solving for x and y in a radical?

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How do you solve for x and y in this radical

2/3 square root of 72x^3y^4

2007-11-15 04:11:41 · 3 answers · asked by the log 1 in Science & Mathematics ➔ Mathematics

2/3 square root of 72x^3y^4
= 2/3 sqrt(36*x^2y^4*2x)
2/3 *6xy^2sqrt(2x)
= 4xy^2sqrt(2x)

2007-11-15 04:21:39 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋

72^2/3 x^(2/3)(3)y^(4)(2/3)
=(4)9^(2/3)x^2y^(8/3)

2007-11-15 12:18:32 · answer #2 · answered by someone else 7 · 0⤊ 1⤋

4xyÂ²â(2x)

2007-11-15 12:15:02 · answer #3 · answered by Dave 6 · 1⤊ 0⤋