PLEASE SOMEONE HELP ME!!!!! ILL GIVE BRAINLIEST1. A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.04x2 + 8.3x + 4.3, 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?A. 208.02 mB. 416.30 mC. 0.52 mD. 208.19 m2. A catapult launches a boulder with an upward velocity of 184 ft/s. The height of the boulder (h) in feet after t seconds is given by the function h= -16t^2 + 184t + 20. How long does it take the boulder to reach its maximum height? What is the boulder's maximum height.A. Reaches a maximum height of 11.6 feet after 5.75 seconded.B. Reaches a maximum height of 549 feet after 11.5 seconds.C. Reaches a maximum height of 549 feet after 5.75 seconds.D. Reaches a maximum height of 23.2 feet after 11.6 seconds.3. A ball is thrown into the air with an upward velocity of 32 ft/s. Its height (h) in feet after t seconds is given by the function h= -16t^2 + 32t + 6. How long does it take the ball to reach its maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary.A. Reaches a maximum height of 22 feet after 1.00 second.B. Reaches a maximum height of 22 feet after 2.00 seconds.C. Reaches a maximum height of 44 feet after 2.17 seconds.D. Reaches a maximum height of 11 feet after 2.17 seconds.

Question
Answer:
1. When it hit the ground=>y = 0 
=>By y = –0.04x^2 + 8.3x + 4.3 
=>0.04x^2 - 8.3x - 4.3 = 0 
=>x = [-(-8.3) +/- √{(-8.3)^2 - 4 x 0.04 x (-4.3)}]/[2 x 0.04] 
=>x = [8.3 +/- 8.34]/[0.08] 
=>x = 208.02 m

Your answer is A. 208.02 m

2. A. Reaches a maximum height of 11.6 feet after 5.75 seconded.

3. B. Reaches a maximum height of 22 feet after 2.00 seconds.

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