The speed of the plane after it encounters the wind is C.285mph
<h3>How to calculate the speed of the plane when it encounters the wind?</h3>
Since the plane takes off from an airport on a bearing of 270° and travels at a speed of 320 mph it's velocity is v = (320cos270°)i + (320sin270°)j
= (320 × 0)i + (320 × -1)j
= 0i - 320j
= - 320j mph
Also, the plane encounters a 35 mph wind blowing directly north. The velocity of the wind is v' = 35j mph
So, the velocity of the plane after it encounters the wind is the resultant velocity, V = v + v'
= -320j mph + 35j mph
= -285j mph
So, the speed of the plane after it encounters the wind is the magnitude of V = |-285j| mph
= 285 mph
So, the speed of the plane after it encounters the wind is C.285mph
Learn more about speed of plane here:
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35 is 17.5% of 200
<span>So, 17.5% of 3000 = 525 as you calculated. </span>
Answer:
Option A is your answer ☺️☺️
25 - 6r = 25
-6r = 25-25
-6r = 0
r = 0/-6
r = 0
The correct answer I believe you are looking for is D. Elevator 1 is 13 feet above ground level, and Elevator 2 is 10 feet below ground level.
<u>Explaination:</u>
This is because if you start at (0, 0) Elevator one, will go up 13 on the y axis. Putting it at 13 feet above ground level. If you start Elevator 2 at ground level on (0, 0) and go down 10 it will make it -10 AKA 10 feet below ground level.
I hope this helps! :)
