My job is to investigate how many squares would be needed to make any cross shape like this build up in the same way.

Borders: Part 1 My job is to investigate how many squares would be needed to make any cross shape like this build up in the same way. Below are diagrams of the cross shape pattern to the 8th sequence: Here is my table of results, with the number of black, red and total squares Black Squares 4 8 2 6 20 24 28 32 Red Squares 5 3 25 41 61 85 13 Total Squares 5 3 25 41 61 85 13 45 Now so I can find an equation which will tell me how many squares there will be in each sequence, I will find the differences for the black squares first. To find the equation I will need bits of information, the equation Is an+b, where a is the difference and b is the 0th term, e.g. the first term here is 4, to find the 0th terms you need to go back one step (take the difference from the first terms to find the 0th term) and n is the term you want to find, e.g. if you want to find the 1st term n=1 if you want to find the second term n=2. a=4, b=0. Now I have the equation, 4n+0, now let's test it! We know the first terms is 4, so 4xn(1)+0, 4x1=4 so the answer is 4 and that is the 1st term. Now ill do the same with the red squares To find this equation I will need different bits of information, the equation is an2+bn+c, A is 1/2 of the second difference, in this case a = 2 B is complicated, to find b you will need to find c first and substitute a and c into the

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  • Level: AS and A Level
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In this piece of coursework, I will use three different methods (Change of Sign, Newton-Raphson, and the Rearrangement Method) to find the roots of a series of different equations. The number of roots found differs from method to method.

Pure Mathematics 2: Coursework Assessment In this piece of coursework, I will use three different methods (Change of Sign, Newton-Raphson, and the Rearrangement Method) to find the roots of a series of different equations. The number of roots found differs from method to method. A comparison of the three methods will be made at the end of the report. Change of Sign Method Equation: x3 + x2 - 2x + 3 = 0 Using the above equation, the function f(x) = x3 + x2 - 2x + 3 may be generated. This may be expressed in graphical form (see page 2) This may be solved using the method of Interval Bisection (see page 3) <--------Interval-------> Mid Point Height at A Height at B Height at M Curve a b Mid Point f(a) f(b) f(m) y=x^3+x^2-2*x+3 -3 -2 -2.5 -9 3 -1.375 -2.5 -2 -2.25 -1.375 3 .171875 -2.5 -2.25 -2.375 -1.375 .171875 -0.005859375 -2.375 -2.25 -2.3125 -0.005859375 .171875 0.606201172 -2.375 -2.3125 -2.34375 -0.005859375 0.606201172 0.306060791 -2.375 -2.34375 -2.359375 -0.005859375 0.306060791 0.151584625 -2.375 -2.359375 -2.3671875 -0.005859375 0.151584625 0.073235035 -2.375 -2.3671875 -2.37109375 -0.005859375 0.073235035 0.033781111 -2.375 -2.37109375 -2.373046875 -0.005859375 0.033781111 0.013984211 -2.375 -2.373046875 -2.374023438 -0.005859375 0.013984211 0.004068256 -2.375 -2.374023438

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Investigation on Boyles law

Investigation on Boyles law Apparatus Hypothesis My hypothesis of this experiment is that if a fixed (set) mass of air is made smaller then the molecules will be closer together, this will cause the molecules to bounce around more causing more pressure. I calculate that if you half the volume of the liquid then the pressure inside the air column will double. This will be caused by the number of molecules per cm3 inside the glass column being doubled. Therefore the results should be inversely proportional. The equation for inverse proportion is: V ? 1/p - This equation will form a reciprocal graph, which should look something like the above diagram. Method For this experiment we have to keep certain constants maintained during this experiment or it could make the test unfair and alter the results. We have to keep temperature the same for this test to work. We used a foot pump and pumped the air into the column, which increased the pressure. We then took readings of pressure and volume every five seconds and noted down the results. Then after we had taken about thirty results we re-tested them again to insure that the results were totally accurate and no error had occurred. From our results we had taken we plotted a graph with pressure (lb/in2) on the vertical axis and volume (cm3) on the horizontal axis. From all these points we drew a line of best fit. Analysis

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  • Level: AS and A Level
  • Subject: Maths
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Exponential decay:The objective of this experiment is to get the trend line of heart rate and to determine the decay constant of this decay.

Heart Rate .0 Introduction .1 Objective The objective of this experiment is to get the trend line of heart rate and to determine the decay constant of this decay. .2 Theory In a radioactive decay process, the rate at which the heartbeat decrease is proportional to the heartbeat after excising(N). where the decay constant is positive number called the decay constant related to differences between hearts.The solution to this equation is: Here is the heartbeat at time t, and = N (0) is the heartbeat after excising. 2.0 Experimental Heart rate decrease that occurs rapidly when after exercise which is entirely due to increase in cardiac vagal activity, and the heartbeat decrease follow the exponential decay. By recording the number of heartbeat in a fixed period of time, we can get the curve of it and determine the decay constant. 2.1 Apparatus 7 people, stop watch, playground. 2.2 Procedures . Everyone put their hands on carotid artery. 2. Timekeeper calculate the time use. 3. Once the timekeeper call "start!", students can begin to count the heartbeat. 4. Repeat 3 times and get 3 different groups of data. 5. All the students run 5 minutes continuously. 6. Repeat the procedure 1-3 every half a minutes for 7 times. 2.3 Data To get a obvious and simple curve, for each item, I just calculate the average value. Base heart rate Heart rate after t mins(after

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  • Level: AS and A Level
  • Subject: Maths
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Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically

Maths Core 3: Coursework Introduction Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically, also I will show each method working, and each method failing. The three methods are: * Change Of Sign Method * Rearranging f(x)= 0 Into The Form x=g(x) * Newton-Raphson Method Change of Sign Method Change of Sign Method Working I am going to solve the equation f(x) = 0, where f(x) =. The graph of y=f(x) is shown here. In order to calculate the value of the root in [0.3, 0.4] to 3 decimal places I must check whether the value is closer to 0.309 than 0.310 meaning that the value 0.3095 needs to be calculated. If this value is negative then the root is 0.309 to 3 decimal places however if the value is positive then the root would have to be given as 0.310. The value of f(x) when x=0.3095 is 0.00614708 meaning that I can deduce that the value of the root in [0, 1] to 3 decimal places is 0.310. The error bounds are 0.3095 and 0.310. Change of Sign Method Failing Consider the equation f(x) =0, where f(x) = I shall try to locate the roots by making a table of values that correspond to this function of x. Rearranging f(x) =0 into x=g(x) Rearranging f(x) =0 into x=g(x) failing I am going to solve the equation f(x) =0 where f(x) =. The

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  • Subject: Maths
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C3 Coursework - different methods of solving equations.

C3 Coursework - Numerical Solutions Decimal Search There a numerous ways to solve a problem and in finding the unknown. Some methods give you the exact and precise answer but usually are harder and more complex. The Decimal search method enables you to get a very close approximate to the real solution but more easily. The way this method works is by looking between two numerical values (for example 1 and 2) and then As a demonstration in applying this method, I will be attempting to solve this equation using the Decimal Search method and going through the method step by step: Below is what this function looks when plotted on a graph: We know that the solution for F(x) = 0 is the point on the X axis where the sign changes from a positive to a negative. So if we zoom in a little bit further, from this graph we can tell where the solution lies, somewhere between 0 and 10 Now that we know the solution is roughly between these two values, I will use excel to solve the problem with firstly taking increments in x, the size of 1. So when I substitute the incremented values of x between -10 and 0 into the equation, I get the following results: x F(x) -1 20 -2 9 -3 0 -4 -13 -5 -56 -6 -125 -7 -226 -8 -365 -9 -548 You can tell that the sign changed between -3 and -4. So I set these as my initial values. The fact that the solution lies between -3 and -4 can

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Fractals. In order to create a fractal, you will need to be acquainted with complex numbers. Complex numbers on a graph are characterized by the coordinates of (x,y)

Ruzeb Chowdhury June 09, 2012 Fractals Usually when one speaks of fractals to an audience, the audience refers to a piece of art that consists of repeating shapes and self-similar patterning (“same as near as far”). However, they do not realize that fractals were made and recognized due to a mathematical concept that was developed by many people such as Gottfried Leibniz, Georg Cantor, Waclaw Sierpiński, Gaston Julia, and Benoit Mandelbrot. Mandelbrot was the one to coin the term “fractal”, but the other mathematicians paved the way for Mandelbrot’s findings. Also, some people may think that there is just one way to make a fractal, which is by using math. This is not totally true, as there ways to make the same fractal in a different approach, such as by using the L-system. A fractal created with the Julia set under constant -.74434 - .10772i. The mathematics behind fractals, such as the Koch Curve, is rather simple, yet it becomes a complicated, yet ingenious method when referring to the Mandelbrot Set. In order to create a fractal, you will need to be acquainted with complex numbers. Complex numbers on a graph are characterized by the coordinates of (x,y) in which the x-coordinate is any rational number, whereas the y-coordinate contains an imaginary number, denoted by “i”. An

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  • Level: AS and A Level
  • Subject: Maths
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Numerical integration coursework

Numerical integration coursework Problem For this coursework, I am going to use my knowledge of numerical methods to produce an approximation to an area which does not have an analytic solution. I will be finding the approximation, to an appropriate degree of accuracy, of the integral shown above. On the graph below is the area that I will be approximating underneath the curve of y= from x=0 to x=2. Note that throughout my method I worked in radians. This problem is appropriate for numerical solution as I chose my graph to be a polynomial curve involving a square root so that there would be no analytical solution. Due to the fact that I cannot yet integrate functions like this approximating methods will have to be used. According to the numerical methods module the three approximation methods to be used are: Mid-point rule- this method was adopted because it is used to approximate the area underneath the graph by dividing it up into individual rectangles. Trapezium rule- this method was adopted because it is used to approximate the area underneath the graph, this is done so by dividing it up into individual trapeziums. Simpson’s rule- I have realised that out of the two, the mid-point rule and the trapezium rule, the mid-point rule is more accurate, Simpson’s rule is therefore just an average that uses this fact by weighting the average towards the mid-point rule to

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  • Level: AS and A Level
  • Subject: Maths
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Analysing; The Reaction of Hydrogen Peroxide and Iodide ions

Skill 3: Analysing; The Reaction of Hydrogen Peroxide and Iodide ions Table of Results Experiment Volume of KI used (dm3) Volume of water used (dm3) Concentration of I- ions Time taken for colour to appear (S) A 5.0 x 10-3 0.020 0.02 9:56 B 0.010 0.015 0.04 5:36 C 0.015 0.010 0.06 4:07 D 0.020 5.0 x 10-3 0.08 3:27 E 0.025 0.000 0.1 2:40 We calculate the concentration of I- ions by following these stages (taking the hypothetical value of experiment A) Number of moles of KI contained = volume x concentration = 5.0 x 10-3 x 0.1 = 5.0 x 10-4 moles = Number of I- ions contained in the solution Concentration of solution = number of mols Volume of solvent = 5.0 x 10-4 / (25/1000) = 0.02 mol/dm3 Thus I have calculated the other values, which appear on the table Analysis Derived from the rate equation Log (1/time) = n log (volume of KI) + constant We can calculate the rate of reaction, by determining first, n, which can be obtained by calculating the gradient of the line on the following page. Nevertheless, we would need to convert most of the data into logs, Experiment Log Volume of KI (dm3) Time taken in seconds Log (1/time) A -2.30 596 -2.77 B -2.00 336 -2.53 C -1.82 247 -2.39 D -1.70 207 -2.31 E -1.60 60 -2.20 Thus, we could derive the gradient of the graph, by drawing it, as seen upon the nest page. We

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  • Level: AS and A Level
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Investigating the Quadratic Function.

Investigating the Quadratic Function Type 1 Dan Plant Mr. Maly 1 IB Mathematics Thursday, May 10, 2007 Investigating the Quadratic Function Type 1 . Based on three resulting graphs, it can be determined that they are in fact all the same shape, of a parabola, however have varied locations. Furthermore, the locations of the graphs are specifically translated positively or negatively vertical according to the constant added or subtracted to the variable (x2). These three graphs may be generalized by the following statement; adding a positive or negative constant h will shift the graph vertically h units in the function y = (x2)+h. (Refer to attached graphs) 2. Based on the three resulting graphs, it can be determined that they are in fact all the same shape, of a parabola, however they have varied locations. Furthermore, the locations of the graphs are specifically translated positively or negatively horizontal according to the constant added or subtracted directly to the variable x. These three graphs may be generalized by the following statement; replacing x in a function x2 with the expression (x - h), where h is positive or negative the graph will be translated to left or right respectively h units. (Refer to attached graphs) 3. The vertex of the equation y = x2 is (0,0). Therefore, the vertex of y = (x - 4)2 + 5 is (4,5) based on previous conclusions above. Using

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  • Subject: Maths
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