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2 years ago

What is the answer to this?

Mathematics

1 answer:

kherson [118]2 years ago

7 0

s = standard version amount

h = high quality version amount

we know that there were 1090 downloads of the song, meaning s + h = 1090.

we also know that the total amount of MBs downloaded was 3353 MBs, and since the standard is 2.1 MBs and the high quality is 4.9MBs, then 2.1s + 4.9h = 3353.

$\begin{cases} s+h=1090\\ 2.1s + 4.9h = 3353\\[-0.5em] \hrulefill\\ h = 1090 - s \end{cases}\qquad \stackrel{\textit{substituting on the 2nd equation}}{2.1s+4.9(1090-s) = 3353} \\\\\\ 2.1s + 5341 - 4.9s = 3353\implies -2.8s + 5341 = 3353 \\\\\\ -2.8s=-1988\implies s = \cfrac{-1988}{-2.8}\implies \boxed{s = 710}$

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Solve for x -4(2x-3)= -36

ki77a [65]

Answer:

x=6

Step-by-step explanation:

-4(2x-3)=-36

-8x+12=-36

-8x=-48

x=6

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2 years ago

Does the order of information matter when writing a proof? (geometry)

n200080 [17]

Answer:

Yes it does matter

Step-by-step explanation:

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2 years ago

Read 2 more answers

Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

1 year ago

The library has DVDs on crime, adventure and science fiction.

kozerog [31]

Answer:

The library has 1,056 crime DVDs.

Step-by-step explanatiion:

Crime : Adventure : Science

11x : 6x : 10x

6x+384=10x

384=4x

x=96

The number of crime DVDs is 11x.

We found that x=96.

So, the number of crime DVDs the library has is equal to 11*96.

11*96=1,056.

Hope this helps!

6 0

2 years ago

Find the mean, median, mode, and range of this data: 49, 49, 54, 55, 52, 49, 55. If necessary, round to the nearest tenth.

krok68 [10]

Answer:

Mean = 51.4.

Mode = 49.

Median = 52.

Range = 6.

Step-by-step explanation:

Mean = Sum of all observations / Number of observations.

Mean = (49+49+54+55+52+49+52)/7

Mean = 360/7

Mean = 51.4 (to the nearest tenth).

Mode = The most repeated values = 49 (repeated 3 times).

Range = Largest Value - Smallest Value = 55 - 49 = 6.

Median = The central value of the data.

First, arrange the data in the ascending order: 49, 49, 49, 52, 54, 55, 55.

It can be seen that the middle value is 52. Therefore, median = 52!!!

7 0

3 years ago

What is the answer to this?​

What is the answer to this?