Give an example of a function with both a removable and a non-removable discontinuity.

(x+5) (x=3)

(X+5) (x+1)

Step-by-step explanation:

A removeable discontinuity is always found in the denominator of a rational function and is one that can be reduced away with an identical term in the numerator. It is still, however, a problem because it causes the denominator to equal 0 if filled in with the necessary value of x. In my function above, the terms (x + 5) in the numerator and denominator can cancel each other out, leaving a hole in your graph at -5 since x doesn't exist at -5, but the x + 1 doesn't have anything to cancel out with, so this will present as a vertical asymptote in your graph at x = -1, a nonremoveable discontinuity.

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24