Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions: y > 2x + 3 y is less than negative 3 over 2 times x minus 4 part a: describe the graph of the system, including shading and the types of lines graphed. provide a description of the solution area. (6 points) part b: is the point (−4, 6) included in the solution area for the system? justify your answer mathematically. (4 points)

Part A: The solution area is to the left of x=-2, over the right line y=2x+3 and below the right line y=-(3/2)x-4.

Part B: No, the point (-4,6) is not included in the solution area for the system.

Step-by-step explanation:

System of inequalities:

(1) y > 2x + 3

(2) y < -(3/2)x - 4

Part A

1) The first inequality represents the region over the line:

y=2x+3

This is a linear equation and its graph is a right line. We must graph the right line, and then shadow the area above this right line. To graph this right line we need two points, because through two points only goes one right line. Giving two arbitrary values to x, for example:

x=0→y=2(0)+3→y=0+3→y=3; Point=(x,y)→Point=(0,3)

x=1→y=2(1)+3→y=2+3→y=5; Point=(x,y)→Point=(1,5)

1) The second inequality represents the region below the line:

y= -(3/2)x - 4

This is a linear equation and its graph isaright line. We must graph the right line, and then shadow the area below this right line. To graph this right line we need two points, because through two points only goes one right line. Giving two arbitrary values to x, for example:

x=0→y=-(3/2)(0)-4→y=0-4→y=-4; Point=(x,y)→Point=(0,-4)

x=2→y=-(3/2)(2)-4→y=-3-4→y=-7; Point=(x,y)→Point=(2,-7)

The two lines intersect at the point (-2,-1). The solution area is to the left of x=-2, over the right line y=2x+3 and below the right line y=-(3/2)x-4.

Part B

No, the point (-4,6) is not included in the solution area for the system, because this point is not below the right line y=-(3/2)x-4

answer: add the !

step-by-step explanation: then we will answer

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.