Let n be the number of sides of a polygon.
To solve for the sum of the interior angles:
(n-2) x 180°
To solve for the measure of each interior angle of the polygon
[(n-2) x 180°] / n
To solve for the measure of each exterior angle:
360° / n
The sum of the exterior angles is always 360°
See attachment for answers.
See, please, suggested decision, if it is possible check it.
Answer:
By calculating the time taken to reach we get to know that the The time taken by the other person is less than yourself
Step-by-step explanation:
Given that the other person and yourself left at the same time and the speed of the other person is 60 miles per hour and the speed of yourself is 40 miles per hour.
We know that the ![speed=\frac{distance}{time}](https://tex.z-dn.net/?f=speed%3D%5Cfrac%7Bdistance%7D%7Btime%7D)
Let the time taken to reach the destination at a distance of d by the other person is T1.
Then ![60=\frac{d}{T1}](https://tex.z-dn.net/?f=60%3D%5Cfrac%7Bd%7D%7BT1%7D)
![T1=\frac{d}{60}](https://tex.z-dn.net/?f=T1%3D%5Cfrac%7Bd%7D%7B60%7D)
Let the time taken to reach the destination at a distance of d by the yourself is T2.
Then ![40=\frac{d}{T2}](https://tex.z-dn.net/?f=40%3D%5Cfrac%7Bd%7D%7BT2%7D)
![T2=\frac{d}{40}](https://tex.z-dn.net/?f=T2%3D%5Cfrac%7Bd%7D%7B40%7D)
Now the time difference is = T2-T1 =
= ![-\frac{d}{120}](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%7D%7B120%7D)
Therefore T1 > T2 which means the time taken to reach the destination by yourself is greater than that of other person .
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.