If y varies jointly as the cube of x, w, and the square of z, what is the effect on w when x, y, and z are doubled?

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If y varies jointly as the cube of x, w, and the square of z, what is the effect on w when x, y, and z are doubled?

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Answer 1 · 2017-06-01T16:51:40

$w$ $w$ decreases by a factor of 16.

That is, $w_{new} = \frac{1}{16} w$ $w_"new" = 1/16w$

Explanation:

Based on this description, we can create the following equation:

$y = k x^{3} w z^{2}$ $y = kx^3wz^2$

Where $k$ $k$ is a constant which doesn't really matter for this particular problem.

The problem asks what happens to $w$ $w$ when the other variables are doubled, so let's solve for $w$ $w$ to see what it is originally.

$y = k x^{3} w z^{2}$ $y = kx^3wz^2$

$\frac{y}{k x^{3} z^{2}} = w$ $y/(kx^3z^2)=w$

Now, the problem asks us to double everything except $w$ $w$ and see what happens.

$\frac{(2 y)}{k {(2 x)}^{3} {(2 z)}^{2}} = w_{new}$ $((2y))/(k(2x)^3(2z)^2)=w_"new"$

$\frac{2 y}{k (8 x^{3}) (4 z^{2})} = w_{new}$ $(2y)/(k(8x^3)(4z^2))=w_"new"$

$\frac{2}{8 \cdot 4} (\frac{y}{k x^{3} z^{2}}) = w_{new}$ $2/(8*4)(y/(kx^3z^2))=w_"new"$

$\frac{1}{16} (\frac{y}{k x^{3} z^{2}}) = w_{new}$ $1/16(y/(kx^3z^2))=w_"new"$

$\frac{1}{16} w = w_{new}$ $1/16w = w_"new"$

So $w$ $w$ decreases by a factor of 16 when the other 3 variables are doubled.

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If y varies jointly as the cube of x, w, and the square of z, what is the effect on w when x, y, and z are doubled?

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