Construct a regular hexagon with vertex a inscribed in the given circle

Im not sure if this is right but.

Make the diameter by putting the line on points A and C. Then put the middle of the tool on point A and make the width to point C and draw a circle. Do the same for point B (other side of point A) without changing the width. Then plot the lines where the circles intersect and you have a hexagon :)

there is one photo without the hexagon drawn out, and one with it drawn, so you can see it all. there's one small hexagon and one long one.

i dont understand the question? ?

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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