Question 1056249
y varies inversely as the square of x.
<pre>
{{{y}}}{{{""=""}}}{{{k/x^2}}}
</pre>
If y = 81 when x = 9, 
<pre>
Substitute that in

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

Solve for k

{{{81}}}{{{""=""}}}{{{k/81}}}

Multiply both sides by 81

{{{6561}}}{{{""=""}}}{{{k}}}

Substitute that in 

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

{{{y}}}{{{""=""}}}{{{6561/x^2}}}  <-- that's the formula
</pre>
find y when x is 3
<pre>
Substitute 3 for x

{{{y}}}{{{""=""}}}{{{6561/3^2}}}

{{{y}}}{{{""=""}}}{{{6561/9}}}

{{{y}}}{{{""=""}}}{{{729}}}

Edwin</pre>