![\bf \sqrt{n}< \sqrt{2n+5}\implies \stackrel{\textit{squaring both sides}}{n< 2n+5}\implies 0\leqslant 2n - n + 5 \\\\\\ 0 < n+5\implies \boxed{-5 < n} \\\\\\ \stackrel{-5\leqslant n < 2}{\boxed{-5}\rule[0.35em]{10em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}2}](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%7Bn%7D%3C%20%5Csqrt%7B2n%2B5%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsquaring%20both%20sides%7D%7D%7Bn%3C%202n%2B5%7D%5Cimplies%200%5Cleqslant%202n%20-%20n%20%2B%205%20%5C%5C%5C%5C%5C%5C%200%20%3C%20n%2B5%5Cimplies%20%5Cboxed%7B-5%20%3C%20n%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B-5%5Cleqslant%20n%20%3C%202%7D%7B%5Cboxed%7B-5%7D%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D0%5Crule%5B0.35em%5D%7B3em%7D%7B0.25pt%7D2%7D)
namely, -5, -4, -3, -2, -1, 0, 1. Excluding "2" because n < 2.
The second cylinder is taller correct?
8 in < 11 in
If the height is also larger, the base is most likely larger too.
Wouldn't the largest cylinder have a Larger surface area?
givens
pi = 3.14
r = 2
h = 4
Formula
v = pi r^2 h
Sub and Solve
V = 3.14 * 2^2 * 4
V = 3.14 * 4 * 4
V = 3.14 * 16
V = 50.24
4 significant figures.
we can start counting the numbers after all the zeros in front. the zeros in front doesn't count in. but the zeros after the first number that is 1 or larger counts.
therefore we only need to look at 3010
To do this i add them uo and easily subtract woth a hand held calculator
