# simplify

• Oct 22nd 2009, 06:18 AM
andyboy179
simplify
this is my question:

-3----4
2a X 3a=?
----------------------7

the --- are there because i can't get the numbers to stay above other numbers without there being anything else there.
• Oct 22nd 2009, 06:27 AM
Quote:

Originally Posted by andyboy179
this is my question:

-3----4
2a X 3a=?
----------------------7

the --- are there because i can't get the numbers to stay above other numbers without there being anything else there.

HI

i think you need to revise your post . If you want to write a fraction ,

\frac{numerator}{denomintator} , then wrap it up with the math tags .
• Oct 22nd 2009, 06:28 AM
andyboy179
its not a fraction
• Oct 22nd 2009, 06:30 AM
andyboy179
• Oct 22nd 2009, 06:35 AM
Quote:

Originally Posted by andyboy179

nope , it should be $\displaystyle 6A^7$ , 2x3=6 ..

btw , to write in latex , 2A^3\times 3A^4 , then wrap with the tags .
• Oct 22nd 2009, 06:40 AM
andyboy179
Quote:

nope , it should be $\displaystyle 6A^7$ , 2x3=6 ..

btw , to write in latex , 2A^3\times 3A^4 , then wrap with the tags .

thanks, $\displaystyle 3^4\times 3^- 2\times 3^7$
• Oct 22nd 2009, 06:45 AM
Quote:

Originally Posted by andyboy179
thanks, $\displaystyle 3^4\times 3^- 2\times 3^7$

3^{-2} for the term in the middle .
• Oct 22nd 2009, 06:47 AM
andyboy179
Quote:

3^{-2} for the term in the middle .

kk, would the answer to that be $\displaystyle 3^9$
• Oct 22nd 2009, 06:49 AM
Quote:

Originally Posted by andyboy179
kk, would the answer to that be $\displaystyle 3^9$

yup , $\displaystyle 3^9$
• Oct 22nd 2009, 06:50 AM
andyboy179
Quote:

yup , $\displaystyle 3^9$

thanks
• Oct 22nd 2009, 06:52 AM
andyboy179
$\displaystyle 2a^3 b\times 3ab^- 3$ would it = $\displaystyle 6a^4 b^4$
• Oct 22nd 2009, 06:57 AM
Amer
Quote:

Originally Posted by andyboy179
$\displaystyle 2a^3 b\times 3ab^- 3$ would it = $\displaystyle 6a^4 b^4$

$\displaystyle 2a^3b\times 3ab^{-3} = (2\times 3) a^{3+1} b^{1-3} =6a^4b^{-2}$
• Oct 22nd 2009, 07:02 AM
andyboy179
Quote:

Originally Posted by Amer
$\displaystyle 2a^3b\times 3ab^{-3} = (2\times 3) a^{3+1} b^{1-3} =6a^4b^{-2}$