The table shows a set of values for [tex]$x$[/tex] and [tex]$y$[/tex].

\begin{tabular}{|c|c|c|c|c|}
\hline
[tex]$x$[/tex] & 1 & 2 & 3 & 4 \\
\hline
[tex]$y$[/tex] & 16 & 4 & [tex]$\frac{16}{9}$[/tex] & 1 \\
\hline
\end{tabular}

[tex]$y$[/tex] is inversely proportional to the square of [tex]$x$[/tex].

a) Find an equation for [tex]$y$[/tex] in terms of [tex]$x$[/tex].

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



Answer :

Let's analyze the relationship between [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and find the necessary values step-by-step.

### Part a: Find an equation for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex].

Given that [tex]\( y \)[/tex] is inversely proportional to the square of [tex]\( x \)[/tex], we can express this relationship mathematically as follows:
[tex]\[ y = \frac{k}{x^2} \][/tex]
where [tex]\( k \)[/tex] is a constant of proportionality.

We can determine the value of [tex]\( k \)[/tex] using any given pair of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values from the table. Let's use the first pair [tex]\((1, 16)\)[/tex]:

When [tex]\( x = 1 \)[/tex] and [tex]\( y = 16 \)[/tex]:
[tex]\[ 16 = \frac{k}{1^2} \][/tex]
[tex]\[ k = 16 \][/tex]

Now that we have determined [tex]\( k = 16 \)[/tex], we can write the equation for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y = \frac{16}{x^2} \][/tex]

### Part b: Find the positive value of [tex]\( x \)[/tex] when [tex]\( y = 25 \)[/tex].

We need to find [tex]\( x \)[/tex] such that when [tex]\( y = 25 \)[/tex], the equation [tex]\( y = \frac{16}{x^2} \)[/tex] holds true. Plug [tex]\( y = 25 \)[/tex] into the equation and solve for [tex]\( x \)[/tex]:

Starting with the equation:
[tex]\[ 25 = \frac{16}{x^2} \][/tex]

Rearranging to solve for [tex]\( x^2 \)[/tex]:
[tex]\[ 25 x^2 = 16 \][/tex]
[tex]\[ x^2 = \frac{16}{25} \][/tex]

Taking the square root of both sides to solve for [tex]\( x \)[/tex]:
[tex]\[ x = \sqrt{\frac{16}{25}} \][/tex]
[tex]\[ x = \frac{\sqrt{16}}{\sqrt{25}} \][/tex]
[tex]\[ x = \frac{4}{5} \][/tex]
[tex]\[ x = 0.8 \][/tex]

Thus, the positive value of [tex]\( x \)[/tex] when [tex]\( y = 25 \)[/tex] is:
[tex]\[ x = 0.8 \][/tex]