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If y varies inversely as the cube of x and directly as the square of z and y = -6 when x=3 and z =9, how do you find y when x =6 and z= -4?

1 Answer

May 31, 2015

If y varies inversely as the cube of x and directly as the square of z
then we can write:
$XXXXX$ $color(white)("XXXXX")$ $\frac{y \cdot x^{3}}{z^{2}} = c$ $(y*x^3)/(z^2) = c$ for some constant $c$ $c$

If
$XXXXX$ $color(white)("XXXXX")$ $(x, y, z) = (3, - 6, 9)$ $(x,y,z) = (3,-6,9)$
is a solution for this equation, then

$XXXXX$ $color(white)("XXXXX")$ $\frac{(- 6) \cdot 3^{3}}{9^{2}} = c$ $((-6)*3^3)/(9^2) = c$
$XXXXX$ $color(white)("XXXXX")$ $= \frac{162}{81} = c$ $=162/81 = c$
$XXXXX$ $color(white)("XXXXX")$ $c = 2$ $c= 2$

When $(x, z) = (6, - 4)$ $(x,z) =(6,-4)$
$XXXXX$ $color(white)("XXXXX")$ $\frac{y \cdot x^{3}}{z^{2}} = 2$ $(y*x^3)/(z^2) = 2$
becomes
$XXXXX$ $color(white)("XXXXX")$ $\frac{y \cdot 216}{16} = 2$ $(y*216)/16 = 2$

$XXXXX$ $color(white)("XXXXX")$ $y = \frac{16 \cdot 2}{216}$ $y = (16*2)/(216)$

$y = \frac{4}{27}$ $y = 4/27$

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