Please help me on this

Mathematics

1 answer:

Leni [432]4 years ago

4 0

5 7/8 is the answer to the problem

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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.

tatyana61 [14]

Answer:

(a) $g(n)=0.05\cdot 2^n$

(b) $g^{-1}(n)=\log_{2}20g(n)$

Step-by-step explanation:

Part A

The paper's thickness = 0.05mm

When the paper is folded, its width doubles (increases by 100%).

The thickness of the paper grows exponentially and can be modeled by the function:

$g(n)=0.05(1+100\%)^n\\\\g(n)=0.05\cdot 2^n$

Part B

$g(n)=0.05\cdot 2^n\\2^n=\dfrac{g(n)}{0.05}\\ 2^n=20g(n)\\$Changing to logarithm form, we have:\\\log_{2}20g(n)=n\\$Therefore:\\g^{-1}(n)=\log_{2}20g(n)$

Part C

If the thickness of the paper, g(n)=384,472,300,000 mm

Then:

$g^{-1}(n)=\log_{2}20g(n)\\g^{-1}(n)=\log_{2}20\times 384,472,300,000\\=\dfrac{\log 20\times 384,472,300,000}{\log 2} \\g^{-1}(n)=42.8 \approx 43\\n=43$