Using the law os cosines formula b^2 = a^2 + c^2 - 2*a*c*cos(B)
a = 17, b = 8, c = 16
8^2 = 17^2 + 16^2 - 2*17*16* cos(B)
64 = 289 + 256 - 544 * cos(B)
544*cos(B) = 289 + 256 - 64
544 * cos(B) = 481
cos (B) = 481/544
B = arccos(481/544)
B = 27.8 degrees
Answer:
4=No
-3=No
2=Yes
0=No
Step-by-step explanation:
enter enter all the factors into the equation and see which one equals -28 on both sides at the end.
Answer:

Step-by-step explanation:
See the attached picture for explanation.
Answer:
100+150DHS
Step-by-step explanation:
100 base and 150 is the rate per a day
Answer:
Required center of mass 
Step-by-step explanation:
Given semcircles are,
whose radious are 1 and 4 respectively.
To find center of mass,
, let density at any point is
and distance from the origin is r be such that,
where k is a constant.
Mass of the lamina=m=
where A is the total region and D is curves.
then,

- Now, x-coordinate of center of mass is
. in polar coordinate 




![=3k\big[-\cos\theta\big]_{0}^{\pi}](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B-%5Ccos%5Ctheta%5Cbig%5D_%7B0%7D%5E%7B%5Cpi%7D)
![=3k\big[-\cos\pi+\cos 0\big]](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B-%5Ccos%5Cpi%2B%5Ccos%200%5Cbig%5D)

Then, 
- y-coordinate of center of mass is
. in polar coordinate 




![=3k\big[\sin\theta\big]_{0}^{\pi}](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B%5Csin%5Ctheta%5Cbig%5D_%7B0%7D%5E%7B%5Cpi%7D)
![=3k\big[\sin\pi-\sin 0\big]](https://tex.z-dn.net/?f=%3D3k%5Cbig%5B%5Csin%5Cpi-%5Csin%200%5Cbig%5D)

Then, 
Hence center of mass 