It’s 6.72 you add the two numbers
Answer: PQ=15
Step-by-step explanation:
XY=(PQ+SR)/2
PQ+SR=2*XY
PQ+35=2*25
PQ+35=50
PQ=15
Answer:
Ipguifiyfohvlbouviyfiyfyivojviyc itcitditcyivuy if it go
Put the argument value where the variable is, then evaluate.
For f(x), you want f(2).
f(2) = 2² + 1 = 4+1 = 5
For g(x), you want g(1).
g(1) = 3·1 +1 = 4
For [f(2) - g(1)] you want the difference of the above values.
[f(2) - g(1)] = [5 - 4] = 1
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)