PLEASE IM TIMED A small rocket is launched from a height of 6 feet off of the ground at an initial upward velocity of 64 feet per second. Its height is given by the equation h=-16t^2+64t+6, where h is the height of the rocket in feet and t is the time since liftoff in seconds. At what time is the rocket 54 feet in the air?

the answer to the question is a

step-by-step explanation:

it has the same slope and the same y-intercept.

y + 11 = -7/9(x - 18)

y + 11 =-7/9x + 14

b

y = -7/9x + 3

y-intercept (0, 3)

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.

this is the formula: v = the value of investment at the end of the time period

p = the principal amount (the initial amount invested)

r = the annual interest rate

n = the annual frequency of compounding (how many times a year interest is added)

t = the number of years the money is invested

^ means raise to the power of