Winning Heights in the Men’s Olympic High Jumps
Introduction:
The Olympics are an international sporting event that is held every four years where people from around the world send their best athletes to compete and see who the best of the best is when it comes to sports. The Olympics date back to Ancient Greece where their basic events included track and field, which are like men’s high jump which is the topic of this report. In this problem we are looking at the data collected from the gold medalists in the men’s high jump in from the years 1932-1980, excluding the years 1940 and 1944 where the Olympics were not held due to World War II. Below is the data for the gold medalists in the men’s high jump from the years 1932-1980 (again excluding 1940 and 1944). Year| 1932| 1936| 1948| 1952| 1956| 1960| 1964| 1968| 1972| 1976| 198| Height (cm)| 197| 203| 198| 204| 212| 216| 218| 224| 223| 225| 236|

Data:
The problem asks for the points in the table to be plotted and I chose to put them into a scatter plot because that is the most logical way of graphing the data shown in the table above. Because the data is not continuous you would not make a line graph. The graph is shown below:

The graph shows that the data collected is for the most part an upward trend. You can tell this because most of the data goes up in a positive trend the majority of the time.
The problem asks for a function to describe the way that the graph moves up, for the most part, in a positive way and I believe the best way to show this trend would be to use a line graph. I feel like this would be the best representation of showing how the points on the graph have a positive trend because you can see the average trend between all the points in the graph. What I did to find the slope was to take the x and y coordinates of both the first and the last plots on the graph and plug them into the equation m=y2-y1x2-x1 where y and x are the x and y components of...

...Jonghyun Choe
March 25 2011
Math IB SL
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 1
1 1
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
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 order to obtain a general statement for the pattern, two different statements will be needed to combine to form one final statement. This means that there will be two different statements, one that...

...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 equations, we can conclude that any equations which have their constants follow a arithmetic series will have the answer for the variable y as 2 and variable x as -1....

...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 (Type 1)
Check for the following:
• I explicitly defined variables and parameters the first time I used them, even if they were
already defined in...

...Alma Guadalupe Luna
MathIA (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 be understood...

...Lacsap’s Fractions
The aim if this IA is to investigate Lacsap’s Fractions and to come up with a general statement for finding the terms.
When I noticed that Lacsap was Pascal spelt backwards I decided to look for a connection with Pascal’s triangle.
Pascal’s triangle is used to show the numbers of ‘n’ choose ‘r’(nCr). The row number represents the value of ‘and the column number represents the ‘r’ value. Eg. Row 3, colomn 2 = 3C2 = 2.
I noticed that all the numerators of the fractions in Lascap’s fraction (3,6,10,15) are also found in Pascal’s triangle. So I tried to see if I would get the denominator of the fractions by using the row as ‘n’ and the colomn (or element) as ‘r’. This did not work out because Lascap’s triangle does not have a row with only one element like Pascal’s does. To solve this I just added 1 to each row number. This gives me the formula[pic].
|(Row number +1)C2 |Numerator |
|(2+1)C2 |= 3 |
|(3+1) C2 |= 6 |
|(4+1)C2 |=10 |
|(5+1)C2 |=15 |
Now that we have found an equation to solve to numerator of the fractions, we now have to try and work out how to solve the denominator....

...EUROPEAN SCHOOL
Mathematics Higher Level
Portfolio
Type 1
SHADOW FUNCTIONS
Candidate Name: Emil Abrahamyan
Candidate Number: 006343-021
Supervisor: Avtandil Gagnidze
Session Year: 2013 May
Candidate Name: Emil Abrahamyan
Candidate Number: 006343-021
Mathematics Higher Level
Type 1: Shadow Functions
SHADOW FUNCTIONS
The Aim of the Investigation:
The overall aim of this investigation is to investigate different polynomials with different powers
and create shadow function for each one. Afterwards identify the real and imaginary components
of complex zeros from the key points along the x-axis using the method of shadow functions and
their generators.
Technology Used:
Technology that had been used is shown below
1)
Autograph (Version 3.3)
Graphing Display Calculator TI-84 Plus Texas Instruments
2)
Defining terms:i
Quadratic, cubic, quartic functions are members of the family of polynomials.
A quadratic function is a function of the form
constants and
A cubic function is a function of the form
are constants and
A quartic function is a function of the form
where
are constants and
Complex numbers is any number of the form
where ,
where
are
where
are real and
.
The vertex of parabola is point where the parabola crosses its axes of symmetry.
ii
nd
Urban, P., Martin, D., Haese, R., Haese, S., Haese M. and Humphries, M. (2008) Mathematics HL (Core). 2 ed.;
Adelaide Airport: Haese & Harris...

...In Lacsap’s Fractions, En(r) refers to the (r+1)th term in the nth row. The numerator and denominator are found separately, therefore to find the general statement, two different equations, one for the numerator and one for the denominator, must be found. Let M=numerator and let D=denominator so that En(r) = M/D.
To find the numerator for any number of Lacsap’s Fractions, an equation must be made that uses the row number to find the numerator. Because the numerator changes depending on the row, the two variables (row number and numerator) must be compared. To find this equation, the relationship between the row number and numerator must be found, put it graph form, and the equation for the graph will be the equation needed.
Row Number, n | Numerator, N |
1 | 1 |
2 | 3 |
3 | 6 |
4 | 10 |
5 | 15 |
Numerator
Numerator
Row Number
Row Number
The equation for the numerator can be derived by using quadratic regression on a graphing calculator. The equation is; y = .5x2 + .5x. This translates into; M=.5n2+.5n, where n=row number, and M=numerator. This means that any numerator from a certain row number can be found by using this equation. For example, to find the numerator of the sixth row, “6” needs to be substituted in for n.
M= .5n2 + .5n
M= .5(6)2 + .5(6)
M= .5(36) + .5(6)
M= 18 + 3
M= 21
The Numerator for row six is 21
They method to find the equation for the denominator is similar, but slightly more difficult...

...Cindy Hwang
IB Math 11 SL 1-3
IB Math Portfolio
Gold medal heights
Aim:
The aim of this task is to consider the winning height for the men’s high jump in the Olympic Games.
Introduction:
The Olympic Games which are held in every four years have an event called Men’s High Jump and usually an athlete tries to jump over a bar which is set up in a certain range from1 meter to 3 meters. The table 1 shows the record of the gold medalists during the Olympic Games held during 1932 to 1980. Note: The Olympic Games were not held in 1940 and 1944.
Table 1
Year | 1932 | 1936 | 1948 | 1952 | 1956 | 1960 | 1964 | 1968 | 1972 | 1976 | 1980 |
Height(cm) | 197 | 203 | 198 | 204 | 212 | 216 | 218 | 224 | 223 | 225 | 236 |
Variables:
Independent Variables:
The heights that are achieved by the gold medalists
Dependent Variables:
The years of Olympic Games
Figure 1:
From the figure 1, the independent variables are the x axis which shows the years of the Olympic Games, and y axis is the dependent variables which represents the heights that are achieved by the gold medalists. Also it shows that it is not constant.
Linear Regression
To create a certain equation, you draw the best fit line on the graph.
The difference between the red graph and the linear function is that the red does not have a predictable pattern. When the best fit is drawn it...

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