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2 years ago

Where do i graph? please send picture of your own for me to copy

Mathematics

1 answer:

melomori [17]2 years ago

7 0

Answer:

Step-by-step explanation:

Compound inequality has been given as,

12(2x - 10) + 24 ≥ 48 Or 6x + 15 - 9x ≥ 6

By simplifying it further,

12(2x - 10) + 24 ≥ 48

$\frac{12(2x - 10)}{12}+\frac{24}{12}\geq \frac{48}{12}$

(2x - 10) + 2 ≥ 4

(2x - 10) + 2 - 2 ≥ 4 - 2

2x - 10 ≥ 2

2x - 10 + 10 ≥ 2 + 10

2x ≥ 12

$\frac{2x}{2}\geq \frac{12}{2}$

x ≥ 6

6x + 15 - 9x ≥ 6

(6x - 9x) + 15 ≥ 6

(-3x) + 15 ≥ 6

-3x + 15 - 15 ≥ 6 - 15

-3x ≥ -9

3x ≤ 9

x ≤ 3

We have to plot the inequality on a number line

x ≥ 6 Or x ≤ 3

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3 years ago

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

tresset_1 [31]

Because I've gone ahead with trying to parameterize $S$ directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over $S$ straight away, let's close off the hemisphere with the disk $D$ of radius 9 centered at the origin and coincident with the plane $y=0$ . Then by the divergence theorem, since the region $S\cup D$ is closed, we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV$

where $R$ is the interior of $S\cup D$ . $\vec F$ has divergence

$\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z$

so the flux over the closed region is

$\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0$

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

$\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0$

$\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}$

Parameterize $D$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k$

with $0\le u\le9$ and $0\le v\le2\pi$ . Take the normal vector to $D$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$