A rocket is launched from atop a 65-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

7.6 seconds

Step-by-step explanation:

To solve this problem we can use the following equation:

S = So + 113*t + at^2/2

Where S is the final position, So is the inicial position, Vo is the inicial speed, a is the acceleration and t is the time.

In this problem, we have that S = 0, So = 65, Vo = 113 and a = -32 ft/s2 (The acceleration of gravity)

So we have that:

0 = 65 + 113t - 16t2

16t2 - 113t - 65 = 0

Using Bhaskara's formula, we have:

Delta = 113^2 + 4*65*16 = 16929

sqrt(Delta) = 130.11

t = (113 + 130.11) / (2*16) = 7.597 seconds

Rounding to nearest tenth, we have t = 7.6 seconds

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6