Question Progress

Homework Progress

[tex]$y$[/tex] is inversely proportional to the square of [tex]$x$[/tex]. A table of values for [tex]$x$[/tex] and [tex]$y$[/tex] is shown.

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$y$ & 4 & 1 & $\frac{4}{9}$ & $\frac{1}{4}$ \\
\hline
\end{tabular}
\][/tex]

a) Express [tex]$y$[/tex] in terms of [tex]$x$[/tex].

b) Work out the positive value of [tex]$x$[/tex] when [tex]$y = 25$[/tex].



Answer :

Sure! Let's work through the problem step-by-step.

### Part (a)

Given that [tex]\( y \)[/tex] is inversely proportional to the square of [tex]\( x \)[/tex], we can write this relationship as:

[tex]\[ y = \frac{k}{x^2} \][/tex]

where [tex]\( k \)[/tex] is a constant.

To find the constant [tex]\( k \)[/tex], we'll use one of the given pairs of values from the table. Let’s use the values [tex]\( x = 1 \)[/tex] and [tex]\( y = 4 \)[/tex]:

[tex]\[ 4 = \frac{k}{1^2} \][/tex]

Simplifying this equation:

[tex]\[ k = 4 \][/tex]

So, the relationship between [tex]\( y \)[/tex] and [tex]\( x \)[/tex] can be written as:

[tex]\[ y = \frac{4}{x^2} \][/tex]

This equation expresses [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex].

### Part (b)

Now, we need to find the positive value of [tex]\( x \)[/tex] when [tex]\( y = 25 \)[/tex].

Using the equation:

[tex]\[ y = \frac{4}{x^2} \][/tex]

we substitute [tex]\( y = 25 \)[/tex]:

[tex]\[ 25 = \frac{4}{x^2} \][/tex]

Rearranging to solve for [tex]\( x^2 \)[/tex]:

[tex]\[ 25x^2 = 4 \][/tex]

[tex]\[ x^2 = \frac{4}{25} \][/tex]

Taking the positive square root of both sides:

[tex]\[ x = \sqrt{\frac{4}{25}} \][/tex]

[tex]\[ x = \frac{2}{5} \][/tex]

Therefore, the positive value of [tex]\( x \)[/tex] when [tex]\( y = 25 \)[/tex] is:

[tex]\[ x = \frac{2}{5} \][/tex]