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
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For this case we can make the following rule of three:
23000 -------> 100%
x --------------> 180%
Clearing the value of x we have:
x = (180/100) * (23000)
x = 41400
Answer:
you would expect about:
a) 41,400 responses
84 is no less than 7 times K, 84 divided by 7= 12,
K=12
