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2 years ago

Convert the following to slope intecept form, 4x+6y=−12

Mathematics

2 answers:

dolphi86 [110]2 years ago

7 0

6y=-4x-12
y = -2/3x -2

Dvinal [7]2 years ago

5 0

Answer:y= -2/3 x -2

Step-by-step explanation:

you find the slope of the line and... I can't remember I did this so long ago

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Saint Petersburg, Russia and Anchorage, Alaska, both have approximately the same latitude of 60 degrees. That means you can reac

user100 [1]

Answer:

Distance from Anchorage (longitude 150 degrees W) to Saint Petersburg (longitude 30 degrees E) = 10,002.47 KM

Step-by-step explanation:

As earth is a sphere of radius 6371 km.

Therefore at 60 degree latitude, the radius of circle can be calculated as shown below:

= 6371 * Cos 60

= 3185.5 KM

Saint Petersburg is nearly 180 degree about Anchorage by flying directly east.

So Distance from Anchorage (longitude 150 degrees W) to Saint Petersburg (longitude 30 degrees E) can be calculated as shown below:

= (180 * 2 * 3.1416 * 3185.5) / 360

= 10,002.47 KM

5 0

2 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

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