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3 years ago

Y=x+4 y=-2x-2 explain the awnser plz

Mathematics

1 answer:

Sliva [168]3 years ago

6 0

Answer:

x = -2 and y = 2

Step-by-step explanation:

We have the following system of linear equations;

y=x+4

y=-2x-2

To solve the above system, we shall be equating the right hand sides of the two equations since we have y's on the left hand side of both equations;

x + 4 = -2x - 2

add 2x on both sides of the equation;

x + 4 + 2x = -2x - 2 + 2x

3x + 4 = -2

subtract 4 on both sides of the equation;

3x + 4 - 4 = -2 - 4

3x = -6

x = -2

We can use the first equation y=x+4 to determine the value of y;

y = -2 + 4

y = 2

The solution to the system of linear equations is thus;

x = -2 and y = 2

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10 months ago

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3 years ago

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ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$