possible homework solutions....
a. (1/2)^23x
b. (1/2)^24x
c. (1/2)^25x
d. (1/2)^26x
e. (1/2)^27x
f. (1/2)^28x
g. (1/2)^30x
h. (1/2)^32x
i. none of these
2007-02-24 08:53:22 · 3 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
[(1/2)^2x][(1/4)^3x][(1/8)^4x]
=[(1/2)^2x][(1/2)^6x][(1/2)^12x]
=[(1/2)^20x]
2007-02-24 09:05:01 · answer #1 · answered by epbr123 5 · 0⤊ 0⤋
The and is unreadable
[(1/2)^2x][(1/4)^3x][(1/8... =
2^-2x * 2^-6x * ... =
2^[-2x + - 6x + ...] =
Th
2007-02-24 17:04:27 · answer #2 · answered by Thermo 6 · 0⤊ 0⤋
[(1/2)^2x][(1/4)^3x][(1/8)^3x]... =?
(does the ... mean that this goes on forever?)
if not then
[(1/2)^2x][(1/4)^3x][(1/8)^3x]
=[(1/2)^2x][(1/2)^6x][(1/2)^9x]
=[(1/2)^17x]
answer is i
2007-02-24 17:01:56 · answer #3 · answered by Glynn R 1 · 0⤊ 0⤋