Math Coursework-T-Total Investigation

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Math Coursework-T-Total Investigation

Part 1

Grid 1

 Let n be 23

T number = n

T-total= (n-19)+(n-18)+(n-17)+(n-9)+n

           = 5n-63

To get the formula 5n-63, let n be 23( the T number), and calculate the T-total in terms of n, as shown above.

                                                                                 

                                                                         

Part 1 continued

Grid 2

Let n be 35

T number = n

T-total=(n-19)+(n-18)+(n-17)+(n-9)+n

           = 5n-63  

I moved the T-shape to other position on the grid and I get the same result, 5n-63, so it proves that the relationship between the T-number and the T-total can be represented by the formula 5n-63.

Part 2

T-shape1

T-shape2

                                                                       

Grid 1

T-shape 1: Let n be 25

T number = n

T- total= (n-21)+(n-20)+(n-19)+(n-10)+n

            =5n-70

T-shape 2: Let n be 78

T number = n

T-total= (n-21)+(n-20)+(n-19)+(n-10)+n

           =5n-70

The grid size is 10x10. I move the T-shapes to different position on the grid but I get the same result, 5n-70.

                                                                                   

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T-shape 1

T-shape 2

Grid 2

T-shape1: Let n be 15

T number = n

T total= (n-9)+(n-8)+(n-7)+(n-4)+n

          = 5n-28

T-shape2: Let n be 10

T number = n

T total= (n-9)+(n-8)+(n-7)+(n-4)+n

          = 5n-28

The grid size is 4x4. Again, I get the same results moving the T-shapes to different positions on the grid. I find out that the size of the grid is a very important factor leading to this result. The formula is always equal to 5n-(7x ...

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