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Doss [256]
2 years ago
11

Given line segment AB with endpoints A(-1,7) and B(11, -1) Find the length of AB.

Mathematics
1 answer:
ser-zykov [4K]2 years ago
6 0

Step-by-step explanation:

some to check if it correct

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G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u

So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
3 years ago
Statistics urgent<br> Use n=6 and p=0.85 to complete
Bas_tet [7]

Answer:

I cant see anything on that picture. Wish I could help bruv. I wish I could help. It's a shame innit.

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labwork [276]
D

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galben [10]
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If it's perimeter, then its 23. If it's multiplying, then it's 810.

Step-by-step explanation: Please brainliest if possible!

4 0
2 years ago
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