Two cars raced at a race track. the faster car traveled 20 mph faster than the slower car. in the time that the slower car traveled
165 miles, the faster car traveled 225 miles. if the speeds of the cars remained constant, how fast did the slower car travel during
the race?
distance
(mi)
165
-
slower car
no
it
faster car
55 mph
60 mnh

complete question:

Two cars raced at a race track. The faster car traveled 20 mph faster than the slower car. In the time that the slower car traveled  165 miles, the faster car traveled 225 miles. If the speeds of the cars remained constant, how fast did the slower car travel during  the race?

The slower car traveled 55 mph

Step-by-step explanation:

Let

Speed of the slower car = a mph

Speed of the faster car = (a + 20) mph

The speed of the car remains constant when the both cars covers different distance.

slower car

speed = distance/time

time = distance/speed

time = 165/a

faster car

time = distance/speed

time = 225/(a + 20)

t = t

165/a = 225/(a + 20)

cross multiply

165(a + 20) = 225a

165a + 3300  = 225a

collect like terms

3300 = 225a - 165a

60a = 3300

divide both sides by 60

a = 3300/60

a = 55 mph

The slower car traveled 55 mph

first, you divide the triangle in two, tô form a retangle triangle. after, using pythagoream theorem:

18^2 = 14^2 + (x/2)^2

(x^2/4) + 196 = 324

x^2/4 = 324 - 196

x^2/4 = 128

x^2 = 512

x = √512

x = √2^9

x = 16√2 cm

i hope i you.