## Algebra 1

$\dfrac{25y^8}{49x^{10}}$
(i) $a^{-m} = \dfrac{1}{a^m}, a\ne0$ (ii) $(ab)^m=a^mb^m$ (iii) $(a^m)^n=a^{mn}$ Use the rules above to obtain: $=\dfrac{1}{\left(-\dfrac{7x^5}{5y^4}\right)^2}$ The negative sign goes away since we are raising everything to an even power. $\\=\dfrac{1}{\left(\dfrac{7^2(x^5)^2}{5^2(y^4)^2}\right)} \\=\dfrac{1}{\left(\dfrac{49x^{2\cdot5}}{25y^{4\cdot2}}\right)} \\=\dfrac{1}{\left(\dfrac{49x^{10}}{25y^{8}}\right)} \\=1 \cdot \dfrac{25y^8}{49x^{10}} \\=\dfrac{25y^8}{49x^{10}}$