Give me a second to take a look at it
H = -16t² + 64t + 80 (:16)
H = -t² + 4t + 5
/\ = b²-4ac
/\ = 4²-4.(-1).5
/\ = 16+20
/\ = 36
Yv = -/\ / 4a
Yv = -36/4.(-1)
Yv = -36/-4
Yv = 9
Answer:
The work done to pump all of the kerosene from the tank to an outlet is
Step-by-step explanation:
The work is defined by:
(1)
The force here will be the product between the volume and the kerosene weighing, so we have :
![dF=\pi R^{2}dy*50](https://tex.z-dn.net/?f=dF%3D%5Cpi%20R%5E%7B2%7Ddy%2A50)
This force will be in-lbs.
Where R is the radius (3 feet)
Then using (1), we have:
Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.
![W=450\pi \int_{0}^{8}dy(8-y)](https://tex.z-dn.net/?f=W%3D450%5Cpi%20%5Cint_%7B0%7D%5E%7B8%7Ddy%288-y%29)
![W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)](https://tex.z-dn.net/?f=W%3D450%5Cpi%28%5Cint_%7B0%7D%5E%7B8%7D%208dy-%5Cint_%7B0%7D%5E%7B8%7D%20ydy%29)
![W=450\pi(8*8 -\frac{8^{2}}{2})](https://tex.z-dn.net/?f=W%3D450%5Cpi%288%2A8%20-%5Cfrac%7B8%5E%7B2%7D%7D%7B2%7D%29)
![W=450\pi(64 -\frac{64}{2})](https://tex.z-dn.net/?f=W%3D450%5Cpi%2864%20-%5Cfrac%7B64%7D%7B2%7D%29)
Therefore, the work done to pump all of the kerosene from the tank to an outlet is
I hope it helps you!
98.4252<span> inches is the exact so about 98 inches</span>