The table shows a set of values for $x$ and $y$. | $x$ | 1 | 2 | 3 | 4 | |-----|---|---|---|---| | $y$ | 9 | $\frac{1}{4}$ | 1 | $\frac{9}{16}$ | $y$ is inversely proportional to the square of $x$. (a) Find an equation for $y$ in terms of $x$. (b) Find the positive value of $x$ when $y = 16$.

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The table shows a set of values for $x$ and $y$ - Edexcel - GCSE Maths - Question 13 - 2017 - Paper 1

Question 13

The table shows a set of values for $x$ and $y$. | $x$ | 1 | 2 | 3 | 4 | |-----|---|---|---|---| | $y$ | 9 | $\frac{1}{4}$ | 1 | $\frac{9}{16}$ | $y$ is inversely ... show full transcript

Worked Solution & Example Answer:The table shows a set of values for $x$ and $y$ - Edexcel - GCSE Maths - Question 13 - 2017 - Paper 1

Step 1

(a) Find an equation for y in terms of x.

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Answer

Since $y$ is inversely proportional to the square of $x$ , we can write this relationship as: $y = \frac{k}{x^2}$ where $k$ is a constant.

To find $k$ , we can use one of the values from the table. For example, taking $x = 1$ when $y = 9$ :

$9 = \frac{k}{1^2} \implies k = 9.$

Thus, the equation for $y$ in terms of $x$ is: $$y = \frac{9}{x^2}.$

Step 2

(b) Find the positive value of x when y = 16.

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Answer

Substituting $y = 16$ into the equation we derived:

$16 = \frac{9}{x^2}$

Multiplying both sides by $x^2$ gives:

$16x^2 = 9.$

Now, divide by 16:

$x^2 = \frac{9}{16}.$

Taking the positive square root:

$x = \frac{3}{4}.$

Thus, the positive value of $x$ when $y = 16$ is: $$x = \frac{3}{4}.$

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