An easy approach


#1

they simply want to ask which one has more complexity than O(n^2)

  1. (15^10) * n + 12099 = const *n +const =const *n => O(n)

2)n^1.98, here 1.98<2 , i.e. (n^2) > (n^1.98)

  1. n^3 / (sqrt(n))

4)(2^20) * n = const * n => O(n)