Suppose a railroad rail is 5 kilometers and it expands on a hot day by 12 centimeters in length. approximately how many meters would the center of the rail rise above the ground?

  about 17 meters

Step-by-step explanation:

We can use the Pythagorean theorem to put an upper bound on the height of the bump in the rail. This assumes half the expanded rail length (d+e) is the hypotenuse of a right triangle whose legs are the bump height (b) and the 2500 meter distance (d) from the center of the rail to its end.

The Pythagorean theorem relates these distances this way:

  b^2 + d^2 = (d+e)^2

Expanding the square on the right, we can simplify the expression to find b.

  b^2 = (d^2 +2de +e^2) -d^2

  b^2 = e(2d +e)

  b = √(e(2d +e))

Using lengths in meters, we can fill this in to calculate b.

  b = √(.06(2·2500 +.06)) = √300.0036

  b ≈ 17.32 . . . . meters

Comment on this solution

We don't expect rails to tear loose from the rail bed and rise up to a height matching that of a 3-story building. That is why there are typically expansion joints and shorter rail lengths used in the construction of railways.

The height is a little lower if we take physics into account and distribute the stress in the rail along its length. No doubt the final curve is somewhat more complicated than the triangle we have assumed.

If it were an ellipse, the height might only be 9.4 meters, with the steepest rise occurring near the ends of the rail. The math for this model is beyond the scope of this answer.

1.4

2.3

step-by-step explanation:

the formula for the constant of variation is

y= kx

divide each side by x

y/x =kx/x

y/x =k

we know a and y

1.   y/x = k

4/1 =4

check the other values to make sure they are the same

8/2 = 4   12/3 =4   and 16/4 =4  

the constant of variation is 4

2.   y/x = k

3/1 =3

check the other values to make sure they are the same

6/2 = 3   9/3 =3   and 12/4 =3  

the constant of variation is 3

555.99.

step-by-step explanation:

hope this