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)

Answer:
Option 3
Step-by-step explanation:
When it says 12 less, that means -12.
Hope I helped. :)
9514 1404 393
Answer:
Step-by-step explanation:
The measure of an inscribed angle (QTR) is half the measure of the arc it intercepts. The measure of an arc is the same as the measure of the central angle it intercepts. So, we have ...
∠QSR = 2×∠QTR
∠QSR = 2×39°
∠QSR = 78°
__
Sides SQ and SR are radii of circle S, so are the same length. That means triangle QRS is an isosceles triangle and the base angles SQR and SRQ are congruent. The sum of angles in a triangle is 180°, so we have ...
∠QSR + 2(∠SQR) = 180°
78° + 2(∠SQR) = 180° . . . . fill in the value we know
2(∠SQR) = 102° . . . . . . . . . subtract 78°
∠SQR = 51° . . . . . . . . . . . . .divide by 2
Answer:
the question is incomplete, so I looked for a similar one:
<em>After the release of radioactive material into the atmosphere from a nuclear power plant, the hay was contaminated by iodine 131 ( half-life, 8 days). If it is all right to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay? </em>
iodine's half life (we are given x, we need to find b):
0.5A₀ = A₀eᵇˣ
x = 8 days
we eliminate A₀ from both sides
0.5 = eᵇ⁸
ln 0.5 = ln eᵇ⁸
-0.69315 = b8
b = -0.69315 / 8 = -0.08664
since the farmers need to wait until only 10% of the iodine remains (we already calculated b, now we need to find x):
0.1A₀ = A₀eᵇˣ
0.1 = eᵇˣ
ln 0.1 = bx
where b = -0.08664
x = ln 0.1 / -0.08664 = -2.302585 / -0.08664 = 26.58 days
The only thing you should do is to divide 3,072 by 96 and the answer is 32.
So each of them should read 32 books :)))
I hope this is helpful
have a nice day