Answer:
0.08
Step-by-step explanation:
Please use parentheses to indicate which operations must be done first. I 'm choosing to believe that you meant 7^(3x-1) and 5^(x-1).
Taking the common log of both sides, we get:
(3x-1)log 7 = (x-1)log 5
Then 3x·log 7 - log 7 = x log 5 - log 5
Grouping the x terms:
x(3·log 7 - log 5) = log 7 - log 5. We combine log 7 and log 5, obtaining log (7/5).
Then x(3·log 7 - log 5) = log (7/5).
Solving for x:
log (7/5)
x = ---------------------
3·log 7 - log 5
0.1461
x = ------------- = 0.08
1.0363
This agrees with the 3rd answer choice.
Answer:
Visible light has higher energy than infrared light because visible light has higher frequencies.
Step-by-step explanation:
Just took the test hope you do well! (:
Also pls brainliest me
Answer: HI! its B
Step-by-step explanation:
have a great day tell your sis i said hi!
16(1/4x-1/2)>24-2x
Use the distributive property on the left side:
4x -8 > 24-2x
Add 2x to each side:
6x -8 > 24
Add 8 to each side:
6x >32
Divide both sides by 6:
x > 32/6 reduces to 16/3
Not sure if there is a typo in any of your answers, but I do not see the correct answer listed.
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)
