I think the brainlest answer here would be my answer because I need points it’s 57^2/45^4x
if the sphere has a diameter of 5, then its radius is half that, or 2.5.
![\bf \textit{volume of a sphere}\\\\ V=\cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=2.5 \end{cases}\implies V=\cfrac{4\pi (2.5)^3}{3}\implies V=\cfrac{62.5\pi }{3} \\\\\\ V\approx 65.44984694978736\implies V=\stackrel{\textit{rounded up}}{65.45}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20sphere%7D%5C%5C%5C%5C%20V%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D2.5%20%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B4%5Cpi%20%282.5%29%5E3%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B62.5%5Cpi%20%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20V%5Capprox%2065.44984694978736%5Cimplies%20V%3D%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B65.45%7D)
Answer:
See in attachment and Mark as brainleist please.
Hope it helps
Answer:
346 2/3 yards
Step-by-step explanation:
A football field is 100 yards long with each end zone being 10 years long, meaning the total length is 120 yards.
The width is 53 1/3 yards.
The perimeter is all the sides added, and a rectangle's, the shape of the field's, formula is p = 2l + 2w.
So if we plug it in and add it all up:
120 + 120 + 53 1/3 + 53 1/3
Perimeter = 346 2/3