Section 1
Decimals are a part of our everyday life in some way, when we put fuel in our cars to buying meat from the butcher. Mastering this critical mathematical concept is a necessity (Stephanie Welch, 2010). A decimal is a proper fraction, which is a number less than 1. It is a part of a whole number. Since our numbering system is based on the powers of 10, it is called a decimal system. Decem in Latin means ten (The Maths Page, 2012, Lesson 3). Decimal fractions are represented as the numbers found between two whole numbers. The decimal fraction shows part of a whole number and is written after the decimal place.

Some key understandings in learning about decimals would be-

* the idea that there are numbers between two consecutive whole numbers, for example between 6 and 7 is 6.54. * the place value system can be extended to the right to show the numbers between two whole numbers * to record a number you write the whole number followed by a decimal point then the part of the number * the numbers to the right of the decimal point have decreasing values in powers of ten ie. 1/10, 1/100, 1/1000 and so on. * decimal numbers can be partitioned just like whole numbers (0.84 = 8/10+4/100 or 84/100 or 840/1000)

Prior to learning about decimal numbers students must have a clear understanding of place value, ordering and rounding whole numbers. Without this secure understanding and ability to work with whole numbers, students will not have the prerequisite skills and understanding to move into decimal numbers. Many of the misconceptions students have with decimals arise from the lack of confidence and skills with whole numbers.

When representing whole numbers and parts of whole numbers the decimal place is a separator between the whole number and the smaller part of the whole number. A major misconception students have with decimals is the idea that the decimal place separates...

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IB Math: Studies Statistics
Portfolio:
What is the relationship between the numbers of goals the top sixteen players of the FIFA world cup 2014 score with the height of the players?
Due date: Friday November 23, 2013
School Name: Franklin Delano Roosevelt
Course: IB Math Studies
Name: Valerie Philco
What is the relationship between football player’s height that is participating in FIFA and has scored more than fourteen goals with the number of goals they have scored?
Introduction
The FIFA world cup is one of the most celebrated soccer tournaments around the world today. Not only does it serve as entertainment to everyone that watches the tournament but to participate teams have to have play a game against every team of there area teams classified this year include country’s like Argentina, Spain, Bosnia, Brazil, USA, Italy, France and many more. This being said for teams to win games they have to score more goals than other teams in this investigation the purpose is to verify if there is a correlation between the height of soccer players and the amount of goals they score.
Statement of task
The main purpose of this investigation is to determine whether there is a relationship between the height of the player and the amount of scores that they scored through out the period of qualifications for the 2014 FIFA world cup Brazil. The type of...

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ANALYSIS
Physics has a lot of topics to cover. In the previous experiments, we discussed Forces, Kinematics, and Motions. In this experiment, the focus is all about Friction. Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction like fluid friction which describes the friction between layers of a viscous fluid that are moving relative to each other; dry friction which resists relative lateral motion of two solid surfaces in contact and is subdivided into static friction between non-moving surfaces, and kinetic friction between moving surfaces; lubricated friction which is a case of fluid friction where a fluid separates two solid surfaces; skin friction which is a component of drag, the force resisting the motion of a fluid across the surface of a body; internal friction is the force resisting motion between the elements making up a solid material while it undergoes deformation and sliding friction.
When surfaces in contact move relative to each other, the friction between the two surfaces converts kinetic energy into heat. This property can have dramatic consequences, as illustrated by the use of friction created by rubbing pieces of wood together to start a fire. Kinetic energy is converted to heat whenever motion with friction occurs, for example when a viscous fluid is stirred. Another important consequence of many types of friction can be wear,...

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The case between Beauty and Stylish involves concept of a valid contract, pre-contractual statements, express term and misrepresentation.
A valid contract is established between Beauty and Stylish when an offer is accepted and there is intention for both parties to create legal relations. An offer refers to the expression of willingness of the offerer to be contractually bound by an agreement if his or her offer is properly accepted. It has to be clear and certain in terms. It must also be communicated to the offeree before it is being accepted. In addition, the acceptance has to be unqualified, unconditional and made by a positive act. In the case of Beauty and Stylish, a positive act refers to the signing of the contract. All terms of the offer must be accepted without any changes and cannot be subjected to any condition, taking effect only upon fulfillment of that condition. When Beauty and Stylish enter into the agreement, they must intend to bind and bound legally to each other by their agreement. This is the intention to create legal relations between two parties. In the meanwhile, this contract must possess consideration. A contract must therefore be a two-sided affair, with each side providing or promising to provide something of value in exchange for what the other is to provide.
Every contract, whether oral or written, contain terms. The terms of a contract set out the rights and duties of the parties. Terms are the promises and undertakings given by each...

...1 1
1 32 1
1 64 64 1
1 107 106 107 1
1 1511 159 159 1511 1
The aim of this task is to find the general statement for En(r). Let En(r) be the element in the nth row, starting with r = 0.
First to find the numerator of the sixth row, the pattern for the numerator for the first five rows is observed. Since the numerator is the same in each row (not counting the first and the last number in each row), I can observe the numerator in the middle of each row. The numerators from row 1 to row 5 are 1,3,6,10,15
Table 1: A table showing the relationship between the row number and the numerator. The table also shows the relationship between the numerators in each row.
Row | Numerator | 1st differences | 2nd differences |
1 | 1 | 2 | 1 |
2 | 3 | | |
| | 3 | |
3 | 6 | | 1 |
| | 4 | |
4 | 10 | | 1 |
| | 5 | |
5 | 15 | | |
The difference between the numerator in row 1 and row 2 is 2, row 2 and row 3 is 3, row 3 and 4 is 4 and row 4 and 5 is 5. The second difference for each row number is 1; this shows that the equation for the numerator is a geometric sequence. So I try to find the equation of the sequence by using the quadratic formula, y = ax2 + bx + c, where y = the numerator and x = the row number.
6 = a(3)2 + b(3) +0
6 = 9a +3b
6 = 9a + 3(-2a + 1.5)
6 = 9a – 6a + 4.5...

...MATHS PORTFORLIO SL TYPE I
CIRCLES
In this portfolio I am investigating the positions of points in intersecting circles. (These are shown on the following page.
The following diagram shows a circle C1 with centre O and radius r, and any point P.
The circle C2 has centre P and radius OP. Let A be one of the points of intersection of C1 and C2. Circle C3 has the centre A, and radius r. The point Pʹ is the intersection of C3 with (OP). This is shown in the diagram below.
As shown on the assignment sheet, r=OA. We therefore need to find the values of OPʹ when r=OA=1 for the following of the values of OP: OP=2, OP=3 and OP=4.
We first of all extract the triangle OPA from the above diagram and since we have the values for all the three sides we can finds the angle AOP which will later on help to get the value of length OPʹ. The circle C2 and triangle OPA are shown below with all side of OPA indicated.
OP=AP since they are the radii of the same circle, C2. Having all the three sides, we can now calubulate the angle AOP using the cosine rule. Angle AOP is calculated below:
Cos AOP=(2^(2 _ ) 2^(2 _ ) 1^2)/(-2×2×1)
Cos AOP=0.25
∴ AOP =COS-10.25
=75.52248781
≈75.5˚...

...MathPortfolio SL TYPE I
LACSAP’S FRACTIONS
Introduction
This assignment requires us to solve patterns in numerators and denominators in LACSAP’S FRACTIONS, and the first five rows look like:
Figure 1: Lacsap’s Fractions
1 1st row
1 3/2 1 2nd row
1 6/4 6/4 1 3rd row
1 10/7 10/6 10/7 1 4th row
1 15/11 15/9 15/9 15/11 1 5th row
Then, let’s look at each part of the question.
Part 1: Numerator of the Sixth Row
Describe how to find the numerator of the sixth row.
For the first part of the question, we need to describe how to find the numerator of the sixth row. To begin with, let’s make all the numerators look the same in a row:
Figure 2: Lacsap’s Fractions
1/1 1/1 1st row
3/3 3/2 3/3 2nd row
6/6 6/4 6/4 6/6 3rd row
10/10 10/7 10/6 10/7 10/10 4th row
15/15 15/11 15/9 15/9 15/11 15/15 5th row
Then, we can take out the denominator and go down by the row and just look at the numerator for this part of question, which will look like:
t1=1; t2=3; t3=6; t4=10; t5=15
We can investigate that the numerator starts with 1 in the first row, and the numerators are the same in one row. Each numerator after 1 is the sum of 1+2+…+n (n=row number) which forms...

...especially triangles. Angles are mostly what decide the shape of triangles. This activity was about grouping similar angles from a set of parallel line with another line intersecting both of them. This activity had an important connection to figuring out the final shadows equation because we put the problems in terms of triangles and triangles are heavily linked with angles. After all, triangles do mean three angles.
Compared to math class back in elementary school, IMP is totally different. When I used to just solve given equations from a book and memorize methods to solve various problems. Now I actually have to create my own equation and work with that. But honestly speaking I preferred doing math the old way. It felt more normal and more like math. Since I started IMP, math has begun to feel more like science class, with different experiments and no real rules to solving the problems in front of you. At least now I have kind of adjusted to this new way but before it was quite overwhelming.
Since starting IMP, I have started disliking math more, but not really in a bad way. I think because I don't like it as much I spend more time doing the homework assignments trying to understand the material so I will not have problems and fall behind. So my goals have changed compared to what they used to be in eighth grade. Now I am more determined to do really well because I am in a class...

...STELLAR NUMBERS
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 stellar numbers 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 applying the following formulae:
Now it is possible to find the nth number by using the formula, going back to the example where n is 592, so we replace n by 592 and solve the equation as follows: so the 592nd term will contain 175528 dots....