HELP ME PLEASE ASAP,

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.

ha!

i just did this!

18. g

20. h 30 degree turn.

21. d 90 degree angles

and also 19 is a. a sphere not a cirlcle

hope i !

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