A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+124x+127

Step-by-step explanation:

so when the height is zero, that's when the rocket hits the ground.  The equation will have two zeros,  but one would be for a negative time,  or before the rocket is launched.    so we only one the one after the rocket is launched.

use the quadratic formula to solve for  x  ( which is time)

-124 +- sqrt [ - 4*(-16)*(127) ] / 2 (-16)

-124 +- sqrt [15376 + 8128 ] / -32

-124 +- sqrt [23504 ] / -32

-124 +- 153.310 / -32

if we take the positive part it will become the negative time so skip that one try the negative part 1st

-124 - 153.310 / -32

-277.310 / -32

8.6659 seconds to hit the ground.    ( time of flight)  aka   TOF

8.67 seconds  ( rounded to nearest 100th)

answer: 193

explanation:

200-7 = 193

-1/4 > -1 2/3. -1/4 is greater than -1 2/3 because it is closer to 0 on the number line.