If it was one year it would be 2200 and you just keep adding 200 for each year.
again, let's assume daily compounding means 365 days per year.



what's their difference? well

First you compare the denomonators and see if they have any common multiples, then what you do to one side you do to the other so you multiply so you get the same number on both denomonators.finally you multipybthe same number you did ti he botto. to the top.p.s. your welcome