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:
brainly.com/question/3387746
#SPJ1
Answer:
GH = 9 units
Step-by-step explanation:
Given HI = 4b - 5 and HI 3, then
4b - 5 = 3 ( add 5 to both sides )
4b = 8 ( divide both sides by 4 )
b = 2
Thus
GI = 5b + 2 = 5(2) + 2 = 10 + 2 = 12
GH = GI - HI = 12 - 3 = 9
Answer:
k;p
Step-by-step explanation:
Not a robot! I don't think.
Y in the beginning goes up to 3.
Y in the end goes down to -2 before shooting back up in an infinite sense.
Increasing: The beginning and the end the line on the graph. (Also the jump in the middle, the round part.)
Decreasing: The middle of the graph. (The jump, downward slope.)
Constant, Y at the near end going in a straight line from 9-12 at a -2.
End behavior: Decide for yourself. Is the line going up without fault at the end an appearing continuous or a discontinuous line?
