SOLUTION: For the following problems y varies inversely with the square of x. If y is 4 when x is 5, find y when x is 2.

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yx^(2)=yx^(2)
4(5^2)=y(2^2)
4(25)=y(4)
y=25

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y varies inversely with x^2
y = c*(1/x^2), where c is the proportanality constant
y = 4 when x = 5
4 = c*(1/(5*5))
therefore c = 100
y = 100*(1/x^2)
when x = 2 then y = 100*(1/(2*2)) = 25

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y varies inversely with the square of x is equivalent to y = k/x^2
y = k/x^2
Substituing y=4 and x=5: 4 = k/25 => k = 100;
Thus, y = 100/x^2
x = 2 => y = 100/2^2 = 100/4 = 25