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

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

An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.

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

Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.

Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
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 William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.

To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.

For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false

For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true

For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true

For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false

For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false

For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false

Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.

4 0

3 years ago

If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find ƒ(-1) + 5g(-3). -40 -38 7

EastWind [94]

$f(x) = x^2 + 1 \\ 
g(x) = 3x + 1$

You can do this in parts.
$f(-1) = (-1)^2 + 1 = 1 + 1 = 2$
g(-3) = 3(-3) + 1 = -9 + 1 = -8
So, 5g(-3) = 5(-8) = -40.
So f(-1) + 5g(-3) = 2 - 40 = -38

Or you can do this as one big chunk.
f(-1) + 5g(-3)
$=[(-1)^2+1] + 5[3(-3) + 1] \\
= (2) + 5(-9 + 1) \\
= 2 + 5(-8) \\
= 2 - 40 \\ = -38$