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)
![V=\pi [28x+\frac{1}{x}+xln^2x-12xlnx-6lnx]](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5B28x%2B%5Cfrac%7B1%7D%7Bx%7D%2Bxln%5E2x-12xlnx-6lnx%5D)
and by evaluating in the limits we have
Hope this helps
regards
In order to solve this problem, we must solve an equation.
First, we need to define our unknown.
Let's call our number "n".
So, the sum of the number and 6 is the same as saying n+6
Then we need to multiply it by 2 (since it's twice the sum)
2(n+6)
This whole thing is equal to three times of the difference of the number and 8
n-8 is equal to the difference between the number and 8, then we need to multiply it by three.
3(n-8)
now, we set both sides equal and solve
The number is 36
This is similar to the other question you posted. Follow the same steps as before.
First find g(41).
g(41) = sqrt{x - 5}
g(41) = sqrt{41 - 5}
g(41) = sqrt{36}
g(41) = 6
We now find f(6).
f(6) = -7(6) + 1
f(6) = -42 + 1
f(6) = -41
Answer:
(fºg)(41) = -41
Did you follow?
0.20(150) = 30 basketballs were in the shipment. Since the shipment presumably comprised only footballs and basketballs, the remainder of the shipment must consist of footballs.
150 - 30 = 120 footballs were in the shipment.
