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Nastasia [14]
4 years ago
15

A cake has a circumference of 25 1/7 inches. What is the area of the cake? Use 22/7 to approximate pi

Mathematics
1 answer:
Novosadov [1.4K]4 years ago
7 0
To calculate for area given the circumference we proceed as follows:
C=2πr
where:
r=radius
thus given that C=25 1/7 in=176/7
thus plugging in our formula to solve for r we get:
176/7=2πr
thus
r=4.002~4.00 in
hence the area will be:
A=πr²
A=π×(4²)
A=50 2/7 in²


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Answer:

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And when we apply the limit we got that:

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Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

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We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

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\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

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We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

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\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

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