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geniusboy [140]
3 years ago
12

7. A student can type 174 words in 3 min and 348 words in 6 mon. how many words can the student type in 8 min?

Mathematics
1 answer:
julsineya [31]3 years ago
8 0
I think D ..........................
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Write 3 numbers that make this inequality true: x>4​
Trava [24]

Answer:

8, 10, 24

(Basically every number above 4)

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20℅ of your employees work part time What fraction of the employees work part time
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1/5
A fifth of your employees ..
4 0
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0.0176165803 is estimated to what
bezimeni [28]
It most common to round to the hundredths place, this is two spaces after the decimal. 

So look two spaces right - 0.17. If the third digit (in this case 6) is higher than 5 you round the number up if its not, keep it the same. 

"6" is greater than five, so round it to about (or approximately) 0.18.
5 0
3 years ago
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
emmasim [6.3K]
\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
5 0
3 years ago
What is the greatest value of 2,463.9051
RoseWind [281]

Hey There @Bre18016,

The answer is \boxed{2}

The greatest value of 2,463.9051 would be the thousands place (2) simply as it is the biggest number out of the other places.

For instance, if we had the number 300, 3 would be the greatest value.

Or let's say we had 10,000 the 1 would be the greatest value.

Furthermore, you could look at the first digit in the entire number to deter mine the greatest value.

6 0
2 years ago
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