Algebra

39.) two planes left st. louis for los angeles at the same time. after 4 hrs they were 700 mi apart. the slower plane traveled at 350 mi/hr. what was the speed of the faster plane?

525 mi/hr

Step-by-step explanation:

Let the speed of the faster plane be x miles per hour

Since we know both planes left St. Louis at the same time, we also know that after 4hours they are 700mi apart, and by definition we know that:

Eq(1): Distance = Velocity * Time <=> d = Vt

Also we know that the slower plane traveled at 350 mi/hr thus plugging all known values in Eq(1) for the Slowest plane we obtain the distance covered in 4 hours as follow:

d = Vt <=> d = 350 * 4 <=> d = 1400miles  

We also know that after 4 hours the planes are 700mil apart.

Thus the Fastest plane has covered a distance as

1400mil + 700mil = 2100mil

Finally recalling Eq(1) again but now for the faster plane, we now kow the distance and the time thus:

d = Vt <=> V = d/t <=> V = 2100/4 <=> V = 525 mi /hr is the speed of the faster plane.

215.92 dollars is the answer

correct answer:

b.(19,-2)