How do you find an equation of variation: y varies inversely as the square of x, and y = 0.15 when x= 0.1?

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Answer 1 · 2016-04-05T11:46:35

$y= (0.0015)/x^2$

Or if you prefer $y=15/(10000x^2)$

Or as scientific notation: $y=15/(x^2)xx10^(-4)$

Splitting the question down into its component parts:

y varies inversely as: $->y=1/?$

the square of $->y=1/(?^2)$

$x" "-> y=1/(x^2)$

But we need a constant of variation
Let the constant of variation be $k$ then we have:

$y=kxx1/(x^2)$
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$color(blue)("Determine the value of "k)$

Known condition: at $y=0.15"; "x=0.1$

So by substitution we have

$0.15=kxx1/((0.1)^2)$

$=>k=0.15xx(0.1)^2$

$k=0.15xx0.01$

To calculate this directly think of $0.01" as "1/100$

Then we have:

$k=0.15/100 = 0.0015$

So the equation becomes:

$y= (0.0015)/x^2$

Or if you prefer $y=15/(10000x^2)$
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