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:
- \( x = 0 \) (the starting point)
- \( -\frac{25}{36}x + \frac{16}{3} = 0 \)
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.
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At \( x = 12 \): \[ h = -\frac{25}{36}(12^2) + \frac{16}{3}(12) = -\frac{25}{36}(144) + 64 = -100 + 64 = -36 \text{ (not 0)} \]
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At \( x = 24 \): \[ h = -\frac{25}{36}(24^2) + \frac{16}{3}(24) = -\frac{25}{36}(576) + 128 = -400 + 128 = -272 \text{ (not 0)} \]
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At \( x = 36 \): \[ h = -\frac{25}{36}(36^2) + \frac{16}{3}(36) = -\frac{25}{36}(1296) + 192 = -900 + 192 = -708 \text{ (not 0)} \]
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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.