Answer:
2
Step-by-step explanation:
For a quadratic function the average rate of change on an interval is the rate of change at the midpoint of the interval. The rate of change of a function is given by its derivative.
The derivative of f(x) = x^2 is f'(x) = 2x. The midpoint of the interval is (4+(-2))/2 = 1. Then the average rate of change is ...
f'(1) = 2(1) = 2
The average rate of change of f(x) on [-2, 4] is 2.
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<em>Alternate solution</em>
The average rate of change is the slope of the line between the end points of the interval:
m = (y2 -y1)/(x2 -x1)
m = (f(4) -f(-2))/(4 -(-2)) = (20 -8)/(6) = 2
The average rate of change on [-2, 4] is 2.
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The attached graph shows the points on the curve and a line with slope 2 between them. It also shows the various slope calculations.
For question 11, you essentially need to find when h(t) = 0, since that is when the height of the ball reaches 0 (ie touches the ground).
For question 12, it is asking for a maximum height, so you need to find when dh/dt = 0 and taking the second derivative to prove that there is maximum at t. That will find you the time at which the ball will hit a maximum height.
Rinse and repeat question 12 for question 13
Total budget = $13,500
Donations = 12%
Taxes = 12%
Food = 26%
Clothing = 20%
Housing = 10%
and the rest are 5 %
Answer: I believe it’s 24
Step-by-step explanation: if not sorry
Step-by-step explanation:
You need to solve for x.
You can do that by either setting both of the equations equal to each other. or Solve each one separately and subtract the EG equation from the EW to get GW