Answer and Step-by-step explanation:
Since we have given that
1. m∠1 + m∠5 = 180° and m∠1 + m∠4=180° - Given
2. m∠1 + m∠5 = m∠1 + m∠4 - Substitution
3. m∠5 = m∠4 -
Subtraction property of equality
4. Ray YZ is parallel to Ray UV - If alternate interior angles equal, then line are ||
Answer:
50 Free Throws
Step-by-step explanation:
Answer: 50 Free Throws
25% = 0.25
0.25*200=50
<em>skylar sucks at basketball</em>
The option A is correct. A circle is the shape of the cross section taken parallel to the base of a cylinder.
According to the statement
We have given that the cylinder and we have to tell that the which shape required for the cross section taken parallel to the base of a cylinder.
So, For this purpose,
We know that the
A cylinder is a three-dimensional solid, one of the most basic geometric shapes.
In this shape the surface formed by the points at a fixed distance from a line segment, which is known as the axis of the cylinder.
As the bases of cylinder are two identical circles which are parallel to the curved surface.
The curved surface when we open will be circle rather than the circle.
So, Due to this reason the answer is a circle.
Therefore, a circle is the shape of the cross section taken parallel to the base of a cylinder.
Learn more about base of cylinder here
brainly.com/question/76387
#SPJ1
We have
![\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)](https://tex.z-dn.net/?f=%5Csqrt%5Bk%5D%7B%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%7D%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%5Cright%29%7Dk%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%5Cright%29%2B%5Cln%5Cleft%28%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%5Cright%29%2B%20%5Ccdots%20%2B%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%5Cright%29%7Dk%5Cright%29)
and as k goes to ∞, the exponent converges to a definite integral. So the limit is
![\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bk%5Cto%5Cinfty%7D%20%5Csqrt%5Bk%5D%7B%5CGamma%5Cleft%28%5Cdfrac1k%5Cright%29%20%5CGamma%5Cleft%28%5Cdfrac2k%5Cright%29%20%5Ccdots%20%5CGamma%5Cleft%28%5Cdfrac%20kk%5Cright%29%7D%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Clim_%7Bk%5Cto%5Cinfty%7D%20%5Cfrac1k%20%5Csum_%7Bi%3D1%7D%5Ek%20%5Cln%5Cleft%28%5CGamma%5Cleft%28%5Cfrac%20ik%5Cright%29%5Cright%29%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cint_0%5E1%20%5Cln%5Cleft%28%5CGamma%28x%29%5Cright%29%5C%2C%20dx%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Cexp%5Cleft%28%5Cdfrac%7B%5Cln%282%5Cpi%29%7D2%7D%5Cright%29%20%3D%20%5Cboxed%7B%5Csqrt%7B2%5Cpi%7D%7D)
Answer:
B
Step-by-step explanation:
The result says y=-x+5 which can be rearrange d to x+y=5
