SOLUTION: Kim starts to walk 3 mi to school at
7:30 A.M. with a temperature of 0°F. Her brother Bryan
starts at 7:50 A.M. on his bicycle, traveling 10 mph faster
than Kim. If they get to
Algebra ->
Functions
-> SOLUTION: Kim starts to walk 3 mi to school at
7:30 A.M. with a temperature of 0°F. Her brother Bryan
starts at 7:50 A.M. on his bicycle, traveling 10 mph faster
than Kim. If they get to
Log On
Question 341165: Kim starts to walk 3 mi to school at
7:30 A.M. with a temperature of 0°F. Her brother Bryan
starts at 7:50 A.M. on his bicycle, traveling 10 mph faster
than Kim. If they get to school at the same time, then how
fast is each one traveling?
How do I solve this word problem because I am at a lost. I am totally pulling my hair out with this one. Please help me somebody!!!! Answer by galactus(183) (Show Source):
You can put this solution on YOUR website! I don't see what the temperature has to do with it other than it was a cold walk to school.
Since d=rt, Kim travels 3=rt.
Bryan starts 1/3 hour later and travels 10 mph faster, but he travels the same distance: 3=(r+10)(t-1/3)
There are two equations with two unknowns. Can you finish now?.
