two cars leave the same point at noon. one car travels north and the other car travels east. suppose the northbound car is traveling 40mph, and after two hours, the two car are 100 miles apart.how fast is the other car going?

recall that speed is distance travelled over time.

let x = speed of the other car.
then we set-up the equation.
when you draw the directions they're travelling, you will form a right triangle. thus recall that in a right triangle, the hypotenuse (in this the the distance between them which is 100 miles) is given by the Pythagorean theorem:
c^2 = a^2 + b^2
we equate distances:
100^2 = (40*2)^2 + (2x)^2
10000 = 80^2 + 4x^2
10000 - 6400 = 4x^2
3600 = 4x^2
x^2 = 900
x = 30 mph

hope this helps~ :)

thanks

To find out how fast the eastbound car is going, we can use the Pythagorean theorem, which relates the lengths of the sides of a right triangle. In this case, the two cars are moving at right angles to each other, and after two hours, they are 100 miles apart.

Let's consider the northbound car's path as the vertical leg of the triangle and the eastbound car's path as the horizontal leg. The distance the northbound car travels in two hours can be calculated by multiplying its speed (40 mph) by the amount of time (2 hours), giving us 80 miles.

Using the Pythagorean theorem, we have:

(80)^2 + (x)^2 = (100)^2

where x represents the speed of the eastbound car.

Simplifying the equation:

6400 + (x)^2 = 10000

Subtracting 6400 from both sides:

(x)^2 = 3600

Taking the square root of both sides:

x = 60 mph

Therefore, the eastbound car is traveling at a speed of 60 mph.