = k / 3 so that k = 4 * 3 = 12. The equation is therefore y = 12 / sqrt(x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: z= k(x^3 + y^2) 1 = k (2^3 + 3^2) 1 = k (8 + 9) 1 = k (17) 1/17 = k k = 1/17 z = (x^3 + y^2)/17 confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The proportionality is z = k (x^3 + y^2). If x = 2, y = 3 and z = 1 we have 1 = k ( 2^3 + 3^2) or 17 k = 1 so that k = 1/17. The proportionality is therefore z = (x^3 + y^2) / 17. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*OK ********************************************* Question: * 2.5.20 / 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt (32) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = k sqrt (l) where k = (2pi/sqrt(32)) confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The equation is T = k sqrt(L), with k = 2 pi / sqrt(32). So we have T = 2 pi / sqrt(32) * sqrt(L). ** **** What equation relates period and length? **** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok Since the k = constant of proportionality = 2pi/sqrt(32), that part of the equation represents the length, while T = period. ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: r (resistance) = k * length / diameter^2 r = kl/d^2 1.24 ohms = k(432) / 4mm^2 k = 1.24 * 4^2 / 432 k = 1.24 * 16 / 432 k = .046 1.44 = .046(l) / 3^2 l = 9 * 1.44 / .046 l = 12.96 /.046 l = 281.739 confidence rating #$&*3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * We have R = k * L / D^2. Substituting we obtain 1.24 = k * 432 / 4^2 so that k = 1.24 * 4^2 / 432 = .046 approx. Thus R = .046 * L / D^2. Now if R = 1.44 and d = 3 we find L as follows: First solve the equation for L to get L = R * D^2 / (.046). Then substitute to get L = 1.44 * 3^2 / .046 = 280 approx. The wire should be about 280 ft long. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating #$&*OK"