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4 years ago

Pls help me and can someone help be my touter

Mathematics

1 answer:

ELEN [110]4 years ago

6 0

Wouldn't it be 6 cm, 8 cm, 10 cm?

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Number patterns 54, 18, 6, 2, 2/3, what are the next 2 terms?

nikitadnepr [17]

Answer:

2/9, 2/27

Step-by-step explanation:

54, 18, 6, 2, 2/3

Take the second term and divide by the first

18/54 = 1/3

We multiply by 1/3

Check by taking the third term and dividing by the second

6/18 = 1/3

This works

Take the last term and multiply by 1/3

2/3*1/3 = 2/9

2/9*1/3 =2/27

4 0

3 years ago

I HAVE LOTS OF POINTS COMING FOR THE END OF ALL MY QUESTIONS SO PLEASE HELP OUT

Irina18 [472]

Answer:

1/5

Step-by-step explanation:

I think you go to the number 1 on the bottom and you go up 5 and there's your answer?

7 0

3 years ago

Read 2 more answers

Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)

zlopas [31]

We can express the recurrence,

$\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}$

in matrix form as

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}$

By substitution,

$\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}$

and continuing in this way we would find that

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

Diagonalizing the coefficient matrix gives us

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

which makes taking the $(n-2)$ -th power trivial:

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

So we have

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}$