o 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=−2536x2+1623x
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To determine how far the firework will travel before reaching the ground, we need to find the value of x that makes the equation true, i.e., when the height is zero (0). The equation you've provided is:
0 = -2536x^2 + 1623x
This is a quadratic equation in the form of ax^2 + bx + c = 0, where a = -2536, b = 1623, and c = 0. To solve for x, we would usually use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
However, since the height is zero when the firework hits the ground, and we're looking for the value of x (which likely represents time) when that happens, we can also solve for x by factoring, or simply by observing that since the constant term c is missing (c = 0), one of the solutions for x is 0. This represents the launch time. We're interested in the second time the firework hits the ground. For that, we can solve the linear term:
-2536x^2 + 1623x = 0
x(-2536x + 1623) = 0
From the factored form, we either have x = 0 or -2536x + 1623 = 0. Since we've already determined that x=0 is the launch time, let's solve the other part for x:
-2536x + 1623 = 0
2536x = 1623
x = 1623 / 2536
Let's divide 1623 by 2536 to find the non-zero solution for x:
x ≈ 0.640093
So, the second solution to the equation, which represents the time at which the firework will hit the ground after being launched, is approximately 0.640093 time units (seconds, if we're using standard units).
Please note, this answer assumes that x represents time and the coefficients of the equation you provided are based on the physics of projectile motion but without accounting for air resistance or other potential real-world factors. Additionally, if the equation also accounts for initial height above the ground, the provided quadratic would not directly determine the ground hit time; the initial position term is not present in the equation you've given.