The answer is 3 ft
The surface area (SA) of a cylinder with radius r and height h is:
SA = 2πr² + 2πrh = π(2r² + 2rh)
We have:
SA = 126π ft²
h = 3 * d
d = 2r
h = 3 * 2r = 6r
126π = π(2r² + 2rh)
126 = 2r² + 2rh
126 = 2r² + 2r * 6r
126 = 2r² + 12r²
126 = 14r²
r² = 126/14
r² = 9
r = √9
r = 3
The solutions appear to be
{π/2, 2π/3, 4π/3}.
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
Answer: 
Step-by-step explanation:
Given
The length of the rectangular prism is
width of the rectangular prism is 
height of the rectangular prism is 
The surface area of the rectangular prism is ![2[wl+lh+hw]](https://tex.z-dn.net/?f=2%5Bwl%2Blh%2Bhw%5D)
Put the values
![\Rightarrow 2[8\times 12+15\times 12+15\times 8]\\\Rightarrow 2[96+180+120]=2\times 396\\\Rightarrow 792\ m^2](https://tex.z-dn.net/?f=%5CRightarrow%202%5B8%5Ctimes%2012%2B15%5Ctimes%2012%2B15%5Ctimes%208%5D%5C%5C%5CRightarrow%202%5B96%2B180%2B120%5D%3D2%5Ctimes%20396%5C%5C%5CRightarrow%20792%5C%20m%5E2)
The surface area of the rectangular prism is 
Short answer: No and yes.
Yes First.
If the lower right point (looks like an C perhaps or an F) and the middle point and the upper left point lie on the same line and if D the middle point and the upper right point lie n the the same line, then yes. You have angle side angle. The lower right angle is given as equal to the upper left angle. The angles at C are vertically opposite angles and they are equal and you have been given 2 enclosed sides that are equal.
No
If the points I've described are not on the same line, then the triangles are not congruent. If I can, I'll edit this a second time.
Please Note:
You have done something that looks a bit illegal to me. I wouldn't just trade pictures. It only costs you another 5 points to issue another question. I do think, however, that it is very clever. In answering about 1700 questions, I've never seen this done. That's commendable.