A standard piece of paper is 0.05 mm thick. Let's imagine taking a piece of paper and folding the paper in half multiple times. We'll assume we can make "perfect folds," where each fold makes the folded paper exactly twice as thick as before - and we can make as many folds as we want. Write a function g g that determines the thickness of the folded paper (in mm) in terms of the number folds made, n n. (Notice that g ( 0 ) = 0.05 g(0)=0.05.)

For each fold, the thickness increases by a factor of 100%. With that in mind, as consecutive folds are made, thickness will grow exponentially. If the initial thickness is 0.05, the thickness 'g' as a function of the number of folds 'n' can be described by:

[tex]g(n) = 0.05*(1+100\%)^n\\g(n) =0.05*2^n[/tex]

Assuming infinite folds are possible, the thickness is given by g(n) =0.05*2ⁿ.

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Sam1s Sam1s · Answer 2 · 2021-12-15T06:22:55+01:00

Answer:

g(n) =0.05*2ⁿ

Step-by-step explanation:

For each overlap, the thickness increments by a calculate of 100%.
With that in intellect, as sequential folds are made, thickness will develop exponentially.
If the initial thickness is 0.05, the thickness 'g' as a function of the number of folds 'n' can be described by:

Thus, the correct answer is G(n)=0.05*2ⁿ

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