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.

its c or d   take a lucky guess my dude

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)