[(x^(2)y^(-3) over x^(-4)y^(-5))]^(1/2)
2007-08-05 18:48:04 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
[(x^(2)y^(-3) over x^(-4)y^(-5))]^(1/2)
=[x^(2-(-4))y^(-3-(-5))]^(1/2)
=[x^(6)y^(-8)]^(1/2)
=x^(3)y^(-4)
2007-08-05 20:09:54 · answer #1 · answered by happy li 1 · 0⤊ 0⤋
N = (x ²)(1 / y ³)
D = (1 / x^4)(1 / y^5)
N / D = (x^6) y ²
(N / D)^(1/2) = x ³ y
2007-08-05 20:14:18 · answer #2 · answered by Como 7 · 0⤊ 0⤋
a million. while a potential is being raised to a diverse potential you multiply the exponents, which might make it x^12 (3x4=12) 2. confident your answer m^a million/3 is actual. You divide the staggering exponent by using the backside. 3. a damaging # situations yet another damaging # is a superb. i do no longer think of you meant to *** the third 3, staggering? If it have been -3x-3x-3 the respond could be a damaging, -27. 4.The ? could be 2. while the exponent of 6 is divided by using the exponent of you subtract them to get 4. 5. In clinical notation the type could be : 9 x10^10 6. 4.3 x 10^ -4 desire I helped.
2016-12-11 11:29:15 · answer #3 · answered by ? 4 · 0⤊ 0⤋
i think the answer is x^(3)y
2007-08-05 18:53:36 · answer #4 · answered by smm123 1 · 0⤊ 0⤋