Y varies directly as x and inversely as the square of z. y=10 when x=80 and z=4. How do you find y when x=36 and z=2?

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Answer 1 · 2016-09-01T03:16:44

#y=18#

As #y# varies directly as #x#, we have #ypropx#. Also it varies inversely as square of #z#, which means #yprop1/z^2#.

Hence, #ypropx/z^2# or

#y=k×x/z^2#, where #k# is a constant.

Now when #x=80# and #z=4#, #y=10#, so

#10=k×80/4^2=k×80/16=5k#

Hence #k=10/5=2# and #y=2x/z^2#.

So when #x=36# and #z=2#,

#y=2×36/2^2=72/4=18#