Answer:
a_n=-\frac{1}{4 a_{n-1}
Step-by-step explanation:
The recursive formula for the geometric sequence is given by:
a_n = a_{n-1} \cdot r
where,
r is the common ratio terms
-16, 4, -1, ...
This is a geometric sequence.
Here, and
Since,
ans so on .....
Substitute the given values we have;
⇒
Therefore, the recursive formula for the following geometric sequence is,
Answer: x = 8
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I'm going to use the notation log(2,x) to indicate "log base 2 of x". The first number is the base while the second is the expression inside the log (aka the argument of the log)
log(2,x) + log(2,(x-6)) = 4
log(2,x*(x-6)) = 4
x*(x-6) = 2^4
x*(x-6) = 16
x^2-6x = 16
x^2-6x-16 = 0
(x-8)(x+2) = 0
x-8 = 0 or x+2 = 0
x = 8 or x = -2
Recall that the domain of log(x) is x > 0. So x = -2 is not allowed. The same applies to log(2,x) as well.
Only x = 8 is a proper solution.
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You can use the change of base rule to check your work
log base 2 of x = log(2,x) = log(x)/log(2)
log(2,(x-6)) = log(x-6)/log(2)
So,
(log(x)/log(2)) + (log(x-6)/log(2)) = 4
(log(8)/log(2)) + (log(8-6)/log(2)) = 4
(log(8)/log(2)) + (log(2)/log(2)) = 4
(log(2^3)/log(2)) + (log(2)/log(2)) = 4
(3*log(2)/log(2)) + (log(2)/log(2)) = 4
3+1 = 4
4 = 4
The answer is confirmed
Answer:
30
Step-by-step explanation:
Just insert the value of x into the equation.
x = 3 right?
6^2 + 3 - 3^2 because x = 3
36 + 3 - 9 = 39 -9 = 30
:D
1/2 is written as a decimal as 0.50
