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Devansh Singh
Apr 17, 2020

Floor, ceil problem

in Video Ideas

For what values of k, ceil ((6k+1) pi /6) =2+floor ((6k-1) pi/6) where k is a positive integer greater than equal to one?

4 comments
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Faizan Ahmed
Apr 18, 2020

It is possible for every integer.


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Faizan Ahmed
Apr 18, 2020

The equation is this whose graph is drawn ceil ((6k+1) pi /6) – 2 – floor ((6k-1) pi/6) = 0

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DEVANSH SINGH IET Lucknow Student
Apr 20, 2020

@Faizan Ahmed -It is not possible for k=39, 46 and many more.

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Jotadiolyne Dicci
Jun 12, 2020

This is my answer !



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4 comments