If y varies inversely as #x^2# and y = 10 when x=2 , find y when x = 0.5?

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Answer 1 · 2016-07-02T21:44:32

#y =160#

Write it as an inverse proportion first

#yprop 1/x^2#

Change into an equation by multiplying by a constant.

#y = k/x^2#

Find the value of the constant

#k = x^2 xx y = 2^2 xx 10 = 40#

Use the constant in the equation

#y = 40/x^2#

Solve for y when x = 0.5

#y = 40/0.5^2 = 160#