Asha makes a conjecture that the perimeter of a square whose sides are an odd integer is always even. Which choice is the best proof of her conjecture?

CLEAR

CHECK

Every time you add four odd numbers, the sum is an even

number.

Let m and n both represent odd numbers that are the side

lengths of a square. And let m +n be an even number.

Therefore, m+n=n+m, which shows that the sum

of two odd numbers is even. And if two odd numbers are

even, then four odd numbers are also even.

Let n be an odd number. Because 4n is a multiple of 2, it

is an even number.

Look at these different examples: 5+5

9+9+9+9= 36. So if the side of a square is add,

the perimeter must be even.

well the 1 and -1 are always after the 0 and the 0 is inthe middle. s thats what i would guess. but i dont know what a slope of 2 is sorry

step-by-step explanation:

answer: b)

step-by-step explanation:

its a dot plot because since the score is not high numbers.

- r3ktforgood ☕

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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