An observation deck extends 201 feet out above a valley. The deck sits 151 feet above the valley floor. If an object is dropped from the observation deck, its height h in feet, after t seconds, is given by h=-16t^2+151 . How long will it take for the object to be feet above the valley floor?

Step-by-step explanation:

The question is incomplete. Here is the complete question.

An observation deck extends 201 feet out above a valley. The deck sits 151 feet above the valley floor. If an object is dropped from the observation deck, its height h in feet, after t seconds, is given by h=-16t^2+151 . How long will it take for the object to be 6 feet above the valley floor?

Given the height of an object dropped from the observation deck modeled by the equation h(t) = -16t²+151. In order to know how long it will take for the object to be feet above the valley floor, we will substitute the height of the valley value into the equation;

h(t) = -16t²+151

If h = 6 feet

6 =  -16t²+151.

6-151 = -16t²

-145 = -16t²

t² = -145/-16

t² = 145/16

t² = 9.0625

t = √9.0625

t = 3.01seconds

Hence it will take 3.01secs take for the object to be 6 feet above the valley floor.

#15. square

step-by-step explanation:

because quadrilatural (idk how to spell it) means somthing that is not 3d or like, equal sides (or somthing like that   you know what i meanz)

hope this halps!

bai!

the answer is $42.96

step-by-step explanation: