Two small planes start from the same point and fly in opposite directions. The first plane is flying 50 mph slower than the second plane. In 3 h, the planes are 900 mi apart. Find the rate of each plane.

6 and 50

Step-by-step explanation:

so in 1 hr the planes are 300 mi apart and if one plane is 50 mph slower then the planes are

300/50=6 and 50

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this