# Someone help if you get a chance

• June 14th 2006, 12:02 PM
Aarxn
Someone help if you get a chance
if any of you get time help me with this problem, thankssss

http://img231.imageshack.us/img231/1427/ughmath6lp.jpg
• June 14th 2006, 12:30 PM
Quick
Quote:

Originally Posted by Aarxn
if any of you get time help me with this problem, thankssss

http://img231.imageshack.us/img231/1427/ughmath6lp.jpg

The image is fuzzy, but I think it says $\frac{21a^2b}{7a^2b}+\left(\frac{a^2}{b^2}\right)^ 0$

In review, $\frac{a^3}{a^2}=a^{(3-2)}=a^1=a$ and $a^0=1$ when $a\neq0$.

$\frac{21a^2b}{7a^2b}+\left(\frac{a^2}{b^2}\right)^ 0$

$\frac{21a^{(2-2)}b^{(1-1)}}{7}+1$

$\frac{21a^0b^0}{7}+1$

$\frac{21\cdot1\cdot1}{7}+1$

$\frac{21}{7}+1$

$3+1$

$4$