<em>60%</em>
- <em>Step-by-step explanation:</em>
<em>Hi there ! </em>
<em>5 ........... 100%</em>
<em>3 .................x%</em>
<em>x = 3×100/5</em>
<em>= 300/5</em>
<em>= 60%</em>
<em>Good luck ! </em>
After all the calculations you should get an answer of 75
Answer:
The proof is derived from the summarily following equations;
∠FBE + ∠EBD = ∠CBA + ∠CBD
∠FBE + ∠EBD = ∠FBD
∠CBA + ∠CBD = ∠ABD
Therefore;
∠ABD ≅ ∠FBD
Step-by-step explanation:
The two column proof is given as follows;
Statement 
Reason
bisects ∠CBE Given
Therefore;
∠EBD ≅ ∠CBD Definition of angle bisector
∠FBE ≅ ∠CBA Vertically opposite angles are congruent
Therefore, we have;
∠FBE + ∠EBD = ∠CBA + ∠CBD Transitive property
∠FBE + ∠EBD = ∠FBD Angle addition postulate
∠CBA + ∠CBD = ∠ABD Angle addition postulate
Therefore;
∠ABD ≅ ∠FBD Transitive property.
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.
Answer: 20$
Step-by-step explanation: You would add 2(5)+10=y so 10+10=y 10 + 10 = 20
In this case (5,20). Hope this helps!!
