It should be acute because if A and B are complementary then B and A added together should be 90 so 90-30=60 and 60 would be an acute angle
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
the answer is 32.........
Answer:
![f^{-1}(x)=\frac{x+1}{6}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D%5Cfrac%7Bx%2B1%7D%7B6%7D)
Step-by-step explanation:
So we have the function:
![f(x)=6x-1](https://tex.z-dn.net/?f=f%28x%29%3D6x-1)
To find the inverse function, switch x and f(x), change f(x) to f⁻¹(x), and solve for f⁻¹(x). Therefore:
![f(x)=6x-1](https://tex.z-dn.net/?f=f%28x%29%3D6x-1)
Switch:
![x=6f^{-1}(x)-1](https://tex.z-dn.net/?f=x%3D6f%5E%7B-1%7D%28x%29-1)
Add 1 to both sides:
![x+1=6f^{-1}(x)](https://tex.z-dn.net/?f=x%2B1%3D6f%5E%7B-1%7D%28x%29)
Divide both sides by 6:
![f^{-1}(x)=\frac{x+1}{6}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D%5Cfrac%7Bx%2B1%7D%7B6%7D)
And that's our answer :)
The solutions to system of equations are the point of intersection of the graphs of the equations.
The point of intersection of the graphs of the given equation is (0, 2). Therefore, point (1, 1) is not a solution to the system of equation.