The coordinate plane below represents a town. Points A through F are farms in the town.
1 answer:
We have
A( 2,-3) B(-3,-4) C(-4,2) D(2,4) E(3,1) F(-2,3)
using a graph tool
see the attached figure N 1
Part A;There are many systems of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-2, 3) but is not satisfied by ( 2,-3) (-3,-4), (2, 4) and (3, 1) can serve.
An example of such system of equation is
x < 0
y > 0
The system of the equation above represents 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.
see the attached figure N 2
Part B:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(-4,2) into the system we have:
-4 < 0
2 > 0
as can be seen, the two inequalities above are true, hence point C is a solution to the set of inequalities.
Also, substituting F(-2,3) into the system we have
:-2 < 0
3 > 0
as can be seen, the two inequalities above are true, hence point F is a solution to the set of inequalities.
Part C:
Given that chicken can only be raised in the area defined by y < 5x - 3.
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( 2,-3): -3 < 5(2) - 3 ⇒ -3 < 10 - 3 ⇒ -3 < 7 which is true
For point B(-3,-4): -4 < 5(-3) - 3 ⇒ -4 < -15 - 3 ⇒ -4 <-18 which is false
For point C(-4,2): 2 < 5(-4) - 3 ⇒ 2 < -20 - 3 ⇒ 2 <-23 which is false
For point D(2,4): 4 < 5(2) - 3 ⇒ 4 < 10 - 3 ⇒ 4 <7 which is true
For point E(3,1): 1 < 5(3) - 3 ⇒ 1 < 15 - 3 ⇒ 1 <12 which is true
For point F(-2,3): 3 < 5(-2) - 3 ⇒ 3 < -10 - 3 ⇒ 3 < -13 which is false
Therefore
the farms in which chicken can be raised are the farms at point A, D, and E
see the attached figure N 3.
A( 2,-3) B(-3,-4) C(-4,2) D(2,4) E(3,1) F(-2,3)
using a graph tool
see the attached figure N 1
Part A;There are many systems of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-2, 3) but is not satisfied by ( 2,-3) (-3,-4), (2, 4) and (3, 1) can serve.
An example of such system of equation is
x < 0
y > 0
The system of the equation above represents 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.
see the attached figure N 2
Part B:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(-4,2) into the system we have:
-4 < 0
2 > 0
as can be seen, the two inequalities above are true, hence point C is a solution to the set of inequalities.
Also, substituting F(-2,3) into the system we have
:-2 < 0
3 > 0
as can be seen, the two inequalities above are true, hence point F is a solution to the set of inequalities.
Part C:
Given that chicken can only be raised in the area defined by y < 5x - 3.
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( 2,-3): -3 < 5(2) - 3 ⇒ -3 < 10 - 3 ⇒ -3 < 7 which is true
For point B(-3,-4): -4 < 5(-3) - 3 ⇒ -4 < -15 - 3 ⇒ -4 <-18 which is false
For point C(-4,2): 2 < 5(-4) - 3 ⇒ 2 < -20 - 3 ⇒ 2 <-23 which is false
For point D(2,4): 4 < 5(2) - 3 ⇒ 4 < 10 - 3 ⇒ 4 <7 which is true
For point E(3,1): 1 < 5(3) - 3 ⇒ 1 < 15 - 3 ⇒ 1 <12 which is true
For point F(-2,3): 3 < 5(-2) - 3 ⇒ 3 < -10 - 3 ⇒ 3 < -13 which is false
Therefore
the farms in which chicken can be raised are the farms at point A, D, and E
see the attached figure N 3.
You might be interested in