Answer:
m <2=m <8(Alternative exterior angles)
Therefore m <8=55°
I believe it's B because we don't know how big the jar is and it's asking for. a statistical question.
The volume generated by rotating the given region
about OC is
<h3>
Washer method</h3>
Because the given region (
) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.
solution
We first find the value of x and y
![y=2(x)^{\frac{1}{4} }](https://tex.z-dn.net/?f=y%3D2%28x%29%5E%7B%5Cfrac%7B1%7D%7B4%7D%20%7D)
![x=(\frac{y}{2} )^{4}](https://tex.z-dn.net/?f=x%3D%28%5Cfrac%7By%7D%7B2%7D%20%29%5E%7B4%7D)
![y=2x](https://tex.z-dn.net/?f=y%3D2x)
![x=\frac{y}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7By%7D%7B2%7D)
![\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cpi%20%7D%20%5C%2C%20%28R_%7Bo%5E%7B2%7D%20%7D%20%20-%20R_%7Bi%5E%7B2%7D%20%7D%20%29%20%20%20%20%20%20%20dy)
![R_{o} = x = \frac{y}{2}](https://tex.z-dn.net/?f=R_%7Bo%7D%20%3D%20x%20%3D%20%5Cfrac%7By%7D%7B2%7D)
![R_{i} = x= (\frac{y}{2}) ^{4}](https://tex.z-dn.net/?f=R_%7Bi%7D%20%3D%20x%3D%20%28%5Cfrac%7By%7D%7B2%7D%29%20%5E%7B4%7D)
![a=0, b=2](https://tex.z-dn.net/?f=a%3D0%2C%20b%3D2)
![v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy](https://tex.z-dn.net/?f=v%3D%20%5Cint%5Climits%5E2_o%20%7B%5Cpi%20%7D%20%5C%2C%20%5B%28%5Cfrac%7By%7D%7B2%7D%29%5E%7B2%7D%20-%20%28%28%5Cfrac%7By%7D%7B2%7D%29%20%5E%7B4%7D%20%29%5E%7B2%7D%20%29%20%20dy)
![v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy](https://tex.z-dn.net/?f=v%3D%20%5Cpi%20%5Cint%5Climits%5E2_o%3D%20%5B%5Cfrac%7By%5E%7B2%7D%20%7D%7B4%7D%20-%20%5Cfrac%7By%5E%7B8%7D%20%7D%7B2%5E%7B8%7D%20%7D%7D%20%20%5D%20dy)
![v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]](https://tex.z-dn.net/?f=v%3D%20%5Cpi%20%5B%5Cint%5Climits%5E2_o%20%7B%5Cfrac%7By%5E%7B2%7D%20%7D%7B4%7D%20%7D%20%5C%2C%20dy%20-%20%5Cint%5Climits%5E2_o%20%7B%5Cfrac%7By%7D%7B2%5E%7B8%7D%20%7D%20%5E%7B8%7D%20%7D%20%5C%2C%20dy%20%5D)
![v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi](https://tex.z-dn.net/?f=v%3D%5Cpi%20%5B%5Cfrac%7B1%7D%7B4%7D%20%5Cfrac%7By%5E%7B3%7D%20%7D%7B3%7D%20%20%5Cint%5Climits%5E2_0%20-%20%5Cfrac%7B1%7D%7B2%5E%7B8%7D%20%7D%20%20%5Cfrac%7By%5E%7Bg%7D%20%7D%7Bg%7D%20%5Cint%5Climits%5E2_o%5C%5Cv%3D%20%5Cpi%20%5B%5Cfrac%7B1%7D%7B12%7D%20%282%5E%7B3%7D%20-0%29-%5Cfrac%7B1%7D%7B2%5E%7B8%7D%2A9%20%7D%20%282%5E%7Bg%7D%20-0%29%5D%5C%5Cv%3D%20%5Cpi%20%5B%5Cfrac%7B2%7D%7B3%7D%20-%5Cfrac%7B2%7D%7Bg%7D%20%5D%5C%5Cv%3D%20%5Cfrac%7B4%7D%7Bg%7D%20%5Cpi)
A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095
We are required to find an inequality which best represents the relationship between the number of hours gardening g and the total charge c
The inequality which best represents the relationship between the number of hours gardening g and the total charge c is c ≥ 15 + 12g
At least means greater than or equal to (≥)
fixed charge = $15
charges per hour = $12
Total charge = c
Number of hours = g
The inequality:
<em>Total charge ≥ fixed charge + (charges per hour × Number of hours</em>
c ≥ 15 + (12 × g)
c ≥ 15 + (12g)
c ≥ 15 + 12g
Therefore, the inequality which best represents the relationship between the number of hours gardening g and the total charge c is c ≥ 15 + 12g
Read more:
brainly.com/question/11067755