A homing device left its home to deliver a message. On takeoff, the device encountered a tailwind of 10 mph and made the delivery in 15 minutes. The return trip into the wind took 30 minutes. How fast could the device fly in still air?

  30 miles per hour

Step-by-step explanation:

The relationship between time, speed, and distance is ...

  time = distance/speed

For some distance d and some speed s (in miles per hour), the travel times (in hours) are ...

  1/4 = d/(s+10)

  1/2 = d/(s -10)

Dividing the second equation by the first gives ...

  (1/2)/(1/4) = (d/(s-10))/(d/(s +10))

  2 = (s +10)/(s -10) . . . . simplify

  2s -20 = s +10 . . . . . . multiply by (s -10)

  s = 30 . . . . . . . . . . . add 20-s

  The device flies 30 miles per hour in still air.

__

Additional comment

The distance to the delivery point is ...

  d = 1/4(s +10) = 1/4(30 +10) = 10 . . . . miles to the delivery point

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

(2,20) means that the 2 is 2 hours and 20 is 20$. so there both 20$ for 2 hours only the same price. like it says. so, the dunk tank and the bounce house are the same price (20$) for the same amount of time (2 hours).

hope i