Given: Lines a and b are parallel and line c is a transversal. Prove: Angle2 is supplementary to Angle8 Horizontal and parallel lines a and b are cut by transversal c. On line a where it intersects line c, 4 angles are created. Labeled clockwise, from uppercase left, the angles are: 1, 2, 4, 3. On line b where it intersects line c, 4 angles are created. Labeled clockwise, from uppercase left, the angles are: 5, 6, 8, 7. What is the missing reason in the proof? Statement Reason 1. a || b, is a transv 1. given 2. ∠6 ≅ ∠2 2. ? 3. m∠6 = m∠2 3. def. of congruent 4. ∠6 is supp. to ∠8 4. def. of linear pair 5. ∠2 is supp. to ∠8 5. congruent supplements theorem corresponding angles theorem alternate interior angles theorem vertical angles theorem alternate exterior angles theorem

the answer to your question is b.

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.