$793.50, All you do is multiply $15.87 by 50
The square root of x is less than 4. There is an open circle on 4 indicating its exclusion from the set. Square roots cannot result in negative numbers so the set ends at 0.
Answer:
Option D.
![A =96\pi\ cm^2](https://tex.z-dn.net/?f=A%20%3D96%5Cpi%5C%20cm%5E2)
Step-by-step explanation:
The area of the circular bases is:
![A_c = 2\pi(a) ^ 2](https://tex.z-dn.net/?f=A_c%20%3D%202%5Cpi%28a%29%20%5E%202)
Where
is the radius of the circle
Then
![A = 2\pi(4) ^ 2](https://tex.z-dn.net/?f=A%20%3D%202%5Cpi%284%29%20%5E%202)
![A = 32\pi\ cm^2](https://tex.z-dn.net/?f=A%20%3D%2032%5Cpi%5C%20cm%5E2)
The area of the rectangle is:
![A_r=b * 2\pi r](https://tex.z-dn.net/?f=A_r%3Db%20%2A%202%5Cpi%20r)
Where
![b=8\ cm](https://tex.z-dn.net/?f=b%3D8%5C%20cm)
b is the width of the rectangle and
is the length
Then the area of the rectangle is:
![A_r=8 * 2\pi (4)](https://tex.z-dn.net/?f=A_r%3D8%20%2A%202%5Cpi%20%284%29)
![A_r=64\pi\ cm^2](https://tex.z-dn.net/?f=A_r%3D64%5Cpi%5C%20cm%5E2)
Finally the total area is:
![A = A_c + A_r\\\\A = 32\pi\ cm^2 + 64\pi\ cm^2\\\\](https://tex.z-dn.net/?f=A%20%3D%20A_c%20%2B%20A_r%5C%5C%5C%5CA%20%3D%2032%5Cpi%5C%20cm%5E2%20%2B%2064%5Cpi%5C%20cm%5E2%5C%5C%5C%5C)
![A =96\pi\ cm^2](https://tex.z-dn.net/?f=A%20%3D96%5Cpi%5C%20cm%5E2)
Brand A is. Just divide 100 (100% of a pickle) by each number, so 100 divided by 240 and 100 by 325. Then you multiply them by the fraction of 100, which 1/3 is 33.3% and 1/2 is 50%. It comes very close but A is the lowest.
Ok I almost have the answer. Do you want me to explain how I did it?