Note: Consider "÷" sign instead of "=".
Given:

To find:
The quotient.
Solution:
We have,

It can be written as

Splitting the middle terms, we get


Taking out the common factor (x-1), we get

Cancel the common factors.

The quotient is
.
Therefore, the correct option is D.
The domain (input values) of the cosine function is all negative and positive angle measures.
Let the function be f(x) = cos(x)
The domain of cos(x) is -∞ < x < ∞
The range is -1 ≤ f(x) ≤ 1
Hence, domain of cos(x) is all (+) and (-) angle measures.
Same goes with sine function as well
For function f(x) = sin(x)
The domain is -∞ < x < ∞ and range -1 ≤ f(x) ≤ 1
However for f(x) = tan(x) the same is not applicable.
1st multiply -6y by 4y=-24y^2
2nd, multiply -6y by 3=-18y
3rd, multiply -7 by 4y=-28y
4th, multiply -7 by 3=-21
5th, write all the terms in standard form: -24y^2-18y-28y-21
Combine like terms:-18y-28y=-46y
Put in standard form: -24y^2-46y-21
Final Answer: -24y^2-46y-21
Answer:
The answer is actually 1/3
Step-by-step explanation: