Using calculus, it can be shown that if a ball is thrown upward with an initial velocity of 16 ft/s from the top of a building 7
04 ft high, then its height h above the ground t seconds later will be h = 704 + 16t − 16t2. During what time interval (in seconds) will the ball be at least 32 ft above the ground? (Enter your answer using interval notation.)
1 answer:
Answer:
[0, 7]
Step-by-step explanation:
We want the height to be greater than or equal to 32 ft, so ...
704 +16t -16t^2 ≥ 32
t^2 -t -42 ≤ 0 . . . . . . . . . . . subtract 32, divide by -16
(t -7)(t +6) ≤ 0
This inequality will be true for values of t between -6 and +7. Since we're only concerned with times t ≥ 0, the appropriate solution interval is ...
0 ≤ t ≤ 7 . . . . [0, 7] in interval notation
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Step-by-step explanation:
2n+5=35
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