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# 4 squre root 81

When I start working on this problem, do I divide the square root by the power of four because the 4 is on top of the root thing?

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Take the fourth root. The square root, denoted by a 2 on top of the "root thing" or more commonly without a number at all, is the second root.

Usually the easiest way to calculate the fourth root is to calculate the square root of the square root. So in this case the fourth root of 81 is the square root of the square root of 81. The square root of 81 is 9, the square root of 9 is 3.

Posted on Apr 18, 2013

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Posted on Jan 02, 2017

Check out this web site, http://www.mathsisfun.com/square-root.html
You can also use the google calculator function. go to google if you want the square root of 5 type: square root 5=
and google will solve it. for use with a calculator the other poster is right, enter the number ie:5 then use the x2 button or sqrt button.

Posted on Oct 05, 2009

SOURCE: we are trying to add

( 3 + 4 ) ^ 3 / ( 1 7 - 2 0 0 sqrt / 2 ) = or
( 3 + 4 ) ^ 3 / ( 1 7 - 2 0 0 sqrt ) / 2 = depending on what you meant.

Posted on Nov 01, 2010

SOURCE: Ok. When i need to

Posted on Mar 03, 2011

First off let me say there are 2 ways to answer this question depending on what you are taking the square root of.

((sqrt(206) - 900 )/ 5) / 4 =

sqrt(206) = 14.35

14.35-900 = -885.64

-885.64/5 = -177.13

-177.13/4 = -44.28 this would be the answer.

Or if you are takeing the sqrt of 206 - 900 which equals -694 then this is a different type of problem.

then you would have

sqrt(-694) = 26.34i

26.34i/5 = 5.26i

5.26/4 = 1.31i

but that is if you are using imaginary numbers that is what the "i" stands for, you cannot take a sqrt of a negative number directly because anytime you multiply 2 numbers together the result must be positive. So hopefully that helps you out, Just remeber where you put your parenthases is very important, like in this case your are potentially asking 2 different questions. Good luck

Posted on Jun 30, 2011

SOURCE: how would I get the

type sqrt(28-4)*sqrt(2)

Posted on Sep 16, 2011

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