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DavidM64550

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  • On Dara O Briain: School Of Hard Sums:

    Posted on 22nd of May 2012

    I got the time t equation as (Dr = distance run): t= [Dr+2*SQRT(Dr^2-150Dr+6250)]/(run speed)so differentiating to get the minimum gives: 1+(2Dr-150)/SQRT(Dr^2-150Dr+6250)=0(independent of run speed) giving Dr=60.56624 (squaring and solving quadratic)phew!

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  • Dara O Briain: School Of Hard Sums

    Posted on 17th of May 2013

    Really enjoy the show. But, whilst working out the 'word sum', I noticed that the solution given on the programme is in fact non-unique: the sum 8542 + 0915 = 09457 is also a solution. Just thought I'd share...