Hello!
E) 3 and 6
interior angles are the angles on the inside of the parallel angles. In this case, 3, 4, 5, and 6 are interior angles.
<u>alternate</u> exterior angles are 2 interior angles on opposite sides of the line passing through the middle.
in this case, the alternate exterior angles are: 3 and 6 / 5 and 4
choosing from your answer choices, you'll see that the only option that matches either of the 2 alternate interior angles I listed is: E) 3 and 6
I hope this helps, and have a nice day!
31. t= 10/9
43. x= -1/14 (the “-“ can go either on the top or bottom)
45. z= 0.21
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.
<u>Step by Step Explanation:</u>
16. x - 4 = 4
Isolate the variable (x). Note the equal sign, what you do to one side, you do to the other. Add 4 to both sides:
x - 4 (+4) = 4 (+4)
x = 4 + 4
x = 8
17. a/4 = 4
Isolate the variable (a). Note the equal sign, what you do to one side, you do to the other. Multiply 4 to both sides:
(a/4)(4) = (4)(4)
a = (4)(4)
a = 16
18. y * 20 = 30
Isolate the variable (y). Note the equal sign, what you do to one side, you do to the other. Divide 20 from both sides.
(y *20)/20 = (30)/20
y = 30/20
y = 3/2 or 1.5
19. x/5 = 20
Isolate the variable (x). Note the equal sign, what you do to one side, you do to the other. Multiply 5 to both sides
(x/5)(5) = (20)(5)
x = (20)(5)
x = 100
~
N would be 6 for this one
