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3 years ago

HELP ASAP

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

7 0

Answer

$\boxed{5 \sqrt{2} \: \: \:units}$

Step by step explanation

Let the points be A and B

A ( - 3 , - 4 ) ⇒( x₁ , y₁ )

B ( 4 , - 5 )⇒( x₂ , y₂ )

Now, let's find the distance between these points :

Distance = $\mathsf{ \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } }$

Plug the values

⇒ $\mathsf{ \sqrt{ {(4 - ( - 3))}^{2} + {( - 5 - ( - 4))}^{2} } }$

Calculate

⇒ $\mathsf{ \sqrt{ {(4 + 3)}^{2} + {( - 5 + 4)}^{2} } }$

⇒ $\mathsf{ \sqrt{ {(7)}^{2} + {( - 1)}^{2} } }$

Evaluate the power

⇒ $\mathsf{ \sqrt{49 + 1} }$

Add the numbers

⇒ $\mathsf{ \sqrt{50} }$

Simplify the radical expression

⇒ $\mathsf{5 \sqrt{2} \: \: units}$

Hope I helped!

Best regards!!

Nadya [2.5K]3 years ago

5 0

Answer:

d=5√2 unit

Step-by-step explanation:

distance between two points d=√(x2-x1)²+(y2-y1)²

two pints (-3,-4) and (4,-5)

d=√(4-(-3)²+(-5-(-4)²

d=√(4+3)²+(-5+4)²

d=√49+1

d=√50

d=√25*2

d=5√2

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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

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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

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butalik [34]

$\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k$

Let

$\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k$

The curl is

$\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)$

where $\partial_\xi$ denotes the partial derivative operator with respect to $\xi$ . Recall that

$\vec\imath\times\vec\jmath=\vec k$

$\vec\jmath\times\vec k=\vec i$

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The cross product reduces to

$\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k$

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

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$4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}$

$\implies g(x,z)=4\sin(xz^2)+h(z)$

Now

$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)$

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$2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}$

$\dfrac{\mathrm dh}{\mathrm dz}=0$

$\implies h(z)=C$

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$f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C$

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