I don't know what I'm doing.
2006-06-21 10:27:10 · 8 answers · asked by t g 1 in Science & Mathematics ➔ Mathematics
3sqrt(11) a^(9/2) b^(5/2)
2006-06-21 10:36:56 · answer #1 · answered by scott_d_webb 3 · 0⤊ 1⤋
99 = 3 x 3 x 11
a^9 = a^4 x a^4 x a
b^5 = b^2 x b^2 x b
sqrt(99a^9b^5) = 3a^4b^2sqrt(11ab)
2006-06-21 17:56:12 · answer #2 · answered by kindricko 7 · 0⤊ 0⤋
3a^4b^2 sqrt (11ab)
Try to make pairs out of all the factors. 99 = 3 * 3 * 11
a^9 = a^4*a^4*a
b^5 = b^2*b^2*b
Whatever you can make a pair of goes outside the sq. rt. and whatever is leftover goes back under the sq. rt.
2006-06-21 17:41:15 · answer #3 · answered by scotsgirl 2 · 0⤊ 0⤋
sqrt(99a^9b^5)
sqrt(9a^8b^4 * 11ab)
sqrt(9) * sqrt(a^8) * sqrt(b^4) * sqrt(11ab)
3 * (a^8)^(1/2) * (b^4)^(1/2) * sqrt(11ab)
3 * a^(8 * (1/2)) * b^(4 * (1/2)) * sqrt(11ab)
3 * a^(8/2) * b^(4/2) * sqrt(11ab)
3a^4b^2sqrt(11ab)
ANS : (3a^4b^2)sqrt(11ab)
2006-06-21 22:23:11 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
its really simple
(99a^9b^5)^0.5 =
3a^4b^2(11ab)^0.5
get it?
2006-06-21 17:37:13 · answer #5 · answered by hanumistee 7 · 0⤊ 0⤋
6.246 squared
2006-06-21 17:31:04 · answer #6 · answered by Hermione G. 2 · 0⤊ 0⤋
3*a^4*b^2*sq.rt(11*a*b)
2006-06-21 17:33:28 · answer #7 · answered by Vivek 4 · 0⤊ 0⤋
ummmm.... 0
2006-06-21 17:30:47 · answer #8 · answered by DAVE 3 · 0⤊ 0⤋