Answer:
(f - g)(x) = -x² + 3x + 5
General Formulas and Concepts:
<u>Pre-Algebra</u>
<u>Algebra I</u>
- Function Notation
- Combining Like Terms
Step-by-step explanation:
<u>Step 1: Define</u>
f(x) = 3x + 5
g(x) = x²
(f - g)(x) is f(x) - g(x)
<u>Step 2: Find (f - g)(x)</u>
- Substitute: (f - g)(x) = 3x + 5 - x²
- Rewrite: (f - g)(x) = -x² + 3x + 5
We can verify this with the graph
(-6, 3) NO
(-3, 3) NO
(-5, 0) NO
(-4, 2) YES
(0,0) NO
(-6, 6) YES
(-3, -6) NO
(-4.0) YES
the area where the solutions are in the yellow area
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)

The correct answer is B 9.74_98.1
Nope- false
The sum of the square roots of two sides of a _right_ triangle is equal to the hypotenuse.