1 km = 0.62137... mile (rounded)
7 km = 4.3296... miles (rounded)
Pretty sure it’s y= 5 + 1.95x
I don’t think that the number of people matters because the cost doesn’t vary based on the number of people
You can make 13.5 three quarter servings
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.
C)
2n + 3
first term = 3
second term = 5 (2 + 3 = 5)
third term = 7 ( 4 + 3 = 7)
d)
term 15 that has value greater than 30
cause
2(14) + 3 = 28 + 3 = 31 greater than 30
