i tried getting this answer but i am way off can someone help
2006-11-11 07:37:31 · 7 answers · asked by danni c 1 in Science & Mathematics ➔ Mathematics
4a^4b^-1
Subtract exponents when dividing a^5/a = a^(5-1) = a^4; b/b^2 = b^(1-2) = b^-1
2006-11-11 07:38:55 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋
Okay. Regroup the numbers first, then the as then the bs
(48/12)(a^5/a)(b/b^2)
4*a^4(1/b)
4a^4/b
2006-11-11 16:32:55 · answer #2 · answered by mom 7 · 1⤊ 0⤋
(48a^5b)/(12ab^2)
This is the same as
(48/12)(a^5/a)(b/b^2)
(4)(a^4)(1/b)
4a^4/b
2006-11-11 15:46:49 · answer #3 · answered by kindricko 7 · 1⤊ 0⤋
(48a^5b)/(12ab^2) divide num & den by 12
4a^5b/(ab^2) divide by a
4a^4b/b^2 divide by b
4a^4/b
2006-11-11 17:51:50 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
4(a^(5b-1))/b^2
2006-11-11 15:43:13 · answer #5 · answered by trackstarr59 3 · 1⤊ 0⤋
4a^4/b
2006-11-11 17:28:21 · answer #6 · answered by J 6 · 0⤊ 0⤋
4a^4/b
2006-11-11 16:25:54 · answer #7 · answered by Anonymous · 0⤊ 0⤋