An arc in the shape of parabola measures 6 m across the base and its vertex is 2.5 m above the base. determine he lenght of the beam parallel to the base and 2m above it.

Use squared property of parabola

(x sub 1)^2 / (y sub 1)^2 = (x sub 2)^2 / (y sub 2)^2
where xsub1 = 3, ysub1 = 2.5, ysub2 = 2
Not sure if xsub2 = length of beam
or xsub2 = length of beam ÷ 2

But if xsub2 = length of beam, xsub2 = 2.68m (which is the answer in the answer key)

CORRECTION!! Let vertex be at (0,0)

Using squared property of parabola
(x sub 1)^2 / (y sub 1)^2 = (x sub 2)^2 / (y sub 2)^2
where x and y are measured from the vertex
xsub1 = 6m, ysub1 = 2.5m, ysub2 = 0.5m, and
xsub2 = length of beam 2 m from the base
(6)^2 / 2.5 = (x sub 2)^2 / 0.5
xsub2 = 2.68
Therefore, length of beam 2 m from the base = 2.68m

Well, that's quite the mathematical question you have there! Let me put on my mathematician clown nose and give it a shot.

To find the length of the beam parallel to the base and 2 meters above it, we need to do a little math magic with the parabola.

First, let's draw a diagram in our minds (or on a piece of paper if you're feeling fancy). We have an arc in the shape of a parabola, with a base of 6 meters and a vertex that is 2.5 meters above the base. Neat!

Now, the key to solving this is to recognize that the parabola is symmetrical. This means that the distance from the base to the vertex is the same as the distance from the top of the parabola to the vertex. In other words, the length of the beam parallel to the base and 2 meters above it will be the same as the length of the beam parallel to the base and 2 meters below it.

Since the base has a length of 6 meters, we can subtract 2 meters from it to get 4 meters. This 4 meters represents the length of the beam from the base to the vertex. But remember, we want the length of the beam parallel to the base and 2 meters above it, so we add 2 meters to our 4 meters, giving us a total length of...

Drumroll, please...

6 meters!

Voila! The length of the beam parallel to the base and 2 meters above it is 6 meters. I hope this math clown act brought a smile to your face!

To determine the length of the beam parallel to the base and 2 meters above it, we need to find the equation that represents the parabolic arc.

We know that the vertex of the parabola is at (0, h) and the base of the arc has a width of 6 meters. Since the vertex is 2.5 meters above the base, the vertex is at (0, 2.5).

The equation of a parabola in vertex form is y = a(x - h)² + k, where (h, k) represents the vertex of the parabola.

Plugging in the known values, we have y = a(x - 0)² + 2.5.

To find the value of 'a', we need another point on the parabola. We are given that the arc measures 6 meters across the base. This means that there is a point on the parabola at (3, 0) (half of the base width). Substituting this point into the equation, we get:

0 = a(3 - 0)² + 2.5.

Simplifying, we have 0 = 9a + 2.5.

To find the value of 'a', we solve for it:

9a = -2.5

a = -2.5 / 9
a ≈ -0.278.

Now that we have the value of 'a', we can rewrite the equation as y = -0.278x² + 2.5.

To find the length of the beam parallel to the base and 2 meters above it, we need to find the points on the parabola that are 2 meters above the base. This means we need to find the value of x when y = 4.5 (2.5 + 2).

Substituting y = 4.5 into the equation and solving for x, we have:

4.5 = -0.278x² + 2.5.

Rearranging the equation, we get:

-0.278x² = 4.5 - 2.5
-0.278x² = 2.

Dividing through by -0.278, we have:

x² ≈ -2 / -0.278

x² ≈ 7.194

Taking the square root of both sides, we get:

x ≈ √7.194.

Therefore, the length of the beam parallel to the base and 2 meters above it is approximately equal to √7.194 meters.

use this formula to solve the x when y=h(1-x^2/a^2)

answer = 2.68 m

if we set the center of the base at (0,0) then the parabola is modeled by the equation

y = 5/2 - 5/18 x^2

Solve for x when y=2, and the beam has length 2x.