course Mth158 I can't remember if I have submitted this assignment or not. f̐Ēҙصassignment #021
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15:00:41 2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.
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RESPONSE --> y=k/x y=4 x=9 y=k/sqrt(x) 4=k/sqrt(9) 4=k/3 12=k y=12/x confidence assessment: 3
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15:00:53 ** 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 --> whooah! self critique assessment: 2
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15:04:01 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 --> y=kx z=k(x^3+y^2) 1=k(2^3+3^2) 1=k(8+9) 1=17k 1/17=17k/17 1/17=k z=1/17(x^3+y^2) confidence assessment: 3
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15:04:14 ** 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 --> good good! self critique assessment: 2
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