Question 459321
If y varies inversely as x^2, and y=2 when x=4, what is the constant of variation? 
:
Let k = constant of variation
:
{{{k/x^2}}} = y
:
Replace x & y:
{{{k/4^2}}} = 2
{{{k/16}}} = 2
multiply both sides by 16
k = 16(2)
k = 32 is the constant of variation
and
y = {{{32/x^2}}} is the equation