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3 years ago

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Mathematics

2 answers:

Alenkinab [10]3 years ago

8 0

Answer:

Step-by-step explanation:

To solve this, you can add 2/5 b to both sides of the equation. You now have 3 + 5/5 (or 1) b = 11. => 3 + b = 11 => Now, subtract 3 from both sides, and you have b = 11 - 3, and then b = 8.

taurus [48]3 years ago

6 0

Step-by-step explanation:

3 + 2/5 b =11 - 2/5 b (this is the given question is it?)

2/5 b + 2/5 b= 11 - 3 (like terms together, you transpose -2/5b to the other side as shown.)

2b+2b = 8 ( at that stage you would find t

5 hat 5 is the common value to go into 5 itself, as shown.

4b = 8 ( you just add 2b + 2b to have 4b)

4b = 8 ( at that point you cross multiply

5 1 as shown)

4b = 8 × 5 ( simple math as shown)

4b = 40 ( you multiply 8 × 5 to obtain 40)

b =10 you can prove that. Thank you.

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anastassius [24]

Answer:

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Fudgin [204]

The third one
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OLga [1]

Answer:

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Step-by-step explanation:

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rjkz [21]

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

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
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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

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
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So the surface integral is equivalent to

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

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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