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3 years ago

All my question are about math and this test determines my high school classes help if you can please

Mathematics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

y = x+2

Step-by-step explanation:

When x = 0, y =2

So, y = 0 + m

2 = 0 + m

2 = m

When x = 1, y =3

y = x + m

3 = x + 2

-2 -2

1 = x

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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

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

$=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0$

$\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}$

8 0

4 years ago

Maria's plant grew 3 1/4 inches every month for 2 1/2 months. How many inches did Maria's plant grow during that time?

Vadim26 [7]

Answer: 8.125 inches

Step-by-step explanation: Ok so if the plant grew 3.25 inches every month, and there were 2.5 months, you need to multiply 3.25 by 2.5 to get the total amount of growth. So that would be 8.125

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3 years ago

Fred is making a bouquet of carnations and roses. The carnations cost $5.25 in all. The roses cost $1.68 each. How many roses di

olga nikolaevna [1]

Answer:

8 roses

Step-by-step explanation:

First start by subtracting $18.69 by $5.25 because you already know how much the carnations cost and how much the bouquet cost in all. When you subtract 18.69 by 5.25 you should get $13.44. If each rose cost $1.68 then divide 13.44 by 1.68 to get the amount of roses in the bouquet.

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3 years ago