Answer:
The average value of 
over the interval ![[-2,5]](https://tex.z-dn.net/?f=%5B-2%2C5%5D)
is 
.
Step-by-step explanation:
Let suppose that function is continuous and integrable in the given intervals, by integral definition of average we have that:
(1)
(2)
By Fundamental Theorems of Calculus we expand both expressions:
(1b)
(2b)
We obtain the average value of over the interval ![[-2, 5]](https://tex.z-dn.net/?f=%5B-2%2C%205%5D)
by algebraic handling:
![F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)](https://tex.z-dn.net/?f=F%285%29%20-%20F%283%29%20%2B%5BF%283%29-F%28-2%29%5D%20%3D%2040%20%2B%20%28-30%29)
The average value of over the interval is .
Answer: 6. d
7. b
8. d
9. c
Step-by-step explanation:
Answer:
Step-by-step explanation:
If you call "5x-2x^2+1" an "equation," then you must equate 5x-2x^2+1 to 0:
5x-2x^2+1 = 0
This is a quadratic equation. Rearranging the terms in descending order by powers of x, we get:
-2x^2 + 5x + 1 = 0. Here the coefficients are a = -2, b = 5 and c = 1.
Use the quadratic formula to solve for x:
First find the discriminant, b^2 - 4ac: 25 - 4(-2)(1) = 25 + 8 = 33
Because the discriminant is positive, the roots of this quadratic are real and unequal.
-b ± √(discriminant)
Applying the quadratic formula x = --------------------------------
2a
we get:
-5 ± √33 -5 + √33
x = ----------------- = --------------------- and
2(-2) -4
-5 - √33
---------------
-4
The answer is 32.72 degree.
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