One airplane is approaching an airport from the north at 199 km/hr. a second airplane approaches from the east at 162 km/hr. find the rate at which the distance between the planes changes when the southbound plane is 26 km away from the airport and the westbound plane is 24 km from the airport.

Because the Pythagorean Theorem states that a^2 + b^2 = c^2 you can say a is the distance to the north plane, and b is the distance to the east plane, while c is the distance between them. You can solve for c by plugging in and getting 2*sqrt(313).If you take the derivative of both sides, you get 2a*a' + 2b*b' = 2c*c' . by plugging in a, a' (which is the speed of the southbound plane), b, b'(speed of the westbound plane), and c, you then get 10348 + 7776 = 4*sqrt(313)*c' simplifying you get 4531=sqrt(313)c' and 4531/sqrt(313) = c' or the rate c the distance between the planes changes, so the answer is 4531/(sqrt(313)).

Y= (2x) /5 - 6step by step(2x)/5 - y = 6- y = -2/5 (x - 15)answer: || y = (2( x - 15 )) /5