Answer:

Step-by-step explanation:
when adding logs, apply the log rule: 
∴ 
when subtracting logs, apply the log rule: 

This can be written as:

"No greater than" does not necessarily mean less than, but it just means it cannot go over. So equal to or less than.
"The difference of a number and 2" can be represented as "n - 2"
If you were to solve this inequality, you would end up getting:

or

Hope this helps! :)
Answer: see proof below
<u>Step-by-step explanation:</u>
Given: A + B + C = π → C = π - (A + B)
→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))
→ sin C = sin (A + B) cos C = - cos(A + B)
Use the following Sum to Product Identity:
sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]
cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]
Use the following Double Angle Identity:
sin 2A = 2 sin A · cos A
<u>Proof LHS → RHS</u>
LHS: (sin 2A + sin 2B) + sin 2C




![\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]](https://tex.z-dn.net/?f=%5Ctext%7BFactor%3A%7D%5Cqquad%20%5Cqquad%20%5Cqquad%202%5Csin%20C%5Ccdot%20%5B%5Ccos%20%28A-B%29%2B%5Ccos%20%28A%2BB%29%5D)


LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C 
Answer:
[-1, 9], [-2, 10], [-3, 11], [1, 7]
Step-by-step explanation:
Starting from the y-intercept of [0, 8], I did <em>rise</em><em>\</em><em>run</em><em> </em>by moving one block <em>north</em><em> </em>over <em>one</em> block <em>west</em><em> </em>for the first three coordinates, but retracing back to the y-intercept, I moved one block <em>south</em><em> </em>over <em>one</em> block <em>east</em> for the last coordinate. It does not matter where you go, as long as you follow the pattern within this <em>rate</em><em> </em><em>of</em><em> </em><em>change</em><em> </em>[<em>slope</em>].
I am joyous to assist you anytime.