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katrin [286]
3 years ago
8

Which statements are true about the ordered pair (7, 19) and the system of equations? {2x−y=−5x+3y=22 Select each correct answer

. The ordered pair (7, 19) is a solution to the first equation because it makes the first equation true. The ordered pair (7, 19) is a solution to the second equation because it makes the second equation true. The ordered pair (7, 19) is not a solution to the system because it makes at least one of the equations false. The ordered pair (7, 19) is a solution to the system because it makes both equations true.
Mathematics
2 answers:
IRISSAK [1]3 years ago
7 0
(7,19) is a solution for only the first equation (2x-y= -5), so it is not a solution to the system.
2x-y=-5 we substitute the x and y for 7, and 19 making the equation 2 x 7-19, 2 x 7 is 14 so we then put the 14 and 19 together, 14-19, and 14-19 is equal to -5 so the first equation is true. For the second equation which is x+3y=22 we do the same thing as the first equation, we substitute x and y for 7, and 19, substituting these numbers in we get 7+3 x 19, 3 x 19 is 57, so 7+57, which equals 64 and is no where near 22. So, the ordered pair (7,19) is a solution to the first equation because, it makes the first equation true, and the ordered pair (7,19) is not a solution to the system because, it makes at least one of the equations false.


Cloud [144]3 years ago
5 0

this would be your only answer

The ordered pair  (7, 19) is a solution to the first equation because it makes the first equation true.

hope this helps

i did the test so i'm sure

just saying:)

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Looks like we have

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\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(z^2x)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial z}=z^2+y^2+x^2

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Then by the divergence theorem,

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(x^2+y^2+z^2)\,\mathrm dV

Compute the integral in spherical coordinates, setting

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

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\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=-u\,\vec k

Then we have

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1\left(\frac{u^3}3\sin^3v\,\vec\jmath+u^2\sin^2v\,\vec k\right)\times(-u\,\vec k)\,\mathrm du\,\mathrm dv

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