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2 years ago

The system has......

Mathematics

2 answers:

Flura [38]2 years ago

7 0

Answer:

No solution

Step-by-step explanation:

SIZIF [17.4K]2 years ago

5 0

The system has one solution.

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If there are 100 tickets to a concert and 200 fans who would like to go to the concert, each placing a slightly different value

Setler79 [48]

It's more efficient to hold a random drawing because the probability is higher.
$\frac{100}{200} = \frac{1}{2}$
For every 2 person, 1 person gets to go to the concert.

Answer is 1/2 aka 0.5 aka 50%

Hope this helps. - M

3 0

2 years ago

How to find area of a rectangle?

anyanavicka [17]

Formula

Area = L*w

Hint : To find the area of a rectangle , multiply the length times the width

I hope that's help !

4 0

2 years ago

HELP DUE IN 10 MINS!!

fredd [130]

Answer:

D. No, they are not similar.

Step-by-step explanation:

<u>Check the ratio of the two given sides:</u>

MV/VT = 21/49 = 3/7
LV/VU = 8/28 = 2/7

Ratios are different so the triangles are not similar

Correct choice is D

5 0

2 years ago

2cos<img src="https://tex.z-dn.net/?f=2cos%5E%7B2%7D%20x-3cosx%2B1%3D0" id="TexFormula1" title="2cos^{2} x-3cosx+1=0" alt="2cos^

sattari [20]

Answer:

Step-by-step explanation:

7 0

2 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

2 years ago