Answer:
[tex]\large\text{$\tt{1.)\:8^{-1}}=\boxed{\tt{\frac{1}{8}}}$}[/tex]
[tex]\large\text{$\tt{2.)\:(2abc)^0}=\boxed{\tt{1}}$}[/tex]
[tex]\large\text{$\tt{3.)\:\frac{18x^8y^3}{6x^3y^2}=\frac{18x^8x^{-3}y^3y^{-2}}{6}=\frac{18x^{8-3}y^{3-2}}{6}=\frac{18x^5y}{6}}=\boxed{\tt{3x^5y}}$}[/tex]
[tex]\large\text{$\tt{4.)\:6(xy)^0=6(1)}=\boxed{\tt{6}}$}[/tex]
[tex]\large\text{$\tt{5.)\:(\frac{1}{3})^{-2}=\frac{1}{(\frac{1}{3})^2}=\frac{1}{\frac{1^2}{3^2}}=\frac{1}{\frac{1}{9}}=1\times9}=\boxed{\tt{9}}$}[/tex]
[tex]\large\text{$\tt{or}$}[/tex]
[tex]\large\text{$\tt{(\frac{1}{3})^{-2}=\frac{1^{-2}}{3^{-2}}=\frac{3^{2}}{1^2}=\frac{9}{1}}=\boxed{\tt{9}}$}[/tex]
[tex]\large\text{$\tt{6.)\:8^0+8=1+8}=\boxed{\tt{9}}$}[/tex]
[tex]\large\text{$\tt{7.)\:3^{-2}\:{\cdot}\:2^{-1}=\frac{1}{3^2}\:{\cdot}\:\frac{1}{2^1}=\frac{1}{9}\:{\cdot}\:\frac{1}{2}}=\boxed{\tt{\frac{1}{18}}}$}[/tex]
[tex]\large\text{$\tt{8.)\:2^0\:{\cdot}\:3^4=1\:{\cdot}\:81}=\boxed{\tt{81}}$}[/tex]
[tex]\large\text{$\tt{9.)\:(\frac{2}{3})^3=\frac{2^3}{3^3}}=\boxed{\tt{\frac{8}{27}}}$}[/tex]
[tex]\large\text{$\tt{10.)\:10^0+1=1+1}=\boxed{\tt{2}}$}[/tex]
Note:
- Any number or variable raised to zero is 1.
- If a numerator or a whole number has a negative exponent, you're going to put it to the denominator to become positive, and vice versa.
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