<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B3%20%7D%20%20%5Ctimes%20%20%5Cfrac%7B5%7D%7B7%7D%20%20%5Ctimes%20%20%5Cfra

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3 years ago

c%7B4%7D%7B5%7D%20%5Ctimes%20%20%5Cfrac%7B3%7D%7B4%7D%20%20" id="TexFormula1" title=" \frac{2}{3 } \times \frac{5}{7} \times \frac{4}{5} \times \frac{3}{4} " alt=" \frac{2}{3 } \times \frac{5}{7} \times \frac{4}{5} \times \frac{3}{4} " align="absmiddle" class="latex-formula">

Mathematics

1 answer:

VikaD [51]3 years ago

5 0

Answer:

$\frac{2}{3}*\frac{5}{7}*\frac{4}{5}*\frac{3}{4}\\=\frac{2}{7}$

Step-by-step explanation:

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Answer the question with explanation;

PSYCHO15rus [73]

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

$\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0$

Therefore, by the series divergence test, the series $\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}$ diverges.

EDIT: To VectorFundament120, if $(x_n)_{n\in\mathbb N}$ is a sequence, both $\lim x_n$ and $\lim_{n\to\infty}x_n$ are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.