2007-10-11 19:46:14 · 6 answers · asked by Michelle S 1 in Science & Mathematics ➔ Mathematics
6t-18/t^2-9
first factor numerator and denominator
6(t-3) / (t+3)(t-3)
the (t-3) cancels
6 / (t+3) is the answer
2007-10-11 19:48:37 · answer #1 · answered by ignoramus 7 · 0⤊ 0⤋
= 6t -18 / t^2 - 9
= (6[t - 3]) / ([t + 3][t - 3])
= 6/(t + 3)
Answer: 6/(t + 3)
2007-10-12 03:30:49 · answer #2 · answered by Jun Agruda 7 · 2⤊ 0⤋
Assume question is:-
( 6 t - 18 ) / ( t ² - 9 ) and NOT as written.
6 (t - 3) / [ (t - 3) (t + 3)]
6 / (t + 3)
2007-10-12 09:55:35 · answer #3 · answered by Como 7 · 0⤊ 0⤋
Um..you factorise the top bit into 6(t-3) and the denominator into (t-3)(t+3) and then you cancel out the (t-3) for both, and you get 6/(t+3). hope this helped! =)
2007-10-12 02:53:05 · answer #4 · answered by anthea p 1 · 0⤊ 0⤋
6(t-3)/(t-3)(t+3)
6/(t+3)
6=t+3
t = 3
easy
2007-10-12 03:02:21 · answer #5 · answered by alkesh831 2 · 0⤊ 0⤋
I'm just filling this in for my 2 points. Sorry.
2007-10-12 02:49:46 · answer #6 · answered by ? 2 · 1⤊ 0⤋