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.

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!

0.6

39mph

step-by-step explanation: