Answer:
B) all real values of x except x=5
Step-by-step explanation:
If [tex]f(x)=x^2-25[/tex] and [tex]g(x)=x-5[/tex], then [tex]\bigr(\frac{f}{g}\bigr)(x)=\frac{f(x)}{g(x)}=\frac{x^2-25}{x-5}=\frac{(x-5)(x+5)}{(x-5)}[/tex].
Given that [tex]x-5[/tex] exists in both the numerator and denominator, this creates a hole on the graph of the function where [tex]x=5[/tex] since [tex]5-5=0[/tex]all real values of x except x=5 is correct.
Review the attached graph for more information
