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2 years ago

Need these 2 answered. 25 points! Please show work, Thanks! :)

Mathematics

2 answers:

adoni [48]2 years ago

5 0

Answer:

2(x+3)(x+5)
-4(x+2)(x-4)

Step-by-step explanation:

$2x^2+16x +30\\\\\mathrm{Factor\:out\:common\:term\:}2:\\\quad 2\left(x^2+8x+15\right)\\\\\mathrm{Factor}\:x^2+8x+15:\\\quad \left(x+3\right)\left(x+5\right)\\\\\\=2\left(x+3\right)\left(x+5\right)$

$-4x^2+8x+32\\\\\mathrm{Factor\:out\:common\:term\:}-4:\\\quad -4\left(x^2-2x-8\right)\\\\\mathrm{Factor}\:x^2-2x-8:\\\quad \left(x+2\right)\left(x-4\right)\\\\=-4\left(x+2\right)\left(x-4\right)$

sasho [114]2 years ago

3 0

Hello!

1. -4x2 + 8x + 32
-4(x2 - 2x - 8)
-4(x2 + 2x - 4x - 8)
-4(x(x + 2) - 4(x + 2))
-4(x + 2)(x - 4)
Answer A.

2. 2x2 + 16x + 30
2(x2 + 8x + 15)
2(x2 + 5x + 3x + 15)
2(x(x + 5) + 3(x + 5))
2(x + 5)(x + 3)
Answer C.

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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

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

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

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

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

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2 years ago

Consider the infinite geometric series 2(p-5) + 2(p-5)² + 2(p-5)³

nasty-shy [4]

10(p+3)-4(p+10)-6(p+15)

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2 years ago

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Anika [276]

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