4.2 with a line above the 2 (repeating numer) into a fraction

kokomocutie88r1
2022-07-26
Answered

4.2 with a line above the 2 (repeating numer) into a fraction

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Monica Dennis

Answered 2022-07-27
Author has **13** answers

Step 1

4.2 with a line above the 2 (repeating numer)

$=4+0.2222222222\dots $

$=4+\frac{2}{10-1}$

$=4+\frac{2}{9}$

$=\frac{38}{9}$

4.2 with a line above the 2 (repeating numer)

$=4+0.2222222222\dots $

$=4+\frac{2}{10-1}$

$=4+\frac{2}{9}$

$=\frac{38}{9}$

PoentWeptgj

Answered 2022-07-28
Author has **6** answers

Step 1

$4+0.22222222\phantom{\rule{0ex}{0ex}}4+\frac{2}{10-1}\phantom{\rule{0ex}{0ex}}4+\frac{2}{9}$

$4+0.22222222\phantom{\rule{0ex}{0ex}}4+\frac{2}{10-1}\phantom{\rule{0ex}{0ex}}4+\frac{2}{9}$

asked 2022-10-15

Prove that $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\ge \frac{3}{2}$

For $a\ge b\ge c>0$. Prove that

$a=100;b=1;c=1/100$ it's wrong ???

$\frac{a}{a+b}}+{\displaystyle \frac{b}{b+c}}+{\displaystyle \frac{c}{c+a}}\ge {\displaystyle \frac{3}{2}$

$<=>\sum \frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac{1}{2}+\frac{c}{c+a}-\frac{1}{2}\ge 0$

$<=>\sum \frac{a-b}{2(a+b)}\ge 0$

$<=>x{y}^{2}+y{z}^{2}+x{z}^{2}-{x}^{2}y-{y}^{2}z-{z}^{2}x\ge 0$

$<=>(y-x)(z-y)(z-x)\ge 0$

And we have $y-x\le 0;z-x\le 0;z-y\ge 0$

we are done !?

For $a\ge b\ge c>0$. Prove that

$a=100;b=1;c=1/100$ it's wrong ???

$\frac{a}{a+b}}+{\displaystyle \frac{b}{b+c}}+{\displaystyle \frac{c}{c+a}}\ge {\displaystyle \frac{3}{2}$

$<=>\sum \frac{a}{a+b}-\frac{1}{2}+\frac{b}{b+c}-\frac{1}{2}+\frac{c}{c+a}-\frac{1}{2}\ge 0$

$<=>\sum \frac{a-b}{2(a+b)}\ge 0$

$<=>x{y}^{2}+y{z}^{2}+x{z}^{2}-{x}^{2}y-{y}^{2}z-{z}^{2}x\ge 0$

$<=>(y-x)(z-y)(z-x)\ge 0$

And we have $y-x\le 0;z-x\le 0;z-y\ge 0$

we are done !?

asked 2022-06-21

What is the difference between a simple "fraction" and a "common fraction"?

I have read about a common fraction in this statement (written in a text book):

Ratio is the simplest form of a common fraction, in which the numerator denotes the antecedent and the denominator denotes the consequent.

The Wordweb dictionary defines a common fraction as

The quotient (defined as "The ratio of two quantities to be divided" / "The number obtained by division") of two integers.

From this definition, a common fraction pretty much seems to be just the same as a fraction. So what is the difference?

I have read about a common fraction in this statement (written in a text book):

Ratio is the simplest form of a common fraction, in which the numerator denotes the antecedent and the denominator denotes the consequent.

The Wordweb dictionary defines a common fraction as

The quotient (defined as "The ratio of two quantities to be divided" / "The number obtained by division") of two integers.

From this definition, a common fraction pretty much seems to be just the same as a fraction. So what is the difference?

asked 2022-08-10

Decompose a fraction into a sum of fractions (with complex variable)

Here is equality:

$\frac{1}{z\mathrm{ln}4+(z\mathrm{ln}4{)}^{2}/2+O({z}^{3})}=\frac{1}{z\mathrm{ln}4}-\frac{1}{2}+O(z)$

I can't understand why is it correct?

Here is equality:

$\frac{1}{z\mathrm{ln}4+(z\mathrm{ln}4{)}^{2}/2+O({z}^{3})}=\frac{1}{z\mathrm{ln}4}-\frac{1}{2}+O(z)$

I can't understand why is it correct?

asked 2022-06-04

Property of fractions

Given two fractions $\frac{h}{k}$ and $\frac{{h}^{{}^{\prime}}}{{k}^{{}^{\prime}}}$ both in reduced form. I am unable to find a case when $\frac{h+{h}^{{}^{\prime}}}{k+{k}^{{}^{\prime}}}$ does not lie in the interval $[}\frac{h}{k},\frac{{h}^{{}^{\prime}}}{{k}^{{}^{\prime}}}{\textstyle ]$. Is there such a case ?

PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.

Given two fractions $\frac{h}{k}$ and $\frac{{h}^{{}^{\prime}}}{{k}^{{}^{\prime}}}$ both in reduced form. I am unable to find a case when $\frac{h+{h}^{{}^{\prime}}}{k+{k}^{{}^{\prime}}}$ does not lie in the interval $[}\frac{h}{k},\frac{{h}^{{}^{\prime}}}{{k}^{{}^{\prime}}}{\textstyle ]$. Is there such a case ?

PS: I was able two prove no such case exists for consecutive terms of Farey series. But can't prove in general.

asked 2022-06-16

Why does $\sqrt{x}/y=\sqrt{x/y/y}$

Sorry for the awkward title, hard to to sum a mathematical problem with words alone.

Having said that, I recently learned that the root of any value, x, and then that over value y, is identical to the root of $x/y/y$ , as in $\sqrt{x/y/y}$

For e.g, $\sqrt{35}/7=\sqrt{5/7}$ and I cannot logically deduce why. So far, I know that:

$35=7\times 5\phantom{\rule{0ex}{0ex}}\frac{\sqrt{35}}{7}=\frac{\sqrt{7}\times \sqrt{5}}{7}$

And somehow this being equal to:

√(5/7)

So it seems that we divided the numerator, √35 by √5 to give √7. However, 7/√7 is not √7! Yet, apparently it is?

So, a fraction is of course equal to itself if scaled up or down by the same amount, in that 10/2 is the same as 5/1, or 5. So scaling down √35 by √5 makes sense, as it only leaves √7, but it seems that the same cannot be said for the denominator, 7. 7/√7 = √7!? Yet, the two expressions really are the same, √35/7 and √(5/7).

Thanks for any help in advance.

P.S This works for any values, so it's no coincidence.

Sorry for the awkward title, hard to to sum a mathematical problem with words alone.

Having said that, I recently learned that the root of any value, x, and then that over value y, is identical to the root of $x/y/y$ , as in $\sqrt{x/y/y}$

For e.g, $\sqrt{35}/7=\sqrt{5/7}$ and I cannot logically deduce why. So far, I know that:

$35=7\times 5\phantom{\rule{0ex}{0ex}}\frac{\sqrt{35}}{7}=\frac{\sqrt{7}\times \sqrt{5}}{7}$

And somehow this being equal to:

√(5/7)

So it seems that we divided the numerator, √35 by √5 to give √7. However, 7/√7 is not √7! Yet, apparently it is?

So, a fraction is of course equal to itself if scaled up or down by the same amount, in that 10/2 is the same as 5/1, or 5. So scaling down √35 by √5 makes sense, as it only leaves √7, but it seems that the same cannot be said for the denominator, 7. 7/√7 = √7!? Yet, the two expressions really are the same, √35/7 and √(5/7).

Thanks for any help in advance.

P.S This works for any values, so it's no coincidence.

asked 2021-12-16

Convert $\frac{3}{20}$ to percentage.

asked 2022-05-11

What is the property that allow the transformation $\frac{16{a}^{3}}{8ac}=\frac{16}{8}\cdot \frac{{a}^{3}}{a}\cdot \frac{1}{c}$?