Answer:
Step-by-step explanation:
a^2 + b^2 = c^2
where
a and b are the legs
and
c is the hypotenuse
.
the problem gives the "legs" because the LONGEST leg is always the hypotenuse.
a^2 + b^2 = c^2
1071^2 + 1840^2 = c^2
1147041 + 3385600 = c^2
4532641 = c^2
Answer:
-17y + 16x
Step-by-step explanation:
-3(3y - 2x) + 2(5x - 4y) = -9y + 6x + 10x - 8y = -17Y + 16X.
If it is helpful, plz give me Brainliest.
Answer:
-2/7
Step-by-step explanation:
After the new test, she now has taken 8 tests.
Set it up just like when finding the average:
(87+79)/8
Solve for the new test average
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
