Answer: 11.25 secs
Step-by-step explanation:
So in this sense the rockets origin or the ground is modeled at h=0 so the time required if used on a table shows that h=0 between the values of 11 and 12. So if you plug and chug decimal values between these two values you get exactly 0 at t=11.25 so it takes approximately 11.25 seconds for the rocket to return to the ground
V=xyz where x,y,z are the three dimensions...
V=(1 4/5)(1 4/5)1
V=(9/5)(9/5)=81/25
V=3 6/25 ft^3
12 = 7+5 Hope this helps:)
The answer would be (B/A)*100%
