Amodel rocket is launched from a roof into a large field the path of the rocket can be modeled by the equation y equals -0.05x^2 + 9.6 x + 5.4 where x is the horizontal distance in meters from the starting point on the roof and y is the height in meters of the rocket above the ground how far horizontally from its starting point will the rocket land round your answer to the nearest hundredth

•82.03 m
•6.50 m
• 90.00 m
• 88.53 m

The answer to this question is going to be 88.53 because you shift it horizontally

it's 160 or 160. 56

Step-by-step explanation:

since you want the height horizontally from the starting height, and that height was 5.4 metres,

-.06x^2 + 9.6x + 5.4 = 5.4

the 5.4 drops out, so

x(-.06x + 9.6) = 0

x = 0 , the starting time

or

x = 9.6/.06 = 160

it will land after 160 m

stop cheating

step-by-step explanation:

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft