• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
• Level: GCSE
• Subject: Maths
• Word count: 2327

# T-Totals. To figure out an equation for different grid sizes, I have to find the relationship between grid sizes and the T total. I will now let S= Grid Size.

Extracts from this document...

Introduction

Page

 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

T-Totals

Part 1

By using algebra I can find the equation for T Total for grid size 9

T Total= n-19+n-18+n-17+n-9+n= 5n-63

T Total= 5n-63 is the relationship between the T total and the T number. This means for example that if the T number is 20 my formula predicts a T Total of 5× 20– 63= 37 which agrees with my earlier calculations.

Part 2

Equation for different grid sizes

To figure out an equation for different grid sizes, I have to find the relationship between grid sizes and the T total. I will now let S= Grid Size.

I get this T

T Total= n+n-S+n-2S+n-2S+1+n+2S-1= 5n-7S

This means that the equation is T total= 5n-7S where S is the grid size so if for example the T number is 24 and the grid size is 11 then the T total will be (5× 24– 11× 7)= 43. I can check whether this works.

 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 82 83 84 85 86 87 88

The totals are the same so my formula seems to work.

Translation

I will now try and figure out the relationship between translation and the T Total.

Middle

79

The resulting T total is 22+23+24+32+41= 142.

Using my equation for a translated t total, New T total= T+5x+5(yS), for a T total of 37, a grid size of 9 and translation of  then the equation will be  T=37+15+90= 142 which the same as the T total on the grid, so my formula works.

Another way of doing this is by representing the number in each square by a letter.

Again I will use to represent the vector movement, S for the grid size T for the new T Total.

a+x+yS+ b+x+yS+ c+x+yS+ d+x+yS+ e+x+yS=

a+b+c+d+e+5x+5yS= New T Total

a+b+c+d+e is equal to the old T Total which I will call T so the formula is

T+5x+5yS= New T Total

This gives the same equation as in the previous result so I know works as I showed it in the previous result.

Rotation

I will now try and work out the relationship between the T number and the T total when it is rotated about the original T number.

n= T number

S= Grid size

T= New T total

I will try a 90º clockwise rotation

n+n+1+n+2+n+2-S+n+2+s= 5n+7

T=5n+7

This is the T for a 180º clockwise rotation

n+n+S+n+2S-1+n+2S+n+2S+1=7n+7S

T= 7n+7S

This is the T for a 270º clockwise rotation

n+n-1+n-2+n-2-S+n-2+S= 7n-7

I can show that these formulas work by drawing a table.

 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 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

The formula for the 90º clockwise rotation is T= 5n+7 so the T number is 20 so the T total would be (5×20) +7= 107. From the table I can see that the T total is 25+26+27+17+37= 107 so my formula works.

The formula for the 180º clockwise rotation is T= 7n+7S so the T number is 20 so the T total would be (7×20) + (7×10)= 210. From the table I can see that the T total is 25+35+45+46+44= 210 so my formula works.

The formula for the 270º rotation clockwise is T= 5n-7 so the T number is 20 so the T total would be (5×20) -7= 93. From the table I can see that the T total is 25+24+23+13+33= 107 so my formula works.

I will now try and figure out the relationship between the T number and the T total in a 90º clockwise rotation about a point on the table.

T number= n

New T total= T

Grid Size= S

Horizontal distance from the T number

To the point of rotation= x

Vertical distance from the T number

To the point of rotation = y

= Distance from the point of rotation

= Rotation Point

= The T number

To figure it out I would have to

I can prove this by drawing a 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 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Conclusion

="c4">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

82

83

84

85

86

87

88

89

90

91

92

93

94

95

96

97

98

99

100

I will now make a T for a rotation of 270º clockwise in the same format as the last one

n+yS+x+xS-y+n+yS+x+xS-y-1+n+yS+x+xS-y-2+n+yS+x+xS-y-2-S+ n+yS+x+xS-y-2+S=

5n+5yS+5x+5xS-5y-7= T

T= 5(n+yS+x+xS-y)-7

To show that it works, I made an example.

 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 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Limits on T

 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 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108

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. ## T-Total Maths

In the one I just done the sign is addition, but in the other one it is subtraction. They are reflected e.g. 3600 with 1800so the reflected each other but the sign changed with the reflection. Grid Number Formular To prove the formular 7 by 7 T=5N-49 T= N+N-7+N-15+N-14+N-13 8

2. ## Objectives Investigate the relationship between ...

195 +140 You can clearly see from the above table of results, the T-total increases by an increment of '+140' every time. I will now find an algebraic formula for finding the T-total of any 180� rotated T-shape. To find this algebraic formula, I will find out a way to

1. ## T-Total. I will take steps to find formulae for changing the position of the ...

Equation T number/T total(translated) 25/62 5 x 25 - 63 - 5 x 5 20/37 69/282 5 x 69 - 63 - 5 x 4 65/262 My formula has worked. Moving the T shape down: for this I will be using a 8 x 8 grid but my formula will have a variable grid size.

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

of 15 the T-Total is 33, again working out the formula leads to t = 5x - 42 Again this is the same "magic number" as found by the predictions for a 6x6 grid found earlier, therefore we can state that: 5x - 7g can be used to find the T-Total (t)

1. ## T totals - translations and rotations

The number directly above this is also 8 places back on my grid so it is N-8-8= N-16. The two remaining numbers in my grid are N-16-1 and N-16+1. Thus my T-total is: N+ (N-8) + (N-16) + (N-16-1) + (N-16+1)

2. ## Maths Coursework T-Totals

- a multiple of 7 with a value dependent on the grid size. We should now try and find the rule that governs the "magic number" that has to be taken from 5x to gain t. If we say g is the grid size (e.g.

1. ## Maths- T-Totals

and you get 5. Therefore the formula of this is 5n � C = T-Total/ 5n � C = 37, (if n=20 (T-Number) ) 20 x 5= 100, 100 -37= 63. Therefore nth term= 5n-63=37 Grid 1: 9 by 9 - Diagonally ( )

2. ## t totals gcse grade A

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 Using a basic t-shape I am trying to find a formula to

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