# Algebra 2

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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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