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March 30, 2017

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A bug is placed on the road. t seconds after it is put down, the function is p(t) = -16tsquared + 64t, where p(t) is measure in feet. At what moment will the bug be 14 feet from where it began?

  • Algebra 2 - ,

    p(t) same as f(x) basically just put in 14 for 7 in the equation but don't put it in for the t in p(t) solve out and that is your answer

    p(t)=-16(14)^2 + 64(14)

  • Algebra 2 - ,

    oh whoops i read the question wrong scratcch that

  • Algebra 2 - ,

    oh okay here it is sorry about the above;

    p(t)= - 16t^2 +64t

    14= - 16t^2 + 64t bring 14 to other side and factor or use the equation -b = or minus square root of (b - 4AC)and divide by 2A and you should get what t is

  • Algebra 2 - ,

    14= - 16t^2 + 64t

    16 t^2 - 64 T + 14 = 0

    8 t^2 -32 t + 7 = 0

    t = [32 +/-sqrt (1024-224)]/16

    = [32 +/- sqrt(800)]/16
    = [32 +/- 20sqrt 2]/16
    = [ 32 +/- 28.3]/16
    = 3.76 on the way down
    and
    = .231 on the way up

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