Question 754534
h(base 1)t = -2t^2+ 20t + 50 and h(base 2) = 2t +66 .
a) How long after their launch will the two projectiles be at the same height?
Solve:
-2t^2 + 20t + 50 = 2t+66
-2t^2 + 18t - 16 = 0
t^2 - 9t + 8 = 0
(t-8)(t-1) = 0
t = 1 second (1st time they are at the same height)
t = 8 seconds (2nd time they are at the same height)
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b) Over what interval of time since the launch is the first projectile higher than the second?
1sec < t < 8 sec
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{{{graph(400,400,-10,10,-10,100,-2x^2+20x+50,2x+66)}}}
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cheers,
Stan H.
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