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2 years ago

Wyd for 50 points answer fast

Mathematics

2 answers:

Vika [28.1K]2 years ago

6 0

Answer: Chillin hbu?

Step-by-step explanation:

Dmitrij [34]2 years ago

5 0

Answer:

Hey, I'm chugging a bottle of Coke, hbu?

Step-by-step explanation:

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Help me with this!!

Neporo4naja [7]

Answer:

Step-by-step explanation:

so we see the two line of the same length of 14, which tells us that the lengths of 5x and 7x-8 have to be the same length. so we know we can set them equal to each other. Often times that's the key to math problems , finding where you can set things equal.

5x = 7x-8 ( subtract 5x from both sides and then add 8 to both sides )

8 = 2x ( divide both sides by 2 )

4 = x

nice this looks right too , both sides add to 20 :)

5 0

2 years ago

Please Help! Shouldn't be that hard.

AysviL [449]

The answer is a 1:3 and b 2:3

5 0

2 years ago

solve the following quadratic equation by completing the square. if you need to round to the nearest tenth. x² - 20x = -75

lesantik [10]

The Answer should be either x=15 or x=5 I got both.

5 0

3 years ago

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

2 years ago

The slope of the line below is 2. Which of the following is the the point-slope form of the line?

Jobisdone [24]

Answer: y + 1 = 2 (x -1)

Step-by-step explanation:

Point slope form is:

y - y1 = m (x - x1)

Point on the line is (1, -1) and slope is 2

m = 2

y1 = -1

x1 = 1

y - (-1) = 2 (x - 1) or y + 1 = 2 (x -1)

5 0

2 years ago