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3 years ago

Help me below.........................................

Mathematics

2 answers:

bagirrra123 [75]3 years ago

7 0

−12+6h+4+−3hCombine Like Terms:=−12+6h+4+−3h=(6h+−3h)+(−12+4)= 3h+−8
Answer: =3h - 8

Afina-wow [57]3 years ago

6 0

-12+6h+4-3h = -8 + 3h

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Which numbers are solutions to the inequality -3x-3<6? Check all that apply

bogdanovich [222]

Answer:

0, 2, and 4.

Step-by-step explanation:

-3x-3≤6

Add 3 on both sides of the inequality.

-3x≤9

Divide both sides of the inequality by -3.

x≥3

The solutions possible for x can be 0, 2, and 4.

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3 years ago

7) y = 5x +3, y = 2x +21 What is the equal value method for this equation?

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5x+3=2x+21

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4 years ago

Solve for the inverse g(x) = 2x^2-6/9

oksian1 [2.3K]

Answer:f(x)=x2−6x−9 f ( x ) = x 2 - 6 x - 9. Replace f(x) f ( x ) with y y . y=x2−6x−9 y = x 2 - 6 x - 9. Interchange the variables. x=y2−6y−9 x = y 2 - 6 y - 9.

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3 years ago

What is the radius of the circle who’s equation is x^2 + y^2 = 9

Svetllana [295]

Answer:

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4 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

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