Cu part two REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS REMOVE THE BOTS ✌️❤️❤️SPREAD THE WORD SPREAD THE WORD SPREAD THE WORD ✌️❤️❤️. CRASH THE SEVERS

Step-by-step explanation:

yes bot su.cks

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

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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