Question 1085676

if  {{{y}}} varies {{{directly}}} as the square of {{{x}}},we have 

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

and if {{{y=0.86}}} when {{{x=0.1}}}, we have:

{{{0.86=k*(0.1)^2}}}

{{{0.86=k*0.01}}}

{{{0.86/0.01=k}}}

{{{86/1=k}}}

{{{k=86}}}

and your equation is: {{{y=86x^2}}}