If y varies inversely as the square of x and y=5 when x=2.5, what is the value of Y when X=9?

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Answer 1 · 2016-04-04T12:53:44

#y=31.25/81#

As #yprop1/x^2#, we have #y=k/x^2# or #x^2y=k#.

As when #x=2.5#, #y=5#,

#2.5^2xx5=k# or

#k=6.25xx5=31.25#

Therefore #x^2y=31.25#

Hence when #x=9#, #9^2xxy=31.25#

and #y=31.25/81#