Arobot begins its journey at the origin, point 0, and travels along a straight line path at a constant rate. fifteen minutes into its journey the robot is at a(5, 8) . (a.) if the robot does not change speed or direction, where will it be 3 hours into its journey (call this point b)? (b.) the robot continues past point b for a certain period of time until it has traveled an additional 3/4 the distance it traveled in the first 3 hours and stops. i. how long did the robot's entire journey take? ii. what is the robot's final location? c. what was the distance the robot traveled in the last leg of its journey?

A) Point B(60,96)
B)i. 315 mins (or 5.25 hours [or 5 1/4 hours])
ii. Ended at coordinates (105,168)
C) it travelled a distance of the origin to coordinates (45,72). It travelled 45 in the x direction and 72 in the y direction.

C is confusing question. Assuming that means after it stopped and continued again, the last leg of the journey was just the slope 72/45; or

First find the slope
Slope=rise/run
Slope=8/5
During 15 mins the robot goes at the slope of 8/5.
3 hours is 180 mins.
180/15=12
8/5*12=96/60
Remember!! Slope=rise/run, now UNDO it
96/60 (undo it) becomes point B (60,96)

An additional 3/4 of slope 96/60 is:
(96/4)*3=72
(60/4)*3=45
72/45=3/4 of 96/60
Dividing 72 by 45 and 96 by 60 will yield the same number (therefore the slope is the same therefore that checks that this is correct)
The time is took to do 96/60 was 3 hours, or 180 mins. (180/4)*3=135
The new total time of journey is 180+135=315 mins
Final location is just adding (45,72) and (60,96) to get (105,168)

)because it's going in order of the math sign