Answer:
-7/27
Step-by-step explanation:
-14/54
Hope this helped :)
The image isn’t loading for me , try reuploading
Answer:
+8, -4, +2
Step-by-step explanation:
The pattern rule is every other number is negative and it alternates. It would be plus, minus, plus, minus, and keep that going. The second part to the pattern rule is the next number is half of the previous. So 32 was half of 64, 16 was half of 32 so now 8 is half of 16, 4 is half of 8, and 2 is half of 4. Keep the + - + - + - pattern too.
342-72=270
270-72=198
198/2=99
Therefore the width is 99
The <em>directional</em> derivative of 
at the given point in the direction indicated is 
.
<h3>How to calculate the directional derivative of a multivariate function</h3>
The <em>directional</em> derivative is represented by the following formula:
(1)
Where:
- Gradient evaluated at the point 
.
- Directional vector.
The gradient of is calculated below:
![\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial r}(r_{o},s_{o}) \\\frac{\partial f}{\partial s}(r_{o},s_{o}) \end{array}\right]](https://tex.z-dn.net/?f=%5Cnabla%20f%20%28r_%7Bo%7D%2C%20s_%7Bo%7D%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20r%7D%28r_%7Bo%7D%2Cs_%7Bo%7D%29%20%20%5C%5C%5Cfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20s%7D%28r_%7Bo%7D%2Cs_%7Bo%7D%29%20%5Cend%7Barray%7D%5Cright%5D)
(2)
Where 
and 
are the <em>partial</em> derivatives with respect to 
and 
, respectively.
If we know that 
, then the gradient is:
![\nabla f(r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{s}{1+r^{2}\cdot s^{2}} \\\frac{r}{1+r^{2}\cdot s^{2}}\end{array}\right]](https://tex.z-dn.net/?f=%5Cnabla%20f%28r_%7Bo%7D%2C%20s_%7Bo%7D%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7Bs%7D%7B1%2Br%5E%7B2%7D%5Ccdot%20s%5E%7B2%7D%7D%20%5C%5C%5Cfrac%7Br%7D%7B1%2Br%5E%7B2%7D%5Ccdot%20s%5E%7B2%7D%7D%5Cend%7Barray%7D%5Cright%5D)
![\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{1+1^{2}\cdot 3^{2}} \\\frac{1}{1+1^{2}\cdot 3^{2}} \end{array}\right]](https://tex.z-dn.net/?f=%5Cnabla%20f%20%28r_%7Bo%7D%2C%20s_%7Bo%7D%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B3%7D%7B1%2B1%5E%7B2%7D%5Ccdot%203%5E%7B2%7D%7D%20%5C%5C%5Cfrac%7B1%7D%7B1%2B1%5E%7B2%7D%5Ccdot%203%5E%7B2%7D%7D%20%5Cend%7Barray%7D%5Cright%5D)
![\nabla f (r_{o}, s_{o}) = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right]](https://tex.z-dn.net/?f=%5Cnabla%20f%20%28r_%7Bo%7D%2C%20s_%7Bo%7D%29%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B3%7D%7B10%7D%20%5C%5C%5Cfrac%7B1%7D%7B10%7D%20%5Cend%7Barray%7D%5Cright%5D)
If we know that 
, then the directional derivative is:
![\nabla_{\vec v} f = \left[\begin{array}{cc}\frac{3}{10} \\\frac{1}{10} \end{array}\right] \cdot \left[\begin{array}{cc}5\\10\end{array}\right]](https://tex.z-dn.net/?f=%5Cnabla_%7B%5Cvec%20v%7D%20f%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B3%7D%7B10%7D%20%5C%5C%5Cfrac%7B1%7D%7B10%7D%20%5Cend%7Barray%7D%5Cright%5D%20%5Ccdot%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D5%5C%5C10%5Cend%7Barray%7D%5Cright%5D)
The <em>directional</em> derivative of at the given point in the direction indicated is . 
To learn more on directional derivative, we kindly invite to check this verified question: brainly.com/question/9964491
