math. simplify
2007-11-26 08:15:50 · 5 answers · asked by Dee Dee 1 in Science & Mathematics ➔ Mathematics
(m-3)/(m^2)-9
=(m-3)/(m^2)-3^2
as from fomula, a^2-b^2=(a+b)(a-b)
therefore,
=(m-3)/(m+3)(m-3)
=1/(m+3)
2007-11-26 08:32:02 · answer #1 · answered by pink 2 · 0⤊ 0⤋
(m -- 3) / (m^2 -- 9)
= (m -- 3) / (m + 3)(m -- 3)
= 1/(m + 3)
2007-11-26 16:20:20 · answer #2 · answered by sv 7 · 1⤊ 0⤋
m^2 - 9 can be factored into (m+3)(m-3)
Then cancel out the (m-3) in the numerator and denominator and you are left with 1/(m+3)
2007-11-26 16:20:48 · answer #3 · answered by dirtyrubberduck 4 · 0⤊ 1⤋
= (m - 3) / [ (m - 3) (m + 3) ]
= 1 / (m + 3)
2007-11-26 17:34:52 · answer #4 · answered by Como 7 · 1⤊ 1⤋
The denominator is a difference of squares. This can be rewritten as:
(m-3)
---------------
(m+3)(m-3)
The (m-3) terms cancel and you have:
1
------
(m+3)
2007-11-26 16:19:27 · answer #5 · answered by Puzzling 7 · 1⤊ 1⤋