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2 years ago

6

At a family reunion, a family friend bring 20 lb of ice and 30 cup. Which expression show the greatest common factor , using dis

tribute property of 20 and 30?

1 answer:

Zepler [3.9K]2 years ago

3 0

Answer:

10(2+3)

Step-by-step explanation:

We are given that

Ice=20 lb

Cup=30

We have to find the expression which shows the greatest common factor using distributive property of 20 and 30.

We know that distributive property of integers a,b and c

$a\cdot(b+c)=a\cdot b+a\cdot c$

$20=2\times 10$

$30=3\times 10$

$10(2+3)=10\times 2+10\times 3$

Hence, the highest common factor of 20 and 30=10

Expression:10(2+3)

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1 year ago

Evaluate the surface integral:S

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Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

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where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
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So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
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$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

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2 years ago