ABCD is a square. Two equilateral triangles AED and BFC have been constructed on the sides AD and BC respectively. Prove that triangle ABF is congruent to triangle CDE.

Check below please.

Step-by-step explanation:

Hi,

Let's plot the figure, 1 square and 2 equilateral triangles.

1) Let's remember all the angles we already know, from the square and the equilateral triangle from their respective definition.

In other words:

Statement                               Reason                          

  Given

2) We have two triangles ABF and CDE

3) The Side, Angle Side Congruence Theorem assures us that both triangles are congruent. When there are two known legs (4 cm and 4 cm) of each  triangle, and their respective formed angle is also known (150º). Therefore, these two triangles are both congruent.

Statement                                                   Reason

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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