Answer:
Conditional probability is probability of a second event given a first event has already occurred.
Step-by-step explanation:
Answer:
Both A and B are possible.
Step-by-step explanation:
It's both!
This is a might tricky. First you have to find the altitude. You have to determine if this is a real triangle.
Sin(22) = opposite / hypotenuse. The hypotenuse = 111. The angle is 22
opposite = hypotenuse * sin(22)
opposite = 111 * sin(22)
opposite = 41.58 and this is the altitude.
What have you learned?
Since 42 is larger than 41.58 you have 2 solutions to the triangle. One of the angles is acute, and the other one is obtuse. They are supplementary angles.
Sin(C) / c = Sin(A) / a
Sin(C) = c * Sin(22) / 42
Sin(C) = 111*sin(22)/ 42
Sin(C) = .99003
Sin(C) = 81.9
So that's your first answer. The second answer comes from Finding the supplement to this angle
supplement + 81.9 = 180
supplement = 180 - 81.9
supplement = 98.1
Answer:
14.598
Step-by-step explanation:
When you round that number to the nearest hundredth, you'll get 14.6. Hope this helps.
The scale factor is 1/2 or .5 because 200*.5=100 and 100*.5=50 and so on
Answer:
Step-by-step explanation:
I will work with radians.
![$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$](https://tex.z-dn.net/?f=%24%5Cfrac%20%7B%5Ccos%5E2%20%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%20%5Cright%29%2B%5Csin%28-x%29-%5Csin%5E2%20%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%20%5Cright%29%2B%5Ccos%20%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D-x%20%5Cright%29%7D%20%7B%5B%5Csin%28%5Cpi%20-x%29%2B%5Ccos%28-x%29%5D%20%5Ccdot%20%5B%5Csin%282%5Cpi%20%2Bx%29%5Ccos%282%5Cpi-x%29%5D%7D%24)
First, I will deal with the numerator
Consider the following trigonometric identities:
Therefore, the numerator will be
Once
Now let's deal with the numerator
![[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]](https://tex.z-dn.net/?f=%5B%5Csin%28%5Cpi%20-x%29%2B%5Ccos%28-x%29%5D%20%5Ccdot%20%5B%5Csin%282%5Cpi%20%2Bx%29%5Ccos%282%5Cpi-x%29%5D)
Using the sum and difference identities:
Therefore,
![[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]](https://tex.z-dn.net/?f=%5B%5Csin%28%5Cpi%20-x%29%2B%5Ccos%28-x%29%5D%20%5Ccdot%20%5B%5Csin%282%5Cpi%20%2Bx%29%5Ccos%282%5Cpi-x%29%5D%20%5Cimplies%20%5B%5Csin%28x%29%2B%5Ccos%28x%29%5D%20%5Ccdot%20%5B%5Csin%28x%29%5Ccos%28x%29%5D)
![\implies [p+4] \cdot [p \cdot 4]=4p^2+16p](https://tex.z-dn.net/?f=%5Cimplies%20%5Bp%2B4%5D%20%5Ccdot%20%5Bp%20%5Ccdot%204%5D%3D4p%5E2%2B16p)
The final expression will be
