Determine whether the conjecture is true or false. Give a counterexample for any false conjecture.

Given: segments RT and ST; twice the measure of .
Conjecture: S is the midpoint of segment RT.
Question 9 options:

False; lines do not have midpoints.

True

False; point S may not be on .

False; could be the segment bisector of .

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! : )

cretdit to source

can you make your screen more see able?

then i might be able to .

answer: group b.

step-by-step explanation: