If x is inversely proportional to y, and x = 60 when y = 0.5, how do you find x when y = 12?

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Answer 1 · 2016-03-31T16:02:54

$y = \frac{30}{x}$ $y=30/x$

At $y = 12; x = 2 \frac{1}{2} \equiv \frac{5}{2}$ $y=12"; "x= 2 1/2 -= 5/2$

Let $k$ $k$ be a constant. Then:

$y = k \times \frac{1}{x} = \frac{k}{x}$ $y=kxx1/x = k/x$

It is given that:

$y = \frac{k}{x} \to 0.5 = \frac{k}{60}$ $y=k/x" "->" "0.5=k/60$

Multiply both sides by 60 giving

$(0.5 \times 60) = k \times (\frac{60}{60})$ $(0.5xx60) = kxx (60/60)$

But $\frac{60}{60} = 1 and 0.5 \times 60 = 30$ $60/60=1" and "0.5xx60=30$

$\Rightarrow k = 30$ $=>k=30$

So the relationship is: $y = \frac{30}{x}$ $y=30/x$
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For $y = 12$ $y=12$ we have

$12 = \frac{30}{x}$ $12=30/x$

$x = \frac{30}{12} = 2 \frac{6}{12} \to 2 \frac{1}{2}$ $x=30/12 = 2 6/12 -> 2 1/2$