Two ships leave port at 10 a. m. one travels at a bearing of 50° at 25 mph, and the other travels at a bearing of 120° at 20 mph. how far apart are the ships at noon? round your answer to the nearest tenth.

The first ship travels (25 mi/h)*(2 h) = 50 mi.
The second ship travels (20 mi/n)*(2 h) = 40 mi.
The angle between their directions of travel is 120° -50° = 70°.

The law of cosines can be used to find the distance (d) between the ships.
  d² = 50² + 40² - 2*50*40*cos(70°)
  d² = 2500 +1600 -4000*cos(70°)
  d² ≈ 2731.919
  d ≈ √2731.919 ≈ 52.3 . . . . miles

At noon, the ships are about 52.3 miles apart.

The answer is c because you turn 17.5% into a decimal which gives you .175 then you times that by $361.35 which gives you 63.23625 but then you round that which gives you 63.24. then you at 63.24 pulse 361.35 and you round it which gives you $425