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2 years ago

Simplify the radical of 10+10

Mathematics

2 answers:

Marianna [84]2 years ago

8 0

10+10=20 the simplified is 2

mina [271]2 years ago

3 0

It should be 20. I hope this helps!!

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What is the equation in slope-intercept form of a line that is perpendicular to y=13x+2 and passes through the point (3, 1)? Ent

geniusboy [140]

Answer:

y = -1/13x + 1 3/13

Step-by-step explanation:

y = 13x + 2

Perpendicular

y = -1/13x + b

Finding b:

1 = -1/13 * 3 + b

1 = -3/13 + b

1 3/13 = b

y = -1/13x + 1 3/13

Please let me know if I'm wrong :)

8 0

2 years ago

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Simplify the expression: 7–3p+4p

trapecia [35]

Answer:

7 + p

Step-by-step explanation:

Solve:

7 - 3p + 4p

Simplify using like terms

Wrote problem using addition

7 + ( -3p + 4p)

7 + 1p or 7 + p

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6 0

3 years ago

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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

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