Answer:
The inverse for log₂(x) + 2 is - log₂x + 2.
Step-by-step explanation:
Given that
f(x) = log₂(x) + 2
Now to find the inverse of any function we put we replace x by 1/x.
f(x) = log₂(x) + 2
f(1/x) =g(x)= log₂(1/x) + 2
As we know that
log₂(a/b) = log₂a - log₂b
g(x) = log₂1 - log₂x + 2
We know that log₂1 = 0
g(x) = 0 - log₂x + 2
g(x) = - log₂x + 2
So the inverse for log₂(x) + 2 is - log₂x + 2.
The wording of this question is a bit confusing... You can't write a sequence in sigma notation, but rather a series or sum. I think the question is asking you to write the sum of the sequence,

which would be

in sigma notation.
To do this, notice that the denominator in each term is a power of 2, starting with
and ending with
. So in sigma notation, this series is

The answer for this is 3 and 4
The answer is it is yes, it is sometimes true, y can =0
Answer: The answer is 11 1/24 or in decimal form 11.04
Step-by-step explanation: