If Z varies directly with x and inversely with y^2 when x=2 and y=5, z=8 what is the value of z when x=4 and y=9?

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Answer 1 · 2016-05-24T16:37:12

#z=4.9383#

Direct Variation is #y= kx#
Inverse Variation is #y = k/y#

If #z# varies directly to #x# the equation would be #z=kx#

If #z# varies inversely to #y^2# the equation would be #z=k/y^2#

Combining these two equations we get #z=(kx)/y^2#

We use the first set of values to solve for the constant of variation #k#

#x=2#
#y=5#
#z=8#

#8=(2k)/5^2#

#8=(2k)/25#

#8(25)=(2k)/cancel(25)cancel(25)#

#200=(2k)#

#200/2=(cancel2k)/cancel2#

#100 = k#

Now use the constant #k# with the second values to solve for #z#

#x=4#
#y=9#
#z=?#
#k=100#

#z=((100)(4))/9^2#

#z = 400/81#

#z=4.9383#