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Ilya [14]
3 years ago
11

0.0007 in standard form

Mathematics
2 answers:
Flauer [41]3 years ago
7 0

Answer:

very simple 0.0007 = 7 / 10000 lol

Step-by-step explanation:

hope this is helpful!

zhannawk [14.2K]3 years ago
5 0
—————>>>>>>>. 7x10^-4
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A=1/2h(a+b) <br> what is h
Ilya [14]

Answer:

h = \frac{2 A}{(a + b)}

Step-by-step explanation:

A =  \frac{1}{2} h(a + b) \\  \\2 A =   h(a + b) \\  \\  \frac{2 A}{(a + b)}  = h \\  \\  \huge \red{ \boxed{h = \frac{2 A}{(a + b)} }}

6 0
4 years ago
I need a little more help ok
Elenna [48]

Answer:

84+(x+9)=180

Step-by-step explanation:

84º+(x+9)º = 180º, so x+93 = 180, so x=87. Hope this helps!

6 0
3 years ago
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fredd [130]

Answer:

A

Step-by-step explanation:

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6 0
3 years ago
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Calculus find the functions and state the domain
Sergio039 [100]

Answer:

check image below

Step-by-step explanation:

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8 0
3 years ago
Et f(x, y, z) = z tan−1(y2)i + z3 ln(x2 + 5)j + zk. find the flux of f across s, the part of the paraboloid x2 + y2 + z = 11 tha
nignag [31]
Since the surface is closed, and the vector field is rather complicated, you can use the divergence theorem. The flux of \mathbf f(x,y,z) across S is given by a surface integral, which the divergence theorem asserts is equivalent to a volume integral:

\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\iiint_R(\nabla\cdot\mathbf f)\,\mathrm dV

where R denotes the space with boundary S. We have

\nabla\cdot\mathbf f(x,y,z)=\dfrac{\partial z\tan^{-1}(y^2)}{\partial x}+\dfrac{\partial z^3\ln(x^2+5)}{\partial y}+\dfrac{\partial z}{\partial z}=0+0+1=1

So in fact the flux across S happens to be equal (in magnitude) to the volume encased by S.

\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\int_{x=-3}^{x=3}\int_{y=-\sqrt{9-x^2}}^{y=\sqrt{9-x^2}}\int_{z=2}^{z=11-x^2-y^2}\mathrm dz\,\mathrm dy\,\mathrm dx

Convert to cylindrical coordinates, setting

\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta

\displaystyle\iint_S\mathbf f\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\int_{\zeta=2}^{\zeta=11-r^2}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=3}r(11-r^2-2)\,\mathrm dr
=\displaystyle2\pi\int_{r=0}^{r=3}(9r-r^3)\,\mathrm dr=\dfrac{81\pi}2
8 0
3 years ago
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