What this question is asking us is
A) From -40C to 5 0c How much did it rise
The formula for this would be
Risen Temperature - Initial Temperature
The risen temparauture is 5 so we plug the 5 in first then the 4 so it would look like
5-(-4)=9
And remeber when you have 2 negatives it becomes a positie, 5+4=9
The asnwer for A is 9 degress
B) The tempurater then fell by 7 degress from 5 so we just have to do
5 - 7
which is -2 Celcius
Answer:
a,c,b
Step-by-step explanation:
If (-1, -1) is an extremum of
, then both partial derivatives vanish at this point.
Compute the gradients and evaluate them at the given point.
![\nabla f = \left\langle y - \dfrac1{x^2}, x - \dfrac1{y^2}\right\rangle \implies \nabla f (-1,-1) = \langle-2,-2\rangle \neq \langle0,0,\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Cleft%5Clangle%20y%20-%20%5Cdfrac1%7Bx%5E2%7D%2C%20x%20-%20%5Cdfrac1%7By%5E2%7D%5Cright%5Crangle%20%5Cimplies%20%5Cnabla%20f%20%28-1%2C-1%29%20%3D%20%5Clangle-2%2C-2%5Crangle%20%5Cneq%20%5Clangle0%2C0%2C%5Crangle)
![\nabla f = \langle 2x+2,0\rangle \implies \nabla f(-1,-1) = \langle0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Clangle%202x%2B2%2C0%5Crangle%20%5Cimplies%20%5Cnabla%20f%28-1%2C-1%29%20%3D%20%5Clangle0%2C0%5Crangle)
![\nabla f = \langle y, x-2y\rangle \implies \nabla f(-1,1) = \langle-1,1\rangle \neq\langle0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Clangle%20y%2C%20x-2y%5Crangle%20%5Cimplies%20%5Cnabla%20f%28-1%2C1%29%20%3D%20%5Clangle-1%2C1%5Crangle%20%5Cneq%5Clangle0%2C0%5Crangle)
![\nabla f = \left\langle y + \frac1{x^2}, x + \frac1{y^2}\right\rangle \implies \nabla f(-1,1) = \langle0,0\rangle](https://tex.z-dn.net/?f=%5Cnabla%20f%20%3D%20%5Cleft%5Clangle%20y%20%2B%20%5Cfrac1%7Bx%5E2%7D%2C%20x%20%2B%20%5Cfrac1%7By%5E2%7D%5Cright%5Crangle%20%5Cimplies%20%5Cnabla%20f%28-1%2C1%29%20%3D%20%5Clangle0%2C0%5Crangle)
The first and third functions drop out.
The second function depends only on
. Compute the second derivative and evaluate it at the critical point
.
![f(x,y) = x^2+2x \implies f'(x) = 2x + 2 \implies f''(x) = 2 > 0](https://tex.z-dn.net/?f=f%28x%2Cy%29%20%3D%20x%5E2%2B2x%20%5Cimplies%20f%27%28x%29%20%3D%202x%20%2B%202%20%5Cimplies%20f%27%27%28x%29%20%3D%202%20%3E%200)
This indicates a minimum when
. In fact, since this function is independent of
, every point with this
coordinate is a minimum. However,
![x^2 + 2x = (x + 1)^2 - 1 \ge -1](https://tex.z-dn.net/?f=x%5E2%20%2B%202x%20%3D%20%28x%20%2B%201%29%5E2%20-%201%20%5Cge%20-1)
for all
, so (-1, 1) and all the other points
are actually <em>global</em> minima.
For the fourth function, check the sign of the Hessian determinant at (-1, 1).
![H(x,y) = \begin{bmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} -2/x^3 & 1 \\ 1 & -2/y^3 \end{bmatrix} \implies \det H(-1,-1) = 3 > 0](https://tex.z-dn.net/?f=H%28x%2Cy%29%20%3D%20%5Cbegin%7Bbmatrix%7D%20f_%7Bxx%7D%20%26%20f_%7Bxy%7D%20%5C%5C%20f_%7Byx%7D%20%26%20f_%7Byy%7D%20%5Cend%7Bbmatrix%7D%20%3D%20%5Cbegin%7Bbmatrix%7D%20-2%2Fx%5E3%20%26%201%20%5C%5C%201%20%26%20-2%2Fy%5E3%20%5Cend%7Bbmatrix%7D%20%5Cimplies%20%5Cdet%20H%28-1%2C-1%29%20%3D%203%20%3E%200)
The second derivative with respect to
is -2/(-1) = 2 > 0, so (-1, -1) is indeed a local minimum.
The correct choice is the fourth function.
Answer:
The reflection of a horizontal line has the same result as a rotation of 90 degrees clockwise.
Step-by-step explanation:
When an image is rotated 90 degrees counterclockwise the shape is still congruent but in a different quadrant. The same is true when an image is reflected across a horizontal line.
Answer: x=-1
Step-by-step explanation: