Answer:
yes
Step-by-step explanation:
each input is put with only one output therefore it is a function
Answer:
x = 27
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define equation</u>
4x + 10 = 3x + 37
<u>Step 2: Solve for </u><em><u>x</u></em>
- Subtract 3x on both sides: x + 10 = 37
- Subtract 10 on both sides: x = 27
<u>Step 3: Check</u>
<em>Plug in x to verify it's a solution.</em>
- Substitute: 4(27) + 10 = 3(27) + 37
- Multiply: 108 + 10 = 81 + 37
- Add: 118 = 118
And we have our final answer!
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.
The base value is the current value
Good luck! I hope this helps!
Answer:
chance is 50 percent or 1/2 so 3 times
Step-by-step explanation:
The die" has 6 sides with 3 odd numbers and 3 even numbers. If rolled 6 times, the chance would be 50 percent to roll an even number.
