Answer:
<h2><u><em>
3.5 Km</em></u></h2>
Step-by-step explanation:
the length of the hypotenuse is 4 Km, it is probable that you are looking for the value of the cathetus a.
it is a right triangle and we use Pythagoras
a² = 4² - 2²
a² = 16 - 4
a² = 12
a = √12
a = 3.46 (round 3.5)
X/y = 2 so x = 2y
z/x = 4 so z = 4x but x = 2y so z = 4(2y) = 8y
now you have
x = 2y, z = 8y
x + y + z 2y + y + 8y 11y 11
------------ = -----------------------= ------------- = ----------- = 5 1/2
x 2y 2y 2
answer
C. 5 1/2
Answer:
$690
Step-by-step explanation:
The amount she makes in a week=her weekly salary + the commission on sales.
Amount=330 + 7.5% of 4800.
Amount = 330+ 360=690.
The she would make if she sold $4,800 of merchandise is $690.
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
Answer:
Expenditure x (in hundreds of dollars) = $2,000
Step-by-step explanation:
Given:
Profit p (in hundreds of dollars)
Expenditure x (in hundreds of dollars)
p(x) = -0.5x² + 20x + 23
Find:
Expenditure for advertising .
Computation:
x = -(Cofficient of x) / 2(Cofficient of x²)
x = -(20) / 2(-0.5)
x = 20
Expenditure x (in hundreds of dollars) = 20 × $100
Expenditure x (in hundreds of dollars) = $2,000
