A skyrocket is launched from a 8-foot-high platform with an initial speed of 220 feet per second. The polynomial 16t^ ^ 2+220t+8 gives the height in feet that the skyrocket will rise in t seconds. How high will the rocket rise if it has a 7-second fuse? The equation above is - 16t ^ 2 + 220t + 8

A) 748
B) 764
C) 788
D) 882

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.

robert puts 2 in each row

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!