Prove that the sum of 2 even integers is an even integer

Let a and b represent even integers. Then by definition of an even integer, we can say that:  a = 2n and b = 2m

Proof by contradiction:

a + b is not an even integer.

Using substitution, 2n + 2m is not an even integer.

Using distibutive property, 2(n + m) is not an even integer.

Consider that (n + m) = k, then we have 2k is not an even integer.

This is a False statement since 2 times any integer is an even integer.

Therefore, a + b is an even integer, so the sum of two even integers is an even integer.  QED

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.

14.29%

step-by-step explanation: