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?

Question

Kennedy Adams · Accepted Answer

If y varies inversely as the cube of x and directly as the square of z
then we can write:
#color(white)("XXXXX")##(y*x^3)/(z^2) = c# for some constant #c# If
#color(white)("XXXXX")##(x,y,z) = (3,-6,9)#
is a solution for this equation, then #color(white)("XXXXX")##((-6)*3^3)/(9^2) = c#
#color(white)("XXXXX")##=162/81 = c#
#color(white)("XXXXX")##c= 2# When #(x,z) =(6,-4)#
#color(white)("XXXXX")##(y*x^3)/(z^2) = 2#
becomes
#color(white)("XXXXX")##(y*216)/16 = 2# #color(white)("XXXXX")##y = (16*2)/(216)# #y = 4/27#

Ethan Adkins · Answer

undefined To find y when x = 6 and z = -4, we can use the inverse variation formula.

First, let's find the constant of variation (k) by substituting the given values into the equation y = k * (x^3) / (z^2).

-6 = k * (3^3) / (9^2)

Simplifying this equation, we get:

-6 = k * 27 / 81

Next, we can solve for k:

k = -6 * 81 / 27

k = -18

Now that we have the value of k, we can substitute it into the equation and solve for y when x = 6 and z = -4:

y = -18 * (6^3) / (-4^2)

Simplifying this equation, we get:

y = -18 * 216 / 16

y = -243

Avery Alexander · Answer

undefined To find $ y $ when $ x = 6 $ and $ z = -4 $, you can use the concept of inverse and direct variation.

Given that $ y $ varies inversely as the cube of $ x $ and directly as the square of $ z $, we can set up the equation:

$$ y = k \times \frac{z^2}{x^3} $$

First, we need to find the constant of variation, $ k $, using the initial values provided:

$$ -6 = k \times \frac{9^2}{3^3} $$

Solving for $ k $:

$$ k = \frac{-6 \times 3^3}{9^2} $$

$$ k = -6 \times \frac{27}{81} $$

$$ k = -2 $$

Now that we have $ k $, we can use it to find $ y $ when $ x = 6 $ and $ z = -4 $:

$$ y = -2 \times \frac{(-4)^2}{6^3} $$

$$ y = -2 \times \frac{16}{216} $$

$$ y = -2 \times \frac{2}{27} $$

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