Which ordered pairs are in the solution set of the system of linear inequalities? y > Negative one-halfx

y < One-halfx + 1

On a coordinate plane, 2 straight lines are shown. The first solid line has a negative slope and goes through (0, 0) and (4, negative 2). Everything above the line is shaded. The second dashed line has a positive slope and goes through (negative 2, 0) and (2, 2). Everything below the line is shaded.
(5, –2), (3, 1), (–4, 2)
(5, –2), (3, –1), (4, –3)
(5, –2), (3, 1), (4, 2)
(5, –2), (–3, 1), (4, 2)

b

step-by-step explanation:

the answer is  

a.

4,624π mm2

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.