Answer:
58 minutes
Step-by-step explanation:
Kelvin and Ornella completed the course in 1 hour 52 minutes
60 minutes + 52 minutes
= 112 minutes
Kelvin time is 54 minutes
Kelvin time is 4 minutes less than ornella
Hence Ornella time is 58 minutes
Answer:
You are differentiating with respect to x
with the formula X^n = nX^n-1
dy/dx= 12x2 - 24x + 14
Answer:
The given function is differentiable at y = 1.
At y = 1, f'(z) = 0
Step-by-step explanation:
As per the given question,
![f(z)\ = (x^{2}+y^{2}-2y)+i(2x - 2xy)](https://tex.z-dn.net/?f=f%28z%29%5C%20%3D%20%28x%5E%7B2%7D%2By%5E%7B2%7D-2y%29%2Bi%282x%20-%202xy%29)
Let z = x + i y
Suppose,
![u(x,y) = x^{2}+y^{2}-2y](https://tex.z-dn.net/?f=u%28x%2Cy%29%20%3D%20x%5E%7B2%7D%2By%5E%7B2%7D-2y)
![v(x,y) = 2x - 2xy](https://tex.z-dn.net/?f=v%28x%2Cy%29%20%3D%202x%20-%202xy)
On computing the partial derivatives of u and v as:
![u'_{x} =2x](https://tex.z-dn.net/?f=u%27_%7Bx%7D%20%3D2x)
![u'_{y}=2y -2](https://tex.z-dn.net/?f=u%27_%7By%7D%3D2y%20-2)
And
![v'_{x} =2-2y](https://tex.z-dn.net/?f=v%27_%7Bx%7D%20%3D2-2y)
![v'_{y}=-2x](https://tex.z-dn.net/?f=v%27_%7By%7D%3D-2x)
According to the Cauchy-Riemann equations
![u'_{x} =v'_{y} \ \ \ \ \ \ \ and\ \ \ \ \ \ u'_{y} = -v'_{x}](https://tex.z-dn.net/?f=u%27_%7Bx%7D%20%3Dv%27_%7By%7D%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20and%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20u%27_%7By%7D%20%3D%20-v%27_%7Bx%7D)
Now,
![(u'_{x} =2x) \neq (v'_{y}=-2x)](https://tex.z-dn.net/?f=%28u%27_%7Bx%7D%20%3D2x%29%20%5Cneq%20%20%28v%27_%7By%7D%3D-2x%29)
![(u'_{y}=2y -2) \ = \ (- v'_{x} =-(2-2y) =2y-2)](https://tex.z-dn.net/?f=%28u%27_%7By%7D%3D2y%20-2%29%20%5C%20%3D%20%5C%20%28-%20v%27_%7Bx%7D%20%3D-%282-2y%29%20%3D2y-2%29)
Therefore,
holds only.
This means,
2y - 2 = 0
⇒ y = 1
Therefore f(z) has a chance of being differentiable only at y =1.
Now we can compute the derivative
![f'(z)=\frac{1}{2}[(u'_{x}+iv'_{x})-i(u'_{y}+iv'_{y})]](https://tex.z-dn.net/?f=f%27%28z%29%3D%5Cfrac%7B1%7D%7B2%7D%5B%28u%27_%7Bx%7D%2Biv%27_%7Bx%7D%29-i%28u%27_%7By%7D%2Biv%27_%7By%7D%29%5D)
![f'(z) =\frac{1}{2}[(2x+i(2-2y))-i(2y-2+i(-2x))]](https://tex.z-dn.net/?f=f%27%28z%29%20%3D%5Cfrac%7B1%7D%7B2%7D%5B%282x%2Bi%282-2y%29%29-i%282y-2%2Bi%28-2x%29%29%5D)
![f'(z) = i(2-2y)](https://tex.z-dn.net/?f=f%27%28z%29%20%3D%20i%282-2y%29)
At y = 1
f'(z) = 0
Hence, the required derivative at y = 1 , f'(z) = 0
P
=
12
5 this is what I got for that
Answer:
x = - 4
Step-by-step explanation:
Given a parabola in standard form
f(x) = ax² + bx + c ( a ≠ 0 )
Then the equation of the axis of symmetry is
x = - ![\frac{b}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7Bb%7D%7B2a%7D)
f(x) = 2x² + 16x - 19 ← is in standard form
with a = 2, b = 16
Then the equation of the axis of symmetry is
x = -
= - 4