This is an investigation about stellar numbers, it involves geometric shapes which form special number patterns. The simplest of these is that of the square numbers (1, 4, 9, 16, 25 etc…) The diagram below shows the stellar triangular numbers until the 6th triangle.

The next three numbers after T5 would be: 21, 28, and 36.
A general statement for nth triangular numbers in terms of n is:

The 6-stellar star, where there are 6 vertices, has its first four shapes shown below:

The number of dots until stage S6: 1, 13, 37, 73, 121, 181
Number of dots at stage 7: 253
Expression for number of dots at stage 7:
Since the general trend is adding the next multiple of 12 (12, 24, 36, 48 etc…) for each of the stars, so for S2 it would be 1+12=13, and for S3 it would be 13+24=37

General statement for 6-stellar star number at stage Sn in terms of n:

For P=9:

Since S1 must equal 1 then we can prove this formula by showing that:

So the first six terms are: 1, 19, 55, 109, 181, 271
Therefore the equation for the 9-Stellar star at

For P=5:

Since S1 must equal 1 then we can prove this formula by showing that:

So the first six terms are: 1, 11, 31, 61, 101, 151
So the expression for 5-Stellar at

General Statement for P-Stellar numbers at stage Sn in terms of P and = For P-Stellar number equals 10:

For P-Stellar number equals 20:

The General Statement works for all number fro 1 to positive infinity. The equation was arrived at since the sum of arithmetic series can be found using , since the difference is always 2P then we can replace 2P with d, and since u1 is always equal to 1, we can replace it with 1 every time. The at the end of the equation serves the purpose of making the difference between the numbers in the series constant. This form of the equation will allow for only one variable to change, which will be . One of the things the student realized while solving this...

...STELLARNUMBERS
In order to develop this mathematics SL portfolio, I will require the use of windows paint 2010 and the graphic calculator fx-9860G SD emulator, meaning that I will use screenshots from this software with the intention of demonstrating my work and process of stellarnumbers sequences.
Triangular numbers are those which follow a triangular pattern, these numbers can be represented in a triangular grid of evenly spaced dots.
The sequence of triangular numbers is shown in the diagrams above. The first stage has 1 dot; the second stage has 3 dots (1+2); the third stage has 6 dots (1+2+3); the fourth stage has 10 dots (1+2+3+4); the fifth stage has 15 dots(1+2+3+4+5); the sixth stage has 21 dots (1+2+3+4+5+6) ; the seventh stage has 28 dots(1+2+3+4+5+6+7) and the eighth stage has36 dots(1+2+3+4+5+6+7+8). As it could be noticed, there is a sequence where in every stage the number of dots is obtained by adding up all the positive integers that correspond to the previous stages and every time one more number is added.
In terms of n, where n matches up to the stage number, it is accurate to establish an equation so that when trying to find the number of dots in stage 592, it is easy and fast by simply...

...patterns evolve, such as the square numbers, triangular numbers, and much more. The StellarNumbers are mostly used in astronomy and astrology. StellarNumbers are figurate numbers based on the number of dots that can fit into a midpoint to form a star shape. The points of the star determine the number of points plotted around the midpoint.
Triangularnumbers is a figurate number system that can be represented in the form of a triangular grid of points where the first row contains a single element and each subsequent row contains one more element than the previous one. The triangular numbers from that pattern are 1 followed by 1+2 followed by 1+2+3 and so on. From the pattern of the triangular numbers, this infinite serious starts with 1, 3, 6, 10, 15… With this pattern, calculated by counting, the next three terms would be 21, 28, and 36. To derive a formula from this pattern, we can see that x is repeated and the number goes up each time by one. After using the rules of the sequences and a few checks, the final formula results inx(x+1)2 where x is any natural number. I found this formula with the calculator with steps show below. To prove this formula, there is the typical guess and check formula where the...

...
StellarNumbers
Results
1. Triangular Numbers
Observation of the number pattern of polynomial type or different pattern needed.
Identifying the order of the general term by using the difference between the succeeding numbers.
Students are expected to use mathematical way of deriving the general term for the sequence.
Students are expected use technology GDC to generate the 7th and 8th terms also can use other graphic packages to find the general pattern to support their result
The general term in this stage is
2. 6- stellarnumbers
Stage
No of dots
1
1
2
13
3
37
4
73
5
121
6
181
Students are expected to make the stellar shape for the next to stages and count the no of dots to get the 6-stellarnumber in 5th and 6th stage
Diagrams can be hand made or using technology
Communication or observation of the number pattern has to be given
From the observation, the expression of the terms of this sequence has to be identified
Expression for the 7th term
General expression 6 – stellar shape
or
p n (n-1) +1
Other stellar shapes – based on the no of...

...The aim of this task is to investigate geometric shapes, which lead to special numbers. The simplest example of these are square numbers, such as 1, 4, 9, 16, which can be represented by squares of side 1, 2, 3, and 4.
Triangular numbers are defined as “the number of dots in an equilateral triangle uniformly filled with dots”. The sequence of triangular numbers are derived from all naturalnumbers and zero, if the following number is always added to the previous as shown below, a triangular number will always be the outcome:
1 = 1
2 + 1 = 3
3 + (2 + 1) = 6
4 + (1 + 2 + 3) = 10
5 + (1 + 2 + 3 + 4) = 15
Moreover, triangular numbers can be seen in other mathematical theories, such as Pascal’s triangle, as shown in the diagram below. The triangular numbers are found in the third diagonal, as highlighted in red.
The first diagrams to be considered show a triangular pattern of evenly spaced dots, and the number of dots within each diagram represents a triangular number.
Thereafter, the sequence was to be developed into the next three terms as shown below.
The information from the diagrams above is represented in the table below.
Term Number (n) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |...

...ID: 22087885
Exam: 986093RR - Business Math, Part 2
1. Ann works 51/2 hours, Mary works 61/3 hours, and John works 41/4 hours. How many combined hours
have they worked?
A. 165/12
*B. 161/12
C. 151/12
D. 155/12
2. You pay the following for supplies: $19.36 for pens; $21.75 for notepads; and $32.26 for computer
paper. What is your total?
A. $73.73
B. $77.37
C. $77.73
*D. $73.37
3. How many liters will a U.S. 6-quart container hold? Round your answer to the nearest hundredth.
A. 6.11 liters
*B. 5.68 liters
C. 5.97 liters
D. 6.37 liters
4. If your paycheck indicates that you worked 7.5 hours each day in an 8-day pay period, how many hours
did you work in all?
A. 40 hours
*B. 60 hours
C. 54 hours
D. 45 hours
5. If the price of upholstery fabric is $12.49 per yard, how much will 16 yards cost?
*A. $199.84
B. $189.84
C. $201.04
D. $199.94
6. Reduce 24/40 to its lowest terms.
A. 2/5
*B. 3/5
C. 4/5
D. 12/20
7. It takes you approximately 1/3 hour to type a memo. If you plan to work for 6 hours, how many memos
can you type?
A. 2
B. 6
C. 9
*D. 18
8. Paul earned a salary of $926.35 and a commission of $136.15. How much did he earn in all?
*A. $1,062.50
B. $780.20
C. $1,052.50
D. $790.20
9. What is 51/4 ÷ 22/7 ?
A. 12
B. 2
C. 1219/64
*D. 219/64
10. To get to a job site, you must travel 45 kilometers. How many miles will you travel? Round your
answer to the...

...MathIAMath Internal Assessment
EF International Academy NY Student Name: Joo Hwan Kim Teacher: Ms. Gueye Date: March 16th 2012
Contents
Introduction Part A Part B Conclusion
Introduction
The aim of this IA is to find out the pattern of the equations with complex numbers by using our knowledge. I used de Moivre’s theorem and binomial expansion, to find out the specific pattern and make conjecture about it. I basically used property of binominal theory with the relationship between the length of the line segments and the roots.
Part A
To obtain the solutions to the equation ) | | Moivre’s theorem, (| | equation, we will get: , I used de Moivre’s theorem. According to de . So if we apply this theorem in to the
(| |
)
(
(| |
)
)
| |
(
)
If we rewrite the equation with the found value of , it shows (| | ( ( ( ( ) )) ))
Let k be 0, 1, and 2. When k is 0, ( ) ( )
√
√
Now I know that if I apply this equation with the roots of
( )
( ) we can
find the answers on the unit circle. I plotted these values in to the graphing software, GeoGebra and then I got a graph as below:
Figure 1 The roots of z-1=0 I chose a root of and I tried to find out the length of two segments from the point Z. I divided each triangle in to two same right angle triangles. By knowing that the radius of the unit circle is 1, with the knowledge of the length...

...IA Task I
Introduction and purpose of task: The purpose of this task is to investigate the positions of points in intersecting circles and to discover the various relationships between said circles. Circle C1 has center O and radius r. Circle C2 has center P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has center A and radius r (therefore circles C1 and C3 are the same size). The point P’ (written P prime) is the intersection of C3 with OP. This is shown in the diagram below.
Analytically find OP’ using r=1 and OP=2, OP=3, and OP=4:
First, I created a line (see the dashed line in the above figure) between AP’ that creates the ΔAOP’. Because P’ is on the circumference of circle C3 and A is the center of circle C3, that means that AP’ is equal to the radius of C3, which is 1. We also know that because line AO connects the circumference of C1 with the center of C1 (O) and the circumference of C3 with the center of C3 (A), the radii of these circles is the same, which means that they are equivalent circles. Therefore, in the ΔAOP’, AO=AP. When a triangle has two equivalent sides, it is an isosceles triangle. By that logic, ∠O=∠P’.
Now, I looked at the triangle that is already drawn in the above figure, ΔAOP. We know that this triangle is also isosceles because OP=AP. By that logic, ∠A=∠O.
Using the law of cosines c^2=a^2+b^2-2abcos(C), which works for...

...Name: Linh Nguyen
IB MathIA
02/06/12
Part A
Consider this 2× 2 system of linear equations: x + 2y = 3
2x - y = -4
We can see patterns in the constants of both equations. In the first equation, the constants are 1, 2, and 3. The common difference between the constants is 1:
3 – 2 = 1
2 – 1 = 1
Based on this, we can set up a general formula for the constant of this equation:
Un = U1 + (n - 1)d Where:
n: The number of the series
d: the common difference in the series.
For the second equation, we also can see that the constants belong to a arithmetic series, which has the common difference of -3:
-1 – 2 = -3
-4 – (-1) = -3
Un = U1 + (n+3)d
Solving the equation:
x + 2y = 3
2x – y = -4
* 2x + 4y = 6
2x – y = -4
* 5y = 10
* y = 2
* x + 4 = 3
* x = -1
Looking at the graph, we can see the intersection point A has the coordinate (-1, 2). At this point, the two lines are equal.
Solving equation with similar formats:
x + 6y = 11 Common difference is 5
6x – 2y = -10 Common difference is -8
* 6x + 36y = 66
* 38y = 76
* y = 2
* x = -1
6x + 7y = 8 Common difference is 1
2x + 4y = 6 Common difference is 2
* 6x + 12y = 18
* 5y = 10
* y = 2
* x = -1
Looking at these systems of linear...