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

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation: