T-total coursework
Part 1 Investigate the relationship between the T-total and the T-number 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 2 3 0 1 2 9 20 21 The T-number is always the number at the bottom of the T, so in this case (the yellow T) it is 20. The sum of all the numbers in the T is the T-total. In this case, the T-number, (which will now be called n) is 20, and the T-total, (which will now be called T) is 37. If the number at the bottom of the T is n, these are the other numbers in terms of n, bearing in mind that the width of the grid is 9 (by 9): n-19 n-18 n-17 0 n-9 2 9 n 21 The number above the T-number is (n-9) because this number is exactly one row above (n). The width of the grid is 9, so by moving up 1 cell from (n), I am decreasing the value by the width of the grid (9). The cell above (n-9) is (n-18) as the value has been decreased by the width of the grid again. The cell to the left of (n-18) is (n-19) as it has been decreased by 1. The cell to the right of (n-18) is (n-17) as it is 1 more than (n-18). When these 5 terms are added together I get: (n) + (n-9) + (n-17) + (n-18) + (n-19) = 5n - 63
The Gradient Function Coursework
The Gradient Function Coursework In this piece of coursework I am going to do research on the gradient of various graphs at various points, in order to find a function, which will determine the gradient of these points without drawing or using approximations. I will only need to know the coordinates of the point as well as the type of graph I am considering, to submit them into the gradient function and determine the gradient at this point. The formulae I will use and produce will have particular parameters. Now I am going to explain them. a: this letter will stand for the coefficient of x in the function y=ax^n and determines how steep the graph will be. n: this letter will be the power to which x is raised in the function y=ax^n and determines the shape of the graph. m: this letter will stand for the gradient at any point of any graph. I can say for example the gradient at the point P(1;1) of the graph y=x is 1. Therefore here m=1. The first range of graphs I am going to investigate will have the function y=ax. I will draw three graphs on the next pages and hope to see a pattern between the gradient and the function of the graphs. I do not need to consider the coordinates of the points at which I will determine the gradient, as the gradient is the same at any point on the graph y=ax. From these three graphs I clearly recognise a pattern. I will show how I noticed
opposite corners
GCSE Maths Coursework Opposite Corners I have been given the task to investigate the differences of the products of the diagonal opposite corners of a square on a 10x10 Grid with the numbers 1 to 100 to start with. I will start with a 2 x 2 square on a 10 x 10 grid and discover the rule for it, then I will progress onto a 3 x 3 square on the same grid. I will then keep on going until I eventually find the rule for any sized square on a 10 x 10 grid. 2x2 Square 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 20 21 22 23 24 25 26 27 28 29 30 (2 x 11) - (1 x 12) = 10 (14 x 25) - (15 x 24) = 10 (8 x 17) - (7 x 18) = 10 (20 x 29) - (19 x 30) = 10 I have discovered that the answer is always 10 I will now use algebra to see if the answer is once again 10. n n+1 n+10 n+11 (n+1)(n+10) - n(n+11) (n2+11n+10) - (n2+11n) 0 As the algebraic equation also gives the answer of 10 I know it must be right. As I believe I can keep on learning throughout the investigation I will now move onto a 3x3 square on the same grid. I predict that once again all answers will be the same. 3 X 3 Square 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 20 21 22 23 24 25 26 27 28 29 30 (3 x 21) - (1 x 23) = 40 (6 x 24) - (4 x 26) = 40 (10 x 28) - (8 x 30) = 40 I believe the answer will always be 40 for a 3 x 3 square on this grid. So I will now use algebra
