A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectil
e is modeled using the equation h(t) = –16t2 + 48t + 190. What is the maximum height of the projectile? 82 feet 190 feet 226 feet 250 feet
2 answers:
Answer:
The answer is 226
Step-by-step explanation:
Answer:
226 feet
Step-by-step explanation:
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Answer:
GH = 4
Step-by-step explanation:
Since G is between F and H then
FG + GH = FH , that is
8 + GH = 12 ( subtract 8 from both sides )
GH = 4
[(1515+1010) / 55] x 33 = 1515
If you calculate separately for pink and red using above formula (909pink, 606red, which equals to the same 1515)
Hope that answers it for you..
-4(-x-10)= 4x• 40
11g+4-6g+7s-10 =16g+6-7s
3(10+9x)+11-3x= 13x+19
The answer is B
Cause they all have the same angle points
