Answer:
x = ![\sqrt{30}](https://tex.z-dn.net/?f=%5Csqrt%7B30%7D)
Step-by-step explanation:
Applying altitude- on - hypotenuse theorem.
(leg of large Δ )² = ( part of hypotenuse below it ) × ( whole hypotenuse )
x² = 3 × (3 + 7) = 3 × 10 = 30 ( take square root of both sides )
x = ![\sqrt{30}](https://tex.z-dn.net/?f=%5Csqrt%7B30%7D)
150 is the answer
right????
Answer:
![V_{E} = -22.5 / 15 = 1.5 miles/minute\\V_{F} = 7.5 / 15 = 0.5 miles/minute](https://tex.z-dn.net/?f=V_%7BE%7D%20%3D%20-22.5%20%2F%2015%20%3D%201.5%20miles%2Fminute%5C%5CV_%7BF%7D%20%3D%207.5%20%2F%2015%20%3D%200.5%20miles%2Fminute)
Step-by-step explanation:
To answer this problem, we must bear in mind that the express train travels 3 times faster than the freight train.
If we call F the distance that the freight train travels and we call E at the distance that the express train travels, we must have the distance between them after 15 minutes it must be 30 miles.
So
![F + E = 30\\](https://tex.z-dn.net/?f=F%20%2B%20E%20%3D%2030%5C%5C)
We also know that:
Since the express train travels 3 times faster than the freight train.
So, like both equations, we have to:
![F + 3F = 30\\F = 30/4\\F = 7.5 miles\\](https://tex.z-dn.net/?f=F%20%2B%203F%20%3D%2030%5C%5CF%20%3D%2030%2F4%5C%5CF%20%3D%207.5%20miles%5C%5C)
Then
miles in the opposite direction.
This is the distance you traveled in 15 minutes.
Therefore, the speed of the express train (
) is the distance traveled between the time it did
![V_{E} = -22.5 / 15 = 1.5 miles/minute\\V_{F} = 7.5 / 15 = 0.5 miles/minute](https://tex.z-dn.net/?f=V_%7BE%7D%20%3D%20-22.5%20%2F%2015%20%3D%201.5%20miles%2Fminute%5C%5CV_%7BF%7D%20%3D%207.5%20%2F%2015%20%3D%200.5%20miles%2Fminute)