(9a^5b^9)/(-3a^2b^6)
Is the answer -3a^3b^3 or do I have to do the whole "complex fraction" thing, OR am I totally doing it wrong?
2006-08-24 10:07:27 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
you got it right.
2006-08-24 10:10:29 · answer #1 · answered by Anonymous · 1⤊ 1⤋
If this is multiplying those 2, the knuckleheads that answered are incorrect.
9*-3 = -27; a^5 times a^2 equals a^7
B^6 * b^9 = b^15
So your final solution is: -27a^7b^15
2006-08-24 17:30:01 · answer #2 · answered by mthtchr05 5 · 0⤊ 0⤋
(9a^5b^9)/(-3a^2b^6)
(9/(-3)) * a^(5 - 2) * b^(9 - 6)
-3a^3b^3
So yes you are correct.
2006-08-24 20:23:36 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
Yes, it is right.
-3a^3.b^3
2006-08-24 17:12:37 · answer #4 · answered by iyiogrenci 6 · 1⤊ 1⤋
Your answer is correct, however, it is not a complex fraction. The problem is an exponential expression.
2006-08-24 17:13:20 · answer #5 · answered by conniehelmuth 2 · 0⤊ 1⤋
yes the answer is rite...well done
add powers of the same base in multiplication and
subtract powers of the same bases in division...
2006-08-24 17:45:34 · answer #6 · answered by cyrus 2 · 0⤊ 0⤋
That looks right to me.
2006-08-24 17:15:06 · answer #7 · answered by lilsis853 2 · 0⤊ 1⤋