Find the x-value of the removable discontinuity of the function.

Answer:
x=2
Step-by-step explanation:
[tex]f(x)=\frac{x^2-4}{x^2-12x+20}[/tex]
Factor the numerator and denominator:
[tex]f(x)=\frac{(x-2)(x+2)}{(x-2)(x-10)}[/tex]
We can remove (x-2) from the function:
[tex]f(x)=\frac{x+2}{x-10}[/tex]
(x-2) is a removable discontinuity.
The x-value of x-2 is x=2.