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Black_prince [1.1K]
2 years ago
14

Un avion sale de Córdoba con destino a Estados Unidos realiza escalas en Chile y en Bolivia la capacidad del avión es de 507 pas

ajeros al salir del aeropuerto tiene 119 asientos vacíos en Chile bajan 33 pasajeros y suben 121 en Bolivia bajan 129 y suben 39 ¿Con cuántos pasajeros aterrizará en Estados Unidos
Mathematics
1 answer:
Dahasolnce [82]2 years ago
7 0

Answer:

386 passengers arrive at Estados Unidos.

386 pasajeros aterrizará en Estados Unidos.

Step-by-step explanation:

English Translation

A plane from Córdoba bound for Estados Unidos and makes stops in Chile and Bolivia. The plane has a capacity for 507 passengers. He leaves the Córdoba airport with 119 empty seats. In Chile 33 passengers go down and 121 go up. In Bolivia 129 go down and 39 go up. How many arrive in Estados Unidos?

- The plane has capacity for 507 passengers.

- 119 seats are empty when it leaves Córdoba.

Meaning that there were (507 - 119) passengers on board = 388 passengers

- In Chile 33 passengers go down and 121 go up.

Number of passenger become

388 - 33 + 121 = 476 passengers

- In Bolivia 129 go down and 39 go up.

Number of passengers become

476 - 129 + 39 = 386 passengers.

Hope this Helps!!!!

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3 years ago
If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the directional derivative of f at (2, 4, 0) in the direction
valentina_108 [34]

Answer:

a) \nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}.

b) Du_{f}(2,4,0) = -\frac{8}{\sqrt{11}}

Step-by-step explanation:

Given a function f(x,y,z), this function has the following gradient:

\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}.

(a) find the gradient of f

We have that f(x,y,z) = x\sin{yz}. So

f_{x}(x,y,z) = \sin{yz}

f_{y}(x,y,z) = xz\cos{yz}

f_{z}(x,y,z) = xy \cos{yz}.

\nabla f(x,y,z) = f_{x}(x,y,z)\mathbf{i} + f_{y}(x,y,z)\mathbf{j} + f_{z}(x,y,z)\mathbf{k}.

\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.

The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.

We have that:

\nabla f(x,y,z) = \sin{yz}\mathbf{i} + xz\cos{yz}\mathbf{j} + xy \cos{yz}\mathbf{k}

\nabla f(2,4,0) = \sin{0}\mathbf{i} + 0\cos{0}\mathbf{j} + 8 \cos{0}\mathbf{k}.

\nabla f(2,4,0) = 0i+0j+8k=(0,0,8)

The vector is v = i + 3j - k = (1,3,-1)

To use v as an unitary vector, we divide each component of v by the norm of v.

|v| = \sqrt{1^{2} + 3^{2} + (-1)^{2}} = \sqrt{11}

So

v_{u} = (\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}})

Now, we can calculate the scalar product that is the directional derivative.

Du_{f}(2,4,0) = (0,0,8).(\frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}) = -\frac{8}{\sqrt{11}}

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