Based on the piecewise function, we have two conditions: if x is equal to 5 and not equal to 5.
Since the given value of x in the question is 3, let's use the first condition that states x is not equal to 5.
[tex]h(x)=\frac{x^2-25}{x-5}[/tex]
Let's plug in x = 3 in the function above.
[tex]h(3)=\frac{3^2-25}{3-5}[/tex]
Then, simplify.
[tex]\begin{gathered} h(3)=\frac{9-25}{-2} \\ h(3)=\frac{-16}{-2} \\ h(3)=8 \end{gathered}[/tex]
Therefore, h(3) = 8.
Similary, for B, since the given value of x is 0, which is not equal to 5 still, let's use the first condition and plug in x = 0.
[tex]\begin{gathered} h(x)=\frac{x^2-25}{x-5} \\ h(0)=\frac{0^2-25}{0-5} \\ h(0)=\frac{-25}{-5} \\ h(0)=5 \end{gathered}[/tex]
Hence, h(0) = 5.
Lastly, for C, the given value of x is 5. With that, we shall use the second condition.
[tex]\begin{gathered} h(x)=4 \\ h(5)=4 \end{gathered}[/tex]
Therefore, h(5) = 4.
