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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.

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  1. anonymous
    • 4 years ago
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    are you supposed to graph?

  2. anonymous
    • 4 years ago
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    \[\cot t = 1/\tan t\] So, cotangent is just an inverse of tangent. Hope this helps.

  3. anonymous
    • 4 years ago
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    not "inverse" but "reciprocal"

  4. anonymous
    • 4 years ago
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    :-) Can be. I am not so sure about the terminology.

  5. anonymous
    • 4 years ago
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    with functions "inverse" means "inverse function" so the inverse of \[f(x)=x+1\] is \[f^{-1}(x)=x-1\] not \[\frac{1}{x-1}\] which is the reciprocal

  6. anonymous
    • 4 years ago
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    no, 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?

  7. anonymous
    • 4 years ago
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    \[h(t)=\cot t\] itervals given, \[\pi/4, 3\pi/4\]

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