A truck starts off 180 miles directly east from the city of Hartville. It travels due south at a speed of 45 miles per hour. After travelling 49 miles, how fast is the distance between the truck and Hartville changing

  11.82 mph

Step-by-step explanation:

The angle between the direction of travel and the direction to the city of Hartville is ...

  arctan(180/49) ≈ 74.77°

The speed from the direction of Hartville is the the actual speed multiplied by the cosine of this angle:

  speed from Hartville = (45 mph)×cos(74.77°) ≈ 11.82 mph

Comment on the solution

There are other approaches you can use to solve this. One is to compute the  change in distance over a small period of time, such as 0.02 hours.

0.01 hours before the time of interest, the distance to the city is ...

  d1 = √(180² +(49-.01(45))²) ≈ 186.4326 . . . miles

0.01 hours after the time of interest, the distance to the city is ...

  d2 = √(180² +(49+.01(45))²) ≈ 186.6690 . . . miles

Then the rate of change of distance is ...

  (d2 -d1)/(t2 -t1) = (186.6690 -186.4326)/0.02 = 11.82 . . . mi/h

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Another is to write the distance equation and differentiate it. (You will find the solution looks very much like the trig solution above.)

  d = √(180² +(45t)²)

  dd/dt = 45²t/√(180² +(45t)²) . . . . . to be evaluated at t=49/45

  rate of change = 45(49/√(180²+49²)) ≈ 11.82 . . . mi/h

total cost earn=$1360

adult sold =45

student sold =80

adult cost twice as student cost so we can say adult sold 45+45=90

so total sold=90+80=170  

to get the cost of 1 meal we divide 1360 by 170

1360/170=8

one student meal cost=8$

one adult meal cost=8*2=16

total students meal cost=8*80=640$

total adult meal cost=16*45=720

640+720=1360