Standard form: ax^2 + bx + c = y
Steps to simplify:
2x^2 - 1 = 3x + 4
~Subtract (3x + 4) to both sides
2x^2 - 1 - 3x + 4 = 3x + 4 - 3x + 4
~Simplify
2x^2 - 3x - 5 = 0
(a = 2, b = -3, c = -5)
Now that we know what these values are, we can use the quadratic formula to solve.

Answer: x = 5/2 or x = -1
Best of Luck!
I believe x = 32 because it is an equalateral triangle meaning each angle is 60 degrees so the equasion is 2x-4=60.
Answer:
The angle measures that are correct are m<2 = 125degrees, m<8 = 55 degrees and m<14 = 100 degrees
Given the following angles from the diagram;
m<5 = 55 degrees
m<9 = 80degrees
From the diagram
m<5 = m<1 = 55 degrees (corresponding angle)
m<1 + m<2 = 180 (sum of angle on a straight line)
Hence;
55 + m<2 = 180
m<2 = 180 - 55
m<2 = 125degrees
Also;
m<5 = m<8 = 55 degrees (vertically opposite angle)
m<9 = m<13 = 80degrees
m<13 + m<14 = 180
Hence;
80 + m<14 = 180
m<14 = 180 - 80
m<14 = 100 degrees
Hence the angle measures that are correct are m<2 = 125degrees, m<8 = 55 degrees and m<14 = 100 degrees
Step-by-step explanation:
Recall that for a home visit, the technician charges $50 regardless on the time spent in the repair.
So, to find out the rate, we should calculate the part that depends on the spent time, and the add 50. So for example, we know that the technician spents 1 hour. So, we multiply 1 times 25 and then add 50. So, 25*1 + 50 = 75, which is the rate for a 1-hour repair.
So, in general, if we know that the number of hours is x, we multiply x times 25 and then add 50. Then a table would like this:
x 25*x 25*x +50
1 25 75
2 50 100
3 75 125
4 100 150
Note that as the time increases by one hour, the fare increases by 25. This is an example of a direct variation, since as the independent variable increases (t