Ray starts walking to school at a rate of 2 mi/h. Ten minutes later, his sister runs after him with his lunch, averaging 6 mi/h. b. Solve the system
to determine how much time it took Ray's sister to catch up to him.
Your answer

Ray walks 15 minutes prior to his sister and walk for 5 minutes to catches up

Step-by-step explanation:

The computation of the time taken is as follows:

Let us assume the Ray time be t

So his sister time would be t - 10

Since Ray distance is 2t

So his sister distance is 6(t - 10) = 6t - 60

As the distance is the same

So, the following equation must be made    

2t = 6t - 60

 -4t = -60

t = 15

Therefore Ray walks 15 minutes prior to his sister and walk for 5 minutes to catches up

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step-by-step explanation:

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of