Ray starts walking to school at a rate of 2 mi/h. Ten minutes later, his sister runs after him with his lunch, averaging 6mi/h.

Complete question :

Ray starts walking to school at a rate of 2 mi/h. Ten minutes later, his sister runs after him with his lunch, averaging 6mi/h. Determine take it takes for sister to catch up

5 minutes

Step-by-step explanation:

Ray's rate = 2mi/ hr

Time walked before sister starts = 10 minutes = (10/60) = 0.16667 hours

Hence, distance already covered = (rate * time) = (2 * 0.16667) = 0.3333 miles

To catchup :

Let t = Time taken to catch ; At t, distance covered must be the same ; thus ;

0.3333 + (2 * t) = 6 * t

0.3333 + 2t = 6t

0.3333 = 6t - 2t

0.3333 = 4t

t = 0.3333 / 4

t = 0.0833333 hours

t = 0.0833333 * 60 = 5 minutes

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.