this is the problem..
(2ab)^2/(3x^3)^3
would the answer be 4a^2b^2/9x^5
2007-06-30 08:05:04 · 8 answers · asked by Pradeep k 2 in Science & Mathematics ➔ Mathematics
You got the right idea in the numerator but the denominator
is 27x^9 which is cubing everything inside the parentheses.
Right?
2007-06-30 08:11:55 · answer #1 · answered by ? 5 · 0⤊ 0⤋
4a^2b^2 / 27x^9
Think of it in this way..
(2ab)^2 / (3x^3)^3
expand it...
= (2ab) * (2ab) / (3x^3) * (3x^3) * (3x^3)
= (2*2 a*a b*b) / (3*3*3 x^3 * x^3* x^3)
= (4 a^2 b^2) / 27 x^(3+3+3)
= (4 a^2 b^2) / 27 x^9
2007-06-30 15:11:39 · answer #2 · answered by Sam 3 · 1⤊ 0⤋
No. (3x^3)^3 = (3^3)[x^(3*3)] = 27x^9
(2ab)^2/(3x^3)^3 = (4/27)(a^2)(b^2)/x^9
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Ideas: You group by like factors: constant, a, b, and x, and then add or subtract exponents.
2007-06-30 15:09:02 · answer #3 · answered by sahsjing 7 · 2⤊ 0⤋
No - when you multiply exponents you add - when raising to powers you multiply
(2ab)^2 = 2^2.a^2.b^2 = 4a^2b^2
(3x^3)^3 = 3x^3.3^x^3.3x^3 = 3.3.3 * x(3+3+3)
=27x^9
2007-06-30 15:12:06 · answer #4 · answered by welcome news 6 · 0⤊ 0⤋
(2ab)^2/( 3x^3)^3=4a^2b^2/27x^9=4a^2b^2/27x^9. answer
you committed a mistake.
2007-06-30 15:15:35 · answer #5 · answered by Anonymous · 0⤊ 0⤋
use result that (x^m)^n = x^(mn)
e.g. (x²)³ = x^6 and (x³)³ = x^9
(2ab)² / (3x³)³ = (4 a² b²) / (27 x^9)
2007-07-04 04:03:47 · answer #6 · answered by Como 7 · 0⤊ 0⤋
No.
It would be 4a^2b^2/(27x^9)
(x^m)^n = x^m*n
2007-06-30 15:10:25 · answer #7 · answered by ironduke8159 7 · 1⤊ 0⤋
[(4)(a^2)(b^2)] / [(27)(x^9)]
2007-06-30 15:17:43 · answer #8 · answered by Anonymous · 0⤊ 0⤋