Two cars leave an intersection at the same time. One drives east while the other travels south at 25 miles per hour faster than the other. After 2 hours, the cars are 250 miles apart. How fast is the southbound car driving?

Step-by-step explanation:

Pythagoras :

c² = a² + b²

c is the Hypotenuse (side opposite of the right angle), which is in our case the distance of the 2 cars.

after 2 hours the east car was driving x miles. and the south car was driving x + 2×25 = x + 50 miles.

x² + (x + 50)² = 250²

x² + x² + 100x + 2,500 = 62,500

2x² + 100x = 60,000

x² + 50x = 30,000

x² + 50x - 30,000 = 0

the solution for a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 50

c = -30,000

x = (-50 ± sqrt(2500 - 4×1×-30000))/(2×1) =

= (-50 ± sqrt(2500 + 120000)/2 =

= (-50 ± sqrt(122500))/2 = (-50 ± 350)/2 =

= -25 ± 175

x1 = -25 + 175 = 150 miles

x2 = -25 - 175 = -200 miles.

negative distances don't make any sense here in our scenario, so the solution is x = 150 miles.

therefore, the east car was going 75 mph (traveled in 2 hours 150 miles).

and the south car was going 75 + 25 = 100 mph (traveled in 2 hours 200 miles).

Therefore, we conclude that  is the resulting object is solid cone.

Step-by-step explanation:

Therefore, from exercise we have a triangle and a vertical line. We know that the triangle is revolved about the vertical line. We know that a right triangle with one side against a vertical line. Based on these conditions, we can conclude that it is solid cone.

Therefore, we conclude that  is the resulting object is solid cone.