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2 years ago

NEED QUICK!!!

Mathematics

2 answers:

OverLord2011 [107]2 years ago

6 0

Answer:

A WiFi connection transmits data via wireless signals, while an Ethernet connection transmits data over cable. ... An Ethernet connection is generally faster than a WiFi connection and provides greater reliability and security.

kakasveta [241]2 years ago

3 0

Answer:

they both connect to a tower of some sort but cable applies to channels of network and wifi applies the a wide web network.

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Nick made his histogram showing the number of sport caps each of his friends own.

pickupchik [31]

Answer:

A. 7

Step-by-step explanation:

2 or 3 caps = 13

6 or 7 caps = 6

13 - 6 = 7

- hope this helps! -

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2 years ago

3) A particular motorcycle is 9 ft long. A

Shalnov [3]

The model is 3 inches, since 3 times 3 is 9 and 1 times 3 is 3.

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3 years ago

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Mary invested $5000 in the bank. How much money does she have at the end of 15 years if the interest rate is 4.5% per annum ? A

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Answer:

350000

Step-by-step explanation:

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3 years ago

Cat being neutred. What age should a cat be neutred? Cuz my cat is on the verge of getting one.. Please help me!

e-lub [12.9K]

for cats: it is generally considered safe for kittens as young as eight weeks old to be spayed or neutered. in animal shelters, surgery is often performed at this time so that kittens can be sterilized prior to adoption.

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3 years ago

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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