All categories

3 years ago

What is the fibonacci sequence who invented it and what is it used for

Mathematics

1 answer:

butalik [34]3 years ago

6 0

Answer:

Read the explanation

Step-by-step explanation:

- The Fibonacci Sequence is the series of numbers:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

- The next number is found by adding up the two numbers before it

# Example:

- The 2 is found by adding the two numbers before it (1+1)

- The 3 is found by adding the two numbers before it (1+2),

- The 5 is (2+3) and so on

- The rule of it is x(n) = x(n-1) + x(n-2) where

# x(n) is term number ⇒ n

# x(n-1) is the previous term ⇒ (n-1)

# x(n-2) is the term before that ⇒ (n-2)

- Example: the 8th term is the 7th term plus the 6th term:

x(8) = x(7) + x(6)

# Note: When we take any two consecutive Fibonacci Numbers,

their ratio is very close to the Golden Ratio φ

- The Golden Ratio φ is approximately 1.618034...

- We can calculate any Fibonacci Number using the Golden Ratio:

x(n) = [φ^n - (1 - φ)^n]/√5

- The answer is a whole number, exactly equal to the addition of the

previous two terms.

# There is an interesting patterns in Fibonacci sequence:

every nth number is a multiple of x(n)

- Example:

* x3 = 2 ⇒ every 3rd number is a multiple of 2 (2, 8, 34, 144, 610, ...)

* x4 = 3 ⇒ Every 4th number is a multiple of 3 (3, 21, 144, ...)

* x5 = 5 ⇒ Every 5th number is a multiple of 5 (5, 55, 610, ...)

- The man who invented it.

His real name was Leonardo Pisano Bogollo, and he lived between

1170 and 1250 in Italy.

- Fibonacci was his nickname

- Fibonacci sequence is it used for:

# Reflects patterns of growth spirals (a spiral curve , shape , pattern ,

object) found in nature

# It is the closest approximation in integers to the logarithmic spiral

series, which follows the same rule as the Fibonacci sequence

You might be interested in

Solve the system of equations. −2x+5y =−35 7x+2y =25

Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

$\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}$

This system of equations can be solved in three different ways:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is $\text{y = mx + b}$ .

Equation 1 is $-2x+5y = -35$ . We need to isolate y.

$\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7$

Equation 1 is now $y=\frac{2}{5}x-7$ .

Equation 2 also needs y to be isolated.

$\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}$

Equation 2 is now $y=-\frac{7}{2}x+\frac{25}{2}$ .

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}$

$\bullet \ \text{For x = 0,}$

$\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5$

Now, we can place these values in our table.

As we can see in our table, the rate of decrease is $-\frac{2}{5}$ . In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract $-\frac{2}{5}$ from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be $y=-\frac{7}{2}x+\frac{25}{2}$ . Therefore, we just use the same process as before to solve for the values.

$\bullet \ \text{For x = 0,}$

$\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}$

$\bullet \ \text{For x = 1,}$

$\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9$

$\bullet \ \text{For x = 2,}$

$\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}$

$\bullet \ \text{For x = 3,}$

$\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2$

$\bullet \ \text{For x = 4,}$

$\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}$

$\bullet \ \text{For x = 5,}$

$\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5$

And now, we place these values into the table.

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1 Equation 2

$\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$ $\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}$

Therefore, using this data, we have one solution at (5, -5).

4 0

3 years ago

Estimate the size of a crowd at an airshow that occupies a rectangular space with dimensions of 500 feet by 750 feet (to the nea

Liono4ka [1.6K]

Answer:

C. 41667 people

Step-by-step explanation:

750 * 500 = 375000 / 9 = 41666.666 = 41667

7 0

3 years ago

Write a polynomial to represent the value of d dimes and n nickels.

nadya68 [22]

Answer: 0.10d + 0.05n
Note: this value expression is in dollars (not cents)
If you want the expression in cents instead of dollars, then it would be 10d+5n

--------------------------------------------------------------------------------

The value of just the dimes only is 0.10d dollars
The value of just the nickels only is 0.05n dollars
Combined the total value is 0.10d+0.05n dollars

3 0

3 years ago

You are building a frame for a window. The window will be installed in the opening shown in the diagram. Determine whether the o

lubasha [3.4K]

Answer:

dfffffdf

Step-by-step explanation:

3 0

3 years ago

Hello please evaluate the expressions when x = 3 & y = 15

Nesterboy [21]

Y-6 = 15-6=9
2y+x= 2(15) + 3= 30+3 = 33

7 0

3 years ago