two cars leave an intersection. one car travels north; the other east. when the car traveling north had gone 15 miles, the distance between the cars was 5 mile more than the distance traveled by the car heading east.how far had the east bound car traveled?
A rt. triangle is formed:
X miles = hor side.
15 miles = ver. side.
x+5 = hyp.
x^2 + 15^2 = (x+5)^2.
x^2 + 225 = x^2 + 10x + 25.
x^2 - x^2 + 10x = 225 - 25 = 200.
10x = 200.
X = 200 / 10 = 20 miles.
To determine the distance traveled by the eastbound car, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's denote the distance traveled by the eastbound car as "x" miles.
According to the problem, when the northbound car traveled 15 miles, the distance between the two cars was 5 miles more than the distance traveled by the eastbound car. This can be expressed as:
(15)^2 = (x)^2 + (x+5)^2
To solve for x, we can simplify the equation by expanding and combining like terms:
225 = x^2 + (x^2 + 10x + 25)
Combine like terms:
225 = 2x^2 + 10x + 25
Rearrange the equation to set it equal to zero:
2x^2 + 10x + 25 - 225 = 0
Simplify:
2x^2 + 10x - 200 = 0
Divide the entire equation by 2 to simplify it further:
x^2 + 5x - 100 = 0
Now, we can solve this quadratic equation. Since it does not factor nicely, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 5, and c = -100. Plugging these values into the quadratic formula, we get:
x = (-5 ± √(5^2 - 4(1)(-100))) / (2(1))
Simplify further:
x = (-5 ± √(25 + 400)) / 2
x = (-5 ± √425) / 2
x ≈ (-5 ± 20.62) / 2
Thus, the two possible values for x are:
x ≈ (-5 + 20.62) / 2 ≈ 7.81
x ≈ (-5 - 20.62) / 2 ≈ -12.81
Since distance cannot be negative, we discard the negative solution. Therefore, the eastbound car traveled approximately 7.81 miles.