Monday

December 22, 2014

December 22, 2014

Posted by **chris** on Wednesday, March 24, 2010 at 9:37am.

- math -
**PsyDAG**, Wednesday, March 24, 2010 at 11:29amC = 2J (rates)

Distance = rate/time

Therefore, time = rate/distance

J/24 - C/24 = 3

Use substitution to solve first for cyclist and then for jogger.

- math -
**Chris**, Wednesday, March 24, 2010 at 11:48amis the answer 12

- math (oops!) -
**PsyDAG**, Friday, March 26, 2010 at 7:52pmSorry, I goofed!

Distance =**rate * time**

Therefore, Time = Distance/rate

24/J - 24/C = 3

Substitute 2j for C.

24/j - 24/2J = 3

Multiply both sides by J.

24 - 12 = 3J

12 = 3j

4 = J

C = 2J = 2 * 4 = 8

Sorry for the goof! This should help.

**Answer this Question**

**Related Questions**

Math - A cyclist and a jogger start from a town at the same time and head for a ...

Math - A cyclist and a jogger start from a town at the same time and head for a ...

maths - a cyclist and jogger are 20 miles apart. the cyclist rides at 17 mph and...

physics - A cyclist maintains a constant velocity of 4.1 m/s headed away from ...

SCIENCE - A cyclist maintains a constant velocity of 5.2 m/s headed away from ...

physics - A cyclist maintains a constant velocity of 7 m/s headed away from ...

physics - A cyclist maintains a constant velocity of 7 m/s headed away from ...

physics - A cyclist maintains a constant velocity of 4.8 m/s headed away from ...

physics - A cyclist maintains a constant velocity of 5.4 m/s headed away from ...

physics - A cyclist maintains a constant velocity of 5.9 m/s headed away from ...