ax² + bx + c = 0
x = (-b ± √(b² - 4ac))/2a
First, rewrite the first equation so that the first coefficient is 1. Divide everything by a.
(ax² + bx + c = 0)/a =
x² + (b/a)x + (c/a) = 0
Isolate (c/a) by subtracting (c/a) from both sides
x² + (b/a)x + (c/a) (-(c/a) = 0 (- (c/a)
x² + (b/a)x = 0 - (c/a)
Add spaces
x² + (b/a)x = -c/a
Take 1/2 of the middle term's coefficient and square it. Remember that what you add to one side, you add to the other.
x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²
Simplify the left side of the equation.
x² + (b/a)x + (b/2a)² = (x + (b/2a))²
(x + b/2a))² = ((b²/4a²) - (4ac/4a²)) -> ((b² - 4ac)/(4a²))
Take the square root of both sides of the equation
√(x + b/2a))² = √((b²/4a²) - (4ac/4a²))
x + b/(2a) = (±√(b² - 4ac)/2a
Simplify. Isolate the x.
x = -(b/2a) ± (∛b² - 4ac)/2a = (-b ± √(b² - 4ac))/2a
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A I am pretty sure. I apologize if I am incorrect.
Answer:
Two angles are said to be linear if they are adjacent angles formed by two intersecting lines.
Step-by-step explanation:
The answer is the first one.
Explanation:
X^3 stays the same because there are no other cubed numbers in the problem
Next you combine the x^2s
The x^2s are +3x^2 and +2x^2
Since they are both positive, you add them: 3x^2 + 2x^2 = 5x^2
Next you do the x values
-x and +6x, also known as 6x - x = 5x
Lastly, you just add in the -2 and get:
X^3 + 5x^2 + 5x - 2
Hello,
f(π/2)=4*sin(π/2)-2=4*1-2=2
f(0)=4*sin(0)-2=0-2=-2
rate=(2+2)/(π/2-0)=8/π
Answer D
