Question 1059370
suppose y varies directly with x and inversely with the square of z ,and x=48 when y=8 and z=3 find x when y=12 and z=2 please help me 
<pre>Based on what's given, we get: {{{y = kx/z^2}}}, with k being the CONSTANT of PROPORTIONALITY
{{{8 = k(48)/3^2}}} ------ Substituting 48 for x, 8 for y, and 3 for z
{{{8 = 48k/9}}}
{{{8 = 16k/3}}}
16k = 24 ------ Cross-multiplying
{{{matrix(1,3, k = 24/16, or, 1.5)}}}

{{{y = kx/z^2}}}
{{{12 = (1.5)(x)/2^2}}} ------ Substituting 12 for y, {{{matrix(1,3, 1.5, for, k)}}}, and 2 for z
{{{1.5x = 12(2^2)}}} ------ Cross-multiplying
{{{highlight_green(matrix(1,5, x, "=", 12(4)/1.5, or, 32))}}}