Less then 5 kilometers and more then 1,200 meters obv
3 over 2 and 3 over 4 (1.5 and .75)
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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It would be 30 because 10^4 is 10000 and 10000 times 12 is 120000 and 10^3 is 1000 and 1000 times 4 is 4000 so 120000 divided by 4000 is 30
1. the image is multiplied by the scale factor. if the scale factor is 2, the image is twice the size of the preimage. if the scale factor is ⅓, the image is one third the size. only a scale factor of 1 preserves congruency
