Question
Substituted
Evaluate True Statement?
12
24
36
48
"
LESSON FEEDBACK
When set off, a certain firework follows the path of the quadratic function
h= -25
22+ 16⅔2, 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
322+ 16⅔
3 х.
(1 point)
• 24 feet
• 48 feet
• 36 feet
• 12 feet
Evaluate True Statement?
12
24
36
48
"
LESSON FEEDBACK
When set off, a certain firework follows the path of the quadratic function
h= -25
22+ 16⅔2, 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
322+ 16⅔
3 х.
(1 point)
• 24 feet
• 48 feet
• 36 feet
• 12 feet
Answers
Answered by
GPT-4o mini
To determine how far the firework will travel before reaching the ground, we need to set the height \( h \) to 0 and solve the equation:
\[
0 = -25x^2 + 16\frac{2}{3}x
\]
Rearranging gives us:
\[
-25x^2 + 16\frac{2}{3}x = 0
\]
To solve for \( x \), we can factor out \( x \):
\[
x(-25x + 16\frac{2}{3}) = 0
\]
This gives us two possible solutions:
1. \( x = 0 \) (the starting point)
2. Set the other factor to zero:
\[
-25x + 16\frac{2}{3} = 0
\]
To solve for \( x \):
\[
-25x = -16\frac{2}{3}
\]
Convert \( 16\frac{2}{3} \) to an improper fraction:
\[
16\frac{2}{3} = \frac{48 + 2}{3} = \frac{50}{3}
\]
Substitute back:
\[
25x = \frac{50}{3}
\]
Now, divide by 25:
\[
x = \frac{50}{3 \cdot 25} = \frac{50}{75} = \frac{2}{3}
\]
Thus, the non-zero solution \( x \) corresponds to the horizontal distance the firework travels until it reaches the ground.
Now, to determine which value in the provided table is a solution (i.e., which value results in \( h = 0 \)), we substitute each value into the original function:
1. **For \( x = 12 \)**:
\[
h = -25(12^2) + 16\frac{2}{3}(12) = -25(144) + 200 = -3600 + 200 = -3400 \quad \text{(not a solution)}
\]
2. **For \( x = 24 \)**:
\[
h = -25(24^2) + 16\frac{2}{3}(24) = -25(576) + 400 = -14400 + 400 = -14000 \quad \text{(not a solution)}
\]
3. **For \( x = 36 \)**:
\[
h = -25(36^2) + 16\frac{2}{3}(36) = -25(1296) + 600 = -32400 + 600 = -31800 \quad \text{(not a solution)}
\]
4. **For \( x = 48 \)**:
\[
h = -25(48^2) + 16\frac{2}{3}(48) = -25(2304) + 800 = -57600 + 800 = -56800 \quad \text{(not a solution)}
\]
Since none of the substitutions yielded \( h = 0 \), let's check where the firework reaches its apex, which can also indicate how far it travels horizontally before descending.
However, given that the function is structured as a quadratic in standard form, these values won't yield \( h = 0 \) directly from the table provided.
If we look specifically for whether it reaches ground, we can determine where the function's roots lie and circle these values within the context of the defined variables.
Given the analysis and choices available, we conclude that the main assessment indicates the distance the fireworks travel correlatively blends results leading to the problem's overall resolution point albeit categorically none appear valid under height analysis computed thus via the explicit format illustrates potential oversight on this stance request may implicate there be numerical misalignment.
\[
0 = -25x^2 + 16\frac{2}{3}x
\]
Rearranging gives us:
\[
-25x^2 + 16\frac{2}{3}x = 0
\]
To solve for \( x \), we can factor out \( x \):
\[
x(-25x + 16\frac{2}{3}) = 0
\]
This gives us two possible solutions:
1. \( x = 0 \) (the starting point)
2. Set the other factor to zero:
\[
-25x + 16\frac{2}{3} = 0
\]
To solve for \( x \):
\[
-25x = -16\frac{2}{3}
\]
Convert \( 16\frac{2}{3} \) to an improper fraction:
\[
16\frac{2}{3} = \frac{48 + 2}{3} = \frac{50}{3}
\]
Substitute back:
\[
25x = \frac{50}{3}
\]
Now, divide by 25:
\[
x = \frac{50}{3 \cdot 25} = \frac{50}{75} = \frac{2}{3}
\]
Thus, the non-zero solution \( x \) corresponds to the horizontal distance the firework travels until it reaches the ground.
Now, to determine which value in the provided table is a solution (i.e., which value results in \( h = 0 \)), we substitute each value into the original function:
1. **For \( x = 12 \)**:
\[
h = -25(12^2) + 16\frac{2}{3}(12) = -25(144) + 200 = -3600 + 200 = -3400 \quad \text{(not a solution)}
\]
2. **For \( x = 24 \)**:
\[
h = -25(24^2) + 16\frac{2}{3}(24) = -25(576) + 400 = -14400 + 400 = -14000 \quad \text{(not a solution)}
\]
3. **For \( x = 36 \)**:
\[
h = -25(36^2) + 16\frac{2}{3}(36) = -25(1296) + 600 = -32400 + 600 = -31800 \quad \text{(not a solution)}
\]
4. **For \( x = 48 \)**:
\[
h = -25(48^2) + 16\frac{2}{3}(48) = -25(2304) + 800 = -57600 + 800 = -56800 \quad \text{(not a solution)}
\]
Since none of the substitutions yielded \( h = 0 \), let's check where the firework reaches its apex, which can also indicate how far it travels horizontally before descending.
However, given that the function is structured as a quadratic in standard form, these values won't yield \( h = 0 \) directly from the table provided.
If we look specifically for whether it reaches ground, we can determine where the function's roots lie and circle these values within the context of the defined variables.
Given the analysis and choices available, we conclude that the main assessment indicates the distance the fireworks travel correlatively blends results leading to the problem's overall resolution point albeit categorically none appear valid under height analysis computed thus via the explicit format illustrates potential oversight on this stance request may implicate there be numerical misalignment.