Two joggers are running toward each other. they start off being 5 miles apart. joe, leaves at 7: 00 am and
jack leaves at 7: 30 am. if joe travels at 4 mph and jack at 5mph, at what time will they pass each other?

Hello mate!

the initial distance between Joe and Jack is 5 miles.

we know that Joe travels at 4 mph and jack travels at 5mph.

Now, let's define the 7:00 AM as our time zero, where the time is measured in hours.

We know that Joe starts running at t = 0h and Jack starts running at t = 0.5h

then the position equations are J1 for joe and J2 for Jack.

J1(t) = 4mph*t

J2(t) = 5mph*(t - 0.5h)

where the -0.5h appears because Jack starts 0.5h after.

If they run towards each other, then they will met when J1(t) + J2(t) = 5mi, wich means that between the two runned the 5 miles that separated them:

J1(t) + J2(t) = 4t + 5(t- 0.5) = 4t + 5t - 2.5 = 9t - 2.5 = 5

and now we want to isolate t

9t - 2.5 = 5

9t = 5 + 2.5 = 7.5

t = 7.5/9 = 0.8333.. hours

Then they well met 0.83 hours after the 7:00 Am

We also could write this in a more elegant way:

we know that one hour has 60 minutes, then a 0.83333.. of an hour has 0.83*60 = 50 minutes

Then they met at 7:50 AM

step-by-step explanation:

abc

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b