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Evaluate the expression Assume t ≠ 0 tᵒ
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for t>0 : 1 for t<0 : -1 because \[t^0 = \pm 1\] depending on the sign of t.
Thank you
t^0 is always 1
I suppose you are right because you would sub -n in as (-n)^0 not -n^0 ... so it will always be 1... I was a bit hastey
Any thing to the pwr o is always 1
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