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Leni [432]
3 years ago
15

Can sumbody help meh wit dis plz

Mathematics
2 answers:
babunello [35]3 years ago
7 0

Answer: B (5x - 4y)

Step-by-step explanation:

Well, let’s take a look at the like terms:

3x and 2x

<u>AND</u>

3y and 7y

So, now let’s solve it!

= 3x + 2x + 3y -7y

Lets add the 3x and 2x first! Keep in mind, it doesn’t matter which like terms are combined first!

= 5x + 3y - 7y

Now, let’s combine the 3y and 7y! Here, they are using subtraction! You can just leave the 5x there!

= 5x - 4y

That is your final answer!

fiasKO [112]3 years ago
3 0

Answer:

5y 4x

Step-by-step explanation:

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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

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