How can you tell from the equation of a rational function if the function has a hole in the graph ( a removable discontinuity) at x, rather than a vertical asymptote? Give an example​

Consider that,

x^2+4x+4 = (x+2)(x+2)

x^2+7x+10 = (x+2)(x+5)

Dividing those expressions leads to

(x^2+4x+4)/(x^2+7x+10) = (x+2)/(x+5)

The intermediate step that happened is that we have (x+2)(x+2) all over (x+2)(x+5), then we have a pair of (x+2) terms cancel as the diagram indicates (see below). This is where the removable discontinuity happens. Specifically when x = -2. Plugging x = -2 into (x+2)/(x+5) produces an output, but it doesn't do the same for the original ratio of quadratics. So we must remove x = -2 from the domain.

answer: 180

step-by-step explanation:

12-2=10

20-2=18

10x18=180