g(θ) = 20θ − 5 tan θ
To find out critical points we take first derivative and set it =0
g(θ) = 20θ − 5 tan θ
g'(θ) = 20 − 5 sec^2(θ)
Now we set derivative =0
20 − 5 sec^2(θ)=0
Subtract 20 from both sides
− 5 sec^2(θ)=0 -20
Divide both sides by 5
sec^2(θ)= 4
Take square root on both sides
sec(θ)= -2 and sec(θ)= +2
sec can be written as 1/cos
so sec(θ)= -2 can be written as cos(θ)= -1/2
Using unit circle the value of θ is 
sec(θ)= 2 can be written as cos(θ)=1/2
Using unit circle the value of θ is 
For general solution we add 2npi
So critical points are

Answer:it is 9 because you subtract 6 from the 8 and then you add 2 and 7 to get 9
Step-by-step explanation:
Answer:
| x + 4 | = -1 No solution
Step-by-step explanation:
3 | x + 4 | + 12 = 9
Subtract 12 from both sides: 3 | x + 4 | = -3
Divide both sides by 3: | x + 4 | = -1
Absolute value cannot be negative, so there is no solution.
X=35/6 i hope i helped :-)