Answer:

Step-by-step explanation:
Let the first term be
, then the terms of the sequence are:
.
Since the fourth term is 108, we have


Hence the sequence becomes:
.
The explicit expression for a geometric expression is given by:

where a=4 and r=3
The required formula is:

Answer:
4ln(2) + 4ln(x)
Step-by-step explanation:
ln(2x)⁴
4ln(2x)
4[ln(2) + ln(x)]
4ln(2) + 4ln(x)
The general formula is:

-- (A)
Where g = 9.8 m/s/s

= 0

= 100
a)
Time=t=3s.
Plug-in the values in (A)
y = 100 -(1/2*9.8*3*3)
y = 100-44.1 = 55.9m
Height of the ball exactly after 3 seconds = 55.9m.
b)
At ground, y=0; Plug-in values in (A):
A=> 0 = 100 - (1/2* 9.8 * [tex]t_{2}[/text])
Therefore t = + 4.52 s and - 4.52 seconds.
As t cannot be negative, therefore:
Time when ball hits the ground =
4.52s
-i
If u can plz give more information I am more than glad to help out