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3 years ago

It's a comon core math question the teacher is nice but.... She dosent like helping students out

Mathematics

1 answer:

Luba_88 [7]3 years ago

7 0

Answer: Well i mean thats a personal question but tell your teacher that she not being much help.

Step-by-step explanation:

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Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

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3 years ago

Prove the identity (n-5)^2 - (6n-35)=(n-10)(n-6)

elena-14-01-66 [18.8K]

Answer:

(n-5)^2 - (6n-35)=(n-10)(n-6)

-----------

n²-10n+25-6n+35 =
n²-16+60 = n²- 10n - 6n + 60 =
n(n-10) - 6(n-10) =
(n-10)(n-6)

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3 years ago

Sandra is opening a savings account which compounds interest quarterly. Her banker gave her the following function to find the a

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3 years ago

Find the Surface Area and Volume of Sphere with radius 12 cm.

Gnom [1K]

$Radius:r=12cm\\\\Surface\ area:\\A=4\pi r^2\to A=4\pi\cdot12^2=4\pi\cdot144=576\pi\ (cm^2)\\\\Volume:\\V=\frac{4}{3}\pi r^3\to V=\frac{4}{3}\pi\cdot12^3=\frac{4}{3}\pi\cdot1728=4\pi\cdot576=2304\pi\ (cm^3)$

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3 years ago

Read 2 more answers

If m∠XWY=20∘,m∠XWZ=40∘, and XY = 16, what is the value of YZ?

Blababa [14]

Answer:

YZ=16

Step-by-step explanation:

Because they are parallel

8 0

3 years ago

It's a comon core math question the teacher is nice but.... She dosent like helping students out​

It's a comon core math question the teacher is nice but.... She dosent like helping students out