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4 years ago

PLEASE HELP NEED BY TOMORROW MORNING AND PLEASE SHOW WORK.

Mathematics

1 answer:

Lostsunrise [7]4 years ago

5 0

-2/9 is -0.222....
-0.222... rounded is -0.22.
so it is -0.22.
when a decimal goes on, u can put a bar over(simpily draw a line on top the first few numbers) or round.

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Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

6 0

2 years ago

How many times does 0.25 go into 41?

marysya [2.9K]

0.25 goes into 41 a total of 164 times. to get this answer you take 41 and divide it by 0.25 to get 164.

8 0

3 years ago

Read 2 more answers

VARVARA [1.3K]

<h3>Solution</h3>

$\begin{aligned}\frac{1}{h-5}+\frac{2}{h+5}&=\frac{16}{h^2-25}\\\frac{(h+5)+2(h-5)}{(h-5)(h+5)}&=\frac{16}{h^2-25}\\\frac{h+2h+5-10}{h^2-25}&=\frac{16}{h^2-25}\\\frac{3h-5}{\cancel{h^2-25}}&=\frac{16}{\cancel{h^2-25}}\\3h-5&=16\\3h&=16+5\\3h&=21\\h&=\frac{21}{3}\\h&=7\end{aligned}$