2 5 3 6 4 7 58 6 9 7 10
There are actually two lines:
2 3 4 5 6 7
5 6 7 8 9 10
It would be 5, then 8/5. The pattern is +2, /2, +3, /3, +4, /4, +5, /5, and so on.
plus 4, minus 3
Starting with the number 4, applying the rule of multiplying by 2 and then subtracting 3 gives the following sequence: 4, 5, 7, 11, 19, 35. This pattern can be calculated as follows: 4 x 2 - 3 = 5, 5 x 2 - 3 = 7, 7 x 2 - 3 = 11, 11 x 2 - 3 = 19, 19 x 2 - 3 = 35.
There are 32 possible subset from the set {1, 2, 3, 4, 5}, ranging from 0 elements (the empty set) to 5 elements (the whole set): 0 elements: {} 1 element: {1}, {2}, {3}, {4}, {5} 2 elements: {1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4,}, {3, 5}, {4, 5} 3 elements: {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5} 4 elements: {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5} 5 elements: {1, 2, 3, 4, 5} The number of sets in each row above is each successive column from row 5 of Pascal's triangle. This can be calculated using the nCr formula where n = 5 and r is the number of elements (r = 0, 1, ..., 5). The total number of subset is given by the sum of row 5 of Pascal's triangle which is given by the formula 2^row, which is this case is 2^5 = 32.
The pattern is: -8, -7, -5, -2, 2, 7, 13, 20 The series is: +1, +2. +3. +4. +5, +6. +7, etc.
It is 2, assuming the pattern is repeated as given. 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1... If the intended pattern is to continue to subtract 1 from the last number, then the 5479th digit of the pattern will be -5470.
It would be 5, then 8/5. The pattern is +2, /2, +3, /3, +4, /4, +5, /5, and so on.
There are 64 subsets, and they are:{}, {A}, {1}, {2}, {3}, {4}, {5}, {A,1}, {A,2}, {A,3}, {A,4}, {A,5}, {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3, 5}, {4,5}, {A, 1, 2}, {A, 1, 3}, {A, 1, 4}, {A, 1, 5}, {A, 2, 3}, {A, 2, 4}, {A, 2, 5}, {A, 3, 4}, {A, 3, 5}, {A, 4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 3, 4}, {2, 3, 5}, {2, 4, 5}, {3, 4, 5}, {A, 1, 2, 3}, {A, 1, 2, 4}, {A, 1, 2, 5}, {A, 1, 3, 4}, {A, 1, 3, 5}, {A, 1, 4, 5}, {A, 2, 3, 4}, {A, 2, 3, 5}, {A, 2, 4, 5}, {A, 3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {A, 1, 2, 3, 4}, {A, 1, 2, 3, 5}, {A, 1, 2, 4, 5}, {A, 1, 3, 4, 5}, {A, 2, 3, 4, 5}, {1, 2, 3, 4, 5} {A, 1, 2, 3,,4, 5} .
3-5, 2-5, 3-5, 2-2, 5-2, 5-2, 3-3, 4-3, 4-4,
3-4, 3-4, 3-4, 4-3, 2-2 or 4-5, 4-5, 4-3, 3-3, 3-3
1 2 3 2 3 4 5 4 3 4 5 6 7 6 5 4 5 6 7 8 9 8 7 6 5 Please this pattern you solve ?
3-4, 3-5, 3-3, 1-3 or 3-4, 3-5, 2-2, 5-3 or 2-2, 3-4, 3-5, 5-3
i0 = 4; in = in-1 - 3
You add 2, then 3, then 4, then 5, then 6...etc
plus 4, minus 3
2-4; 3-4; 3-4; 4-4; 4-3; 4-6 et voila...
1970 could have a 5 on 4.5 or a 5 on 4.75 bolt pattern measure from one cbolt enter to the next bolt center if distance 2 5/8 then 5 on 4.5 bolt pattern if distance 2 3/4 then 5 on 4.75 bolt pattern if distnace 3 then 5 on 5 bolt pattern if distance 3 1/4 then 5 on 5.5 bolt pattern hope this helps