Best answer gets

which of the following completes the proof?

triangles abc and edc are formed from segments bd and ac, in which point c is between points b and d and point e is between points a and c.

given: segment ac is perpendicular to segment bd

prove: δacb ~ δecd

reflect δecd over segment ac. this establishes that then, translate point e' to point a. this establishes that therefore, δacb ~ δecd by the aa similarity postulate.

∠abc ≅ ∠e'd'c'; ∠e'd'c' ≅ ∠abc
∠acb ≅ ∠e'c'd'; ∠d'e'c' ≅ ∠bac
∠acb ≅ ∠e'c'd'; ∠e'd'c' ≅ ∠abc
∠abc ≅ ∠e'd'c'; ∠d'e'c' ≅ ∠bac

the answer is f(x)=2x-1

step-by-step explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

the answer is 2/54

step-by-step explanation:

care to share were the trapezoid shown is ?