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.

answer: b.48 for

step-by-step explanation:

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )