Answer:
D) 140°
Step-by-step explanation:
The total number of degrees in a triangle is 180. Since both sides of the triangle are the radius of the circle, the triangle is isosceles. Therefore the two opposite angles are equal. Thus, the central angle is 180° − 2(20°) = 140°.
Answer:
3/2
Step-by-step explanation:
Okay, I can represent this equation as rise over run
the coordinates 4,6 are subcoordinates 1
y: 6 - 3 = 3
x: 4 - 2 = 2
Since it is RISE over run then Y is the numerator and X is the denominator.
So the slope is 3/2.
Use a ruler for this problem. Take a picture of the measurement & then I could help you.
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:

And the derivative of x:

Now, we can calculate the area of the surface:

We could calculate this integral (not very hard, but long), or use
(1),
(2) and
(3) to get:



Calculate indefinite integral:

And the area:
![A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}](https://tex.z-dn.net/?f=A%3D2%5Cpi%5Cint%5Climits_0%5E%7B10%7Dx%5Csqrt%7B4x%5E2%2B1%7D%5C%2Cdx%3D2%5Cpi%5Ccdot%5Cdfrac%7B1%7D%7B12%7D%5Cbigg%5B%5Cleft%284x%5E2%2B1%5Cright%29%5E%5Cfrac%7B3%7D%7B2%7D%5Cbigg%5D_0%5E%7B10%7D%3D%5C%5C%5C%5C%5C%5C%3D%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5Cleft%5B%5Cbig%284%5Ccdot10%5E2%2B1%5Cbig%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cbig%284%5Ccdot0%5E2%2B1%5Cbig%29%5E%5Cfrac%7B3%7D%7B2%7D%5Cright%5D%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%5CBig%28%5Cbig401%5E%5Cfrac%7B3%7D%7B2%7D-1%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%29%3D%5Cboxed%7B%5Cdfrac%7B401%5E%5Cfrac%7B3%7D%7B2%7D-1%7D%7B6%7D%5Cpi%7D)
Answer D.
= 2x^10y^12
answer is B 2x^10y^12