I^68 is the same as i^4
I is the square root of -1
The square root of -1 to the fourth power is 1.
II. f(x) doubles for each increase of 1 in the x values. Thus, r must be 2, and so we our ar^1 = 6 from ( I ) above becomes f(x) = a*2^x. Applying the restriction ar^1 = 6 results in f(1) = a*2^1 = 6, or a = 3.
Then f(x) = ar^x becomes f(x) = 3*2^2 (Answer A)
Answer:
N^2*5= 3^2*5^5*x^6
Step-by-step explanation:
N=3*5^2*x^3
First lets express N^2
N^2= (N)^2= (3*5^2*x^3)^2=3^2*5^4*x^6 (1)
To express 5*N^2 just multiple (1) by 5:
N^2*5=5* 3^2*5^4*x^6 = 3^2*5^5*x^6