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borishaifa [10]
3 years ago
13

Nzmxmhxmxnmxhsk6xmxhxmx​

Mathematics
1 answer:
serg [7]3 years ago
5 0

Answer:

Yeah

Step-by-step explanation:

Hope that helped

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What is the volume of the triangular prism?
nekit [7.7K]
12 if not just look it up on quiz lit and you’ll get it it’s on safari or use scan and solve
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3 years ago
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Solve for a. a + 16 = 20 Enter your answer in the box. a =
Gala2k [10]
A=4 hope this helped
6 0
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Jasmine had a carrot cake with cream cheese frosting for her birthday .she and her friend each ate 1/4 of the cake and jasmine's
zloy xaker [14]
Hey there!

To solve this problem, you first would need to solve for how much of the cake is left after Jasmine, her friend, and Jasmine's brother got their slices. Then, you would need to split the remaining amount in half to get the answer of what each parent ate. We would need to make all of the denominators the same, so let's make all of the fractions out of 12, a common denominator of 3 and 4. 

\frac{12}{12} - \frac{1}{4} - \frac{1}{4} - \frac{1}{3}
\frac{12}{12} - \frac{3}{12} - \frac{3}{12} - \frac{4}{12}
\frac{12}{12} - \frac{10}{12} = \frac{2}{12}

It can then be concluded that each parent ate \frac{1}{12} of the cake, since there was \frac{2}{12} remaining. 

You can check this answer by adding up all of the parts we used in this problem:

\frac{3}{12} + \frac{3}{12} + \frac{4}{12} +  \frac{1}{12} +  \frac{1}{12} = \frac{12}{12}

Hope this helped you out! :-)
3 0
3 years ago
How much more interest is earned on an investment of $12,585 at 3.5% interest over 5 years if the interest is compounded annuall
Nina [5.8K]

Answer:

It is B.$159.65


Step-by-step explanation:

I got it the test

7 0
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
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
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