Answer:
Sorry, I only know the answer to B.
Step-by-step explanation:
The answer is 0=0, 2=5, 4=10, 6= 15. Sorry if I am late. But I hope this helps, and Have an amazing day!!
I believe it is B. I did not look at the both the given points but one of the points that are unidentified and one of the given points:
(0, -8) and (2, -2)
To find slope use this equation:
(Y2 - Y1)/(X2 + X1)
Input the points into the equation:
(-2 - (-8))/(2 - 0) —> (-2 + 8)/(2 - 0)
You find that it becomes:
6/2
Then simplify it:
3/1
So the answer is A
Answer:
Step-by-step explanation:
If the ratio of the interior angle to the exterior angle for a regular polygon is 5:1, find:a. the size of each exterior angle, b. the number of sides of the polygon, c. the sum of the interior angles
hope that help
mark me brill ?
The first integral has a well-known beta function representation, so the second one should too. The beta function itself is defined as
![B(x,y) = \displaystyle \int_0^1 t^{x-1} (1-t)^{y-1} \, dt](https://tex.z-dn.net/?f=B%28x%2Cy%29%20%3D%20%5Cdisplaystyle%20%5Cint_0%5E1%20t%5E%7Bx-1%7D%20%281-t%29%5E%7By-1%7D%20%5C%2C%20dt)
and satisfies the identity
![\displaystyle B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%28x%2Cy%29%20%3D%20%5Cfrac%7B%5CGamma%28x%29%20%5CGamma%28y%29%7D%7B%5CGamma%28x%2By%29%7D)
Later on, we'll also use the so-called reflection formula for the gamma function; for non-integer z,
![\Gamma(z) \Gamma(1-z) = \dfrac{\pi}{\sin(\pi z)}](https://tex.z-dn.net/?f=%5CGamma%28z%29%20%5CGamma%281-z%29%20%3D%20%5Cdfrac%7B%5Cpi%7D%7B%5Csin%28%5Cpi%20z%29%7D)
as well as the identity
![\dfrac{\Gamma(z+1)}{\Gamma(z)} = z](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CGamma%28z%2B1%29%7D%7B%5CGamma%28z%29%7D%20%3D%20z)
Replace
in both integrals, so that
![\displaystyle \int_0^{\frac\pi2} \sqrt{\sin(x)} \, dx = \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Csqrt%7B%5Csin%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E1%20%5Cfrac%7B%5Csqrt%20x%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%20%5C%2C%20dx)
![\displaystyle \int_0^{\frac\pi2} \frac{dx}{\sqrt{\sin(x)}} = \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%5Csin%28x%29%7D%7D%20%3D%20%5Cint_0%5E1%20%5Cfrac%7Bdx%7D%7B%5Csqrt%20x%20%5Csqrt%7B1-x%5E2%7D%7D)
Now replace
:
![\displaystyle \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx = \frac12 \int_0^1 x^{-\frac14} (1-x)^{-\frac12} \, dx = \frac12 B\left(\frac34, \frac12\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cfrac%7B%5Csqrt%20x%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%20%5C%2C%20dx%20%3D%20%5Cfrac12%20%5Cint_0%5E1%20x%5E%7B-%5Cfrac14%7D%20%281-x%29%5E%7B-%5Cfrac12%7D%20%5C%2C%20dx%20%3D%20%5Cfrac12%20B%5Cleft%28%5Cfrac34%2C%20%5Cfrac12%5Cright%29%20)
![\displaystyle \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}} = \frac12 \int_0^1 x^{-\frac34} (1-x)^{-\frac12} \, dx = \frac12 B\left(\frac14, \frac12\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cfrac%7Bdx%7D%7B%5Csqrt%20x%20%5Csqrt%7B1-x%5E2%7D%7D%20%3D%20%5Cfrac12%20%5Cint_0%5E1%20x%5E%7B-%5Cfrac34%7D%20%281-x%29%5E%7B-%5Cfrac12%7D%20%5C%2C%20dx%20%3D%20%5Cfrac12%20B%5Cleft%28%5Cfrac14%2C%20%5Cfrac12%5Cright%29)
So, the original integral (which I condense here to a double integral) is
![\displaystyle \int_0^{\frac\pi2} \int_0^{\frac\pi2} \sqrt{\frac{\sin(x)}{\sin(y)}} \, dx \, dy = \frac14 B\left(\frac34, \frac12\right) B\left(\frac14, \frac12\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Csqrt%7B%5Cfrac%7B%5Csin%28x%29%7D%7B%5Csin%28y%29%7D%7D%20%5C%2C%20dx%20%5C%2C%20dy%20%3D%20%5Cfrac14%20B%5Cleft%28%5Cfrac34%2C%20%5Cfrac12%5Cright%29%20B%5Cleft%28%5Cfrac14%2C%20%5Cfrac12%5Cright%29)
![\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\Gamma\left(\frac54\right) \Gamma\left(\frac34\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5Cfrac14%20%5Cfrac%7B%5CGamma%5Cleft%28%5Cfrac14%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac12%5Cright%29%5E2%7D%7B%5CGamma%5Cleft%28%5Cfrac54%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%7D)
![\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\frac14 \Gamma\left(\frac14\right) \Gamma\left(\frac34\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5Cfrac14%20%5Cfrac%7B%5CGamma%5Cleft%28%5Cfrac14%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac12%5Cright%29%5E2%7D%7B%5Cfrac14%20%5CGamma%5Cleft%28%5Cfrac14%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%7D)
![\displaystyle = \Gamma\left(\frac12\right)^2 = \frac{\pi}{\sin\left(\frac\pi2\right)} = \boxed{\pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5CGamma%5Cleft%28%5Cfrac12%5Cright%29%5E2%20%3D%20%5Cfrac%7B%5Cpi%7D%7B%5Csin%5Cleft%28%5Cfrac%5Cpi2%5Cright%29%7D%20%3D%20%5Cboxed%7B%5Cpi%7D)
Answer:
The height is 9 cm
Step-by-step explanation:
The area of a triangle is given by
A = 1/2 bh where b is the base and h is the height
72 = 1/2 ( 16) h
72 = 8h
Divide each side by 8
72/8 = 8h/8
9 = h
The height is 9 cm