All categories

3 years ago

What is the value of 53i9

2 answers:

6 0

Your question have 53i9 is that correct. ... You've already chosen the best response. 0. yes that's my question. What is the value of 53i 9 LynFran. 11 months ago ...

Alex787 [66]3 years ago

5 0

Answer:

the value of 5^3 i^9 = 125 i

Step-by-step explanation:

We need to find the value of 5^3 i^9

'i' is a complex number (Imaginary)

The value of 5^3 = 5*5*5= 125

the value of i^2 =-1

i^3 = -i

i^4= 1

$i^9 = i ^{4+4+1}$

$i^9 = i^4 * i^4 * i$

$i^9 = 1*1 * i=i$

So the value of 5^3 i^9 = 125 i

You might be interested in

What number is a factor of 22 but not a multiple of 2

Mrrafil [7]

11 is a factor of 22 that is not a multiple of 2.

7 0

3 years ago

Read 2 more answers

Jacob bought a magazine for $2.80 and three candy bars. Write an expression for how much Jacob paid.

harkovskaia [24]

Answer:

total = 2.8 + 3x

x is the price of the candy bars

4 0

2 years ago

Is 1/4 bigger than 3/4

Leto [7]

<span> The larger the numerator, the larger the fraction.
3/4 is bigger then 1/4.
You could also draw a model :D
Your answer is <em>no.</em>
</span>

7 0

2 years ago

Read 2 more answers

Aubrey was sorting books for the garage sale. she had 831 books and wanted to put them in 5 equal boxes. how many books would fi

lianna [129]

she would need 166 boxes. there is a remainder of 1. one book wont fit in a box

6 0

2 years ago

Read 2 more answers

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

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

3 0

2 years ago