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When set off, a certain firework follows the path of the quadratic function h=−2536x2+1623x , where: h= the height of the firew...Question
When set off, a certain firework follows the path of the quadratic function h=-25/36x62 + 16 2/3x, where :h = the height of the firework in feet and x=the horizontal distance in travels in feet
to determine how far the firework will travel before reaching the ground, determine which value of x in the table is a solution to the equation 0=-25/36x62 + 16 2/3x
Values of x in the table are 12, 24, 36, 48
to determine how far the firework will travel before reaching the ground, determine which value of x in the table is a solution to the equation 0=-25/36x62 + 16 2/3x
Values of x in the table are 12, 24, 36, 48
Answers
Answered by
GPT-4o mini
To determine which value of \( x \) corresponds to the firework reaching the ground, we need to solve the equation:
\[
0 = -\frac{25}{36}x^2 + \frac{50}{3}x
\]
This simplifies to solving:
\[
\frac{25}{36}x^2 - \frac{50}{3}x = 0
\]
Factoring out \( x \):
\[
x \left(-\frac{25}{36}x + \frac{50}{3}\right) = 0
\]
Setting each factor to zero, we have:
1. \( x = 0 \)
2. \(-\frac{25}{36}x + \frac{50}{3} = 0\)
Let's solve the second equation:
\[
-\frac{25}{36}x + \frac{50}{3} = 0
\]
Rearranging gives:
\[
\frac{25}{36}x = \frac{50}{3}
\]
Multiplying both sides by \( \frac{36}{25} \):
\[
x = \frac{50}{3} \cdot \frac{36}{25}
\]
Calculating the right-hand side:
\[
x = \frac{50 \times 36}{3 \times 25} = \frac{1800}{75} = 24
\]
Thus, \( x = 24 \) is the value where the firework will reach the ground.
Checking the table’s values (12, 24, 36, 48), the solution to the equation is:
\[
\boxed{24}
\]
\[
0 = -\frac{25}{36}x^2 + \frac{50}{3}x
\]
This simplifies to solving:
\[
\frac{25}{36}x^2 - \frac{50}{3}x = 0
\]
Factoring out \( x \):
\[
x \left(-\frac{25}{36}x + \frac{50}{3}\right) = 0
\]
Setting each factor to zero, we have:
1. \( x = 0 \)
2. \(-\frac{25}{36}x + \frac{50}{3} = 0\)
Let's solve the second equation:
\[
-\frac{25}{36}x + \frac{50}{3} = 0
\]
Rearranging gives:
\[
\frac{25}{36}x = \frac{50}{3}
\]
Multiplying both sides by \( \frac{36}{25} \):
\[
x = \frac{50}{3} \cdot \frac{36}{25}
\]
Calculating the right-hand side:
\[
x = \frac{50 \times 36}{3 \times 25} = \frac{1800}{75} = 24
\]
Thus, \( x = 24 \) is the value where the firework will reach the ground.
Checking the table’s values (12, 24, 36, 48), the solution to the equation is:
\[
\boxed{24}
\]
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