Aplane flying horizontally at an altitude of 2 mi and a speed of 510 mi/h passes directly over a radar station. find the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station. (round your answer to the nearest whole number.)

P is the plane's position

 R is the radar station's position

 V is the point located vertically of the radar station at the plane's height

 h is the plane's height

 d is the distance between the plane and the radar station

 x is the distance between the plane and the V point

 Since the plane flies horizontally, we can conclude that PVR is a right triangle. Therefore, the pythagorean theorem allows us to know that d is calculated:

 d=(h^2+x^2)^(1/2)

 We are interested in the situation when d=2mi, and, since the plane flies horizontally, we know that h=1mi regardless of the situation.

 We are looking for 

 (dd/dt)=d'.

 d^2=h^2+x^2

 d'=(x*x')/d

 We can calculate that, when d=3mi

 x=(d^2−h^2)^(1/2)=(3^2−2^2)^(1/2)=(5)^(1/2) m

 Knowing that the plane flies at a constant speed of 510mi/h, we can calculate:

 d'=(x*x')/d=(((5)^(1/2))*(510))/3= 380.13 mi/h.

 the rate at which the distance from the plane to the station is increasing when it is 3 mi away from the station is 380 mi/h

a4 - 3a2 - 1

step-by-step explanation:

step   1   :

equation at the end of step   1   :

  ((a4) -   3a2) -   1

step   2   :

trying to factor by splitting the middle term

2.1     factoring   a4-3a2-1  

the first term is,   a4   its coefficient is   1 .

the middle term is,   -3a2   its coefficient is   -3 .

the last term, "the constant", is   -1  

17.08

correct me if i am wrong.