Write each expressionusing positiveexponents, thenevaluate.

Answer:The answer is [tex]\frac{125}{243}[/tex]Step-by-step explanation:* Lets explain how to solve the problem- Any number to power of zero = 1 (except 0)- Ex: [tex]5^{0}=1[/tex]  and  [tex](\frac{1}{7})^{0}=1[/tex]- If the power of a number is negative we can change it to positive  if we reciprocal the number- Ex: [tex]3^{-2}=(\frac{1}{3})^{2}[/tex]  and  [tex](\frac{1}{2})^{-3}=(2)^{3}[/tex]* Lets solve the problem- We want to write each expression  using positive  exponents,  and evaluate# [tex]2^{0}[/tex]∵ Zero is not a positive exponent- Change [tex]2^{0}[/tex] to 1 because 1 has positive exponent 1∴ [tex]2^{0}=1[/tex]# [tex]15^{3}[/tex] - We can written it as [tex](3 × 5)^{3}[/tex]∴ [tex]15^{3}=(5*3)^{3}=(5)^{3}(3)^{3}[/tex]# [tex]9^{-4}[/tex]- Change the negative exponent by reciprocal 9∴ [tex]9^{-4}=(\frac{1}{9})^{4}[/tex]∴[tex](\frac{1}{9})^{4}=\frac{1}{6561}[/tex]- Lets evaluate ∵ 5³ = 125 and 3³ = 27∴ 5³ × 3³ = 3375∴[tex](\frac{1}{9})^{4}=\frac{1}{6561}[/tex]∴ [tex](1)(3375)(\frac{1}{6561})=\frac{125}{243}[/tex]∴ The answer is [tex]\frac{125}{243}[/tex]