Michael and Derrick each completed a separate proof to show that corresponding angles AKG and ELK are congruent. Who completed the proof incorrectly? Explain. Line AB is parallel to EF, transversal GJ crosses line AB at K and crosses line EF at L.

Michael's Proof

Statement Justification
1. line AB ∥ line EF with transversal segment GJ 1. Given
2. angle AKG is congruent to angle AKL 2. Vertical Angles Theorem
3. angle BKL is congruent to angle ELK 3. Alternate Interior Angles Theorem
4. angle AKG is congruent to angle ELK 4. Transitive Property

Derrick's Proof

Statement Justification
1. line AB ∥ line EF with transversal segment GJ 1. Given
2. angle AKG is congruent to angle BKL 2. Vertical Angles Theorem
3. angle BKL is congruent to angle ELK 3. Alternate Interior Angles Theorem
4. angle AKG is congruent to angle ELK 4. Transitive Property

d

step-by-step explanation:

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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.