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We have 10n+100 = O(n).

Proof: Let c := 110 and m:=1 and take arbitrary n >= m. We show

|10n+100| <= c*|n|.
We know
|10*n+100| = 10*n+100 <= 110n = 110|n| = c|n|.

Author: Wolfgang Schreiner
Last Modification: December 14, 1999

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