If the airport is located at the origin, for units of miles and hours, we can write the equations of position for airplanes "a" and "b" in rectangular coordinates as
a = (-30 +250t, 0)
b = (0, -40 +300t)
The distance between these (moving) points can be computed in the usual way using the Pythagorean theorem.
d = √((-30 +250t - 0)² +(0 - (-40 +300t))²)
d = √(2500 -39000t +152500t²)
Then the rate of change of d is the derivative of this.
d'(t) = (-19500 +152500t)/√(2500 -39000t +152500t²)
At the present time (t=0), the rate of change of distance between the planes is
d'(0) = -19500/√2500 = -390
The distance between the planes is decreasing at 390 mi/h.
12ft. how did I get this? I simply plugged in 4 for s and so we have
p=3(4) wich is p=12
Answer:
D=20
Step-by-step explanation:
Use PEMDAS
5x3=15
45+15=60
60-40=20
could you please make this the brainliest
