Because the vertex of the parabola is at (16,0), its equation is of the formy = a(x-10)² + 15
The graph goes through (0,0), thereforea(0 - 10)² + 15 = 0100a = -15a = -0.15
The equation is y = f(x) = -0.15(x - 10)² + 15
The graph is shown below.
Part A
Note that y = f(x).
The x-intercepts identify values where the function or y=0. The x-intercepts occur at x=0 and x=20, or at (0,0) and (20,0).
The maximum value of y occurs at the vertex (10, 15) because the curve is down due to the negative leading coefficient of -0.15.
The curve increases in the interval x = (-∞, 10) and it decreases in the interval x = (10, ∞).
Part B
When x=12, y = -0.15(12 - 10)² + 15 = 14.4When x=15, y = -0.15(15 - 10)² + 15 = 11.25
The average rate of change between x =12 to x = 15 is(11.25 - 14.4)/(15 - 12) = -1.05
This rate of change represents the slope of the secant line from A to B. It approximates the rate at which f(x) decreases in the interval x =[12, 15].
There are 39 boys.
First you divide 4 into 52 because there are 3 boys for 4 girls.
Then you take the quotient, 13, and multiply by 3.
Answer:
The answer is "
"
Step-by-step explanation:
Given:

Find critical points:

differentiate the value with respect of x:
critical points
![\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3](https://tex.z-dn.net/?f=%5Cto%20%28x-e%29%5E2%20e%5E%7B%28e-x%29%7D%20%5Be%2B3-x%5D%3D0%5C%5C%5C%5C%5Cto%20e%5E%7B%28e-x%29%7D%5Cneq%200%20%5C%5C%5C%5C%5Cto%20%28x-e%29%5E2%3D0%5C%5C%5C%5C%20%5Cto%20%5Be%2B3-x%5D%3D0%5C%5C%5C%5C%5Cto%20x%3De%5C%5C%5C%5C%5Cto%20x%3De%2B3%5C%5C%5C%5C%5Cto%20x%3D%20e%2Ce%2B3)
So,
The critical points of 
Answer:
135
Step-by-step explanation:
My method 45 hours - 1 so i could divide by 2 because its 2 hours every 6 problems and 1 hour would be 3 math problems we can add that to the total
44 / 2 = 22, 22 * 6 = 132, 132 + 3 = 135
Alternate method
45 * 3 = 135
Intersects a given segment at a 90 degree angle,and passes through the given segment's midpoint.