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pentagon [3]
4 years ago
11

Recall from class that we found that the Fibonacci sequence, with $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n - 2} + F_{n - 1}$, had a

closed form $$F_n = \frac{1}{\sqrt{5}} \left( \phi^n - \widehat{\phi}^n \right),$$ where $$\phi = \frac{1 + \sqrt{5}}{2} \; \text{and} \; \widehat{\phi} = \frac{1 - \sqrt{5}}{2}.$$
The Lucas numbers are defined in the same way, but with different starting values. Let $L_0$ be the zeroth Lucas number and $L_1$ be the first. If
\begin{align*}
L_0 &= 2 \\
L_1 &= 1 \\
L_n &= L_{n - 1} + L_{n - 2} \; \text{for}\; n \geq 2
\end{align*}
then what is the tenth Lucas number? (Note: We seek a numerical answer.)
Mathematics
2 answers:
diamong [38]4 years ago
8 0

Here is an easy way:

2+1=3

3+1=4

4+3=7

7+4=11

11+7=18

18+11=29

29+18=47

47+29=76

76+47=<u>123</u>

bekas [8.4K]4 years ago
4 0
\begin{bmatrix}L_n\\L_{n-1}\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}L_{n-1}\\L_{n-2}\end{bmatrix}
\begin{bmatrix}L_n\\L_{n-1}\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}^2\begin{bmatrix}L_{n-2}\\L_{n-3}\end{bmatrix}
\begin{bmatrix}L_n\\L_{n-1}\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}^3\begin{bmatrix}L_{n-3}\\L_{n-4}\end{bmatrix}
\vdots
\begin{bmatrix}L_n\\L_{n-1}\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}L_1\\L_0\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}^{n-1}\begin{bmatrix}1\\2\end{bmatrix}

The easiest way to find an integer power of a matrix is to diagonalize the matrix first. If \mathbf A=\begin{bmatrix}1&1\\1&0\end{bmatrix}=\mathbf{SD}\mathbf S^{-1} for some matrix \mathbf S and diagonal matrix \mathbf D, then \mathbf A^n=\mathbf{SD}^n\mathbf S^{-1}.

You have

\begin{bmatrix}1&1\\1&0\end{bmatrix}=\begin{bmatrix}\widehat\phi&\phi\\1&1\end{bmatrix}\begin{bmatrix}\widehat\phi&0\\0&\phi\end{bmatrix}\begin{bmatrix}\widehat\phi&\phi\\1&1\end{bmatrix}^{-1}

which is a result you may have derived in class.

When n=10, you have

\begin{bmatrix}L_{10}\\L_9\end{bmatrix}=\begin{bmatrix}1&1\\1&0\end{bmatrix}^9\begin{bmatrix}1\\2\end{bmatrix}
\begin{bmatrix}L_{10}\\L_9\end{bmatrix}=\begin{bmatrix}\widehat\phi&\phi\\1&1\end{bmatrix}\begin{bmatrix}{\widehat\phi}^9&0\\0&\phi^9\end{bmatrix}\begin{bmatrix}\widehat\phi&\phi\\1&1\end{bmatrix}^{-1}\begin{bmatrix}1\\2\end{bmatrix}
\implies L_{10}=123
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nasty-shy [4]

Answer: 5 + 5w

Step-by-step explanation:

We can use the variable <em>x </em>to describe each additional hour, although the vairable can be substituted for any other vairable.

We know that the main cost for one hour is $5, and that each additional hour will cost $5.

Therefore, the equation to reflect this situation would be 5 + 5w, because the first 5 represents the first hour, and the 5w represents how much each hour costs. If you rent two extra hours, then the cost would end up being 15, because the equation would become: 5 + (5x2) = 15.

The equation is 5 + 5w. Hope this helps, and good luck!

4 0
3 years ago
*
Colt1911 [192]

Answer: 6x + 8y =1,245

Step-by-step explanation:

Hi, to answer this question we have to write an equation for each person:

Jack = The total cost (650) must be equal to the product of the number of shirts he bought (2) and the price per shirt (x); plus the product of the number of pair of jeans he bought (5) and the price per jean (y).

2 x+ 5 y=650

Dylan = The total cost (595) must be equal to the product of the number of shirts he bought (4) and the price per shirt (x); plus the product of the number of pair of jeans he bought (3) and the price per jean (y).

4x+3y = 595

Vertical addition:

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4x+3y = 595

_________

6x + 8y =1,245

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3 years ago
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never [62]

Answer:2 2/15

Step-by-step explanation:

4/5= 12/15

+1 1/3 =5/15

1 and 17/15

17/15= 1 and 2/15 +1= 2 and 2/15

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Answer:

15

Step-by-step explanation:

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3 years ago
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irinina [24]

Answer:B

Step-by-step explanation:

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