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3 years ago

If ST = 12 and RT = 50 find RS. Use number line below

Mathematics

1 answer:

Inga [223]3 years ago

6 0

Answer:

I believe RS is 38.

Step-by-step explanation:

RT is 50, which is the total length of the number line. ST is 12, which is one part of the line. To find RS. I believe you have to subtract 50-12 to get the length of the missing part. which is RS.

50-12 = 38

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iren [92.7K]

Answer:

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Step-by-step explanation:

Given

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Multiply through by 3 to clear the fraction

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17 = 3n ( divide both sides by 3 )

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3 years ago

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Answer:

a. x=2, y=13

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3 years ago

Five more than n is equal to 9

dedylja [7]

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4 years ago

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Neporo4naja [7]

Answer:

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3 years ago

Suppose a > 0 is constant and consider the parameteric surface sigma given by r(phi, theta) = a sin(phi) cos(theta)i + a sin(

Gnom [1K]

$\Sigma$ should have parameterization

$\vec r(\varphi,\theta)=a\sin\varphi\cos\theta\,\vec\imath+a\sin\varphi\sin\theta\,\vec\jmath+a\cos\varphi\,\vec k$

if it's supposed to capture the sphere of radius $a$ centered at the origin. ( $\sin\theta$ is missing from the second component)

a. You should substitute $x=a\sin\varphi\cos\theta$ (missing $\cos\theta$ this time...). Then

$x^2+y^2+z^2=(a\sin\varphi\cos\theta)^2+(a\sin\varphi\sin\theta)^2+(a\cos\varphi)^2$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\cos^2\theta+\sin^2\varphi\sin^2\theta+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi\left(\cos^2\theta+\sin^2\theta\right)+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2\left(\sin^2\varphi+\cos^2\varphi\right)$

$x^2+y^2+z^2=a^2$

as required.

b. We have

$\vec r_\varphi=a\cos\varphi\cos\theta\,\vec\imath+a\cos\varphi\sin\theta\,\vec\jmath-a\sin\varphi\,\vec k$

$\vec r_\theta=-a\sin\varphi\sin\theta\,\vec\imath+a\sin\varphi\cos\theta\,\vec\jmath$

$\vec r_\varphi\times\vec r_\theta=a^2\sin^2\varphi\cos\theta\,\vec\imath+a^2\sin^2\varphi\sin\theta\,\vec\jmath+a^2\cos\varphi\sin\varphi\,\vec k$

$\|\vec r_\varphi\times\vec r_\theta\|=a^2\sin\varphi$

c. The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=a^2\int_0^\pi\int_0^{2\pi}\sin\varphi\,\mathrm d\theta\,\mathrm d\varphi$

You don't need a substitution to compute this. The integration limits are constant, so you can separate the variables to get two integrals. You'd end up with

$\displaystyle\iint_\Sigma\mathrm dS=4\pi a^2$

# # #

Looks like there's an altogether different question being asked now. Parameterize $\Sigma$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+u^2\,\vec k$

with $\sqrt2\le u\le\sqrt6$ and $0\le v\le2\pi$ . Then

$\|\vec s_u\times\vec s_v\|=u\sqrt{1+4u^2}$

The surface area of $\Sigma$ is

$\displaystyle\iint_\Sigma\mathrm dS=\int_0^{2\pi}\int_{\sqrt2}^{\sqrt6}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv$

The integrand doesn't depend on $v$ , so integration with respect to $v$ contributes a factor of $2\pi$ . Substitute $w=1+4u^2$ to get $\mathrm dw=8u\,\mathrm du$ . Then

$\displaystyle\iint_\Sigma\mathrm dS=\frac\pi4\int_9^{25}\sqrt w\,\mathrm dw=\frac{49\pi}3$

# # #

Looks like yet another different question. No figure was included in your post, so I'll assume the normal vector points outward from the surface, away from the origin.

Parameterize $\Sigma$ by

$\vec t(u,v)=u\,\vec\imath+u^2\,\vec\jmath+v\,\vec k$

with $-1\le u\le1$ and $0\le v\le 2$ . Take the normal vector to $\Sigma$ to be

$\vec t_u\times\vec t_v=2u\,\vec\imath-\vec\jmath$

Then the flux of $\vec F$ across $\Sigma$ is

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=\int_0^2\int_{-1}^1(u^2\,\vec\jmath-uv\,\vec k)\cdot(2u\,\vec\imath-\vec\jmath)\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-\int_0^2\int_{-1}^1u^2\,\mathrm du\,\mathrm dv$

$\displaystyle\iint_\Sigma\vec F\cdot\mathrm d\vec S=-2\int_{-1}^1u^2\,\mathrm du=-\frac43$

If instead the direction is toward the origin, the flux would be positive.

8 0

4 years ago