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Yanka [14]
2 years ago
8

Find the length of side x in simplest radical form with a rational a denominator.

Mathematics
1 answer:
Drupady [299]2 years ago
5 0

Step-by-step explanation:

{x}^{2}  =  {7}^{2}   +  {7}^{2}  \\  {x}^{2}  = 98 \\  x =  \sqrt{98}  = 7 \sqrt{2}

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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.

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2 years ago
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Hey guys, I'm relearning some Algebra and was confused about this problem. In the equation, the problem was divided by 3. This m
spayn [35]

Answer and Step-by-step explanation:

The signs didn't really "swap". Instead, the whole function was divided by -1, or we could say the function was divided by -3. That would turn:

-18x² - 15x + 3 = 0

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2x − 2) + (x + 1) + x + (x + 1) = (2x − 9) + (x + 1) + (x + 8) + x,
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Answer:

In this problem, we need to describe the relation between variables, if that relation is functional or not. It's important to say that we assumed that the first variable is independent, and the second is dependent.

<h3>(a)</h3>

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<h3>(b)</h3>

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<h3>(c)</h3>

Gasoline price - Day of the Month: These relation is not functional, becasue time must be the independent variable.

<h3>(d)</h3>

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A number and its fifth part: Notice that the fifth part depends on the number, it's defined by it, so this can be a function.

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A number and its square root: These two variables represent a function, where "a number" represents the domain value and "its square root" represents a range vale.

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