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anonymous
 4 years ago
h(t)=cot t
intervals: pie/4, 3pie/4
totally lost on this one... cant find a formula to plug in the intervals. Help :/ Thanks.
anonymous
 4 years ago
h(t)=cot t intervals: pie/4, 3pie/4 totally lost on this one... cant find a formula to plug in the intervals. Help :/ Thanks.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0are you supposed to graph?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\cot t = 1/\tan t\] So, cotangent is just an inverse of tangent. Hope this helps.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not "inverse" but "reciprocal"

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0:) Can be. I am not so sure about the terminology.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0with functions "inverse" means "inverse function" so the inverse of \[f(x)=x+1\] is \[f^{1}(x)=x1\] not \[\frac{1}{x1}\] which is the reciprocal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, no graphing I just have to plug the intervals in and come up with the answer. answer in the book is, 4/pie how do I denote pie on this site?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[h(t)=\cot t\] itervals given, \[\pi/4, 3\pi/4\]
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