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Anna35 [415]
3 years ago
7

Please help I’ll mark brainliest

Mathematics
2 answers:
tamaranim1 [39]3 years ago
3 0

Answer:

m = v + 2

Step-by-step explanation:

m is Mags, and v is Vector

m (Mags' age) =  v(Vector's age) + 2

Anni [7]3 years ago
3 0

Answer:

m= v + 2

Step-by-step explanation:


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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.

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<span>The square root of -16 is an imaginary number and a complex number. Sqrt(-16)=4i. We use the i to indicate that the number is imaginary since there is no number that can be multiplied by itself to get a negative number (a negative times a negative is a positive, and a positive times a positive is also a positive). So the use of i tells you immediately that it's an imaginary number. You can tell the number is complex because it has both a real and an imaginary part and could be written in the form a+bi, where a is a real number and bi is an imaginary number. In this specific case, the real part (a) is 0 and the imaginary part (bi) is 4i.</span>
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