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

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation:

if the pizza is divided equally into 2 equal halves then you would get

πr^2/2

where r is the radius of the pizza.

happy to

pls mark as brainliest