1/2n+1.5m
2006-12-07 02:28:05
·
answer #1
·
answered by Anonymous
·
1⤊
0⤋
First and foremost, it might pay to ask if you are SURE you typed this problem correctly; did you intend to use two different variable names (n and m)?
If so, then the only like terms are the three 'm' terms, whose coefficients we must add together:
1/2n + 2/3m +1/2m + 1/3m
= (1/2)n + (2/3 + 1/2 + 1/3)m
= (1/2)n + (1 + 1/2)m
= (1/2)n + (3/2)m
On the other hand, if all the variables were supposed to be just 'm', then the coefficients add much more simply, by grouping the 'halves' terms together, then the 'thirds' terms.
1/2m + 2/3m +1/2m + 1/3m
= (1/2 + 1/2)m + (2/3 + 1/3)m
= 1m + 1m
= 2m
Hope this helps!
2006-12-07 02:31:34
·
answer #2
·
answered by Tim GNO 3
·
0⤊
0⤋
2/3m can add with 1/3m
2/3m+1/3m=3/3m
3/3m+1/2m=6/6m+3/6m
=9/6m
=3/2m
you need to make the denominator equal in order to + the equation
3/2m+1/2n or in decimal 1.5m+0.5n
2006-12-07 02:43:04
·
answer #3
·
answered by tru m 1
·
0⤊
0⤋
1/2n + 1 1/2m
2006-12-07 02:30:21
·
answer #4
·
answered by Tag 1
·
0⤊
0⤋
1/2n+1 1/2m
2006-12-07 02:48:28
·
answer #5
·
answered by Ranna Renni 2
·
0⤊
0⤋
3m + 1/2n
2006-12-07 02:38:08
·
answer #6
·
answered by Dom 3
·
0⤊
0⤋
1/2n + 2/3m +1/2m + 1/3m =(1/2n + 1/2m) +(2/3m + 1/3m)
=(1/2n + 1/2m) +(3/3m)
=(1/2n + 1/2m) +m
=[(n+m)/2]+(2m/2)
=(n+3m)/2
or=n/2+3m/2
or=1/2n + 3/2m
or=0.5n+1.5m
that all...................
2006-12-07 03:05:47
·
answer #7
·
answered by San 2
·
0⤊
0⤋
1/2n+(2/3+1/2+1/3)m
2006-12-07 02:44:37
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋
1/2 n + 3/2 m
because 2/3 + 1/3 = 1 and 1 + 1/2 = 3/2
2006-12-07 02:28:31
·
answer #9
·
answered by anton3s 3
·
0⤊
0⤋
if that 1/2n is m the answer would be
= 12/6m or 2/m
2006-12-07 02:31:08
·
answer #10
·
answered by (,") Verns (",) 3
·
0⤊
0⤋
Assuming you mean (1/2)n and so on
(1/2)n+((4+3+2)/6)m
or(1/2)n+(9/6)m
(1/2)n+(3/2)m
2006-12-07 02:29:07
·
answer #11
·
answered by math_teacher_02 2
·
0⤊
0⤋