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QUESTION
If y varies directly as x and inversely as the square of z and if ##y=20## when ##x=50## and ##z=5## how do you find y when ##x=3## and ##z=6##?
##y=kx/z^2## ------------------------------------------ Equation 1
Given ##x=50, y=20 and z=5##
##20=k 50/(5)^2##
##k=10##
Sub ##k=10## into Equation 1
##y=10x/z^2## ------------------------------------------ Equation 2
Sub ##x=3 and z=6## into Equation 2
##y=10(3)/(6)^2## ##=5/6##
