1/7x1/7x1/7= 3/7 true or false

Mathematics

2 answers:

Lostsunrise [7]3 years ago

3 0

Answer:

The answer is false

Tcecarenko [31]3 years ago

3 0

I’m pretty sure the answer is false

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Answer:

I think its D. 70 - 79

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2 years ago

Is it possible to create a regular tessellation using a regular decagon? Explain. Yes, 2 decagons with a common vertex will have

Gala2k [10]

Answer:

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Step-by-step explanation:

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2 years ago

Evaluate the limit of sequence:

mr_godi [17]

Rationalize both the numerator and denominator. Given

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}$

we can rationalize it by introducing conjugates of the numerator and denominator:

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d} \cdot \dfrac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d} \\\\ = \dfrac{\left(\sqrt a\right)^2 - \left(\sqrt b\right)^2}{\left(\sqrt c\right)^2-\left(\sqrt d\right)^2} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b} \\\\ = \dfrac{a-b}{c-d} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b}$

Then the limit is equivalent to

$\displaystyle \lim_{n\to\infty} \frac{(n+3)-n}{(n+1)-n} \cdot \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = 3 \lim_{n\to\infty} \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n}$