Solve the lp relaxation of your model in part (a). if required, round your answers to the nearest whole number. if your answer is zero enter "0".

hope this .

r3t40

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.

step-by-step explanation:

consider the parallelogram shown in the attached image.

in the parallelogram hijk, the length of hk is given as,

from the properties of parallelogram we know that the opposite sides of a parallelogram are parallel to each other and are equal in measure.

therefore, the length of the side ij must be equal to the length of the side hk and that is .

hence, the expression for the length of ij confirms that hijk is a parallelogram.