Question #f65fd

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Question #f65fd

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Answer 1 · 2017-03-01T11:47:05

$k = 2$ $k=2$

Explanation:

Consider each statement separately.

$∙ y \propto x^{2} \leftarrow direct variation$ $• ypropx^2larrcolor(blue)" direct variation"$

$∙ y \propto \frac{1}{z} \leftarrow inverse variation$ $• yprop1/zlarrcolor(blue)" inverse variation"$

Combine the right sides as a product.

$∙ y \propto \frac{x^{2}}{z} \leftarrow joint variation$ $• ypropx^2/zlarrcolor(blue)" joint variation"$

Convert to an equation by introducing k,
$the constant of variation$ $color(blue)" the constant of variation"$

$\Rightarrow y = k \times \frac{x^{2}}{z}$ $rArry=kxxx^2/z$

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

$cross-multiplying$ $color(blue)"cross-multiplying"$ gives.

$k x^{2} = y z$ $kx^2=yz$

$\Rightarrow k = \frac{y z}{x^{2}}$ $rArrk=(yz)/x^2$

To calculate k, substitute the given values for x, y and z

$\Rightarrow k = \frac{6 \times 27}{9^{2}} = \frac{162}{81} = 2$ $rArrk=(6xx27)/9^2=162/81=2$

$"the equation of variation is "color(red)(bar(ul(|color(white)(2/2)color(black)(y=(2x^2)/z)color(white)(2/2)|)))$

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