\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) Use this to solve quadratic equations of the form ax^2 + bx + c = 0
495 =2000(.o45n) where n= number of years
495/(2000(.o45)) = n = 5.5 years
3/4 + 1/2 - 5/8 = ?
6/8 + 4/8 -5/8 =
5/8
45.5x + 15 = 20.5x + 20 Subtract 15 from both sides of the equation
45.5x = 20.5x + 5 Subtract 20.5x from both sides
25x = 5 Divide by 25
x = 5/25 = 1/5
3 classes x 25 studs/class x 2 gloves/stud = 150 gloves total (75 pairs)
-r/3 < 6 Multiply both sides by 3
-r < 18 Multiply by -1 ....this will reverse the <
r > 18
EVERYtime you subtract a NEGATIVE it becomes a POSITIVE
example: 4 -(-3) = 4 + 3 = 7 x -(-2x) = x +2x = 3x
If it is just a line, then you add them together.
a________________b ________c = 15
If it is a right triangle and we assume ab and bc are legs then you use the pythagorean theorem
ab^2 + bc^2 = ac^2
10^2 + 5^2 = ac^2
100 + 25 = ac^2
sqrt(125) = ac
That is the 'standard form' equation for a line.
Here ya go:
