Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 miles per hour. then, 15 minutes later, paul passed the same gas station while traveling west at a constant speed of 60 miles per hour. if both drivers maintained their speeds and both remained on the highway for at least 2 hours, how long after he passed the gas station did paul catch up with mary?

1 hour 15min

Step-by-step explanation:

When Paul passes the gas station Mary si already 12.5 miles away

X=v t = 50mph 0.25h = 12.5 miles. (15min equals a quarter hour)

From that moment to the meeting point we’ll have only one t (time) and one x (final point) for both cars

For Mary we have

X= Xo + v t= 12.5 miles + 50mph t

For Paul we have

X=v t = 60mph t

As x si the same in both equations

60mph t = 12.5 miles + 50mph t

(60mph - 50mph) t = 12.5miles

10mph t = 12.5miles

t = 12.5miles/10mph

t = 1,25 h

4 packs of water and 7 packs of candy bars

step-by-step explanation:

35x4=140

140/20=7

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i