a. x^2/y^ -3
b. x^3y ^-5
c. x^2y^ -4z^-4
d. x^2/y-1/2
e. x^2y^3t-4z-6w
2006-11-07 05:59:26 · 4 answers · asked by alize2882 1 in Science & Mathematics ➔ Mathematics
a) x^2y^3
b) x^3/y^5
c) x^2/(y^4z^4)
d) x^2y^1/2 (assuming you mean y^-1/2 and not y-1/2)
e) x^2y^3t/[y^(4z6w)] (I'm assuming you mean y^(3t-4z-6w))
2006-11-07 06:05:40 · answer #1 · answered by slider 2 · 0⤊ 0⤋
Use the fact that x^-n = 1/(x^n) and 1/(x^-n) = x^n
a. x^2/y^ -3 = x^2y^3
b. x^3y ^-5 = x^3/y^5
c. x^2y^ -4z^-4 = x^2/(y^4z^4)
d. x^2/y-1/2 = x^2(y-1/2)^-1
e. x^2y^3t-4z-6w = x^2y^3(t-4z-6w)^-1
2006-11-07 06:05:22 · answer #2 · answered by ohaqqi 2 · 0⤊ 0⤋
a) x^2y^3
b) x^3/y^5
c) x^2/(yz)^4
d) x^2y^(1/2)
e) x^2y^3w/(t^4z^6)
2006-11-07 06:16:12 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋
1.x^2y^3
2.x^3y^5
3.x^2/y^4z^4
4.x^2y^(1/2)
5.x^2y^3t-4z-6w
2006-11-07 06:10:36 · answer #4 · answered by raj 7 · 0⤊ 0⤋