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3 years ago

Work out the area of a circle

Mathematics

2 answers:

stellarik [79]3 years ago

5 0

Answer:

50.24

Step-by-step explanation:

A= πr2

314/100×4×4=50.24

HACTEHA [7]3 years ago

3 0

Answer:

$Area \: = \pi {r}^{2} \\ r = 4 \\ A = 3.142 \times {4}^{2} \\ A = 3.142 \times 16 \\ A = 50.272$

Area ≈50.3

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Y"-3xy=0 given ordinary point x=0. power series

liberstina [14]

$\displaystyle y=\sum_{n\ge0}a_nx^n$
$\implies\displaystyle y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}$

$y''-3xy=0$
$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0$
$\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge0}a_nx^{n+1}=0$
$\displaystyle2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}-3\sum_{n\ge3}a_{n-3}x^{n-2}=0$
$\displaystyle2a_2+\sum_{n\ge3}\bigg(n(n-1)a_n-3a_{n-3}\bigg)x^{n-2}=0$

This generates the recurrence relation

$\begin{cases}a_0=a_0\\a_1=a_1\\2a_2=0\\n(n-1)a_n-3a_{n-3}=0&\text{for }n\ge3\end{cases}$

Because you have

$n(n-1)a_n-3a_{n-3}=0\implies a_n=\dfrac3{n(n-1)}a_{n-3}$

it follows that $a_2=0\implies a_5=a_8=a_{11}=\cdots=a_{n=3k-1}=0$ for all $k\ge1$ .

For $n=1,4,7,10,\ldots$ , you have

$a_1=a_1$
$a_4=\dfrac3{4\times3}a_1=\dfrac{3\times2}{4!}a_1$
$a_7=\dfrac3{7\times6}a_4=\dfrac{3\times5}{7\times6\times5}=\dfrac{3^2\times5\times2}{7!}$
$a_{10}=\dfrac3{10\times9}a_7=\dfrac{3\times8}{10\times9\times8}a_7=\dfrac{3^3\times8\times5\times2}{10!}a_1$

so that, in general, for $n=3k-2$ , $k\ge1$ , you have

$a_{n=3k-2}=\dfrac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}a_1$

Now, for $n=0,3,6,9,\ldots$ , you have

$a_0=a_0$
$a_3=\dfrac3{3\times2}a_0=\dfrac3{3!}a_0$
$a_6=\dfrac3{6\times5}a_3=\dfrac{3\times4}{6\times5\times4}a_3=\dfrac{3^2\times4}{6!}a_0$
$a_9=\dfrac3{9\times8}a_6=\dfrac{3\times7}{9\times8\times7}a_6=\dfrac{3^3\times7\times4}{9!}a_0$

and so on, with a general pattern for $n=3k$ , $k\ge0$ , of

$a_{n=3k}=\dfrac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}a_0$

Putting everything together, we arrive at the solution

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_0\underbrace{\displaystyle\sum_{k\ge0}\frac{3^k\displaystyle\prod_{\ell=1}^k(3\ell-2)}{(3k)!}x^{3k}}_{n=0,3,6,9,\ldots}+a_1\underbrace{\displaystyle\sum_{k\ge1}\frac{3^{k-1}\displaystyle\prod_{\ell=1}^{k-1}(3\ell-1)}{(3k-2)!}x^{3k-2}}_{n=1,4,7,10,\ldots}$

To show this solution is sufficient, I've attached is a plot of the solution taking $y(0)=a_0=1$ and $y'(0)=a_1=0$ , with $n=6$ . (I was hoping to be able to attach an animation that shows the series solution (orange) converging rapidly to the exact solution (blue), but no such luck.)

6 0

3 years ago

NEED DONE ASAP. IT'S 12AM AND I WANNA GET THIS DONE SO I CAN SLEEP..

mr Goodwill [35]

The options seem a bit mis-formatted but here is the way p is computed:

the decrease is 2000-1600

and it is relative to: 2000 (2003 attendance)

so $p=100 \frac{2000-1600}{2000} = 100 \frac{400}{2000}$

and so that aligns with your Option A. Options B, C, D look incorrect for sure.

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3 years ago

Which of the following operations will change the solution of the system x − y = 2 and 3x − 3y = 2?

Anon25 [30]

The answer would be times 3.
Hope that helped.

8 0

4 years ago

The area of a circle is 28.26 square centimeters. What is its diameter (use 3.14 for ð)?

horsena [70]

Area for circle is πr² so πr² =28.26 and we can sub 3.14 as pi
3.14*r²=28.26 we can divide by 3.14 to get r² on it's own
r²=28.26/3.14
then we root both sides to get r on it's own
28.26/3.14=9 √9=3
and the diameter is double the radius 3*2=6 so the diameter is 6

5 0

3 years ago

Write a system of linear equations for the graph below

Oliga [24]

1q+2x-1 liner #1 liner # is 18% one

5 0

3 years ago