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3 years ago

Simplify 4n 5 + 3.5n 5 + (-2.1n 5).

Mathematics

2 answers:

Mars2501 [29]3 years ago

8 0

Answer:

$5.4n^5$

Step-by-step explanation:

Given:

Expression, $4n^5+3.5n^5+(-2.1)n^5$

Simplify the given polynomial.

$\Rightarrow 4n^5+3.5n^5+(-2.1)n^5$

Sign rule: (+)(-) = -

$\Rightarrow 4n^5+3.5n^5-2.1n^5$

Each term has variable n with degree 5.

Combine the coefficients of variable.

$\Rightarrow (4+3.5-2.1)n^5$

$\Rightarrow 5.4n^5$

Hence, The simplest form of polynomial is 5.4n⁵

lyudmila [28]3 years ago

3 0

4n 5 + 3.5n 5 - 2.1n 5 = n 5 (4 + 3.5 - 2.1) = n 5 * 5.4 = 5.4n 5

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What should be subtracted from -8xy + 2x^2 + 3y^2 to get x^2 + 2

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(-8xy + 2x^2 + 3y^2) - unknown = x^2 + 2

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(-8xy + 2x^2 + 3y^2) - (x^2 - 8xy + 3y^2 - 2)

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2 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$