Two children own two-way radios that have a maximum range of 3 miles. one leaves a certain point at 1: 00 p. m., walking due north at a rate of 6 mi/hr. the other leaves the same point at 1: 15 p. m., traveling due south at 9 mi/hr. for how many minutes after 1: 00 p. m. will they be able to communicate with one another? (round your answer to the nearest minute.)

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

4) Sum the equations and equate to 3

5) Substitute the values we already know

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

6hryt%4578855

false

step-by-step explanation:

false : it can not go to 9 the highest that part can go is 7

here is your proof