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: 0.5 < x < 16.5.
Step-by-step explanation:
Given: Two sides of triangle: 8.0 units and 8.5 units
Measure of third side = x
According to the triangle's inequality,
- Sum of any two sides > third side. (i)
- Difference between the sides < third side. (ii)
If x is the third side, then
x < 8+8.5 [Using (i)]
i.e. x< 16.5
Also, x > 8.5-8 [Using (ii)]
i.e. x> 0.5
Hence, Range of possible sizes for side x would be 0.5 < x < 16.5.
Answer:
4,-6,2,4,8
Step-by-step explanation:
the domain are the one at the top like the numerator of a fraction
Answer:
y = (-1/4)x + 9/4
Step-by-step explanation:
2y = 8x - 6
So, y = 4x - 3 (Slope = 4)
Slope of perpendicular line = -1/4
Equation of line: y-3 = (-1/4)(x-(-3))
y-3 = (-1/4)x - 3/4
y = (-1/4)x + 9/4
The Integers Are Positive They Are Not Negative Numbers. If It Was Negative It Would Be -2,-3,-4. While If It's Positive It Should Be 0,1,2,3,4 And So On.
