Three consecutive even integers add to −6. what are these integers?

Let's consider the smallest even integer = x.

then second even integer = (x + 2).

and third even integer = (x + 4).

So three consecutive even integers would be x, (x + 2), and (x + 4).

The sum of these numbers is -6. Its equation would be as follows :-

x + (x + 2) + (x + 4) = -6

3x + 6 = -6

3x = -6 - 6

3x = -12

x = -4

So first number = x = -4

second number = (x + 2) = -2

third number = (x + 4) = 0

Hence, three consecutive even integers are -4, -2, and 0

answer

a the answer hopes this

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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