Arlington and vernonville are 160 miles apart. a car leaves arlington traveling towards vernonville, and another car leaves vernonville at the same time, traveling towards arlington. the car leaving arlington averages 10 miles per hour more than the other, and they meet after 1 hour and 36 minutes. what are the average speeds of the cars? 1. the car leaving arlington . the car leaving vernonville

Arlington moves at 55miles/hr

Vernon ville moves at 45 miles/hr

Step-by-step explanation:

Let a be the speed of the car leaving Arlington

Let v be the speed of the car leaving

Vernon ville

a = v+10

Distance travelled by the car leaving Arlington after 1hr 36minutes(1.6hrs)

= speed × time = 1.6a

Distance travelled by the car leaving Vernon ville after 1hr 36minutes(1.6hrs)

= speed × time = 1.6v

They meet after travelling this distance each thus covering 160 miles

1.6a + 1.6v = 160 1

a = v +10 (a is faster with 10miles/hr)

Put a in eqn 1

1.6(v+10) + 1.6v = 160

1.6v + 16 + 1.6v = 160

3.2v = 160-16=144

v =45 miles/hr

a =45+10= 55miles/hr

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,

it's d

step-by-step explanation: