What is the value of? $\frac{1}{3}\xf74$

snajat3l
2022-02-07
Answered

What is the value of? $\frac{1}{3}\xf74$

You can still ask an expert for help

limegreenleosp9

Answered 2022-02-08
Author has **12** answers

Step 1

You can work it out using the usual fraction division process, or just through what is happening

If you take one third and cut it in half ( same as dividing by 2), then each piece will be$\frac{1}{6}$ (More pieces, therefore they get smaller)

If you take$\frac{1}{6}$ and cut it in half, the pieces get smaller again. Each piece will be $\frac{1}{12}$

$\frac{1}{3}\xf74=\frac{1}{3}\xf72\xf72=\frac{1}{12}$

A nifty short cut: To divide a fraction in half, either halve the top (if it is even) or double the bottom:

$\frac{2}{3}\xf72=\frac{1}{3}$

$\frac{4}{11}\xf72=\frac{2}{11}$ pretty obvious if you think about it!!

$\frac{5}{9}\xf72=\frac{5}{18}$

$\frac{7}{8}\xf72=\frac{7}{16}$

In the same way: To divide a fraction by 3 in half, either divide the by 3 (if possible) or treble the bottom:

$\frac{6}{11}\xf7=\frac{2}{11}$ share out 6 portions equally.

$\frac{5}{8}\xf73=\frac{5}{24}$

You can work it out using the usual fraction division process, or just through what is happening

If you take one third and cut it in half ( same as dividing by 2), then each piece will be

If you take

A nifty short cut: To divide a fraction in half, either halve the top (if it is even) or double the bottom:

In the same way: To divide a fraction by 3 in half, either divide the by 3 (if possible) or treble the bottom:

liofila3w7

Answered 2022-02-09
Author has **13** answers

Step 1

What you do is the KCF method. Keep, Change, Flip. You would keep the$\frac{1}{3}$ . Then you change the divide sign to a multiply sign.

Then you flip the 4 to$\frac{1}{4}$ .

You do that since$\frac{1}{4}$ is the reciprocal of 4

$\frac{1}{3}\xf74=\frac{1}{3}\times \frac{1}{4}$

What you do is the KCF method. Keep, Change, Flip. You would keep the

Then you flip the 4 to

You do that since

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Prove that if $a+b+c=1$ then $\sum _{cyc}\frac{1}{\sqrt{{a}^{2}+{b}^{2}}}\le \frac{9\sqrt{2}}{2}$

My try:since

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it suffices to prove that

$\sum _{cyc}{\displaystyle \frac{1}{a+b}}\le {\displaystyle \frac{9}{2}}$

But since Cauchy-Schwarz inequality we have

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or

$\sum _{cyc}{\displaystyle \frac{1}{a+b}}\ge {\displaystyle \frac{9}{2}}$

so this try can't works.so How to prove $(1)$

My try:since

$\sqrt{{a}^{2}+{b}^{2}}\ge {\displaystyle \frac{\sqrt{2}}{2}}(a+b)$

it suffices to prove that

$\sum _{cyc}{\displaystyle \frac{1}{a+b}}\le {\displaystyle \frac{9}{2}}$

But since Cauchy-Schwarz inequality we have

$2\sum _{cyc}(a+b)\sum _{cyc}{\displaystyle \frac{1}{a+b}}\ge 9$

or

$\sum _{cyc}{\displaystyle \frac{1}{a+b}}\ge {\displaystyle \frac{9}{2}}$

so this try can't works.so How to prove $(1)$

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