Answer:
Sum=720 x=105 degrees Angle H=110 degrees Angle I= 100 degrees Angle K= 135
Step-by-step explanation:
*A hexagons angles add up to 720
*This is a hexagon
*No matter how the hexagon is shaped, it's still going to add up to 720
- Combine all of the angles (known and unknown) into an equation to equal 720
- 140+105+(x+30)+130+(x-5)+(x+5)=720
- remove parentheses
- 140+105+x+30+130+x-5+x+5=720
- Combine like terms and simplify
- 3x+405=720
- subtract 405 from both sides
- 3x=315
- divide by 3 on both sides
- 3x/3=315/3
- x=105
- Angle H = x+5
- Plug in x
- 105+5=110
- Angle H= 110 degrees
- Angle I = x-5
- pug in x
- 105-5=100
- Angle I= 100 degrees
- Angle K= x+30
- plug in x
- 105+30=135
- Angle K= 135 degrees
(g₀f)(x)=g(f(x))
=g(2x-2)
=5(2x-2)^2-3
=5(4x^2-8x+4)-3
=20x^2-40x+17
The coordinates of the centroid are the average values of the
- and
-coordinates of the points
that belong to the region. Let
denote the bounded region. These averages are given by the integral expressions
![\dfrac{\displaystyle \iint_R x \, dA}{\displaystyle \iint_R dA} \text{ and } \dfrac{\displaystyle \iint_R y \, dA}{\displaystyle \iint_R dA}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cdisplaystyle%20%5Ciint_R%20x%20%5C%2C%20dA%7D%7B%5Cdisplaystyle%20%5Ciint_R%20dA%7D%20%5Ctext%7B%20and%20%7D%20%5Cdfrac%7B%5Cdisplaystyle%20%5Ciint_R%20y%20%5C%2C%20dA%7D%7B%5Cdisplaystyle%20%5Ciint_R%20dA%7D)
The denominator is just the area of
, given by
![\displaystyle \iint_R dA = \int_0^8 \int_{\min(8\sin(2x), 8\cos(2x))}^{\max(8\sin(2x),8\cos(2x))} dy \, dx \\\\ ~~~~~~~~ = \int_0^8 |8\sin(2x) - 8\cos(2x)| \, dx \\\\ ~~~~~~~~ = 8\sqrt2 \int_0^8 \left|\sin\left(2x-\frac\pi4\right)\right| \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_R%20dA%20%3D%20%5Cint_0%5E8%20%5Cint_%7B%5Cmin%288%5Csin%282x%29%2C%208%5Ccos%282x%29%29%7D%5E%7B%5Cmax%288%5Csin%282x%29%2C8%5Ccos%282x%29%29%7D%20dy%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cint_0%5E8%20%7C8%5Csin%282x%29%20-%208%5Ccos%282x%29%7C%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%208%5Csqrt2%20%5Cint_0%5E8%20%5Cleft%7C%5Csin%5Cleft%282x-%5Cfrac%5Cpi4%5Cright%29%5Cright%7C%20%5C%2C%20dx)
where we rewrite the integrand using the identities
![\sin(\alpha + \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)](https://tex.z-dn.net/?f=%5Csin%28%5Calpha%20%2B%20%5Cbeta%29%20%3D%20%5Ccos%28%5Calpha%29%5Ccos%28%5Cbeta%29%20%2B%20%5Csin%28%5Calpha%29%5Csin%28%5Cbeta%29)
Now, if
![8(\cos(2x) - \sin(2x)) = R \sin(2x + \alpha) = R \sin(2x) \cos(\alpha) + R \cos(2x) \sin(\alpha)](https://tex.z-dn.net/?f=8%28%5Ccos%282x%29%20-%20%5Csin%282x%29%29%20%3D%20R%20%5Csin%282x%20%2B%20%5Calpha%29%20%3D%20R%20%5Csin%282x%29%20%5Ccos%28%5Calpha%29%20%2B%20R%20%5Ccos%282x%29%20%5Csin%28%5Calpha%29)
with
, then
![\begin{cases} R\cos(\alpha) = 8 \\ R\sin(\alpha) = -8 \end{cases} \implies \begin{cases}R^2 = 128 \\ \tan(\alpha) = -1\end{cases} \implies R=8\sqrt2\text{ and } \alpha = -\dfrac\pi4](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%20R%5Ccos%28%5Calpha%29%20%3D%208%20%5C%5C%20R%5Csin%28%5Calpha%29%20%3D%20-8%20%5Cend%7Bcases%7D%20%5Cimplies%20%5Cbegin%7Bcases%7DR%5E2%20%3D%20128%20%5C%5C%20%5Ctan%28%5Calpha%29%20%3D%20-1%5Cend%7Bcases%7D%20%5Cimplies%20R%3D8%5Csqrt2%5Ctext%7B%20and%20%7D%20%5Calpha%20%3D%20-%5Cdfrac%5Cpi4)
Find where this simpler sine curve crosses the
-axis.
![\sin\left(2x - \dfrac\pi4\right) = 0](https://tex.z-dn.net/?f=%5Csin%5Cleft%282x%20-%20%5Cdfrac%5Cpi4%5Cright%29%20%3D%200)
![2x - \dfrac\pi4 = n\pi](https://tex.z-dn.net/?f=2x%20-%20%5Cdfrac%5Cpi4%20%3D%20n%5Cpi)
![2x = \dfrac\pi4 + n\pi](https://tex.z-dn.net/?f=2x%20%3D%20%5Cdfrac%5Cpi4%20%2B%20n%5Cpi)
![x = \dfrac\pi8 + \dfrac{n\pi}2](https://tex.z-dn.net/?f=x%20%3D%20%5Cdfrac%5Cpi8%20%2B%20%5Cdfrac%7Bn%5Cpi%7D2)
In the interval [0, 8], this happens a total of 5 times at
![x \in \left\{\dfrac\pi8, \dfrac{5\pi}8, \dfrac{9\pi}8, \dfrac{13\pi}8, \dfrac{17\pi}8\right\}](https://tex.z-dn.net/?f=x%20%5Cin%20%5Cleft%5C%7B%5Cdfrac%5Cpi8%2C%20%5Cdfrac%7B5%5Cpi%7D8%2C%20%5Cdfrac%7B9%5Cpi%7D8%2C%20%5Cdfrac%7B13%5Cpi%7D8%2C%20%5Cdfrac%7B17%5Cpi%7D8%5Cright%5C%7D)
See the attached plots, which demonstrates the area between the two curves is the same as the area between the simpler sine wave and the
-axis.
By symmetry, the areas of the middle four regions (the completely filled "lobes") are the same, so the area integral reduces to
![\displaystyle \iint_R dA \\\\ ~~~~ = 8\sqrt2 \left(-\int_0^{\pi/8} \sin\left(2x-\frac\pi4\right) \, dx + 4 \int_{\pi/8}^{5\pi/8} \sin\left(2x-\frac\pi4\right) \, dx \right. \\\\ ~~~~~~~~~~~~~~~~~~~~ \left. - \int_{17\pi/8}^8 \sin\left(2x-\frac\pi4\right) \, dx\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_R%20dA%20%5C%5C%5C%5C%20~~~~%20%3D%208%5Csqrt2%20%5Cleft%28-%5Cint_0%5E%7B%5Cpi%2F8%7D%20%5Csin%5Cleft%282x-%5Cfrac%5Cpi4%5Cright%29%20%5C%2C%20dx%20%2B%204%20%5Cint_%7B%5Cpi%2F8%7D%5E%7B5%5Cpi%2F8%7D%20%5Csin%5Cleft%282x-%5Cfrac%5Cpi4%5Cright%29%20%5C%2C%20dx%20%5Cright.%20%5C%5C%5C%5C%20~~~~~~~~~~~~~~~~~~~~%20%5Cleft.%20-%20%5Cint_%7B17%5Cpi%2F8%7D%5E8%20%5Csin%5Cleft%282x-%5Cfrac%5Cpi4%5Cright%29%20%5C%2C%20dx%5Cright%29)
The signs of each integral are decided by whether
lies above or below axis over each interval. These integrals are totally doable, but rather tedious. You should end up with
![\displaystyle \iint_R dA = 40\sqrt2 - 4 (1 + \cos(16) + \sin(16)) \\\\ ~~~~~~~~ \approx 57.5508](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_R%20dA%20%3D%2040%5Csqrt2%20-%204%20%281%20%2B%20%5Ccos%2816%29%20%2B%20%5Csin%2816%29%29%20%5C%5C%5C%5C%20~~~~~~~~%20%5Capprox%2057.5508)
Similarly, we compute the slightly more complicated
-integral to be
![\displaystyle \iint_R x dA = \int_0^8 \int_{\min(8\sin(2x), 8\cos(2x))}^{\max(8\sin(2x),8\cos(2x))} x \, dy \, dx \\\\ ~~~~~~~~ = 8\sqrt2 \int_0^8 x \left|\sin\left(2x-\frac\pi4\right)\right| \, dx \\\\ ~~~~~~~~ \approx 239.127](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_R%20x%20dA%20%3D%20%5Cint_0%5E8%20%5Cint_%7B%5Cmin%288%5Csin%282x%29%2C%208%5Ccos%282x%29%29%7D%5E%7B%5Cmax%288%5Csin%282x%29%2C8%5Ccos%282x%29%29%7D%20x%20%5C%2C%20dy%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%208%5Csqrt2%20%5Cint_0%5E8%20x%20%5Cleft%7C%5Csin%5Cleft%282x-%5Cfrac%5Cpi4%5Cright%29%5Cright%7C%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%5Capprox%20239.127)
and the even more complicated
-integral to be
![\displaystyle \iint_R y dA = \int_0^8 \int_{\min(8\sin(2x), 8\cos(2x))}^{\max(8\sin(2x),8\cos(2x))} y \, dy \, dx \\\\ ~~~~~~~~ = \frac12 \int_0^8 \left(\max(8\sin(2x),8\cos(2x))^2 - \min(8\sin(2x),8\cos(2x))^2\right) \, dx \\\\ ~~~~~~~~ \approx 11.5886](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciint_R%20y%20dA%20%3D%20%5Cint_0%5E8%20%5Cint_%7B%5Cmin%288%5Csin%282x%29%2C%208%5Ccos%282x%29%29%7D%5E%7B%5Cmax%288%5Csin%282x%29%2C8%5Ccos%282x%29%29%7D%20y%20%5C%2C%20dy%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%20%5Cfrac12%20%5Cint_0%5E8%20%5Cleft%28%5Cmax%288%5Csin%282x%29%2C8%5Ccos%282x%29%29%5E2%20-%20%5Cmin%288%5Csin%282x%29%2C8%5Ccos%282x%29%29%5E2%5Cright%29%20%5C%2C%20dx%20%5C%5C%5C%5C%20~~~~~~~~%20%5Capprox%2011.5886)
Then the centroid of
is
![(x,y) = \left(\dfrac{239.127}{57.5508}, \dfrac{11.5886}{57.5508}\right) \approx \boxed{(4.15518, 0.200064)}](https://tex.z-dn.net/?f=%28x%2Cy%29%20%3D%20%5Cleft%28%5Cdfrac%7B239.127%7D%7B57.5508%7D%2C%20%5Cdfrac%7B11.5886%7D%7B57.5508%7D%5Cright%29%20%5Capprox%20%5Cboxed%7B%284.15518%2C%200.200064%29%7D)