Same Maths Coursework

Tags: Bacchae Essay QuestionsShopping At Stores And Shopping Online EssayImprove Your Problem Solving SkillsThesis Statement For After PickingReligious Views On Euthanasia EssayAcademic Paper Writing ServicesMcsl-036 Solved AssignmentResearch Paper On BrandBusiness Plan Goals ExamplesBest Argumentative Research Paper Topics

However, since the second column was n = 7n – 2 this means there is obviously not a pattern of the n = 7n – x being directly related to which column it’s in.

Conclusion of Ex 2: I have noticed that generally, all columns are n = 7n – x where x varies with the start date.

n x 6 gives the sequence: 6 12 18 24 30, but to get the correct sequence subtract four: n = 6n – 4 I will test the formula on 2 other diagonals descending to the left: one more January, one July.

Ex 4.2: Diagonally descending to the left, starting with 9th January.

Ex 1.4 Results for 2001 Months ( n ) Start day 1 Monday 2 Thursday 3 Thursday 4 Sunday 5 Tuesday 6 Friday 7 Sunday 8 Wednesday 9 Saturday 10 Monday 11 Thursday 12 Saturday The matching months are: 1, 10 = same 2, 3, 11 = same 4, 7 = same 5 7 8 9, 12 = same As you can see, once again the results for the months that start on the same day show the same pattern as in Ex1.1 and Ex1.3.

This clearly shows that there is indeed a pattern between the start days of the months each non – leap year.However, to make this a fair test I will check another year, 2003, to be sure of this, as it could be an anomalous result of some kind.Ex1.3 Results for 2003: Months ( n ) Start day 1 Wednesday 2 Saturday 3 Saturday 4 Tuesday 5 Thursday 6 Sunday 7 Tuesday 8 Friday 9 Monday 10 Wednesday 11 Saturday 12 Monday Thus, the months which have the same start day are: 1, 4, 7 = same 2, 8 = same 3, 11 = same 5 6 9, 12 10 These results are very important, as the dates that are grouped together are the same dates as those in the study sample, which we know is not 2003 because of the differing days.Ex 2.2 – Seventh column in January n date 1 2 2 9 3 16 4 23 5 30 As you can see from the table above, the difference is 7 again.The formula here is n = 7n – 5 Ex 2.3 – Second column in July n dates 1 5 2 12 3 19 4 26 The above table shows that the formula is n = 7n – 2 Ex 2.4 – Sixth column in December n dates 1 3 2 10 3 17 4 24 5 31 The above table yields the formula n = 7n – 4 It is interesting to note that in the sixth column the formula was n = 7n – 4 and that in the seventh column the formula was n =7n -5.n 1 3 2 11 3 19 4 27 I notice from the above table that the difference between the terms is always 8.Multiply each n number by 8 and you get: 8 16 24 32.Ex 3.2 The second row in June: n 1 6 3 7 4 8 5 9 6 11 7 12 As you can see from the table above, the formula is quite simple: n = n 5 Ex 3.3 The third row in June: n 1 13 2 14 3 15 4 16 5 17 6 18 7 19 The above table gives another formula: n = n 12 Ex 3.4, Finally the fourth row in June: n 1 20 2 21 3 22 4 23 5 24 6 25 7 26 The final formula is n = n 19 Ex 3.5: These are the formulae for the rows in the month of June: First row: n = n Second row: n = n 5 Third row: n = n 12 Fourth row: n = n 19 I didn’t draw the table for 5th, but predict it as: Fifth row: n = n 26 The general expression is n = n x depending on the start date. Studying diagonal relationships Ex 4.1 Diagonally descending to the left, starting with 2nd January.n date 1 2 2 8 3 14 4 20 5 26 The difference is always 6.If so, then that means there is a pattern of which months start on the same day each year.The results of this first test are as follows: 1, 10 = same 2, 3, 11 = same 4, 7 = same 5 6 8 9, 12 = same Now I must investigate to find out if the pattern is the same in other years.

SHOW COMMENTS

Comments Same Maths Coursework

The Latest from igmk.ru ©