Arock is thrown from the top of a tall building. the distance in feet between the rock and the ground t seconds after thrown is given by d = -16t2 -4t + 412. how long after the rock is thrown is it 407 feet from the ground?

0.448 seconds

Step-by-step explanation:

d = -16t² -4t + 412

find t when d = 407

substituting d = 407 into the equation:

407 = -16t² -4t + 412 (subtract 407 from both sides)

-16t² -4t + 412 - 407 = 0

-16t² -4t + 5 = 0 (multiply both sides by -1)

16t² + 4t - 5 = 0  

solving using your method of choice (i.e completing the square or using the quadratic equation), you will end up with

t = (-1/8) (1 + √21)= -0.70 seconds (not possible because time cannot be negative)

or

t = (-1/8) (1 -√21) = 0.448 seconds (answer)

idk

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