Answer:

Step-by-step explanation:
Radius = r = 3 in.
Height = h = 2 in.
<u>Volume:</u>
![\sf Volume \ of \ the \ cylinder = \pi r^2 h\\\\V = (3.14)(3)^2(2)\\\\V =(3.14)(9)(2)\\\\V = (3.14)(18)\\\\V = 56.5 \ in.^3\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%20Volume%20%5C%20of%20%5C%20the%20%5C%20cylinder%20%3D%20%5Cpi%20r%5E2%20h%5C%5C%5C%5CV%20%3D%20%283.14%29%283%29%5E2%282%29%5C%5C%5C%5CV%20%3D%283.14%29%289%29%282%29%5C%5C%5C%5CV%20%3D%20%283.14%29%2818%29%5C%5C%5C%5CV%20%3D%2056.5%20%5C%20in.%5E3%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
The midpoint would be (5,9)
It's b 240000
I'm sorry if wrong I'm not too good at these things
Answer:
B) -3/2
Step-by-step explanation:
If [x/2]=0 then x/2 is a number such that the least integer greater than or equal to x/2 is 0. We can rewrite this as the inequality x/2≤0. Then, the value of x in C, D and E is wrong because they are positive numbers, then x/2 would be a positive number which contradicts this inequality.
Now, 0 is the least integer that satisfies this inequality, therefore we cannot have that x/2≤-1 since -1 is an integer and -1<0. Then x/2>-1. This discards A as wrong, because for x=-2, x/2=-1, contrary to x/2>-1.
Thus B is the right answer. To verify, if x=-3/2, then x/2=-3/4 and we have that -1<-3/4≤0 as required.