# 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?

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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

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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} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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