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)
The answer is 31.8 cause u move the one that is in 18 behind the 3
40xa^2 + 24 ax + 32 a =
all you have to do is factor out a 8a from each term
aka reverse distribution property
8a (5 a + 3x + 4)
Answer:
YES
Step-by-step explanation:
YES
Seems like you had your minus signs backwards. Try y = 3 and y = -6
