2006-10-30 09:03:24 · 18 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
18.110770276274833253147616333968
according to my calculator.
2006-10-30 09:06:58 · answer #1 · answered by Philip A 1 · 0⤊ 1⤋
328/2 = 164
164/2 = 82
82/2 = 41
Therefore, 328 = 2^3 * 41 or 4*82
The square root of 328 = sqrt(328) = sqrt(4*82) = 2sqrt(82)
which is approximately 18.11
2006-10-30 17:09:38 · answer #2 · answered by MsMath 7 · 4⤊ 0⤋
The desired answer depends on why the question was asked. Guy M got it right - it is 2sqrt(82) Any evaluation of a surd given in decimal form is not as good unless you want to use it to get an approximate answer to a specific question. In this case you have to determine the level of accuracy you want - how many decimal places. Could be anything from 18.1 or 18.11078 plus a few more besides.
It would have been good to know why you had asked the question.
2006-10-30 19:59:54 · answer #3 · answered by saljegi 3 · 0⤊ 0⤋
328 =4*82 sqrt4=2
try the iteration to find sqrt 82,guess 9
x1=9,x(n+1)=(1/2)(xn+82/xn)
x2=(1/2)(9+82/9)=9.055
x3=(1/2)(9.055+82/9.055)
=9.055385146
therefore,sqrt 328 =sqrt4*sqrt82
= 2*9.055385146
=18.11077029
i hope that this helps
2006-10-30 17:54:22 · answer #4 · answered by Anonymous · 0⤊ 0⤋
take your calcolator, type in 328, hit the square root key.
but the hard way, 18x18=324, so you know it's over 18 & less than 19...
2006-10-30 17:13:53 · answer #5 · answered by Anonymous · 0⤊ 0⤋
all yu do is go on google search and type "square root of 328" and you get it!
2006-10-30 17:11:40 · answer #6 · answered by Rose 5 · 0⤊ 0⤋
sqrt(328) = sqrt(4 * 82) = 2sqrt(82)
2006-10-30 19:42:06 · answer #7 · answered by Sherman81 6 · 0⤊ 0⤋
sqrt(328)=sqrt(4*82)=ssqrt(82)=18.111
2006-10-30 17:28:48 · answer #8 · answered by yupchagee 7 · 0⤊ 0⤋
18.11 Try using a calculator
2006-10-30 17:04:55 · answer #9 · answered by Anonymous · 0⤊ 1⤋
to 5 decimal points it is:
18.11077
2006-10-30 17:16:40 · answer #10 · answered by costa 4 · 0⤊ 0⤋
Here explains how is done.
http://www.nist.gov/dads/HTML/squareRoot.html
To be honest i still could not do it :(
I hope you understand better then me.
2006-10-31 05:52:01 · answer #11 · answered by Smile :) 3 · 0⤊ 0⤋