Question

When set off, a certain firework follows the path of the quadratic function h=−25/36x^2+16/23x, where:

h= the height of the firework in feet.
x= the horizontal distance it travels in feet.
To determine how far the firework will travel before reaching the ground, determine which value of x in table is a solution to the equation 0=−25/36x^2+16 2/3x
24 feet
24 feet

12 feet
12 feet

36 feet
36 feet

48 feet
48 feet

Answer 1

To determine how far the firework will travel before reaching the ground, we need to find the value of \( x \) that satisfies the equation \( 0 = -\frac{25}{36}x^2 + \frac{16}{3}x \). This is a quadratic equation in standard form:

\[
-\frac{25}{36}x^2 + \frac{16}{3}x = 0
\]

We can factor out \( x \) from the equation:

\[
x \left(-\frac{25}{36}x + \frac{16}{3}\right) = 0
\]

This gives us two solutions:

1. \( x = 0 \) (the starting point)
2. \( -\frac{25}{36}x + \frac{16}{3} = 0 \)

Now we solve for \( x \) in the second equation:

\[
-\frac{25}{36}x + \frac{16}{3} = 0
\]

Rearranging gives:

\[
-\frac{25}{36}x = -\frac{16}{3}
\]

Multiplying both sides by \(-1\):

\[
\frac{25}{36}x = \frac{16}{3}
\]

Now, multiply both sides by \( \frac{36}{25} \) to solve for \( x \):

\[
x = \frac{16}{3} \cdot \frac{36}{25}
\]

Calculating the right side:

\[
x = \frac{16 \cdot 36}{3 \cdot 25}
\]

Calculating \( 16 \cdot 36 \):

\[
16 \cdot 36 = 576
\]

Now calculating \( 3 \cdot 25 \):

\[
3 \cdot 25 = 75
\]

Thus:

\[
x = \frac{576}{75}
\]

To simplify \( \frac{576}{75} \):

Calculating the GCD of 576 and 75 is 3, so we can simplify:

\[
x = \frac{192}{25} \approx 7.68 \text{ feet}
\]

So now we will check which choice from the given options can be a solution to the original quadratic equation. None of those values are equal to the height calculated. Hence, out of the choices available:

- **12 feet**
- **24 feet**
- **36 feet**
- **48 feet**

We will test which of these might be the points where the height is zero.

Let's evaluate the quadratic function at the provided points.

1. **At \( x = 12 \)**:
\[
h = -\frac{25}{36}(12^2) + \frac{16}{3}(12) = -\frac{25}{36}(144) + 64 = -100 + 64 = -36 \text{ (not 0)}
\]

2. **At \( x = 24 \)**:
\[
h = -\frac{25}{36}(24^2) + \frac{16}{3}(24) = -\frac{25}{36}(576) + 128 = -400 + 128 = -272 \text{ (not 0)}
\]

3. **At \( x = 36 \)**:
\[
h = -\frac{25}{36}(36^2) + \frac{16}{3}(36) = -\frac{25}{36}(1296) + 192 = -900 + 192 = -708 \text{ (not 0)}
\]

4. **At \( x = 48 \)**:
\[
h = -\frac{25}{36}(48^2) + \frac{16}{3}(48) = -\frac{25}{36}(2304) + 256 = -1600 + 256 = -1344 \text{ (not 0)}
\]

Thus, none of the provided answers yield a height of zero when plugged into the quadratic function.

The simplest solution to provide is that since the parabola descends and we solve \( h=0 \): we are not yielding real solutions within given options; hence, **the firework travels before touching the ground is greater even above maximum evaluated above**, that is beyond 48 ft without solutions provided.