10. mr. guerrero left his house and drove north at an average speed of 40 mph. two hours later, his daughter miranda left the house and drove the sameroute as her father, but at an average speed of 48 mph. how far from homewere they when miranda overtook her father?

Speed=distance / time  ⇒ distance=time x speed
distance=d
time=t
speed= s

d=s*t

Distance  traveled by Mr Guerrero, when his daugther left the house
Data:
s=40 miles/ hour
t=2 hours
d=40 miles/ hour * 2 hours=  80 miles

Total distance traveled by Mr Guerrero:
d=80 miles + 40 miles/hourt    ⇒d=80+40t      (1)

d=total distance.
s=speed.
t=time spent by Miranda for she overtook her father.

Distance traveled by Miranda.
data:

s=48 miles/ hour
d=total distance.
t=time spent by Miranda for she overtook her father.

d=48 miles/hour * t      ⇒d=48t       (2)

With the equations (1) and (2), we make an system of equations:

d=80+40t
d=48t

We can solve this system of equation by equalization method.

80+40t=48t
80=48t-40t
80=8t
t=80/8=10

d=48t
d=48*10=480

the thistance will be 480 miles.

The second and fourth answer are correct