Question 1113941

If {{{y}}} varies {{{inversely}}} with the {{{square}}} of {{{x}}}:

{{{y=k/x^2}}}

 and if {{{y=9}}} when {{{x=2}}}, we have

{{{9=k/2^2}}}

{{{9=k/4}}}

{{{9*4=k}}}

{{{k=36}}}

 find {{{y}}} when {{{x=0.1}}}:

{{{y=36/0.1^2}}}

{{{y=36/0.01}}}

{{{y=3600}}}