A rocket launched from a tower. the height of the rocket, y in feet, i related to the time after launch, x in seconds, by the given equation. using the equation find the time that the rocket will hit the ground to the nearest 100th of a second. y = -16x² + 165x + 257

cheater

step-by-step explanation:

lo98765432wqasdfghjkloi87654567890

step-by-step explanation:

20days louise can read300 pg

step-by-step explanation:

1 day 15 pg 300: 15= 20

x day 300 pg

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )