Pretty sure the answer is 29/20 or 1 9/20. You get this by find the common denominator , 60, and then adding the number all together and simplifying
well, if the diameter is 5, thus its radius must be half that, or 2.5, and therefore, the radius of the one four times as much will be (4)(2.5).
Let's simply get their difference, since that'd be how much more is needed from the smaller to larger sphere.
![~\hfill \stackrel{\textit{surface area of a sphere}}{SA=4\pi r^2}\qquad \qquad r=radius~\hfill \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large difference of their areas}}{\stackrel{\textit{radius of (4)(2.5)}}{4\pi (4)(2.5)^2}~~ - ~~\stackrel{\textit{radius of 2.5}}{4\pi (2.5)^2}}\implies 100\pi -25\pi \implies 75\pi ~~ \approx ~~235.62~ft^2](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsurface%20area%20of%20a%20sphere%7D%7D%7BSA%3D4%5Cpi%20r%5E2%7D%5Cqquad%20%5Cqquad%20r%3Dradius~%5Chfill%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20difference%20of%20their%20areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bradius%20of%20%284%29%282.5%29%7D%7D%7B4%5Cpi%20%284%29%282.5%29%5E2%7D~~%20-%20~~%5Cstackrel%7B%5Ctextit%7Bradius%20of%202.5%7D%7D%7B4%5Cpi%20%282.5%29%5E2%7D%7D%5Cimplies%20100%5Cpi%20-25%5Cpi%20%5Cimplies%2075%5Cpi%20~~%20%5Capprox%20~~235.62~ft%5E2)
Answer:
7 track practices.
Step-by-step explanation:
36 miles - 8 miles she already ran= 28 miles left to run.
28 divided by 4 (miles) = 7.
Answer:
176
Step-by-step explanation:
every 11 mm in the drawing equals 4m.
every 1 mm in the drawing equals 4/11 m
Actual size of school yard = 484*4/11 = 176m
