the question is 4/(4√64m4n2)
the given answer is
(4√4n2) / 2mn
but I get 4(4√n2)/ 2mn
4√ = root with 'index=4'
*the index 4 root is for the all
-( 64m4n2)
-(4n2)
-(n2)
I'm not good at writing mathematics symbols sorry. help please SOS
Printable View
the question is 4/(4√64m4n2)
the given answer is
(4√4n2) / 2mn
but I get 4(4√n2)/ 2mn
4√ = root with 'index=4'
*the index 4 root is for the all
-( 64m4n2)
-(4n2)
-(n2)
I'm not good at writing mathematics symbols sorry. help please SOS
Hang on, is itQuote:
Originally Posted by edmundyong
?
yes :) I dunno how to type the 'root' like that
the fourth-root of (64m^4n^2) = 2m fourth-root(2^2 n^2) = 2m sqrt(2n). Then multipy and divide Nr and Dr by sqrt(2n) you will get the answer.
so you basically what you do is reduce the 4th-root(22n2 ) into sqrt(2n) by diving all by 2?
so basically what you do is reduce the 4th-root(22n2 ) into sqrt(2n) by diving all by 2?Quote:
Originally Posted by coolge
No, you take everything to the power of 1/4.Quote:
Originally Posted by edmundyong
But you can learn to do the mathematical symbols.Quote:
Originally Posted by edmundyong
[tex]\frac{4}{\sqrt[4]{64m^4n^2}} [/tex] gives![\frac{4}{\sqrt[4]{64m^4n^2}}](http://latex.codecogs.com/png.latex? \frac{4}{\sqrt[4]{64m^4n^2}} )
Click on the “go advanced” tab. On the toolbar you will seeclicking on that give the LaTeX wraps, [tex] [/tex]. The code goes between them.
okQuote:
Originally Posted by Prove It
so it'll be 4/2m sqrt(2n)
and mutiply all by sqrt(2n)
1 question
- is sqrt(2n) x sqrt(2n) = 2n?
if yes then...the wouldn't the Nr become
4 [sqrt(2n)] / 2m(2n)
=4 [sqrt(2n) / 4mn
???