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makvit [3.9K]
3 years ago
12

The sale price of a bike is $189.50. The sale tax is 5%. What is the total price?

Mathematics
2 answers:
omeli [17]3 years ago
6 0
It is 198.97. So the first thing you have to do is multiply 189.50 by .05. 
Then you add that to 189.50 and thats how you get 198.97


Roman55 [17]3 years ago
3 0
It would be $947.50 because you would multiply $189.50 by 0.05.
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A company invests $10,000 in a mutual fund account that earns a fixed
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Answer:

a(t)=10000+1.08t

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3 years ago
F(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 3)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 18 that l
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\mathbf F(x,y,z)=z\tan^{-1}(y^2)\,\mathbf i+z^3\ln(x^2+3)\,\mathbf j+z\,\mathbf k
\implies\mathrm{div}\mathbf F(x,y,z)=0+0+1=1

By the divergence theorem, the flux of \mathbf F across the *closed* surface \mathcal S combined with the plane z=2 is given by a volume integral over the closed region:

\displaystyle\iint_{\mathcal S}\mathbf F\cdot\mathrm d\mathbf S=\iiint_{\mathcal R}\nabla\cdot\mathbf F\,\mathrm dV

So in fact, to find the flux over \mathcal S alone, we'll need to subtract the flux of \mathbf F over the planar portion, oriented outward. First, compute the volume integral by converting to cylindrical coordinates:

x^2+y^2+z=18
z=2\implies x^2+y^2=16\implies r^2=16\implies r=4

\displaystyle\iiint_{\mathcal R}\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}\int_{z=2}^{z=18-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=128\pi

If the surface does actually contain z=2, then you can stop here; otherwise, continue.

Now, parameterize the part of the *closed* surface in z=2 by

\mathbf s(r,\theta)=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j+2\,\mathbf k

where 0\le r\le4 and 0\le\theta\le2\pi. We get a surface element

\mathrm d\mathbf S=(\mathbf s_r\times\mathbf s_\theta)\,\mathrm dr\,\mathrm d\theta=(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta

We don't need to worry about the first two components of

and so the surface integral over this region is

\displaystyle\iint_{z=2\,\land\,x^2+y^2\le16}\mathbf F\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=4}2r\,\mathrm dr\,\mathrm d\theta=32\pi

Then the total flux over \mathcal S alone is (128-32)\pi=96\pi.
4 0
3 years ago
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DochEvi [55]

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Answer:

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