The answer is actually 5.25 because i took the test and use the other users answer and it was wrong.
Recall that
![\sin(x+y)=\sin x\cos y+\cos x\sin y](https://tex.z-dn.net/?f=%5Csin%28x%2By%29%3D%5Csin%20x%5Ccos%20y%2B%5Ccos%20x%5Csin%20y)
![\sin(x-y)=\sin x\cos y-\cos x\sin y](https://tex.z-dn.net/?f=%5Csin%28x-y%29%3D%5Csin%20x%5Ccos%20y-%5Ccos%20x%5Csin%20y)
Adding these together, you have
![\sin(x+y)+\sin(x-y)=2\sin x\cos y](https://tex.z-dn.net/?f=%5Csin%28x%2By%29%2B%5Csin%28x-y%29%3D2%5Csin%20x%5Ccos%20y)
![\dfrac12\sin(x+y)+\dfrac12\sin(x-y)=\sin x\cos y](https://tex.z-dn.net/?f=%5Cdfrac12%5Csin%28x%2By%29%2B%5Cdfrac12%5Csin%28x-y%29%3D%5Csin%20x%5Ccos%20y)
Replace
![x](https://tex.z-dn.net/?f=x)
with
![2x](https://tex.z-dn.net/?f=2x)
and
![y](https://tex.z-dn.net/?f=y)
with
![3x](https://tex.z-dn.net/?f=3x)
. You end up with
![\dfrac12\sin(2x+3x)+\dfrac12\sin(2x-3x)=\sin2x\cos3x](https://tex.z-dn.net/?f=%5Cdfrac12%5Csin%282x%2B3x%29%2B%5Cdfrac12%5Csin%282x-3x%29%3D%5Csin2x%5Ccos3x)
![\sin2x\cos3x=\dfrac12\sin5x+\dfrac12\sin(-x)=\dfrac12\sin5x-\dfrac12\sin x](https://tex.z-dn.net/?f=%5Csin2x%5Ccos3x%3D%5Cdfrac12%5Csin5x%2B%5Cdfrac12%5Csin%28-x%29%3D%5Cdfrac12%5Csin5x-%5Cdfrac12%5Csin%20x)
and so
![A=\dfrac12](https://tex.z-dn.net/?f=A%3D%5Cdfrac12)
and
![B=-\dfrac12](https://tex.z-dn.net/?f=B%3D-%5Cdfrac12)
.
Answer:
divide both sides by 1m
u2-v2=T/1m
add -u2 both sides of the equation
-v2=T/1m -u2
divide both sides by -1
v2= u2 - T/1m
The answer to your problem is 3^1..
Answer:
< 20
Step-by-step explanation:
hope this help idrk how to explain this one lol