The volume of the cone will be volume
V≈400.19
Answer:
y = 6x + 8?
Step-by-step explanation:
Answer:
The midpoint of the segment is (6,5)
Step-by-step explanation:
To find the midpoint of a segment we use the formula
midpoint = (x1+x2)/2, (y1+y2)/2
= (10+2)/2 , (7+3)/2
= 12/2, 10/2
=6,5
The midpoint of the segment is (6,5)
To solve this problem you must apply the proccedure shown below:
1. You have the following functions given in the problem above:

and

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2. You have that g(x) times f(x) can be written as g(x)*f(x). Therefore, you must multiply f(x) * g(x), as following:
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
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3. Applying the distributive property, you have:
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The answer is:
Answer:

Step-by-step explanation:
This problem can be solved by using the expression for the Volume of a solid with the washer method
![V=\pi \int \limit_a^b[R(x)^2-r(x)^2]dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint%20%5Climit_a%5Eb%5BR%28x%29%5E2-r%28x%29%5E2%5Ddx)
where R and r are the functions f and g respectively (f for the upper bound of the region and r for the lower bound).
Before we have to compute the limits of the integral. We can do that by taking f=g, that is

there are two point of intersection (that have been calculated with a software program as Wolfram alpha, because there is no way to solve analiticaly)
x1=0.14
x2=8.21
and because the revolution is around y=-5 we have

and by replacing in the integral we have
![V=\pi \int \limit_{x1}^{x2}[(lnx+5)^2-(\frac{1}{2}x+3)^2]dx\\](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint%20%5Climit_%7Bx1%7D%5E%7Bx2%7D%5B%28lnx%2B5%29%5E2-%28%5Cfrac%7B1%7D%7B2%7Dx%2B3%29%5E2%5Ddx%5C%5C)
and by evaluating in the limits we have

Hope this helps
regards