Jack and Diane are jogging back and forth along a one-mile path. They started out at 9:00 A. M. from opposite ends of the path. They passed each other in 10 minutes when Diane has gone 1/3 mile. What time will they first meet at one end of the path? You have to assume they keep jogging at the same speeds. Explain :

30 minutes

Step-by-step explanation:

that problem description is imprecise.

I think what is meant here : they each keep jogging at their own same speed.

Diane's speed is 1/3 miles / 10 min.

Jack's speed is 2/3 miles / 10 min.

now, to bring this to regular miles/hour format, we need to find the factor between 10 minutes and an hour (60 minutes) and multiply numerator and denominator (top and bottom of the ratio) by it.

60/10 = 6.

so, we need to multiply both speeds up there by 6/6 to get the miles/hour speeds.

Diane : (1/3 × 6) / hour = 2 miles / hour

Jack : (2/3 × 6) / hour = 4 miles / hour

since Jack is running twice as fast as Diane, she will finish one length in the same time he finishes a round trip (back and forth).

Diane running 1 mile going 2 miles/hour takes her 30 minutes.

Jack running 2 miles (back and forth) going 4 miles/hour will take him also 30 minutes.

so, they will meet at his starting point after 30 minutes.

93.8

step-by-step explanation:

lets say, first number = x

second number = 3x+7

total = 49

(x)+(3x+7)= 49

4x+7 = 49

4x = 42

x = 42/4

x= 10.5

x= 10.5

3x+7 = 3(10.5)+7 = 93.8

hope this

g

step-by-step explanation: