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2 years ago

Find two numbers whose difference is 30 and whose product is a minimum.

Mathematics

1 answer:

guapka [62]2 years ago

3 0

Answer:60-30 I guess

Step-by-step explanation:n60-30=30

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Please help me. I need help

mel-nik [20]

<h3>Answer:</h3>

a. -(3√13)/13

<h3>Step-by-step explanation:</h3>

The cosine can be found from the tangent by way of the secant.

tan(θ)² +1 = sec(θ)² = 1/cos(θ)²

Then ...

cos(θ) = ±1/√(tan(θ)² +1)

The <em>cosine is negative in the second quadrant</em>, so we will choose that sign.

cos(θ) = -1/√((-2/3)² +1) = -1/√(4/9 +1) = -1/√(13/9)

cos(θ) = -3/√13 = -(3√13)/13 . . . . . matches your selection A

3 0

3 years ago

Read 2 more answers

What’s the difference need help

labwork [276]

You would circle the 5

4 0

2 years ago

PLEASE HELP ANYONE.....

irga5000 [103]

Answer:

the awnser should have is. b

8 0

2 years ago

Read 2 more answers

A cell phone company has a basic monthly plan of $40 plus $0.45 for any minutes used over 700. Before receiving his statement, J

Serga [27]

Ok so we know he used over 700 because he was charged more than 40. Then we take the $8.10 and divide 0.45 into it. This then gives you 18 more minutes. Then add 18 to 700 and you get 718 min.

4 0

3 years ago

Evaluate the surface integral:S

rjkz [21]

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

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

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

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

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

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

4 0

2 years ago