Answer:

Step-by-step explanation:
Given (Missing Information):
;
; 
Required
Determine the volume
Using Shell Method:

First solve for a and b.
and
Substitute 8 for y

Take 2/3 root of both sides





This implies that:

For 
This implies that:

So, we have:


The volume of the solid becomes:

Open bracket



Integrate
![V = 2\pi * [{\frac{8x^2}{2} - \frac{x^{1+\frac{5}{2}}}{1+\frac{5}{2}}]\vert^4_0](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B%5Cfrac%7B8x%5E2%7D%7B2%7D%20-%20%5Cfrac%7Bx%5E%7B1%2B%5Cfrac%7B5%7D%7B2%7D%7D%7D%7B1%2B%5Cfrac%7B5%7D%7B2%7D%7D%5D%5Cvert%5E4_0)
![V = 2\pi * [{4x^2 - \frac{x^{\frac{2+5}{2}}}{\frac{2+5}{2}}]\vert^4_0](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B4x%5E2%20-%20%5Cfrac%7Bx%5E%7B%5Cfrac%7B2%2B5%7D%7B2%7D%7D%7D%7B%5Cfrac%7B2%2B5%7D%7B2%7D%7D%5D%5Cvert%5E4_0)
![V = 2\pi * [{4x^2 - \frac{x^{\frac{7}{2}}}{\frac{7}{2}}]\vert^4_0](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B4x%5E2%20-%20%5Cfrac%7Bx%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%7D%7B%5Cfrac%7B7%7D%7B2%7D%7D%5D%5Cvert%5E4_0)
![V = 2\pi * [{4x^2 - \frac{2}{7}x^{\frac{7}{2}}]\vert^4_0](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B4x%5E2%20-%20%5Cfrac%7B2%7D%7B7%7Dx%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%5D%5Cvert%5E4_0)
Substitute 4 and 0 for x
![V = 2\pi * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [{4*0^2 - \frac{2}{7}*0^{\frac{7}{2}}])](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%28%5B%7B4%2A4%5E2%20-%20%5Cfrac%7B2%7D%7B7%7D%2A4%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%5D%20-%20%5B%7B4%2A0%5E2%20-%20%5Cfrac%7B2%7D%7B7%7D%2A0%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%5D%29)
![V = 2\pi * ([{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}] - [0])](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%28%5B%7B4%2A4%5E2%20-%20%5Cfrac%7B2%7D%7B7%7D%2A4%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%5D%20-%20%5B0%5D%29)
![V = 2\pi * [{4*4^2 - \frac{2}{7}*4^{\frac{7}{2}}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B4%2A4%5E2%20-%20%5Cfrac%7B2%7D%7B7%7D%2A4%5E%7B%5Cfrac%7B7%7D%7B2%7D%7D%5D)
![V = 2\pi * [{64 - \frac{2}{7}*2^2^{*\frac{7}{2}}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B64%20-%20%5Cfrac%7B2%7D%7B7%7D%2A2%5E2%5E%7B%2A%5Cfrac%7B7%7D%7B2%7D%7D%5D)
![V = 2\pi * [{64 - \frac{2}{7}*2^7]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B64%20-%20%5Cfrac%7B2%7D%7B7%7D%2A2%5E7%5D)
![V = 2\pi * [{64 - \frac{2}{7}*128]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B64%20-%20%5Cfrac%7B2%7D%7B7%7D%2A128%5D)
![V = 2\pi * [{64 - \frac{2*128}{7}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B64%20-%20%5Cfrac%7B2%2A128%7D%7B7%7D%5D)
![V = 2\pi * [{64 - \frac{256}{7}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%7B64%20-%20%5Cfrac%7B256%7D%7B7%7D%5D)
Take LCM
![V = 2\pi * [\frac{64*7-256}{7}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%5Cfrac%7B64%2A7-256%7D%7B7%7D%5D)
![V = 2\pi * [\frac{448-256}{7}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%5Cfrac%7B448-256%7D%7B7%7D%5D)
![V = 2\pi * [\frac{192}{7}]](https://tex.z-dn.net/?f=V%20%3D%202%5Cpi%20%20%2A%20%5B%5Cfrac%7B192%7D%7B7%7D%5D)
![V = [\frac{2\pi * 192}{7}]](https://tex.z-dn.net/?f=V%20%3D%20%5B%5Cfrac%7B2%5Cpi%20%20%2A%20192%7D%7B7%7D%5D)


Hence, the required volume is:

Answer:
The answer to your question is a. Infinite solution
Step-by-step explanation:
When two lines have the same slope and the same intercepts, this means that these lines are the same so they system of equations will have infinite solutions.
If the lines do not cross, there will be no solution.
If the lines cross in one point there will be one solution
If there are a quadratic and a linear function they will cross in two points.
The ratio is common, so you could write:
a1 * r^4 = a^5 (you do r^4 because there are 4 times you need to multiply by the ratio to get from a1 to a5)
Plug in the values:
3 * r^4 = 48
Divide by 3:
r^4 = 16
Take the fourth root of both sides:
r = 2
The common ratio is 2.
It opens upward and the vertex is (2,-6) here are some plotted points: (0,6),(1,-3),(2,-6),(3,-3),(4,6). Hope that helps. I hate graphing but if you do it enough times you'll get the hang of it!