PDA

View Full Version : simplifying expressions with positive exponents



06-26-2005, 10:17 AM
I have gotten pretty far on myself , but this one is kicking me somewhere. Can anyone help?

(6a^5b)^2/ (8ab^-2 ) x (1) / (2a^-3b)

I am not sure where I am going wrong. :(

soroban
06-26-2005, 02:02 PM
Hello, afreemanny!


(6a^5b)^2/ (8ab^-2 ) x (1) / (2a^-3b)

I am not sure where I am going wrong.
I'm not sure either . . . it's hard to see your work from here.
But it looks like you wrote "cole slaw" instead of "July 4, 1776".
. . . . . . . . . (6a<sup>5</sup>b)<sup>2</sup> . . . 1 . . . . . . .6<sup>2</sup>(a<sup>5</sup>)<sup>2</sup>(b<sup>2</sup>) . . . . 36a<sup>10</sup>b<sup>2</sup>
We have: . ---------- . -------- . = . --------------- . = . -----------
. . . . . . . . . . 8ab<sup>-2</sup> . . 2a<sup>-3</sup>b . . . . . 16a<sup>-2</sup>b<sup>-1</sup> . . . . . 16a<sup>-2</sup>b<sup>-1</sup>

Reduce the fraction and use the Division Rule for exponents:

. . . . . . 9a<sup>12</sup>b<sup>3</sup>
. . . . . . ---------
. . . . . . . . 4

06-26-2005, 08:58 PM
thank you for the help. I know you can't tell how I am doing. I got a good star so far from my teacher. Some things really confuse me though. I appreciate all you do. :lol: :D

06-28-2005, 08:04 PM
my book said this answer was wrong. You can look at what I did with my most recent message. Thanks for all the help :D