Question 795285
1st problem : y varies directly as x and inversely as the square of z. 
y = k*x/z^2
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Solve for "k" using "y=30 when x=40 and z=2{
30 = k*40/2^2 
120 = 40k
k = 3
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Equation for this problem::
y = 3x/z^2
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find y when x=80 and z=4
y = 3*80/4^2 = 3*5 =15
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2nd problem: if y varies as x and y=7 when x=6 find y when x=18
y/18 = 7/6
y = (18/6)7
y = 21
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Cheers,
Stan H.
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