anonymous
  • anonymous
PLEASE HELP ME ON THIS ALGEBRA PROBLEM:-)Rewrite in simplest rational exponent form square root x multiplied by 4 square root x . Show each step of your process.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@Vocaloid @veechan @Gambino @goformit100 @Destinyyyy @dan815
anonymous
  • anonymous
@LiveLaughDie @LiteNing1337 @LexiLuvv2431
anonymous
  • anonymous
Can you use the equation button to correctly type out the equation so its easy to read :) I can help you, I just want to make sure that my information is accurate.

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More answers

anonymous
  • anonymous
okay! thanks
anonymous
  • anonymous
\[\sqrt{x} \times \sqrt[4]{x}\]
anonymous
  • anonymous
and then in simplest rational exponent form :-) @LexiLuvv2431
anonymous
  • anonymous
Okayy, do you have any answer choices? or is this a word problem
anonymous
  • anonymous
its just a question where I need to explain each step and how i got it and the answer of course :)
anonymous
  • anonymous
Okayy so, I'm gonna walk you through this just so you know for next time :) Do you know where to begin?
anonymous
  • anonymous
not really for this one
anonymous
  • anonymous
You're going to rewrite \[\sqrt{x}\] with rational exponents first. so \[\sqrt{x}\] = \[x_{2}^{1}\]
anonymous
  • anonymous
where'd you get the 2 from?
anonymous
  • anonymous
That is x with an exponent of 1/2 if you're confused.
anonymous
  • anonymous
okay thanks!
anonymous
  • anonymous
2 is a rational exponent of x thats were the 2 comes from
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
Next we have \[\sqrt[4]{x}\] You then take the rational exponent from here too which is the 4 aand turn it into a fration so it would look like this \[x_{4}^{1}\] x has the exponent of 1/4
anonymous
  • anonymous
that makes sense
anonymous
  • anonymous
So the final product is |dw:1439933580751:dw|
anonymous
  • anonymous
That will be your answer :)
anonymous
  • anonymous
okay! so thats all?
anonymous
  • anonymous
Thank you so much! :-) @LexiLuvv2431
anonymous
  • anonymous
You're welcome :)

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