According to the given information, segment ab is parallel to segment dc and segment bc is parallel to segment ad.. using a straightedge, extend segment ab and place point p above point b. by the same reasoning, extend segment ad and place point t to the left of point a. angles bcd and pbc are congruent by the alternate interior angles theorem. angles pbc and bad are congruent by the corresponding angles theorem. by the property of equality, angles bcd and bad are congruent. angles abc and bat are congruent by the alternate interior theorem. angles bat and cda are congruent by the corresponding angles theorem. by the property of equality, ∠abc is congruent to ∠cda. consequently, opposite angles of parallelogram abcd are congruent. what properties accurately complete the proof? addition transitive reflexive reflexive substitution reflexive transitive transitive

Transitive and Transitive

Step-by-step explanation:

By the TRANSITIVE Property of Equality, angles BCD and BAD are congruent. Angles ABC and BAT are congruent by the Alternate Interior Theorem.  

Angles BAT and CDA are congruent by the Corresponding Angles Theorem. By the TRANSITIVE Property of Equality, ∠ABC is congruent to ∠CDA.

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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