Math Help please?!?

Question

Simplify.

1.[3^0 * 2/2^-2]-3

2.The cost of living in a certain city has been increasing so much that an item that costs a $1 today would cost 1.08 in t years. Using this relationship, how much did todays on-dollar item cost 9 years ago?

3.(4n^-1)^2


Thanks

Answer 1

1) 3^0 = 1 2/2^-2 = 2(2)^2 = 8

8^-3 = 1/8^3

2) 8 cents a year for 9 years = 72 cents = cost 28 cents
3) (4/n)^2 = 16/n^2

Answer 2

1. [3^0 * 2/2^-2]-3

well, anything to the zero power is 1
2/2^-2 = 2*2^2 = 8

(1*8)^-3 = (1/8^3) = 1/512

2. not too sure, sorry
but I'd try setting it up like this: 1+x*1 = 1.08
but, im not sure where to go from there

3. (4n^-1)^2
can also be written as

(4n^-1)=4/n

(4/n)^2
= 16/n^2

Answer 3

1. -2.5
3. 16n^2