Question 279787
Suppose z varies directly as the square of x and inversely as y. If z=8 when x=4 and y=6, find z when x=6 and y=12
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z = kx^2/y
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Solve for "k":
8 = k*4^2/6
48 = 16k
k = 3
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Equation:
z = 3x^2/y
find z when x=6 and y=12
z = 3*6^2/12
z = 3*3
z = 9
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Cheers,
Stan H.
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