Answer:

*Note c could be written as a/b
Step-by-step explanation:
-sin(-t - 8 π) + cos(-t - 2 π) + tan(-t - 5 π)
The identities I'm about to apply:



Let's apply the difference identities to all three terms:
![-[\sin(-t)\cos(8\pi)+\cos(-t)\sin(8\pi)]+[\cos(-t)\cos(2\pi)+\sin(-t)\sin(2\pi)]+\frac{\tan(-t)-\tan(5\pi)}{1+\tan(-t)\tan(5\pi)}](https://tex.z-dn.net/?f=-%5B%5Csin%28-t%29%5Ccos%288%5Cpi%29%2B%5Ccos%28-t%29%5Csin%288%5Cpi%29%5D%2B%5B%5Ccos%28-t%29%5Ccos%282%5Cpi%29%2B%5Csin%28-t%29%5Csin%282%5Cpi%29%5D%2B%5Cfrac%7B%5Ctan%28-t%29-%5Ctan%285%5Cpi%29%7D%7B1%2B%5Ctan%28-t%29%5Ctan%285%5Cpi%29%7D)
We are about to use that cos(even*pi) is 1 and sin(even*pi) is 0 so tan(odd*pi)=0:
![-[\sin(-t)(1)+\cos(-t)(0)]+[\cos(-t)(1)+\sin(-t)(0)]+\frac{\tan(-t)-0}{1+\tan(-t)(0)](https://tex.z-dn.net/?f=-%5B%5Csin%28-t%29%281%29%2B%5Ccos%28-t%29%280%29%5D%2B%5B%5Ccos%28-t%29%281%29%2B%5Csin%28-t%29%280%29%5D%2B%5Cfrac%7B%5Ctan%28-t%29-0%7D%7B1%2B%5Ctan%28-t%29%280%29)
Cleaning up the algebra:
![-[\sin(-t)]+[\cos(-t)]+\frac{\tan(-t)}{1}](https://tex.z-dn.net/?f=-%5B%5Csin%28-t%29%5D%2B%5B%5Ccos%28-t%29%5D%2B%5Cfrac%7B%5Ctan%28-t%29%7D%7B1%7D)
Cleaning up more algebra:

Applying that sine and tangent is odd while cosine is even. That is,
sin(-x)=-sin(x) and tan(-x)=-tan(x) while cos(-x)=cos(x):

Making the substitution the problem wanted us to:

Just for fun you could have wrote c as a/b too since tangent=sine/cosine.
Answer:
If it went up it would be: 50596.35 and if it went down it would be: 7556.66
Step-by-step explanation:
Here will be the function if it went up:
20000(1+0.0475)^20
Here’s if it went down:
20000(1-0.0475)^20
Answer:
17 m
Step-by-step explanation:
As shown in the figure,
is the ground level,
is the old level of water which was at 20 m below ground level,
is the new level of water which is 5m above of old level. The height of wall of the well, 1.20 m, and the the pelly at 80 cm =0.80 m, has been shown too.
The minimum length of the rope is equal to the distance between the pully and the new water level, indicated by the length
in the figure.
So, 
Hence, Raghu must use a rope having the minimum length 17 m to draw water from the well.
Answer:
-26/5 or -5.2
Step-by-step explanation:
First, you need to expand the parentheses:
50x+500=240
Now, subtract 500 from both sides:
50x=-260
Now, divide both sides by 50
x=-26/5=-5.2
Hope this helps!
I think The answer is a.59