How do you rationalize the denominator and simplify $\frac{\sqrt{77}}{\sqrt{35}}$ $(sqrt (77))/ (sqrt (35))$ ?

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How do you rationalize the denominator and simplify $\frac{\sqrt{77}}{\sqrt{35}}$ $(sqrt (77))/ (sqrt (35))$ ?

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Answer 1 · 2015-04-19T18:49:17

In this way:

$\frac{\sqrt{77}}{\sqrt{35}} = \frac{\sqrt{77}}{\sqrt{35}} \cdot \frac{\sqrt{35}}{\sqrt{35}} = \frac{\sqrt{77 \cdot 35}}{35} =$ $sqrt77/sqrt35=sqrt77/sqrt35*sqrt35/sqrt35=sqrt(77*35)/35=$

$= \frac{\sqrt{7 \cdot 11 \cdot 5 \cdot 7}}{35} = \frac{7 \cdot \sqrt{55}}{35} = \frac{\sqrt{55}}{5}$ $=sqrt(7*11*5*7)/35=(7*sqrt55)/35=sqrt55/5$ .

OR:

$\frac{\sqrt{77}}{\sqrt{35}} = \sqrt{\frac{77}{35}} = \sqrt{\frac{11}{5}} = \frac{\sqrt{11}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{55}}{5}$ $sqrt77/sqrt35=sqrt(77/35)=sqrt(11/5)=sqrt11/sqrt5*sqrt5/sqrt5=sqrt55/5$ .

OR:

$\frac{\sqrt{77}}{\sqrt{35}} = \frac{\sqrt{7} \cdot \sqrt{11}}{\sqrt{7} \cdot \sqrt{5}} = \frac{\sqrt{11}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{55}}{5}$ $sqrt77/sqrt35=(sqrt7*sqrt11)/(sqrt7*sqrt5)=sqrt11/sqrt5*sqrt5/sqrt5=sqrt55/5$ .

Answer 2 · 2015-04-19T21:03:32

$\frac{\sqrt{77}}{\sqrt{35}}$ $sqrt(77)/sqrt(35)$

$= \frac{\sqrt{11} \sqrt{7}}{\sqrt{5} \sqrt{7}}$ $=(sqrt(11)sqrt(7))/(sqrt(5)sqrt(7))$

$= \frac{\sqrt{11}}{\sqrt{5}}$ $=sqrt(11)/sqrt(5)$

$= \frac{\sqrt{11}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}$ $=sqrt(11)/sqrt(5)*sqrt(5)/sqrt(5)$

$= \frac{\sqrt{11} \sqrt{5}}{5}$ $=(sqrt(11)sqrt(5))/5$

$= \frac{\sqrt{55}}{5}$ $=sqrt(55)/5$

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How do you rationalize the denominator and simplify $\frac{\sqrt{77}}{\sqrt{35}}$ $(sqrt (77))/ (sqrt (35))$ ?

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How do you rationalize the denominator and simplify √77√35(sqrt (77))/ (sqrt (35))?

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How do you rationalize the denominator and simplify $\frac{\sqrt{77}}{\sqrt{35}}$ $(sqrt (77))/ (sqrt (35))$ ?