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hammer [34]
2 years ago
14

Compare partial products and regrouping and describe how the methods are alike and different

Mathematics
1 answer:
erastovalidia [21]2 years ago
5 0
Partial products are different in regrouping in terms of how numbers are clustered from a set equation as a whole delivering it individual but naturally to all the numbers involved in the set. 
Regrouping is just like the commutative or associative property of numbers.
<span>Associative property of addition is used when you want to group addends. This is mainly used to cluster set of numbers or in this case, addends. How do you use the associative property when you break apart addends? Simple you group them using the open and closed parentheses or brackets. Take for an example 1 + 1 + 2 = 4. Using the associative property you can have either (1 + 1) + 2 = 4 or 1 + (1 + 2) = 4 clustered into place.  <span>
</span></span>
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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]

Answer:

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
2 years ago
At barlow school ,4/9 of the 873 students are boys
Maksim231197 [3]

Answer:

=Barlow by 181 boys

Step-by-step explanation:

lets calculate the no of boys in Barlow

=4/9×873=388

388 boys

in Willow school

to bet no of boys

3/3-2/3=1/3

1/3×620=207(in a calculator it will give you decimals so i rounded of to the nearest whole no.)

388-207=181

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