HELP ME PLEASE ASAP,,,,,,,
1 answer:
Answer:
a) i) The drawing of Zeno's route is attached
ii) The bearing of Zeno's return journey is 241° from point C to point A
b) Yes, Zeno returns to Port A before 5.15pm
Step-by-step explanation:
a) i) Please find attached the drawing of Zeno's route
ii) From the attached diagram of Zeno's route created with Microsoft Whiteboard, we find the bearing of his return journey as 241° from point C to point A
b) The distance from point C to point A = 10·√5 km
The speed with which Zeno sails as he returns = 10 km/h
The time it takes Zeno to return t = Distance/Speed
∴ The time it takes Zeno to return t = 10·√5 km/(10 km/h) = √5 h ≈ 2.2361 hours ≈ 2 hours 14 minutes and 9.845 seconds
The time Zeno arrives at point A from point A ≈ 3.00 pm + 2 hours 14 minutes and 9.845 seconds = 5:14.1641 p.m. ≈ 5:14 pm.
Therefore, Zeno returns to Port A before 5.15pm.
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Answer: Choice A. 4 times
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Explanation:
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To start things off, we'll compute the AROC from x = 1 to x = 2.
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Answer:
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Step-by-step explanation:
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