course Mth 158 Ubzqassignment #021
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12:39:43 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.
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RESPONSE --> 4 = k / sqrt(9) = 4 = k / 3. confidence assessment: 1
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12:41:53 ** 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 --> I know that 12 / 3 = 4. I didn't think that I was supposed to put a number in place of k because k is a constant. I understand my mistake now. self critique assessment: 2
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12:51:12 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 = k ( x^3 + y^2) 1 = k ( 2^3 + 3^2) k = 1/ 17 so z = 1/17 ( x^3 + y^2) confidence assessment: 1
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12:54:30 ** 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 knew that k = 1/17. I multiplied by 1/ 17 instead of dividing. self critique assessment: 2
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12:56:58 2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)
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RESPONSE --> period= sqrt ( length) * 2 pi / sqrt (32) confidence assessment: 1
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12:57:16 ** 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 got it right. self critique assessment: 3
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12:57:45 **** What equation relates period and length? ****
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RESPONSE --> period = k * sqrt ( length) confidence assessment: 1
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13:09:04 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 --> resistance = length / diameter^2 1.44 = length / 3^2 1.44 = length / 9 then multiply both sides by 9 so 1.44 * 9 = length 12.96 = length. confidence assessment: 1
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13:10:39 ** 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 --> I forgot about the constant. I understand how to solve for the constant and how to find the length. self critique assessment: 2
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