Answer:
![\cos (a-b)=\cos a \cos b+\sin a \sin b](https://tex.z-dn.net/?f=%5Ccos%20%28a-b%29%3D%5Ccos%20a%20%5Ccos%20b%2B%5Csin%20a%20%5Csin%20b)
Step-by-step explanation:
Given : ![\cos (180^{\circ}-q)=-\cos q](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-%5Ccos%20q)
We have to write which identity we will use to prove the given statement.
Consider ![\cos (180^{\circ}-q)=-\cos q](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-%5Ccos%20q)
Take left hand side of given expression ![\cos (180^{\circ}-q)](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29)
We know
![\cos (a-b)=\cos a \cos b+\sin a \sin b](https://tex.z-dn.net/?f=%5Ccos%20%28a-b%29%3D%5Ccos%20a%20%5Ccos%20b%2B%5Csin%20a%20%5Csin%20b)
Comparing , we get, a= 180° and b = q
Substitute , we get,
![\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D%5Ccos%20180%5E%7B%5Ccirc%7D%20%20%5Ccos%20%28q%29%2B%5Csin%20q%20%5Csin%20180%5E%7B%5Ccirc%7D)
Also, we know
and ![\cos 180^{\circ}=-1](https://tex.z-dn.net/?f=%5Ccos%20180%5E%7B%5Ccirc%7D%3D-1)
Substitute, we get,
![\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-1%5Ccdot%20%5Ccos%20%28q%29%2B%5Csin%20q%20%5Ccdot%200)
Simplify , we get,
![\cos (180^{\circ}-q)=-\cos (q)](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-%5Ccos%20%28q%29)
Hence, use difference identity to prove the given result.
1.) 28-3x-4= 2x+12+x
2.) 24-3x= 3x+12
3.)24-6x=12
4.)-6x=-12
5.)x=2
Answer:
10) C
11) A
Step-by-step explanation:
Answer:
what grade is this???
Step-by-step explanation:
Answer:
You are at the top of the Empire State Building on the 102nd floor, which is located 373 m above the ground when your favorite superhero flies over the building parallel to the ground at 70.0% the speed of light.
You have never seen your favorite superhero in real life. Out of curiosity you calculate her height to be 1.57 mm . If the superhero landed next to you, how tall would she be when standing?
Step-by-step explanation:
so this problem is from Einstein's special theory of relativity about length contraction.
the formula is given by:
![L'=L\sqrt{1-\beta^2 }](https://tex.z-dn.net/?f=L%27%3DL%5Csqrt%7B1-%5Cbeta%5E2%20%7D)
where ![\beta =v/c](https://tex.z-dn.net/?f=%5Cbeta%20%3Dv%2Fc)
c is speed of light
and v is velocity of object.
L is called proper length. L' is length that the observer measure in moving frame.
here velocity of object is 70% speed of light. i.e ![v=209 854 721 m / s](https://tex.z-dn.net/?f=v%3D209%20854%20721%20m%20%2F%20s)
so ![L=1.57 m](https://tex.z-dn.net/?f=L%3D1.57%20m)
putting all values
![L'= 1.57m \sqrt{1-(\frac{209854721}{3*10^8})^2 } \\\\L'=1.12m](https://tex.z-dn.net/?f=L%27%3D%201.57m%20%5Csqrt%7B1-%28%5Cfrac%7B209854721%7D%7B3%2A10%5E8%7D%29%5E2%20%7D%20%5C%5C%5C%5CL%27%3D1.12m)