Two sailboats leave a harbor in the Bahamas at the same time. The first sails at 25 mph in a direction 340°. The second sails at 29 mph in a direction 210°. Assuming that both boats maintain speed and heading, after 5 hours, how far apart are the boats? A. 157.4 mi
B. 212 mi
C. 175 mi
D. 244.8 mi

9514 1404 393

  D. 244.8 mi

Step-by-step explanation:

The law of cosines can be used to find the distance between the boats.

The first boat will have traveled (5 h)×(25 mi/h) = 125 mi. The second boat will have traveled (5 h)×(29 mi/h) = 145 mi. The angle between their travel directions is 340° -210° = 130°.

  c² = a² +b² -2ab·cos(C) . . . . . for a=125, b=145, C=130°

  c² ≈ 59951.05

  c ≈ √59951.05 ≈ 244.8 . . . miles

The boats are about 244.8 miles apart after 5 hours.

75 tickets have been sold so far.

step-by-step explanation:

in this question it is clearly mentioned that 45 children's tickets have been sold and this number is 60% of the total tickets sold.

now let's assume the total number of tickets sold be x.

then 60% of x = 45

⇒

⇒ x(.6) = 45

by dividing with .6 on both the sides of the equation

⇒ x = 75

therefore 75 tickets have been sold so far.