Plzz ! the figure below shows      adc  and    bec.complete the two-column proof.be    ac  and  ad    cbcb  ≅  caprove:   adc  ≅    bec

Consider triangles ADC and BEC.

1. BE⊥AC (statement ) and AD⊥BC;

CB≅CA (statement )

Reason: Given.

2. ∠ADC (statement ) and ∠BEC are right angles.

Reason: Perpendicular lines form right (statement ) angles.

3.  ∠BCE (statement ) = ∠ACD  (statement )

Reason: Reflexive Property of Congruence (statement ).

4. ΔADC≅ΔBEC (statement ).

Reason: HA Postulate (statement ).

none cuz there are no "following statements" regarding the question

step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.