Question 970554
The variable z varies jointly with x and y and inversely with the square of t.
z = k*x*y/t^2
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 Also, z=5 when x=2, y=-3, and t=1/2. 
Use these value to solve for "k"::
5 = k*2*-3/(1/2)^2
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5 = k*-6/4
-6k = 20
-k = 10/3
k = 10/3
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Equation::
z = (10/3)*x*y/t^2
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Write a function such that f(y)=t when x=-12/(5z)^2 and z=2/5.
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(2/5) = (10/3)*(12/25z^2)*y/t^2
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Solve for "y"::::
y = (2/5)t^2/[(8/5z^2]
y = (1/4)(t/z)^2
Cheers,
Stan H. 





f(y)=________
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Cheers,
Stan H.
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