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2 years ago

Whole numbers and the counting numbers less than zero are called ______.

Mathematics

1 answer:

Allisa [31]2 years ago

7 0

Answer:

integers

Step-by-step explanation:

integers include whole numbers and negative non fractions

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Paul is taking inventory of unsold ladies’ dresses from last month. The following matrix shows how many of each dress were left

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there were three times as many new subscribers in the 20-29 and the 30-39 age groups combined for the chronicle than in the 20-29 age group for the times.

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Find the inverse, g−1(x), of the following function: g(x) = 4x - 11

Sphinxa [80]

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$g^{-1}$ (x) = $\frac{x+11}{4}$

Step-by-step explanation:

let y = g(x) and rearrange making x the subject, that is

y = 4x - 11 ( add 11 to both sides )

y + 11 = 4x ( divide both sides by 4 )

$\frac{y+11}{4}$ = x

Change y back into terms of x , with x = $g^{-1}$ (x) , thus

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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

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

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