The image of the regular hexagon coincides with the pre-image 6 times during rotation.
A regular hexagon is a polygon that has six equal sides and six equal angles.
In a single rotation which is usually the rotation of an object at 360°, the number of times in which the regular hexagon coincides with its pre-image is 6 times, this is because it has 6 equal sides and 6 equal angles.
Learn more about regular polygon here:
brainly.com/question/1592456
Answer:
66 I think might he wrong
Step-by-step explanation:
I'm not sure if its right
Using the intersecting chord theorem:
15 x 2 = 5 x n
Simplify:
30 = 5n
Divide both sides by 5:
n = 30/5
n = 6 m
8 x n+8 = 16 x n+2
Simplify:
8n +64 = 16n +32
Subtract 8n from both sides:
64 = 8n +32
Subtract 32 from both sides:
32 = 8n
Divide both sides by 8:
n = 32 /8
n = 4
Answer:
20 games
Step-by-step explanation:
Let
x ----> total games played this season
we know that
12 games represent 60% of the games they have played this season
so
using proportion
Find out how many games represent a percentage of 100%
![I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\ = [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\ =0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\ = (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\ ](https://tex.z-dn.net/?f=I%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%2Asin%5E%7Bn-1%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%5B-cos%28x%29%2Asin%5E%7Bn-1%7D%28x%29%5D_0%5E%20%5Cfrac%7B%5Cpi%7D%7B2%7D%2B%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%28x%29%2Asin%5E%7Bn-2%7D%28x%29%2Acos%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D0%20%2B%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bcos%5E2%28x%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7B%281-sin%5E2%28x%29%29%2Asin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7Bn-2%7D%28x%29%7D%20%5C%2C%20dx%20-%20%28n-1%29%2A%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5En%28x%29%20%5C%2C%20dx%5C%5C%0A%0A)
![\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\ = \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\ = \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\ ](https://tex.z-dn.net/?f=%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%5E%7B3%7D%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cfrac%7B2%7D%7B3%7D%20%5Cint%5Climits%5E%7B%20%5Cfrac%7B%5Cpi%7D%7B2%7D%20%7D_0%20%7Bsin%28x%29%7D%20%5C%2C%20dx%20%5C%5C%0A%3D%20%5Cdfrac%7B2%7D%7B3%7D%5C%20%5B-cos%28x%29%5D_0%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D%7D%3D%5Cdfrac%7B2%7D%7B3%7D%20%5C%5C%0A%0A%0A%0A%0A)
