How can you simplify i^99? and also solve (3^0)(8^1)?
A-Best: i^1=i
i^2=-1
i^3=-i
i^4=1
i^5=i
and you see that the values of i repeat every four exponents
so, we know that i^100=i^4 since they are both at the end of a cycle of 4, so i^99=i^3=-i
3^0=1 since any number raised to the zero power is 1; and 8^1=8 since any number raised to the first power is jus the number, so the second problem is just 8
A: i^99=i^3=-i
3^0=1
8^1=8
8X1=8
A: i^99=(i^(2x49)) X i^1
since i^2= -1 and -1 multiplied with -1 for 49 times it becomes -1
= -1 X i = -i
AND
3^0=3^(1-1)=(3^1)/(3^1) = 3/3 = 1
8^1=8
(3^0)(8^)1= 1x8=8
A: (3^0) = 1
(8^1) = 8
therefore 1 * 8 = 8(ans)
A: i^n
*if n is even
if n is divisible by 4...the answer is 1
if n is not divisible by 4...the answer is -1
*if n is odd......(use n-1)
if n-1 is divisible by 4...the answer is i
if n-1 is not divisible by 4...the answer is -i
i^99= -i
99 -1= 98...(not divisible by 4)
(3^0)(8^1)
1(8)
=8
any number raised to zero is 1
