so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)](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%5Ccot%5Cstackrel%7B%5Cstackrel%7Bdegrees%7D%7B%5Cdownarrow%20%7D%7D%7B%5Cleft%28%20%5Cfrac%7B180%7D%7Bn%7D%20%5Cright%29%7D~%20%5Cbegin%7Bcases%7D%20n%3D%5Ctextit%7Bnumber%20of%20sides%7D%5C%5C%20s%3D%5Ctextit%7Blength%20of%20a%20side%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D%5Cstackrel%7Bhexagon%7D%7B6%7D%5C%5C%20s%3D%5Cfrac%7B9%7D%7B2%7D%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%5Cleft%28%20%5Ccfrac%7B9%7D%7B2%7D%20%5Cright%29%5E2%20%5Ccot%5Cleft%28%20%5Ccfrac%7B180%7D%7B6%7D%20%5Cright%29)
![A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%5Ccfrac%7B9%5E2%7D%7B2%5E2%7D%20%5Ccot%2830%5Eo%29%5Cimplies%20A%3D%5Ccfrac%7B243%7D%7B8%7D%5Ccot%2830%5Eo%29%5Cimplies%20A%3D%5Ccfrac%7B243%5Csqrt%7B3%7D%7D%7B8%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Barea%20of%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D%5Cfrac%7B4%7D%7B5%7D%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%5Cleft%28%20%5Ccfrac%7B4%7D%7B5%7D%20%5Cright%29%5E2%5Cimplies%20A%3D%5Ccfrac%7B16%5Cpi%20%7D%7B25%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Below are the choices:
A. 80 mL of the 3.5% solution and 120 mL of the 6% solution
<span>B. 120 mL of the 3.5% solution and 80 mL of the 6% solution </span>
<span>C. 140 mL of the 3.5% solution and 60 mL of the 6% solution </span>
<span>D. 120 mL of the 3.5% solution and 80 mL of the 6% solution
</span>
Let fraction of 3.5% in final solution be p.
<span>p * 3.5 + (1 - p) * 6 = 4.5 </span>
<span>3.5p + 6 - 6p = 4.5 </span>
<span>2.5p = 1.5 </span>
<span>p = 3/5 </span>
<span>3/5 * 200 = 120 </span>
<span>Therefore the answer is B. 120 ml of 3.5% and 80 ml of 6%.</span>
P( both allergic) = 0.17*0.17 = 0.0289
P(at least one is allergic) = 1 - P(neither are allergic) = 1 - 0.83^2 = 0.311
<span>Let x = amt of water evaporated :.05(50) = .08(50-x) 2.5 = 4 - .08x .08x = 4 - 2.5 .08x = 1.5 x = 18.75 lb of water evaporated : ; Check; amt of salt remains the same, only the percentage is different, right? .05(50) = .08(50-18.75) .05(50) = .08(31.25) 2.5 = 2.5</span>
