An air traffic controller spots two airplanes at the same altitude converging to a point as they fly at right angles to each other. One airplane is 70 miles from the point and has a speed of 280 miles per hour. The other is 240 miles from the point and has a speed of 960 miles per hour. (a) At what rate is the distance between the planes changing

DL/dt = 1000 miles/hour

Step-by-step explanation:

Let´s call A Point for the airplane at 70 miles from the point O (converging point), and point B for the airplane at 240 miles from point O.

Then the three-points A, B, and O shape a right triangle with legs (distances from each of the airplane to point O, and hypotenuse L distance between the two airplanes

Then according to Pithagoras´theorem:

L²  =  (AO)²  + (BO)²

At the moment t when the airplanes are far away as 70 and 240 miles per hour

L²  = (70)²  + ( 240)²

L²  = 4900 + 57600

L = √62500

L = 250 miles

In general

L²  = x² + y²

That equation is always valid for a right triangle if the airplanes are approaching keeping the right triangle shape then:

L² = x² + y²        where x and y are the legs ( that legs change in time then):

Tacking derivatives on both sides of the equation

2*L*DL/dt = 2*x*Dx/dt + 2*y*Dy/dt

By substitution:    since Dx/dt = 280 m/h   and Dy/dt = 960 m/h

2*(250)*DL/dt = 2*70*280 + 2*(240)*960

500*DL/dt  =  39200 + 460800

DL/dt  =  500000/ 500

DL/dt = 1000 miles/hour

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

step-by-step explanation: