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##