Is this a irrational or rational number?

Mathematics

2 answers:

Natali [406]3 years ago

7 0

I think its irrational bc if you put it as the squareroot of 329, it results in neverending decimals.

18^2=324+5=329

Ivenika [448]3 years ago

7 0

It would be irrational because the decimal place doesn’t terminate or repeat

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Karo-lina-s [1.5K]

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3 years ago

Can someone explain how I multiply both sides ?

stellarik [79]

Answer:

x=3

Step-by-step explanation:

Given,

1 = 1/(x-2)

Multiply both sides by (x-2),

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Adding 2 on both sides,

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Leya [2.2K]

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

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

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