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)
The answer would most likely be reflection
Answer:
Step-by-step explanation:
<u>The difference in balances is:</u>
- 5000*(1 + 0.04)^3 - 5000*(1 + 0.04*3) =
- 5624.32 - 5600 =
- 24.32
Answer:
−2 is a member of the domain of f.
f(0) > 0
Step-by-step explanation:
Statement 1 is TRUE. Domain value includes set of values of x of the function. -2 is plotted on the x-axis on the graph. Therefore, it is a member of domain of f.
Statement 2 is incorrect. Range includes all possible y-values of the function. On the graph, no y-value plotted is -2. Therefore, -2 is NOT a member of the range of f.
Statement 3 is TRUE.
f(0) is approximately 2.2 on the graph. i.e. at x = 0, y ≈ 2.2.
Therefore, f(0) > 0.
Statement 4 is INCORRECT.
f(2) = 1, that is at x = 2, y = 1 as seen in the graph. Therefore, f(2) is not greater than 2.
