To determine which value of x in the table is a solution to the equation \( 0 = -2536x^2 + 1623x \), we can substitute each value of x into the equation and check if it equals zero.
The equation can be rewritten in the form of a quadratic function: \[ h(x) = -2536x^2 + 1623x \]
Then we can evaluate it for each x value provided in the table (12, 24, 36, 48).
-
For \( x = 12 \): \[ h(12) = -2536(12)^2 + 1623(12) \] \[ h(12) = -2536(144) + 1623(12) \] \[ h(12) = -364224 + 19476 = -344748 \quad \text{(not 0)} \]
-
For \( x = 24 \): \[ h(24) = -2536(24)^2 + 1623(24) \] \[ h(24) = -2536(576) + 1623(24) \] \[ h(24) = -1458176 + 38952 = -1419224 \quad \text{(not 0)} \]
-
For \( x = 36 \): \[ h(36) = -2536(36)^2 + 1623(36) \] \[ h(36) = -2536(1296) + 1623(36) \] \[ h(36) = -3284736 + 58388 = -3226348 \quad \text{(not 0)} \]
-
For \( x = 48 \): \[ h(48) = -2536(48)^2 + 1623(48) \] \[ h(48) = -2536(2304) + 1623(48) \] \[ h(48) = -5840640 + 77808 = -5762832 \quad \text{(not 0)} \]
None of the evaluated heights equal zero, so the autumn provided values do not solve the equation directly. However, to find the value of \( x \) where the firework reaches the ground, we can solve the equation directly for its roots using the quadratic formula or graphing, but based on the context, you should check each substituted x value in the original equation.
Please note that if you have numerical errors or expectations based on the value being verified, it'd be useful to check those calculations accurately or refer to additional numerical results required.
From the values given and evaluations, none of the options provided yield \( h(x) = 0 \). The solutions must be calculated based on the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).