To determine the values of \( x \) for which the firework reaches the ground (i.e., \( h = 0 \)), we need to solve the equation:
\[ 0 = -\frac{25}{36}x^2 + \frac{16}{23}x \]
We can factor out \( x \) from the equation to find the potential solutions:
\[ 0 = x \left(-\frac{25}{36}x + \frac{16}{23}\right) \]
This gives us one solution:
\[ x = 0 \]
Now, we need to set the remaining factor equal to zero to find the other solutions:
\[ -\frac{25}{36}x + \frac{16}{23} = 0 \]
To isolate \( x \), we can rearrange the equation:
\[ \frac{25}{36}x = \frac{16}{23} \]
Next, we multiply both sides by \( \frac{36}{25} \) to solve for \( x \):
\[ x = \frac{16}{23} \cdot \frac{36}{25} \]
Calculating this:
\[ x = \frac{16 \times 36}{23 \times 25} = \frac{576}{575} \]
Next, we will compute this value numerically:
\[ x \approx 1.003 \]
Since we are interested in the values listed in the table (12, 24, 36, 48), we can now simply check which of these \( x \) values provides \( h = 0 \).
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When \( x = 12 \): \[ h = -\frac{25}{36}(12^2) + \frac{16}{23}(12) = -\frac{25}{36}(144) + \frac{192}{23} \] Calculating these: \[ h = -100 + \frac{192}{23} \approx -100 + 8.35 \approx -91.65 \quad (\text{Negative}) \]
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When \( x = 24 \): \[ h = -\frac{25}{36}(24^2) + \frac{16}{23}(24) = -\frac{25}{36}(576) + \frac{384}{23} \] Calculating these: \[ h = -400 + \frac{384}{23} \approx -400 + 16.65 \approx -383.35 \quad (\text{Negative}) \]
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When \( x = 36 \): \[ h = -\frac{25}{36}(36^2) + \frac{16}{23}(36) = -\frac{25}{36}(1296) + \frac{576}{23} \] Calculating these: \[ h = -900 + \frac{576}{23} \approx -900 + 25.04 \approx -874.96 \quad (\text{Negative}) \]
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When \( x = 48 \): \[ h = -\frac{25}{36}(48^2) + \frac{16}{23}(48) = -\frac{25}{36}(2304) + \frac{768}{23} \] Calculating these: \[ h = -1600 + \frac{768}{23} \approx -1600 + 33.39 \approx -1566.61 \quad (\text{Negative}) \]
Summary:
All calculations yield negative heights, which means that they have not yet reached the ground.
To conclude, whilst \( x = 0 \) is the point at which the firework is launched, the quadratic equation indicates that it does intersect the ground at a distance different from the table given values. Therefore, none of the table values are solutions to reaching the ground, as they all result in negative heights.
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