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3 years ago

Please help!!!!!!!!!!!!!!

Mathematics

1 answer:

Vlad1618 [11]3 years ago

5 0

What do you need specifically just that one question

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Who else uses time4learning? am i the only one?

Greeley [361]

Answer:

i think i used it before

Step-by-step explanation:

5 0

3 years ago

Solve using the quadratic frmula 2x^2-5x=3

Mnenie [13.5K]

2x2−5x=3

I think the first step would be to

Subtract 3 from both sides

2x2−5x−3=3−3

2x2−5x−3=0

Then

For this equation: a=2, b=-5, c=-3

2x2+−5x+−3=0

Lastly you would

Use quadratic formula with a=2, b=-5, c=-3

I think your answer would then have to be...

x = 3 or x = −1 /2

4 0

3 years ago

Read 2 more answers

Hay 150 numeros en una rifa ¿ cuantos habria que comprar para tener un 8% de probalidad pra ganarla

krok68 [10]

Answer:

12 numeros

Step-by-step explanation:

para que pueda obtener el 8% de probabilidad, tendrá que comprar el 8% de los números

8% x 150

<u> 8 </u> x 150

100

=12 numeros

espero eso ayude

3 0

3 years ago

A line with a slope of 5 passes through the point (2, 5). What is its equation in

Maru [420]

Answer: y = 5x - 5

Step-by-step explanation:

6 0

3 years ago

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Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago