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3 years ago

I need help as soon as poible π×5×5×7=

Mathematics

1 answer:

Marrrta [24]3 years ago

8 0

Answer:

549.5

Step-by-step explanation:

For a estimate of pi we use 3.14. So 3.14x5x5x7=549.5

For a realistic answer its 549.778714378.....

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Help now i have 20 mins!!!!!!!!!!!

IgorLugansk [536]

Answer:

Hope this solution helps you

7 0

4 years ago

You are saving to buy a new phone you have $150 in your bank account and can save $55 each week assuming you don’t spend any mon

FromTheMoon [43]

Answer:

150+55w

Step-by-step explanation:

150+55w

150 never changes

(55)(w), 55 changes depending on number of weeks

8 0

3 years ago

Read 2 more answers

What is 1/4 of 20? what is an equivalent fraction to 1/2?

Vinvika [58]

1/4 of 20 is 5.. and 5/10, 3/6, 2/4, 6/12 are all equal to 1/2

3 0

3 years ago

What is the zero for 2x2-7=4

rosijanka [135]

2x^2-7=4
subtract 4 from both sdies
2x^2-11=0
therefor
2x^2=11
divide by 2
x^2=5.5
square root both sdies
x=-2.345 or 2.345

6 0

3 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

3 years ago