A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using
this equation, find out the time at which the rocket will reach its max, to the
nearest 100th of a second.
y = -16x2 + 239x + 127

mk./km: lm

step-by-step explanation:

,mlmlm

; mk

lk

5.25 ounces

step-by-step explanation:

divided 3/4 and 6

step-by-step explanation:

bob the banana

statement #3: proof- ang is congruent to ang ecd : angles are congruent.  

step-by-step explanation:

1: the missing angles in statement #3  

a.) acb  

reason- the reason why the answer is acb is because angle ecd and angle acb are both the same size *that's what the word congruent means this is really simple.  

2: the missing word in reason #3  

c.) vertical  

reason- the reason why the answer is vertical is because the angles in that diagram on figure 2 is slanted, which is vertical, and the e and a both lay on vertical lines. along with the other's, but e and a are the 2 main points you want to focus on. and c is kind of like your starting point, because c is for both angles, not just one.  

i hope that was and understanding.