Answer:
360 mi/h
Step-by-step explanation:
The speed for the outbound trip was ...
speed = distance/time = (715 mi)/(2 1/6 h) = 330 mi/h
The inbound speed was ...
(715 mi)/(1 5/6 h) = 390 mi/h
The airspeed of the plane is the average of these two ground speeds, so is ...
(330 +390)/2 = 360 . . . . mi/h
<span>sin θ = opposite/hypotenuse
P(18,24) lies in the first quadrant as both are positive.
x=18
y=24
Hypotenuse = sqrt (18^2 + 24^2)
</span><span>Hypotenuse = 30
</span><span>sin θ = 24/30
</span><span>sin θ = 4/5
</span>θ = sin^-1 (4/5)
<span>θ = 53.13 degrees</span>
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