Answer / ali

In the case of T(n) = n, say O(n), and n is 1000, then on
the old computer 1 minute and the new one, 1/1000 minute
For T (n) =n3, the size will be 1000^3 = 1000,000,000 which
takes 1000,000 minutes on the old computer and 1000 on the
new one
For T(n)=10n Because of the limited space, I won’t write
1000 ZEROs here, instead it’s clear that it takes 10^997
minutes on the old computer, and 10^994 minutes on the new one.

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