Answer:
80 hope this helps!
Step-by-step explanation:
We are given the function <span>f(x)=sqrt of (4sinx+2) and is asked to find the first derivative of the function when x is equal to zero.
</span><span>f(x)=sqrt of (4sinx+2)
f'(x) = 0.5 </span><span>(4sinx+2) ^ -0.5 * (4cosx)
</span>f'(0) = 0.5 <span>(4sin0+2) ^ -0.5 * (4cos0)
</span>f'(0) = 0.5 <span>(0+2) ^ -0.5 * (4*1)
</span>f'(x) = 0.5 (2) ^ -0.5 * (4)
f'(x) = -.1.65
Solution: Let be the number of birds and be the number of cats.
Therefore, there were 14 birds at the shelter on Friday
Answer:
ASA
Step-by-step explanation:
Because I need trust in my life
Answer:
See steps below
Step-by-step explanation:
The function
is a particular case of the general Rosenbrock’s Function.
a)
Since
for all the values of x and equals 0 when x=1 and
for all the values of y and equals 0 only when y=1, we conclude that (1,1) is a minimum.
On the other hand,
f(x,y)>0 for (x,y) ≠ (1,1) so (1,1) is a global minimum.
b)
To confirm that this function is not convex, we will be using the following characterization of convexity
“f is convex if, and only if, the Laplace operator of f for every (x,y) in the domain of f”
Given that the domain of f is the whole plane XY, in order to prove that f is not convex, we must find a point (x,y) at where the Laplace operator is < 0.
The Laplace operator is given by
Let us compute the partial derivatives
and
we have then
if we take (x,y) = (0,1)
hence f is not convex.