course MTH 158
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10:55:29 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.
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RESPONSE --> So y = k / sqrt(x) 4 = k / sqrt(9) 4 = k/3 k = 12 y = 12 / sqrt(x) confidence assessment: 2
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10:55:43 ** 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 self critique assessment: 3
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11:02:05 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 = k x^3 y^2 1 = k 2^3 * 3^2 1 = 72 k k = -71 z = -71 x^3 y^2 confidence assessment: 2
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11:04:01 ** 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 --> I see that I wrongly multiplied x^3 and y^2 instead of adding, as well as subtracting the number by k instead of dividing. self critique assessment: 2
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11:08:30 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)
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RESPONSE --> T = 2 pi / sqrt(32) * sqrt (1) I don't know if I need to simplify anything additionally.. confidence assessment: 2
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11:09:46 ** 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 --> I meant to type sqrt(L) instead of sqrt(1)... when I wrote the problem down I put a lower-case L instead of an upper case L, and it looked like I'd written a 1... self critique assessment: 3
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11:10:55 **** What equation relates period and length? ****
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RESPONSE --> Probably combined variation confidence assessment: 2
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11:34:53 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 = kL/d^2 1.24 = k*432 / 4^2 1.24 = 432k / 16 1.24 = 27k k = 1.24/27 The formula is r = .046 L / d^2 For the length of a wire with resistance 1.44 ohms and a diameter of 3mm, we substitute... 1.44 = .046 L / 3^2 1.44 = .046 L / 9 1.44 = .05 L
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11:35:44 ** 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. **
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RESPONSE --> Ok.. why and how did you change 2.8 to 280? self critique assessment: 2
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