anonymous
  • anonymous
Can someone help me with this radical?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\frac{ 7 }{ \sqrt{45} }\]
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
maybe before you do that, note that \(45=9\times 5\) and so \(\sqrt{45}=\sqrt{9\times 5}=\sqrt{9}\sqrt{5}=3\sqrt{5}\)

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anonymous
  • anonymous
then \[\frac{ 7 }{ \sqrt{45} }=\frac{7}{3\sqrt{5}}\]and now you only need to multiply top and bottom by \(\sqrt{5}\) to remove the radical from the denominator
anonymous
  • anonymous
@julia_copen you got this or you need the steps?
anonymous
  • anonymous
Steps please.
anonymous
  • anonymous
lets start with \[\frac{7}{3\sqrt{5}}\] your job is to get the radical out of the denominator, so multiply top and bottom by \(\sqrt 5\) you get \[\frac{7}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{5}}{3\sqrt{5}\sqrt{5}}=\frac{7\sqrt{5}}{3\times 5}\]
anonymous
  • anonymous
typo there, meant \[\frac{7}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{7\sqrt{5}}{3\sqrt{5}\sqrt{5}}=\frac{7\sqrt{5}}{3\times 5}\]
anonymous
  • anonymous
final answer is \(\frac{7\sqrt{5}}{15}\)
anonymous
  • anonymous
another quick example \[\frac{4}{\sqrt{3}}=\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{3}\]
anonymous
  • anonymous
Thanks so much! Me and my friend were having a hard time trying to understand how to break it down.
anonymous
  • anonymous
yw
anonymous
  • anonymous
can you help me with another?
anonymous
  • anonymous
sure
anonymous
  • anonymous
\[\frac{ 1 }{ \sqrt{75z} }\]
anonymous
  • anonymous
is the \(z\) inside the radical?
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok the idea is to see if the number is the product of some "perfect square" so in this case \(75=25\times 3\) making \[\sqrt{75}=\sqrt{25}\sqrt{3}=5\sqrt{3}\] so star with \[\frac{1}{5\sqrt{3z}}\] and then multiply top and bottom by \(\sqrt{3z}\)
anonymous
  • anonymous
you get \[\frac{1}{5\sqrt{3z}}\times \frac{\sqrt{3z}}{\sqrt{3z}}=\frac{\sqrt{3z}}{15z}\]
anonymous
  • anonymous
Okay I see. It's always difficult in the beginning for me.
anonymous
  • anonymous
you will get used to it, (and then probably forget it because it is not really that useful) but in any case it gets easier
anonymous
  • anonymous
Are you still on?

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