[tex]1.\\(-1.5)^0=1\\/\text{for any real numbers not equal 0 we have}\ a^0=1/\\\\2.\\3g^{-2}b^2=3\cdot\dfrac{1}{g^2}\cdot b^2=\dfrac{3b^2}{g^2}\\\\/a^{-n}=\dfrac{1}{a^n}/\\\\3.\\\dfrac{3}{g^{-2}h^3}=\dfrac{3}{\frac{1}{g^2}\cdot h^3}=\dfrac{3}{h^3}\cdot\dfrac{g^2}{1}=\dfrac{3g^2}{h^3}\\\\4.\\2x^{-2}y^{-2}=2\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y^2}=\dfrac{2}{x^2y^2}\\\\\text{substitute x = 3 and y = -2}\\\\\dfrac{2}{3^2\cdot(-2)^2}=\dfrac{\not2^1}{9\cdot\not4_2}=\dfrac{1}{9\cdot2}=\dfrac{1}{18}\\[/tex]
[tex]\dfrac{1}{2^{-2}x^{-3}y^5}=\dfrac{1}{\frac{1}{2^2}\cdot\dfrac{1}{x^3}\cdot y^5}=\dfrac{1}{y^5}\cdot\dfrac{2^2}{1}\cdot\dfrac{x^3}{1}=\dfrac{2^2x^3}{y^5}\\\\\text{substitute x = 2 and y = -4}\\\\\dfrac{2^2(2^3)}{(-4)^5}=\dfrac{2^{2+3}}{(-4)^5}=\dfrac{2^5}{(-4)^5}=\left(\dfrac{2}{-4}\right)^5=\left(-\dfrac{1}{2}\right)^5=-\dfrac{1}{32}\\\\/a^n\cdot a^m=a^{n+m};\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}/[/tex]
