3 u^2 y^3 z^3 and 14 u^7 y^2
2007-07-31 03:50:31 · 6 answers · asked by Brandon 1 in Science & Mathematics ➔ Mathematics
LCM of 3 and14=42
LCM of the two terms
=42u^7y^3z^3
2007-07-31 03:55:27 · answer #1 · answered by moona 4 · 2⤊ 0⤋
3 u^2 y^3 z^3 = 3 * uu * yyy * zzz
14 u^7 y^2 = 2 * 7 * uuuuuuu * yy
Take the "least" number of each element needed to make both numbers:
You need at least one 2 (otherwise you can't make 14)
You need at least one 3 (otherwise you can't make 3)
You need at least one 7.
You need at least seven u (otherwise, you can't make u^7)
You need at least three y.
You need at least three z.
The "least" common multiple is:
2*3*7*uuuuuuu*yyy*zzz =
42 u^7 y^3 z^3
2007-07-31 04:03:06 · answer #2 · answered by Raymond 7 · 0⤊ 0⤋
if you want to solve this problem i advice you to use a free online math too
2007-08-03 23:44:44 · answer #3 · answered by Iulian 1 · 0⤊ 0⤋
42u^7y^3z^3
2007-07-31 03:54:02 · answer #4 · answered by Dave 6 · 0⤊ 0⤋
42u^7y^3z^3
2007-07-31 03:54:01 · answer #5 · answered by me 2 · 0⤊ 0⤋
u^2y^2
2007-07-31 03:53:50 · answer #6 · answered by suesysgoddess 6 · 0⤊ 0⤋