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International Baccalaureate: Maths
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In this task, we are going to show how any two vectors are at right angles to each other by using patterns with vectors.
and then a direction vector . The line that goes through the set of points has an arrow at the end to show its direction. Two vectors are at right angles (perpendicular) to each other when they satisfy the equation Results/ Analysis By plotting the vector equation:, After plotting this vector equation, the first step is to find various points in time. That is to substitute values for (t) to find the coordinates of the object. t = 0, , , t = 1, , , t = 2, , , t = 3, , , t = 4, , From these points for t it is now possible to plot them on a set of Cartesian axes.
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236 Table 1.1 Processed data from Table 1.0 for the height of the gold medallist in high jump from 1932 until 1980 with 1896 as year 0 Plotted on a graph the data from Table 1.1 results in this graph. Figure 1.0 Graph from Table 1.1 with years x versus Height in centimetres h Figure 1.1 Graph from Table 1.1 with years x versus Height in centimetres h (Identical to Figure 1.0 inserted here again to make reading more convenient)
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Although the data does not tell us the reason why the height increased in 1932 and 1936 and drop down abruptly in 1948, we can assume that the World War II had affected athletics health critically. With this graph and the numbers given in the table I am able to develop a model of function that fits the data points in my graph by using Geogebra software. I chose linear function to model it, because the graph shows a general increase on the heights from 1948-1980 so in my opinion the linear will be the best function for it.
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Stellar numbers. The aim of the current investigation is to consider different geometric shapes, which lead to specific numbers, to formulate the universal formulas
Procedure 1. The first step is to work out the sequence of triangular pattern of evenly spaced dots. The number of dots in each are examples of triangular numbers (1, 3, 6,...). 1 3 6 10 15 * To complete the triangular numbers sequence with three more terms. 21 28 36 * To find the general statement that represents the nth triangular in terms of n. Term Number Number of Dots nth triangular in terms of n n = 1 u1 = 1 u1 = u1 = 1 n = 2 u2 = 3 u2 = u1 + n =
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Pascal's triangle 1 1 1 1 1 1 1 1 1 1 LACSAP'S Fractions. Since I'm trying to find the numerator, I was looking for any similarities between the two triangles. As it so happens, in Pascal's triangle, when r = 2, the diagonal sequence is the same as the numerator order in LACSAP's fractions. This sequence is called the Triangular sequence. Here is a table to show this. R = 2 (Pascal's Triangle) Numerator (LACSAP'S Fractions) 1 1 3 3 6 6 10 10 15 15 Knowing this, in the equation nCr, r would have =2 in this case.
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1st row = 2 2nd row = 2 2 + 2 4 N3 (3rd) 1st row = 3 2nd row = 3 3rd row = 3 3 + 3 + 3 9 N4 (4th) 1st row = 4 2nd row = 4 3rd row = 4 4th row = 4 4 + 4 + 4 + 4 16 When we compare the square numbers to the triangular numbers, a connection is discovered. We can see that: Qualitative Quantitative N1 (triangular numbers) + N2 (triangular numbers) = N2 (square numbers) 1 + 3 = 4 N2 (triangular numbers)
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Stellar Numbers. In this study, we analyze geometrical shapes, which lead to special numbers. We consider various geometrical patterns of evenly spaced dots and derive general statements of the nth special numbers in terms of n.
plus . Lets denote this number by Gn Using this formula, we can calculate the number of dots in the triangular number 6: In the triangular number 7: And in the triangular number 8: and etc. The following table can demonstrate this sequence: - Number of the Triangular Pattern - nth Triangular Number 1 1 2 3 2 3 6 3 4 10 4 5 15 5 6 21 6 7 28 7 8 36 8 The last 3 triangular numbers can be graphically demonstrated as following.
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Tn = T1 + (Sumn - 1) Tn = T1 + [(n-1) �2] x [2 x AP + (n - 1 -1) d] Tn = 1 + [(n-1) �2] x [2 x 2 + (n - 2)] Tn = 1 + [(n-1) �2] x [4 + (n - 2)] Or Tn = n + [(n-1) �2] x [2 x T1 + (n - 1 -1) d] Tn = n + [(n-1) �2] x [2 x 1 + (n - 1 -1)1] Tn = n + [(n-1)
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SL Type 1 PF - Infinite Summation - A general statement has been reached, taking a step further into knowing the infinity.
Calculations of the first 10 terms are implemented via GDC and are shown in the following equations. The data is also organized in Table 1. (Note: All data in this assignment are correct to six decimal points.) 0 1.000000 1 1.693147 2 1.933374 3 1.988878 4 1.998496 5 1.999829 6 1.999983 7 1.999999 8 2.000000 9 2.000000 10 2.000000 Table 1: Values of when , , In order to visualize the set of data, a graph can be plotted based on the values of .
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Analysis of Functions. The factors of decreasing and decreasing intervals (in the y axis) in a polynomial function depend on the turning points of the function.
It is defined as a function for which f(x+a) = f(x), where T is the period of the function. In the case of polynomial functions, clearly the only exception that can't be a periodic function is that there is no definite or constant period like "a". The relation of periodicity, however, holds for any change to x, so it can also be accepted the idea that polynomial functions are periodic functions with no period. There are two types of maximums and minimums, which are relative and absolute. In polynomial functions there can be both of them, but always just one absolute maximum and one minimum (if the degree of the function is odd), or two maximums or minimums (if
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SHORT EXPLANATION: Each *point* is considered as a separate task. Task 3 $ 4 are resolved into one place because both are closely linked. TASK 1 -------------------------------------- Explanation: Note that Ann wins if her die is higher than Bob's die, Bob wins in the event of a tie, and neither wins if Bob's dice is higher than Ann's - we should clearly see that Ann's probability of winning should be higher than Bobs's and the probability of NONE Wins should be also higher than the probability of Bob's win, while probablity of Ann's Win and NONE Wins should be the same.
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Solving the surd was done in the following method. The exact value of the infinite surd is. Only the positive is used as you cannot have a real negative surd. In addition, 1.61803398 which is very close to the value for however the gaps between the values for become smaller and the difference between and is approximately -3.72528E-06 which is a minute change compared to the gap between and the value at which . These results support the answer which was provided algebraically due to the small margin between the calculated and algebraic results.
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Moss's Egg. Task -1- Find the area of the shaded region inside the two circles shown below. The two large circles have a radius of 6cm.
The area of the small circle can therefore be calculated using the formula indicated: , where A equals the area and r is the radius. Thus: A = (32) = 9 � 28.3 cm2 Task -2- The same circles are shown below. Find the area and perimeter of the triangle ABC. a) In order to determine the area of triangle ABC, we must adopt the formula: , where b is the base and h is the perpendicular height. From the information we are given, we know that the base of the triangle is 6 cm in length, as we know that this is the diameter of the small circle and that the base extends to its ends (A to B).
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While the general population may be 15% left handed, MENSA membership is populated to 20% left handed people (Left Handed Facts). Intrigued by the statistics, I decided it would be interesting to do some independent exploratory research in
Surveying solely students that are enrolled in the International Baccalaureate program will ensure that the coursework is consistent in difficulty level and that the GPAs are weighted the same, with 5.0 being the highest attainable grade point average due to the weighted grade scale of the rigorous classes. I plan to collect data via an online survey. The most efficient way that I can reach the greatest number of students is though the social networking website Facebook.com. In creating a Facebook poll, every one of my friends will be able to see the poll and answer quickly and conveniently from their household.
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The linear function does not give us very accurate results about the long run as well, and it is most suitable just for portraying present data, because with this trend, it will show that the population will grow continuously and constantly up to infinity. Points Value of the table (a) Value of the linear curve (b) Difference between the values (b-a) Systematic error percentage ((b-a)/b)*100 50 554.8 531.3 -23.5 -4.4 55 609.0 609.3 0.3 0.05 60 657.5 687.3 29.8 4.3 65 729.2 765.3 36.1 4.7 70 830.7 843.3 12.6 1.5 75 927.8 921.3 6.5 -0.7 80 998.9 999.3 0.4 0.04
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show that the relationship between n and L can be represented by L= an The data begins to increase by a smaller amount about each consecutive n, suggesting that the data may be approaching as asymptote. As these values get very large, they will probably not get much higher than the value of a10, because there already appears to be almost horizontal trend.
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Linear Function: y=ax+b First, I use two points of (1955, 609) and (1985, 1070) from the given data to find the parameters of the function. I chose these two points because they are the only two combinations that are all integers, which will be easier to calculate. 609=1955a+b 1070=1985a+b Finding value a and b by using simultaneous equation: 1070=1985a+b - 609=1955a+b ________________________________ 461= 30a a = ?15.37 b = 609-1955�15.37=29439.35 Which gives the equation of y=15.37x-29439.35 as below: Graph 2: Population of China 1950~1995 with linear function It appears to be rather fit to the coordinates.
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The purpose of this paper is to investigate an infinite summation patter where Ln(a) is a constant and the coefficient of x is an increasing factor to Ln(a).
Sn approaches a horizontal asymptote when y=4. There is a y-intercept at (0,1). Checks: tn = n t(n) S(n) 0.000000 1.000000 1.000000 1.000000 1.609438 2.609438 2.000000 1.295145 3.904583 3.000000 0.694819 4.599402 4.000000 0.279567 4.878969 5.000000 0.089989 4.968958 6.000000 0.024139 4.993096 7.000000 0.005550 4.998646 8.000000 0.001117 4.999763 9.000000 0.000200 4.999962 10.000000 0.000032 4.999995 As n � +?, Sn � +5 There is a horizontal asymptote as n approaches positive infinite (?). As n approaches positive infinite then Sn will approach positive five. N t(n) S(n) 0.000000 1.000000 1.000000 1.000000 1.791759 2.791759 2.000000 1.605201 4.396960 3.000000 0.958711 5.355672 4.000000 0.429445 5.785117 5.000000 0.153892 5.939009 6.000000 0.045956 5.984966 7.000000 0.011763 5.996729 8.000000 0.002635 5.999364 9.000000 0.000525 5.999888 10.000000 0.000094 5.999982 As n � +?, Sn � +6 Sn approaches a horizontal asymptote when y=6.
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How many pieces? In this study, the maximum number of parts obtained by n cuts of a four dimensional object will be analyzed by looking at the patterns of the maximum number of segments made by n cuts on a one dimensional line, the maximum number of regio
it can be seen that the common difference, d, is 1 since the values of S increase by 1 as n increases by 1. If the numbers for the variables are plugged into the equation, an = 2 + (n - 1)1, the equation for an would be an = n + 1, so the rule for S to obtain the maximum amount of cuts per segment would be S = n + 1. This result also shows a linear regression line, so the equation to find the maximum amount of cuts for a one dimensional object would use a linear regression.
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It is of course not perfectly linear because the growth of a population would not be perfectly linear. This data is close to linear but it is not perfectly linear. A linear function is plausible. This is a graph with the data points on it as well as a linear function: y = 15.496x - 29690.2501 The function of the parent linear function is If in the TI-84 calculator you do a linear regression. You first need to put all the data from the years 1950 to 1995 into L1 and the corresponding population data into L2. After you have put the data in the L1 and L2 lists. After the data has been entered into the lists.
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On the middle part an equilateral triangle is drawn. And this continues in the following stages. Below are the values for the first 4 diagrams, . n Nn Ln Pn An 0 3 1 3 1 12 4 0.57735 2 48 0.64150 3 192 0.67001 (5 s.f) Number of sides Initially there is an equilateral triangle at stage n=0, each of those sides is divided into 3 sides. And there is another equilateral triangle created at the center points. Thus from that I can conclude that each side becomes from sides. as the diagrams are given to us, for the first 4 stages I counted the sides.
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tn), it is clear that a quadratic equation is formed by this relation, because the second differences are constant and not zero. n tn First differences Second differences Third differences 1 1 / / / 2 3 2 / / 3 6 3 1 / 4 10 4 1 0 5 15 5 1 0 6 21 6 1 0 7 28 7 1 0 8 36 8 1 0 Let f equal the value of tn . Consider any quadratic equation which can be represented as.
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Stellar Numbers Portfolio. The simplest example of these is square numbers, but over the course of this investigation, both triangular numbers and stellar numbers will be looked at in greater depth.
Complete the triangular numbers sequence with three more terms. The given diagrams for the problem were through and are shown below. The next three terms needed to complete the diagram are through . These are illustrated in the pictures below. The white lines on bottom of each diagram are the dots that need to be added to go from one triangular number to the next. Question 2 Find a general statement that represents the nth triangular number in terms of n When completing different trials in attempts to find the general term for Triangular Numbers, the following information was determined.
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Maths IA Type 2 Modelling a Functional Building. The independent variable in this investigation is the height of the building. The maximum volume of a cuboid under the roof depends on the height of the roof, which is the dependant variable.
The intercept will be modelled at the maximum height of the parabola Let intercept ? The roof is at a maximum turning point ? ? The form of the model quadratic is: To solve for , substitute in the known fixed values (the root) for and (36,0): Substitute all these values into the general form of the quadratic to obtain the general formula of the roof: Modelling at Minimum Height: Below in Fig 1 is a model of the roof structure with the minimum height of 36m, where the fa�ade is designed at the width.
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Logarithms. In this investigation, the use of the properties of logarithms will be used to identify patterns, relationships, and limits of logarithms and sequences.
A geometric sequence is a sequence whose consecutive terms are multiplied by a fixed, non-zero real number called a common ratio. Therefore, to prove that the sequence is indeed, geometric, the common ratio must be found. When the sequence is examined closely, you will notice that the log number remains the same throughout the sequence and that it is the base of the logarithm that changes with every term. Understanding this, you can use the bases to find the common ratio of the sequence by dividing the consecutive terms by the terms preceding them.
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