Answer: y=12.287
Step-by-step explanation:
Check the picture below.
so the volume will simply be the area of the hexagonal face times the height.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\stackrel{\qquad degrees}{\cot\left( \frac{180}{n} \right)}~~ \begin{cases} n=\stackrel{number~of}{sides}\\ s=\stackrel{length~of}{side}\\[-0.5em] \hrulefill\\ n=6\\ s=12 \end{cases}\implies A=\cfrac{1}{4}(6)(12)^2\cot\left( \frac{180}{6} \right) \\\\\\ A=216\cot(30^o)\implies A=216\sqrt{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the hexagon}}{(216\sqrt{3})}~~\stackrel{height}{(10)}\implies 2160\sqrt{3}~~\approx ~~3741.2~cm^3](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B4%7Dns%5E2%5Cstackrel%7B%5Cqquad%20degrees%7D%7B%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7Bn%7D%20%5Cright%29%7D~~%20%5Cbegin%7Bcases%7D%20n%3D%5Cstackrel%7Bnumber~of%7D%7Bsides%7D%5C%5C%20s%3D%5Cstackrel%7Blength~of%7D%7Bside%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D6%5C%5C%20s%3D12%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%2812%29%5E2%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7B6%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D216%5Ccot%2830%5Eo%29%5Cimplies%20A%3D216%5Csqrt%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagon%7D%7D%7B%28216%5Csqrt%7B3%7D%29%7D~~%5Cstackrel%7Bheight%7D%7B%2810%29%7D%5Cimplies%202160%5Csqrt%7B3%7D~~%5Capprox%20~~3741.2~cm%5E3)
Answer:
2x + (-2x + 3) = 3
2x - 2x +3
0 + 3
3
2x-7 + (-2x) = -7
2x -2x - 7
0 -7
-7
-(5x -1) + 5x = 1
-5x +1 + 5x
-5x + 5x + 1
0 + 1
1
Step-by-step explanation: In addition, the parentheses help identify the individual addends. If adding a negative number, it is the same as subtracting. If there is a negative sign outside the parentheses, as in example c, you must distribute it: the signs of the numbers within the parentheses must be reversed when the parentheses are removed.
