Provide the missing reasons for the proof. Given: AB/DB = CB/EB Prove: ∆A/B~∆DBE

Here is their argument. given the obtuse angle x, we make a quadrilateral abcd with ∠dab = x, and ∠abc = 90◦, and ad = bc. say the perpendicular bisector to dc meets the perpendicular bisector to ab at p. then pa = pb and pc = pd. so the triangles pad and pbc have equal sides and are congruent. thus ∠pad = ∠pbc. but pab is isosceles, hence ∠pab = ∠pba. subtracting, gives x = ∠pad−∠pab = ∠pbc −∠pba = 90◦. this is a preposterous conclusion – just where is the mistake in the "proof" and why does the argument break down there?

Rewrite the following quadratic functions in intercept or factored form. show your work. y = 5x^2 + 10x

Which expressions are equivalent to -7+3(-4e-3)? choose all answers that apply: choose all answers that apply: a -4(3e+4)? 4(3e+4) b 12e12e c none of the above