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OpenStudy (anonymous):
Arrange the following functions in increasing order of growth rate: a)2^2^n b)2^n^2 c)n^2^n
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OpenStudy (anonymous):
a, b, c. I'm fairly sure about this, but i'm not a bible.... :P
OpenStudy (anonymous):
ok, then 2^n log(n) > n^2^n, no?
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