MATH SOLVE

4 months ago

Q:
# The coordinate plane below represents a town. Points A through F are farms in the town.graph of coordinate plane. Point A is at 1, 3. Point B is at 3, 1. Point C is at 3, negative 3. Point D is at negative 4, 2. Point E is at negative 1, 5. Point F is at negative 3, negative 3.Part A: Using the graph above, create a system of inequalities that only contains points D and E in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above. (5 points)Part B: Explain how to verify that the points D and E are solutions to the system of inequalities created in Part A. (3 points)Part C: Chickens can only be raised in the area defined by y > 3x β 4. Explain how you can identify farms in which chickens can be raised. (2 points)I REALLY NEED THIS ANSWERED PLEASE

Accepted Solution

A:

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 β 3 > 3 - 4 β 3 > -1 which is true

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

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

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

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

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

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

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 β 3 > 3 - 4 β 3 > -1 which is true

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

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

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

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

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

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