If the expression:
-2t-5 / 9t-6
is to make sense, the values that t can take must be restricted. State the restriction on t
2007-07-21 11:15:33 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
When the denominator of a term is zero, the term "blows up" (goes to plus-or-minus infinity).
So, you won't get a useful answer when the denominator = 0. That is, when...
9*t - 6 = 0
9*t = 6
t = 6/9
t = 2/3
So, when t = 2/3, the denominator is zero, and the expression blows up to plus-or-minus infinity.
2007-07-21 11:19:28 · answer #1 · answered by lithiumdeuteride 7 · 1⤊ 0⤋
It's quite simple. The denominator (9t-8) cannot equal 0. So, 9t-6 /= 0, thus solving for 9t-6=0 we get t=2/3 so t cannot be 2/3
2007-07-21 11:24:40 · answer #2 · answered by mindspin311 2 · 0⤊ 0⤋
the denominator cannot = 0, so 9t - 6 /= 0 or t /= 6/9 /= 2/3
2007-07-21 11:18:56 · answer #3 · answered by dr_no4458 4 · 1⤊ 0⤋
The denominator won't be able to be 0 because of the fact branch by making use of 0 is undefined. as a consequence fixing 7u+8 ? 0 for u supply you the respond: 7u+8 ? 0 7u ? -8 u ? -8/7 The limit is as a consequence u ? -8/7
2016-12-14 15:38:25 · answer #4 · answered by ? 4 · 0⤊ 0⤋
(-2t - 5) / (9t - 6) . . . . . if this cannot be integrated, change its form, divide it
= [ -2/9 + (33/9) / (9t-6) ]
2007-07-21 11:28:48 · answer #5 · answered by CPUcate 6 · 0⤊ 0⤋