
now, the angle of θ, can only have a "y" value that is positive on, well, y is positive at 1st and 2nd quadrants
and "x" is positive only in 1st and 4th quadrants
now, that angle θ, can only have those two fellows, "y" and "x" to be positive, only in the 1st quadrant, and also both to be negative on the 3rd quadrant.
and that those two fellows, can also be both negative in the 3rd quadrant
3/1 = 3, and -3/-1 = 3
so, the solutions can only be "3", when both "y" and "x" are the same sign, and that only occurs on the 1st and 3rd quadrants
The Lagrangian for this function and the given constraints is

which has partial derivatives (set equal to 0) satisfying

This is a fairly standard linear system. Solving yields Lagrange multipliers of

and

, and at the same time we find only one critical point at

.
Check the Hessian for

, given by


is positive definite, since

for any vector

, which means

attains a minimum value of

at

. There is no maximum over the given constraints.
Answer:
we just did this in class a few weeks ago in pretty sure its 3 because that's where the line crosses the y axis but im not 100 percent sure good luck!❤
Answer: Our required probability would be 0.9641.
Step-by-step explanation:
Since we have given that
Number of hours he works a day = 8
So, Number of minutes he worked in a day = 
Number of calls = 220
So, Average 
Standard deviation 
Mean = μ = 2.0 minutes
Standard deviation = σ = 1.5 minutes
Using the normal distribution, we get that

So, the probability that Albert will meet or exceed his quota would be

Hence, our required probability would be 0.9641.
Answer:
the third one
Step-by-step explanation:
the absolute value of -6 is 6