Question #91fad

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Question #91fad

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Answer 1 · 2017-03-01T10:36:46

$y \approx 2.9 to 1 decimal place$ $y≈2.9" to 1 decimal place"$

Explanation:

When dealing with $inverse proportion$ $color(blue)"inverse proportion"$ we know that the relationship involves $\frac{1}{quantity}$ $color(blue)(1/("quantity"))$

$the square of x means x^{2}$ $"the square of x means "x^2$

$\Rightarrow y \propto \frac{1}{x^{2}}$ $rArryprop1/(x^2)$

To convert into an equation, we require to introduce k ,
$the constant of proportion$ $color(blue)"the constant of proportion"$

$\Rightarrow y = k \times \frac{1}{x^{2}}$ $rArry=kxx1/x^2$

$\Rightarrow y = \frac{k}{x^{2}} \to (1)$ $rArry=k/x^2to(1)$

To find k, use the given values of x and y.

$That is y = 7.5 when x = 2.6$ $"That is "y=7.5" when "x=2.6$

$From (1) : k = y x^{2} \leftarrow cross-multiplying$ $"From "(1):k=yx^2larr" cross-multiplying"$

$\Rightarrow k = 7.5 \times {(2.6)}^{2} = 50.7$ $rArrk=7.5xx(2.6)^2=50.7$

$We can now express (1) as y = \frac{50.7}{x^{2}}$ $"We can now express "(1)" as "y=50.7/x^2$

$x = 4.2 \to y = \frac{50.7}{{(4.2)}^{2}}$ $x=4.2toy=50.7/(4.2)^2$

$\Rightarrow y = 2.9 to 1 decimal place$ $rArry=2.9" to 1 decimal place"$

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Question #91fad

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