Can anyone explain how [(1-t)/t]/(t-1) simplifies into -1/t?
2007-07-07 10:14:02 · 3 answers · asked by oscarjr1990 2 in Science & Mathematics ➔ Mathematics
t-1 = (-1)*(1-t)
So...
[(1-t)/t]/(t-1) =
[(1-t)/t]/[(-1)(1-t)] =
Cancel 1-t from the numerator and denominator:
[(1-t)/t]/[(-1)(1-t)] =
(1/t)/(-1) =
-1/t
2007-07-07 10:19:13 · answer #1 · answered by McFate 7 · 1⤊ 0⤋
notes
[a/b] / c = a / (b *c)
1 - t = - (t -1) = -1 * (t -1)
[(1-t) / t] / (t-1) =
= (1-t) / [t * (t-1)]
= - (t - 1) / [ t * (t - 1) ]
= -1 /t
assuming t not equal to 1
2007-07-07 19:31:10 · answer #2 · answered by buoisang 4 · 0⤊ 0⤋
[(1-t)/t]/(t-1)=[(1-t)/t]/[(t-1)/1]
then multiply crossing the denominators with the numerators, and we've got:
(1-t)/[t*(t-1)]
then factorizing the "-" on the numerator, we've got:
[-(t-1)]/[t*(t-1)]
then simplify the t-1 factor and we've got:
-1/t
2007-07-07 17:25:16 · answer #3 · answered by patoo_zz 2 · 0⤊ 0⤋