• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16
17. 17
17
18. 18
18
19. 19
19
20. 20
20
• Level: GCSE
• Subject: Maths
• Word count: 8074

# T-Total Investigation

Extracts from this document...

Introduction

Maths GCSE Coursework 2000 - T-Total Introduction In this investigation I aim to find out relationships between grid sizes and T shapes within the relative grids, and state and explain all generalizations I can find, using the T-Number (x) (the number at the bottom of the T-Shape), the grid size (g) to find the T-Total (t) (Total of all number added together in the T-Shape), with different grid sizes, translations, rotations, enlargements and combinations of all of the stated. Relations ships between T-number (x) and T-Total (t) on a 9x9 grid 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 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 From this we can see that the first T shape has a T number of 50 (highlighted), and the T-total (t) adds up to 187 (50 + 41 + 31 + 32 + 33). With the second T shape with a T number of 80, the T-total adds up to 337, straight away a trend can be seen of the larger the T number the larger the total. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 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 From these Extra T Shapes we can plot a table of results. ...read more.

Middle

We know with will not make a difference to the final answer as proved in question 2. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 As we can see, we have a horizontal translation of the first T-Shape (where v =17) by +4. Where v = 17, t = 63, and where v = 21, t = 83 (both found by using t = 5v - 2g), if be draw up a table in the same format as the one we used for the 9x9 grid, we should be able to find some relationships (from 21 to 17). Middle number (v) T-Total (t) Equation used Difference 21 83 t = (5 x 21) + ( 2 x 11 ) 5 (83 - 78) 20 78 t = (5 x 20) + ( 2 x 11 ) 5 (78 - 73) 19 73 t = (5 x 19) + ( 2 x 11 ) 5 (73 - 68) 18 68 t = (5 x 18) + ( 2 x 11 ) 5 (68 - 63) 17 63 t = (5 x 17) + ( 2 x 11 ) N/a From this we can see that 5 is the "magic" number again as for a grid width of 9 for horizontal translations. From this an obvious relation ship can bee seen that for all grid sizes, a horizontal translation of a T-Shape by +1, makes the T-Total +5 larger, but this is only a prediction. To verify this we can see what the "magic" number is on a grid width of 10. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 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 ...read more.

Conclusion

Using the formula we get; T=5(36+2x10+0)+2 T=5(56)+2 T=282 Thus proving this formula works and it is obvious that it will work in the same fashion as my static (middle number as the centre) rotations, as they both find the position as the new V number then generate the t-total based on that number, therefore I can state; The T-Total of any rotation of a T-Shape with any centre of rotation on any grid size can be found by using the formula of t=5(c+d(g)+b)+y were t is the T-Total, c is the centre of rotation (grid value) d is the horizontal difference of v from the relative centre of rotation , is the grid width, and b is the vertical difference of v from the relative central number. y is to be substituted by the ending required by the type of rotation, these are : Rotation (degrees) Direction Ending (y) 0 Clockwise - 2g 90 Clockwise + 2 180 Clockwise + 2g 270 Clockwise - 2 0 Anti-Clockwise - 2g 90 Anti-Clockwise - 2 180 Anti-Clockwise + 2g 270 Anti-Clockwise + 2 In terms of x (T-Number); The T-Total of any rotation of a T-Shape with any centre of rotation on any grid size can be found by using the formula of t=5(c+d(g)+b)+y were t is the T-Total, c is the centre of rotation (grid value) d is the horizontal difference of x from the relative centre of rotation, g is the grid width, and b is the vertical difference of x from the relative centre of rotation. y is to be substituted by the ending required by the type of rotation, these are : Rotation (degrees) Direction Ending (y) 0 Clockwise - 7g 90 Clockwise + 7 180 Clockwise + 7g 270 Clockwise - 7 0 Anti-Clockwise - 7g 90 Anti-Clockwise - 7 180 Anti-Clockwise + 7g 270 Anti-Clockwise + 7 [PP1]i ?? ?? ?? ?? 1 Philip Price 07/10/01 ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our GCSE T-Total section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE T-Total essays

1. ## Connect 4 - Maths Investigation.

The final rule for Connect 3 is H(4L-6)-6L+8 In order to find out the overall rule for any amount of connects I have to find out the overall rule for another connect. This will make sure that the difference between them is the same.

2. ## T-Totals. Aim: To find the ...

This is what I will find out next. Grid size 6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 There is a

1. ## In this section there is an investigation between the t-total and the t-number.

This is another way to work out the t-total. What we need now is a formula for the relationship between the t-total and the t-number. I have found a formula which is 5t-number-63 = t-total. The question is how did we work out this formula and what can we do with it?

2. ## Given a 10 x 10 table, and a 3 steps stair case, I tried ...

total sum of numbers in a staircase which its bottom left number is in the diagonal line, n is equal to the position of the bottom left number in the diagonal line (e.g. 1st, 2nd, 3rd, etc... see table in next page, the number in blue is the number of which I worked out the position).

1. ## T-totals. I am going to investigate the relationship between the t-total, T, and ...

+ (n +1) + (n + 2) + (n + 2 -g) + (n + 2) + (n + 2 + g) T= 5n + 7 Rotation of 180� n-2g-1 n-2g n-2g+1 n-g n n n+g n+2g-1 n+2g n+2g+1 T= (n) + (n + g)

2. ## T-Shapes Coursework

72 + 82 + 92 + 102] [96]+ [504] 600 3n + 1/2 l {2n + 10(l + 1)} [3 x 32] + [1/2 x 7 x {(2 x 32) + 10(8)}] [96] + [1/2 x 7 x {64 + 80}] [96] + [1/2 x 7 x {144}] [96] + [504] 600 2)

1. ## T totals. In this investigation I aim to find out relationships between grid sizes ...

the values we used in are table, we get the same answers, for example taking x to be 80: t = 80 + 80 - 9 + 80 - 19 + 80 - 18 + 80 - 17 = 337 And x as 52; t = 52 + 52 -

2. ## T-total Investigation

G = 9 T-2G+1 T - 2G T- 2G-1 T- G T I added up all the G terms from inside the T to get 7G the final expression. (2G+1) + (2G) + (2g-1) + (G) = 7G T = 5T - 7G I will now test out this formula to see if it works on a 9by9 grid.

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to