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c=0 -> √8 = √-8 = -√8
Or if you mean √(8-c) = c-8: (8-c)=(c-8)2 (8-c)=(c2-16c+64) c2-16c+64-(8-c)=0 c2-15c+56=0 using quadratic equation => c=8 (√0=√0)
decimal: 301 000 binary : 00000100 10010111 11001000 octal : 1113710 hexadecimal: 04 97 C8 base 36 : 6G94
There are 5 groups of order 8 up to isomorphism. 3 abelian ones (C8, C4xC2, C2xC2xC2) and 2 non-abelian ones (dihedral group D8 and quaternion group Q)
In EXCEL spread sheet you can click and hold on first number, scroll down or across the numbers you want to add to next available blank cell, and then release key and hit the summation toolbar. Or, you can add numbers by command =SUM(C1:C8) for example, where C is column to sum from rows 1 to 8
There are five groups of order 8: three of them are Abelian and the other two are not. These are 1. C8, the group generated by a where a8 = 1 2. C4xC2, the group generated by a and b where a4 = b2 = 1 3. C2xC2xC2, the group generated by a, b and c where a2 = b2 = c2= 1 4. the dihedral group 5. the quaternion group
1.for j = 2 to length[A] c1 n 2. do key ¬ A[j] c2 n-1 3. //insert A[j] to sorted sequence A[1..j-1] 0 n-1 4. i ¬ j-1 c4 n-1 5. while i >0 and A[i]>key c5 Sum (j=2->n) tj 6. do A[i+1] ¬ A[i] c6 Sum (j=2->n) (tj -1) 7. i ¬ i-1 c7 Sum (j=2->n) (tj -1) 8. A[i+1] ¬ key c8 n -1 Sum j=2->n tj evaluates to (n(n+1)/2)-1 and j=2->(tj-1) evaluates to n(n-1)/2 thus the highest order term after droping constants becomes n2 thus the complexity is n2