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Completely simplify and rationalize the denominator:

\[\frac{\sqrt{160}}{\sqrt{252}}\times\frac{\sqrt{245}}{\sqrt{128}}\]

 Jun 22, 2021

Best Answer 

 #1
avatar+502 
+2

\(\frac{\sqrt{160}}{\sqrt{252}}\cdot \frac{\sqrt{245}}{\sqrt{128}}\)

 

\(=\frac{7\sqrt{5}}{8\sqrt{2}}\cdot \frac{2\sqrt{10}}{3\sqrt{7}}\)

 

\(=\frac{2\sqrt{10}\cdot 7\sqrt{5}}{3\sqrt{7}\cdot 8\sqrt{2}}\)

 

\(=\frac{14\sqrt{10}\sqrt{5}}{24\sqrt{7}\sqrt{2}}\)

 

\(=\frac{\sqrt{7}\sqrt{10}\sqrt{5}}{12\sqrt{2}}\)

 

\(=\frac{\sqrt{7}\sqrt{2}\sqrt{5}\sqrt{5}}{2^2\cdot \:3\sqrt{2}}\)

 

\(=\frac{5\sqrt{7}}{3\cdot \:2\sqrt{2}\sqrt{2}}\)

 

\(=\frac{5\sqrt{7}}{12}\)

 

pray that i didn't mess up

 Jun 22, 2021
 #1
avatar+502 
+2
Best Answer

\(\frac{\sqrt{160}}{\sqrt{252}}\cdot \frac{\sqrt{245}}{\sqrt{128}}\)

 

\(=\frac{7\sqrt{5}}{8\sqrt{2}}\cdot \frac{2\sqrt{10}}{3\sqrt{7}}\)

 

\(=\frac{2\sqrt{10}\cdot 7\sqrt{5}}{3\sqrt{7}\cdot 8\sqrt{2}}\)

 

\(=\frac{14\sqrt{10}\sqrt{5}}{24\sqrt{7}\sqrt{2}}\)

 

\(=\frac{\sqrt{7}\sqrt{10}\sqrt{5}}{12\sqrt{2}}\)

 

\(=\frac{\sqrt{7}\sqrt{2}\sqrt{5}\sqrt{5}}{2^2\cdot \:3\sqrt{2}}\)

 

\(=\frac{5\sqrt{7}}{3\cdot \:2\sqrt{2}\sqrt{2}}\)

 

\(=\frac{5\sqrt{7}}{12}\)

 

pray that i didn't mess up

mworkhard222 Jun 22, 2021
 #2
avatar+502 
+2

best answer yayyyy

mworkhard222  Jun 22, 2021
 #3
avatar+121048 
+1

Yep.....good  job,  MWH!!!!!!

 

 

cool cool cool

CPhill  Jun 22, 2021

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