2006-09-25 08:51:14 · 9 answers · asked by cak 1 in Science & Mathematics ➔ Mathematics
6a - 3b + 6z + 3
2006-09-25 08:54:04 · answer #1 · answered by Darunik 3 · 0⤊ 0⤋
3(2a - b + 2z) + 3
= 6a - 3b + 2z + 3
= 6a + 2z - 3b + 3
= 2 (3a + z) - 3 (b + 1)
= 2 -3 ( 3a + z)(b + 1)
=-1 ( 3a + z) (b + 1)
= - ( 3a + z) ( b + 1)
2006-09-25 16:07:03 · answer #2 · answered by ayie 2 · 0⤊ 0⤋
6a -3b +6z +3
In these problems you have to multiply the outside number by the ones in the paretheses then add the outside number. Since you cant add a's, b's, and z's together they stay the same.
2006-09-25 08:57:36 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The answer: 6a - 3b + 6z + 3
2006-09-25 08:56:14 · answer #4 · answered by modeledge 3 · 0⤊ 0⤋
3(2a-b+2z)+3
=6a-3b+6z+3
=3(2a-b+2z+1)
2006-09-25 21:51:08 · answer #5 · answered by KSA 3 · 0⤊ 0⤋
take the 3 inside
6a -3b +6z +3
this is as far as you can go
2006-09-25 08:56:02 · answer #6 · answered by springdaletutor 1 · 0⤊ 0⤋
looks like you've got a whole set of problems that you've posted here. Really- please just try to do some of them yourself.
Learning is not an easy thing. It takes work
2006-09-25 09:23:36 · answer #7 · answered by Morey000 7 · 0⤊ 0⤋
yup he got it
2006-09-25 08:56:01 · answer #8 · answered by t1ch1a 2 · 0⤊ 0⤋
Can you not do your own homework???
2006-09-25 08:59:21 · answer #9 · answered by koreangurl 2 · 1⤊ 0⤋