Given :
A skyrocket is launched from a 8-foot-high platform with an initial speed of 220 feet per second.
The polynomial -16t² + 220t + 8 gives the height in feet that the skyrocket will rise in t seconds.
To Find :
How high will the rocket rise if it has a 7-second fuse?
Solution :
Since, it is given that the rocket has a fuse time of 7 seconds.
Putting value of 7 sec in given equation, we get :
Height = -16×( 7 )² + 220×( 7 ) + 8
Height = 764 m
Therefore, height of rocket in 7 second fuse is 764 m.
Answer:
150
Step-by-step explanation:
i triangles original angle is 180 150+30 is 180
Answer:
f(f(2))=8
Step-by-step explanation:
f(f(2))=5(5*2 - 7) - 7=5*3-7=8
Answer:
2
Step-by-step explanation:
it has 2 in the equation
