Question 1155352
{{{a= km*n^2​}}} and
{{{a=k/y^3}}}

=> {{{a= k(m*n^2/y^3)}}}

 If {{{a=4}}} when {{{m=9​}}}, {{{n=4​}}}, and {{{y=2}}}​, we have 

{{{a= k(9*4^2/2^3)}}}

{{{4=144k/8}}}

{{{32=144k}}}

{{{k=32/144}}}

{{{k=2/9}}}


=>{{{a= (2/9)(m*n^2/y^3)}}}


find {{{a}}} when {{{m=8​}}}, {{{n=3​}}}, and {{{y=5}}}


{{{ a= (2/9)(8*3^2/5^3)}}}

{{{ a= 16/125}}}

{{{a=0.128}}}