It’s simple. The volume of a sphere is V=4/3πr^3.. so you will convert the mm to m okay?? 10mm=1cm .. so do your conversion from here then when you get the radius then you will substitute it into the volume where r is the radius.. so for the first one l changed mm to cm so.. radius is 4cm then l substitute it into the formula so l got 270 after rounding it to the nearest tenth.. pi=22/7 bear in mind. So do the same for the other one
the assumption being that the endpoints are two continuous points in the pentagon, Check picture below.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[2-(-1)]^2+[3-4]^2}\implies d=\sqrt{(2+1)^2+(3-4)^2} \\\\\\ d=\sqrt{9+1}\implies d=\sqrt{10}~\hfill \stackrel{\stackrel{~\hfill \stackrel{\textit{5 sides}}{}}{\textit{perimeter of the pentagon}}}{5\sqrt{10}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B2%7D~%2C~%5Cstackrel%7By_2%7D%7B3%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B%5B2-%28-1%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%282%2B1%29%5E2%2B%283-4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%3D%5Csqrt%7B9%2B1%7D%5Cimplies%20d%3D%5Csqrt%7B10%7D~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7B~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B5%20sides%7D%7D%7B%7D%7D%7B%5Ctextit%7Bperimeter%20of%20the%20pentagon%7D%7D%7D%7B5%5Csqrt%7B10%7D%7D)
A 40% chance
Unless!
*wispers*
It’s a trick question?
The answer is 270 . if you multiply all the yds together you will get the solution to your volume
Where’s the data?................
