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3 years ago

When simplified completely, the product of a monomial and a monomial is _____ a monomial

Mathematics

2 answers:

Evgesh-ka [11]3 years ago

8 0

Answer: Always

Step-by-step explanation:

sukhopar [10]3 years ago

6 0

im gonna say always a monomial ...

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g Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 7yi + xzj + (

Sati [7]

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

where $S$ is any oriented surface with boundary $C$ . We have

$\vec F(x,y,z)=7y\,\vec\imath+xz\,\vec\jmath+(x+y)\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-7)\,\vec k$

Take $S$ to be the ellipse that lies in the plane $z=y+9$ with boundary on the cylinder $x^2+y^2=1$ . Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+9)\,\vec k$

with $0\le u\le1$ and $0\le v\le2\pi$ . Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k$

Then we have

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\big((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+2)\,\vec k\big)\cdot\big(-u\,\vec\jmath+u\,\vec k\big)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(3u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{3\pi}$

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Answer:

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Step-by-step explanation:

I think

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Step-by-step explanation:

Not sure

But Hope It Help

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Alja [10]

Im 100% sure that the answer is:
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