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OpenStudy (danielbarriosr1):
\[\frac{ \sqrt{10} }{ \sqrt{14} }\]
OpenStudy (misty1212):
HI!
OpenStudy (danielbarriosr1):
Hello
OpenStudy (misty1212):
you can start with
\[\frac{\sqrt5}{\sqrt7}\]if you like
OpenStudy (misty1212):
then if you have to rationalize the denominator write
\[\frac{\sqrt5\sqrt7}{\sqrt7\sqrt7}=\frac{\sqrt{35}}{7}\]
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OpenStudy (danielbarriosr1):
I'm doing carnegie and it is telling me I have to multiply by form of 1
OpenStudy (danielbarriosr1):
what is the form of one @misty1212
OpenStudy (misty1212):
lol "form of 1"
i think it means when i multiplied
\[\frac{\sqrt5}{\sqrt7}\times \frac{\sqrt7}{\sqrt7}\] the second part is the "form of 1"
OpenStudy (danielbarriosr1):
Oh
OpenStudy (mathmate):
@danielbarriosr1
If you start with the following identities, everything else will follow:
\(\sqrt{a} \sqrt{b}=\sqrt{ab}\)
\(\dfrac{\sqrt a}{\sqrt b}=\sqrt{\dfrac{a}{b}}\)
Example:
\(\dfrac{\sqrt{25}}{\sqrt{15}}=\sqrt{\dfrac{25}{15}}=\sqrt{\dfrac{5\times 5}{3\times 5}}=\sqrt{\dfrac{5}{3}}\sqrt{\dfrac{5}{5}}=\sqrt{\dfrac{5}{3}}\)
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OpenStudy (danielbarriosr1):
@mathmate so then would it be like @misty1212 said?
OpenStudy (mathmate):
Yes, but in addition to that, I would suggest you to be familiar with the two identities, because they can help you simplify things easier!
OpenStudy (danielbarriosr1):
Oh
OpenStudy (danielbarriosr1):
I got it
OpenStudy (danielbarriosr1):
thanks
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