tahrawright2
tahrawright2 tahrawright2
  • 23-03-2016
  • Mathematics
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Аноним Аноним
  • 23-03-2016
1)
The vertex of the graph is (1,3).

2)
The vertex of the graph is a maximum.

3)
h=0
Then:
-16t²+45t+6=0

we need solve this square equation:
t=[-45⁺₋√(2025+384)] / -32=(-45⁺₋49.08) / -32
We have two solutions:
t₁=(-45+49.08)/-32=2.94 s≈3 s
t₂(-45+49.08)/-32=-0.15 this solution is not valid.

Answer: 3 s.
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