Purpose of Investigation
The purpose of this investigation is to find out the general trends of the Olympic gold medal height each time the event is held. It also could be used to predict the next gold medal height in the upcoming Olympic events. We could know as well what functions can be used to plot the graphs. People could also analyze the pattern of rise or decrease in height of the winning height in the Olympic game.

This investigation also allows future participants to find out information about previous gold medal heights and can make them easier to set targets for their performance in the Olympic Games.

a.

b.The function I used for constructing this graph is Gaussian. Because from the list of functions in the graph-constructing program, the Gaussian function is the most accurate shape when plotted according to the data given which is the statistics of height of gold medalist for men’s high jump in the Olympic. The technology I used to plot all the graphs is Logger Pro 3.50.

c.

The difference is not significant after I adjust it. It can be seen from the graph itself that the shape of it is also similar. The limitation is that it is plotted on a lower values compared to the original and so it couldn’t reach the 1980 mark like the original graph.

d.

With the technology I used, the function I find to have a similar shape is the cubic function which is the red line. There are some period that the cubic function has a higher value. Also there are some points where the two graphs meet. But in the end the cubic is lower than the original and couldn’t reach the 1980 mark. e.According to the original graph, if in the year 1940 and 1944, the estimated height will be 198.2 and 199.6 respectively. f.Estimated winning heights in 2984 and 2016 are 235 and 241 respectively. From the data, the winning...

...Introduction:
The task of this assessment is to consider and investigate geometric shapes that lead to stellar numbers. One of the simplest forms of this would be of square numbers, terms 1, 2, 3, 4, and 5 in terms of geometric shapes lead to the special numbers of 1, 4, 9, 16, and 25. To find these outcomes, simply all that is needed to be done is to square the nth term of the sequence. For example, 12=1, 22=4, 32=9, 42=16, and 52=25, therefore the resulting formula for square numbers would be n2=Tn where T=total amount of n. For this investigation though, the geometric shapes of equilateral triangles are studied. Triangular numbers are the amount of dots that evenly fit inside the triangle. After solving the sequence of the triangular numbers and finding a formula to solve for all equilateral triangles, the knowledge attained from triangular numbers will then be used to find and solve the correlation between triangular numbers and stellar (star) numbers.
Here is a sample of the first 5 terms in the triangular numbers sequence:
The row of numbers above the triangles is the number of each term or n.
The rows of numbers below the triangles are triangular numbers of each term or n.
` n1 n2 n3 n4 n5
Here is a sample of the first 4 stages of p-stellar shapes that are being investigated. P indicates the amount of vertices in the stellar shape. In this case the stellar shapes being investigated are 6-stellar, however...

...
Laurie Scott
SLMath Internal Assessment
Mr. Winningham
9/5/12
Instructions: In this task you will consider a set of numbers that are presented in a symmetrical pattern.
Pascal’s Triangle
|n=0 |1 |
|1 |0 |
|2 |3 |
|3 |6 |
|4 |10 |
|5 |15 |
|6 |21 |
Table 1: Relationship between Row Number and Numerator of Figure 2
[pic]
Figure 3: Graph of the relationship between Row Number and Numerator of Figure 2
In order to find the sixth and seventh rows, a pattern for determining the denominator must be found:
First it is helpful to determine a relationship between the numerator and denominator of the first term in each row:
|Row Number ( n ) |Difference of Numerator and |
| |Denominator (1st term) |
|1 |0 |
|2 |1 |
|3 |2 |
|4 |3 |
|5 |4 |
Table 2: Relationship between Row Number and the difference of the Numerator and...

...Lacsap’s Fractions
IBMathSL
Internal Assessment Paper 1
Lacsap’s Fractions
Lacsap is Pascal spelled backward. Therefore, Pascal’s Triangle can be used practically especially with this diagram.
(Diagram 1)
This diagram is of Pascal’s Triangle and shows the relationship of the row number, n, and the diagonal columns, r. This is evident in Lacsap’s Fractions as well, and can be used to help understand some of the following questions.
Solutions
Describe how to find the numerator of the sixth row.
There are multiple methods for finding the numerator of each consecutive row; one way is with the use of a formula, and another by using a diagonal method of counting illustrated by a diagram.
The following image can be used to demonstrate both techniques to finding the numerator:
(Diagram 2)
This formula uses “n” as the row number and the outcome is the numerator of the requested sixth row.
n2 + n
2
As indicated, inputting the number 6, as the requested sixth row, for n gives the solution of 21.
X = n2 + n
2
X = (6)2 + (6)
2
X = 36 + 6
2
X = 42
2
X = 21
Therefore, as shown, the numerator of the sixth row is 21, and this can be checked for validity by entering each number, 1 through 5, into the formula and making sure that the answer corresponds with the numerator in the above diagram.
Where n = 5:...

...Jonghyun Choe
March 25 2011
MathIBSL
Internal Assessment – LASCAP’S Fraction
The goal of this task is to consider a set of fractions which are presented in a symmetrical, recurring sequence, and to find a general statement for the pattern.
The presented pattern is:
Row 111
Row 21 32 1
Row 3
1 64 64 1
Row 4
1 107 106 107 1
Row 5
1 1511 159 159 1511 1
Step 1: This pattern is known as Lascap’s Fractions. En(r) will be used to represent the values involved in the pattern. r represents the element number, starting at r=0, and n represents the row number starting at n=1. So for instance, E52=159, the second element on the fifth row. Additionally, N will represent the value of the numerator and D value of the denominator.
To begin with, it is clear that in...

...IB Mathematics SL Year 1
Welcome to IB Mathematics. This two-year course is designed for students who have a strong foundation in basic mathematical concepts. The topics covered in this course include:
* Algebra
* Functions
* Equations
* Circular functions
* Trigonometry
* Vectors
* Statistics
* Probability
* Calculus
-------------------------------------------------
Resources:
* Textbook: Mathematics SL 3rd edition. Haese Mathematics 2012
ISBN: 978-1-921972-08-9
* Edmodo: A virtual learning website where students can get homework assignments, ask homework questions, access class documents, and communicate with peers. https://iam.edmodo.com/
* Class code: ybu2s4
* Quest: Virtual learning website where students will complete graded homework assignments. https://quest.cns.utexas.edu/
* Graphing Calculator. Recommended for this course and all subsequent math courses. The Ti-Nspire or Ti-84 plus silver edition are both good choices. http://education.ti.com
* Khan Academy: http://www.khanacademy.org. Will be used to provide students with alternative resources.
* Geometer’s Sketchpad: Graphing software. Will be installed on student computers. http://www.dynamicgeometry.com/
* Additional print, online resources and worksheets may be used....

...Fractions
IBMathSLSLType1
December 11, 2012
Lacsap’s Fractions:
Lacsap is Pascal backwards and the way that Lacsap’s fractions are presented is fairly similar to Pascal’s triangle. Thus, various aspects of Pascal’s triangle can be applied in Lacsap’s fraction.
To determine the numerators:
To determine the numerator (n), consider it in relation to the number of the row (r) that it is a part of.
Consider the five rows below:
Row 111
Row 2 1 32 1
Row 3 1 64 64 1
Row 4 1 107 106 107 1
Row 5 1 1511 159 159 1511 1
The relation between the numerator and the row number can be shown by the equation:
Where the numerator = n
And the row = r
0.5r2 + 0.5r = the numerator (n)
When r=1
0.5(12) + 0.5(1) = 1
When r=2
0.5(22) + 0.5(2) = 3
When r=3
0.5(32) + 0.5(3) =6
When r=4
0.5(42) + 0.5(4) = 10
When r=5
0.5(52) + 0.5(5) = 15
Therefore if you are attempting to find the 6th or 7th row numerators, you simply plug (n) into the...

...MathSL Portfolio – Tips and Reminders Checklist
Notation and Terminology
Check for the following:
• I did not use calculator notation. (I didn’t include things like ‘x^2’ for or Sn for Sn)
• I used appropriate mathematical vocabulary.
Communication
Check for the following:
• The reader will not need to refer to the list of questions in order to understand my work.
• My responses are not numbered.
• I have an introduction, conclusion, title page, and table of contents.
• All graphs are labeled – Each graph has a title, labeled axes, and appropriate scale.
• My graphs and tables are within the body of my work. They are not separate or in an appendix.
• I have explained why I made the choices I did when going through the task.
• I did not include key stroke sequences, e.g. “I pressed the 2nd key, then TRACE…”
• My tables do not straddle pages.
• My tables are labeled well, including my variable definitions in each column.
Use of Technology
Check for the following:
• I used technology to illustrate my points and ideas. I didn’t just “stick in” a graph.
• Each graph or table (or other piece of tech.) is accompanied by explanations and my ideas.
• I did not include too many graphs on the same axes – my graphs are easy to read.
Mathematical Process (Type1)
Check for the following:
• I explicitly defined variables and parameters the first time I used them, even if...

...Alma Guadalupe Luna
Math IA (SL TYPE1)
Circles
Circles
Introduction
The objective of this task is to explore the relationship between the positions of points within circles that intersect.
The first figure illustrates circle C1 with radius r, centre O, and any point P. r is the distance between the centre O and any point (such as A) of circle C1.
Figure 1
The second diagram shows circle C2 with radius OP and centre P, as well as circle C3 with radius r and centre A. An intersection between C1 and C2 is marked by point A. The intersection of C3 with OP is marked by point P’.
Figure 2
Through this investigation I will examine how the r values correlate with the values of OP in determining the length of OP’ when r is held as a constant variable and the value of OP is the variable that is subject to change. I will then venture on to study the inverse, the relationship when the r values becomes the variable that is changed and the OP value is held constant.
r as a Constant
If we let the value of r be equal to 1, we can use that information to find the length of OP’ when OP=2, 3, and 4. The first thing one can deduce is that by using the points A, O, P’, and P two isosceles triangles can be formed; ∆AOP and ∆AOP’. To rationalize this assertion through an analytic approach it should first...

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