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Elanso [62]
3 years ago
5

Please help. I attached a picture.

Mathematics
1 answer:
Serjik [45]3 years ago
3 0
One hundred, if ur teacher asks why say the vertical angle theorem
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Pls pls pls pls help. im so stressed rn
Sloan [31]

Answer:

The function role is +2 for x, and -1 for y.

Step-by-step explanation:

The reason for my answer is because as you can see on the table;

For X, it starts at 0 and counts up by 2. For Y, it starts at 2 and counts back by 1.

6 0
2 years ago
a web music store offers two version of a popular song. the size of the standard version is 2.9 megabytes (MB). the size of the
miv72 [106K]

Answer: 260 downloads of the standard version were there.

Step-by-step explanation:

Let x represent the number of downloads of the standard version.

Let y represent the number of downloads of the high-quality version.

Yesterday, the high-quality version downloaded four times as often as the standard version. It means that

y = 4x

The size of the standard version is 2.9 megabytes (MB). the size of the high-quality version is 4.6 MB. the total size downloaded for the two versions was 5538 MB. It means that

2.9x + 4.6y = 5538- - - - - - - - - - - -1

Substituting y = 4x into equation 1, it becomes

2.9x + 4.6 × 4x = 5538

2.9x + 18.4x = 5538

21.3x = 5538

x = 5538/21.3

x = 260

6 0
3 years ago
Three times the difference between 4 times h and five is 45
Anestetic [448]

3(4h-5) =45

12h-15=45

+15 +15

12h = 60

÷12 ÷12

h = 5

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
3 years ago
Volume of sun is 1.4×10^18 and volume of earth is 1.1×10^12 how many earth can fit in the sun
Advocard [28]
\frac{1.4\cdot10^{18}}{1.1\cdot10^{12}}=
1.(27)\cdot10^6=1272727.(27)

1272727 \hbox{ Earths}

6 0
3 years ago
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