Hey Math Wizards!

[rallysb1tch]

apparently rallysman and I are retarded because we can't figure out how to set up an equation for this problem from my homework. any suggestions?


Two children own two-way radios that have a maximum range of 2 miles. One leaves a certain point at 1 p.m. walking due north at a rate of 4 mi/hr. the other leaves the same point at 1:15 p.m., travleing due south 6 mi/hr. When will they be unable to communicate with one another?

 

[cichlaguapote]

When will they be unable to communicate with one another?

1:28 pm?

Edit: No I think that's wrong.. Holy crap that's hard..

 

[rallysb1tch]

I need a tutor. It's been too damn long since I last took an algebra class!

 

[ewurm]

One starts walking North at 4 miles an hour for 15 minutes, that puts them 1 mile apart. Then the other starts walking south, making their combined speed away from each other 10 mph. They need to travel 1 more mile, so it would take an additional 1/10th of an hour, or 6 minutes. The answer is 1:21

 

[ewurm]

Algebraically speaking, 4mph(1/4hr) + 10mph(X hrs) = 2
4 x 1/4 = 1
1 +10x= 2
10x =1
x= 1/10 hr or 6 minutes

1:15 + 6 minutes = 1:21

 

[rallysb1tch]

Algebraically speaking, 4mph(1/4hr) + 10mph(X hrs) = 2
4 x 1/4 = 1
1 +10x= 2
10x =1
x= 1/10 hr or 6 minutes

1:15 + 6 minutes = 1:21

where is the 10mph coming from?

 

[cookiemonster]

Algebraically speaking, 4mph(1/4hr) + 10mph(X hrs) = 2
4 x 1/4 = 1
1 +10x= 2
10x =1
x= 1/10 hr or 6 minutes

1:15 + 6 minutes = 1:21

Whoa you sounded really smart right there lol!

 

[Nate77]

where is the 10mph coming from?

I would guess if you add 4mph + 6mph =10mph

 

[rallysb1tch]

Math makes me feel inbred or something

 

[Bigairmxman]

yes they are traveling away one at 4 and one at 6 from each other at a rate off 10 mph..which is the same as one standing still and the other walking away at 10