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3 years ago

I NEED HELP PLEASE, THANKS! :)

Mathematics

1 answer:

kompoz [17]3 years ago

8 0

Answer: C

<u>Step-by-step explanation:</u>

Use the multiplication rule for matrices - Multiply 1st row of first matrix with 1st column of second matrix and find their sum. Repeat for each row and column.

A x B

$\left[\begin{array}{cc}1&5\\-3&4\end{array}\right] \times \left[\begin{array}{cc}2&6\\6&-1\end{array}\right]=\left[\begin{array}{cc}1(2)+5(6)&1(6)+5(-1)\\-3(2)+4(6)&-3(6)+4(-1)\end{array}\right]\\\\\\.\qquad \qquad \qquad \qquad \qquad \quad =\left[\begin{array}{cc}32&1\\18&-22\end{array}\right]$

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A quantity and its 2/3 are added together and from thesum 1/3 of the sum is subtracted, and 10 remains.What is the quantity?

musickatia [10]

Answer:

$14\frac{1}{3}$

Step-by-step explanation:

Let's write this out as an equation. Let the unknown quantity be x:

$x+\frac{2}{3} - \frac{1}{3} (x+\frac{2}{3}) = 10\\\\3x+2-x -\frac{2}{3} = 30\\\\9x+6-3x-2=90\\\\6x= 86\\\\x = \frac{86}{6} =\frac{43}{3} =14\frac{1}{3} \\$

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Is 2/3 and 8/12 in equivalent ratio

VLD [36.1K]

Answer: if your question here is if they are the same since it says equivalent then yes

Step-by-step explanation: 2/3 = 8/12

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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

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