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Molodets [167]
2 years ago
8

The Sierpinski Triangle is a type of progression where an equilateral triangle has 1/4 of its area removed to create a new shape

. A series of these triangles is shown below, starting with an area of 64. If we remove 1/4 of the area what fraction of the area remains?
Mathematics
1 answer:
FinnZ [79.3K]2 years ago
7 0
<span>The Sierpinski Triangle is a type of progression where an equilateral triangle has 1/4 of its area removed to create a new shape.
</span><span>starting with an area of 64. If we remove 1/4 of the area 
so the removed area = 64* ( 1/4) = 16
remaining area = 64 - 16 = 48

fraction of the remaining area = 48 / 64 = 3/4
or another way 1 - 1/4 = 3/4

</span>
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What is the value of z in the equation 3z - 7 = 14
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This is simply solving an equation...

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Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:
Vanyuwa [196]

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This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. \lim_{n \to \infty} a_n =A it must hold that for any small \epsilon>0, there must exist n_0\in \mathbb{N} so that starting from that term of the series all of the following terms satisfy that  |a_n-A|n_0 .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with a_n=1 , the other is constant with a_n=3 , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}

If we  denote the first series as b_n=1, we will have that \lim_{k \to \infty} b_k=1.

The second series is denoted as c_k=2^k and we have that \lim_{k \to \infty} c_k=+\infty.

The third sub-series d_k=3 is a constant series and it holds that \lim_{k \to \infty} d_k=3.

Since those limits of sub-series are different, we can never find such n_0\\ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose \epsilon =1, all those terms associated with the third sub-series will be out of this interval (A-1, A+1)=(0, 2).

Therefore, the observed series diverges.

5 0
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lina2011 [118]

Answer:

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