the coordinate plane below represents a town. points a through f are farms in the town. graph of coordinate plane. point a is at negative 3, negative 4. point b is at negative 4, 3. point c is at 2, 2. point d is at 1, negative 2. point e is at 5, negative 4. point f is at 3, 4.

part a: using the graph above, create a system of inequalities that only contains points c and f in the overlapping shaded regions. explain how the lines will be graphed and shaded on the coordinate grid above.

part b: explain how to verify that the points c and f are solutions to the system of inequalities created in part a.

part c: chickens can only be raised in the area defined by y > −4x + 3. explain how you can identify farms in which chickens can be raised.

Part A;

There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.

Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.

An example of such system of equation is

x < 0

y > 0

The system of equation above represent all the points in the second quadrant of the coordinate system.

The area above the x-axis and to the left of the y-axis is shaded.

Part B:

It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.

Substituting D(-4, 2) into the system we have:

-4 < 0

2 > 0

as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.

Also, substituting E(-1, 5) into the system we have:

-1 < 0

5 > 0

as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.

Part C:

Given that chicken can only be raised in the area defined by y > 3x - 4.

To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.

For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true

For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false

For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false

For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true

For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true

For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true

Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.

Step-by-step explanation:

we have that

A (-3,-4)

B (-4,3)

C (2,2)

D (1,-2)

E (5,-4)

F (-3,-13)

using a graph tool

see the attached figure N 1

Part A: Using the graph above, create a system of inequalities that only contains points A and E in the overlapping shaded regions.

A (-3,-4)  E (5,-4)

y<= -3

y>=-5

is a system of a inequalities that will only contain A and E

to graph it, I draw the constant y = -3 and y=-5  and and I shade the region between both lines

see the attached figure N 2

Part B: Explain how to verify that the points A and E are solutions to the system of inequalities created in Part A

we know that

the system of a inequalities is

y<= -3

y>=-5

the solution is all y real numbers belonging to the interval [-5,-3]

therefore

if points A and E are solutions both points must belong to the interval

points A and E have the same coordinate y=-4

and y=-4 is included in the interval 

therefore

both points are solution

Part C: Chickens can only be raised in the area defined by y < −2x + 4. Explain how you can identify farms in which chickens can be raised

step 1

graph the inequality 

y < −2x + 4

see the attached figure N 3

the farms in which chickens can be raised are the points A, B and D

are those that are included in the shaded part

hope this helps!!

answer4 to the second power is 16

and 4 to the third power is 64,

for this case we must apply distributive property to the following expression:

the distributive property states that:

then, when applying the property we have to multiply the -2 for each term within the parenthesis: