Answer:
8/3 or 2 2/3 as a mixed number
Step-by-step explanation:
Given :
An equation, 2cos ß sin ß = cos ß .
To Find :
The value for above equation in (0, 2π ] .
Solution :
Now, 2cos ß sin ß = cos ß
2 sin ß = 1
sin ß = 1/2
We know, sin ß = sin (π/6) or sin ß = sin (5π/6) in ( 0, 2π ] .
Therefore,

Hence, this is the required solution.
Hello!
First, let's write the problem.

Apply the distributive property on the left side of the equation.

Add like terms.

Let's plug that in into the original equation.

Add 4 to both sides.


Divide both sides by 5.

Our final answer would be,

You can feel free to let me know if you have any questions regarding this!
Thanks!
- TetraFish
<h3>
Answer: D) 0</h3>
Work Shown:
f(x) = 4x^(-2) + (1/4)x^2 + 4
f ' (x) = 4*(-2)x^(-3) + (1/4)*2x .... power rule
f ' (x) = -8x^(-3) + (1/2)x
f ' (2) = -8(2)^(-3) + (1/2)*(2)
f ' (2) = 0