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4 years ago

0.81>0.094 is it true

Mathematics

2 answers:

-Dominant- [34]4 years ago

8 0

It is true. Hope it help!

Firlakuza [10]4 years ago

4 0

-My answer-
No it is not true in actuality it should be like this 0.094 > 0.81.

Glad to help you If you have any questions about brainly let me know so I can help okay :)

Sincerely,
teenwolfgirlxD

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Circle P has a circumference of approximately 75 inches. what is the approximate length of the radius, r? use 3.24 for pi. round

zloy xaker [14]

Answer:11.9 inches

Step-by-step explanation:

Circumference(c)=75inches

Pi=3.14

Circumference(c)=2 x π x radius(r)

75=2 x 3.14 x r

75=6.28 x r

Divide both sides by 6.28

75 ➗ 6.28=(6.28 x r) ➗ 6.28

11.9=r

Therefore radius=11.9 inches

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3 years ago

What fractions are equivalent to 1/2

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3 years ago

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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

Cerrena [4.2K]

$\mathbf F(x,y,z)=z\tan^{-1}(y^2)\,\mathbf i+z^3\ln(x^2+3)\,\mathbf j+z\,\mathbf k$
$\implies\mathrm{div}\mathbf F(x,y,z)=0+0+1=1$

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 $(128-32)\pi=96\pi$ .

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3 years ago

Help me with my math

shutvik [7]

Answer:

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3 years ago