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a_sh-v [17]
1 year ago
6

Factor −6x^2 + 18x.

Mathematics
1 answer:
snow_lady [41]1 year ago
5 0

Answer:

6x(-x+3)

Step-by-step explanation:

-6x^2+18x

factor out 6x

6x(-x^2+3)

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The bin's volume is 36cm³
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What is the slope of a line that is perpendicular to the line who’s equation is y=4/5x-3
alexandr402 [8]

This line y = 4/5 x - 3 has slope = 4/5

Perpendicular lines, slope is opposite and reciprocal so slope of a line that is perpendicular to the line y = 4/5 x - 3 is -5/4

Answer

- 5/4

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3 years ago
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What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
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\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

So, the flux is given by

\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA
\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du
\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv
\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv

where t=-\sin v in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi
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3 years ago
Which of the following inequalities is graphed below?
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Im dum but please help me<br>​
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Answer:

336

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