\begin{tabular}{lll}\hline & & [15] \\ Question 2 & (implify the following and give your answers as positive exponents. \\ 2.1 & (1) \\ 2.1 .1 & \( -4\left(x^{3}\right)^{0} \) & (1) \\ 2.1 .2 & \( \left(-4 x^{3}\right)^{0} \) & (3) \\ 2.1 .3 & \( 3^{-1} a^{2} b^{-5} \) & (2) \\ 2.1 .4 & \( \left(-2 y^{2}\right)(-2 y)^{-2} \) & (3) \\ 2.1 .5 & \( x^{4} y^{3} \) & \( x^{5} y^{3} \)\end{tabular}

Let's simplify each expression step by step, ensuring that we express all answers with positive exponents. ### 2.1.1: \( -4\left(x^{3}\right)^{0} \) According to the exponent rule, any non-zero number raised to the power of 0 is 1. Therefore: \[ \left(x^{3}\right)^{0} = 1 \] Thus, we have: \[ -4\left(x^{3}\right)^{0} = -4 \cdot 1 = -4 \] ### 2.1.2: \( \left(-4 x^{3}\right)^{0} \) Again, using the same exponent rule: \[ \left(-4 x^{3}\right)^{0} = 1 \] ### 2.1.3: \( 3^{-1} a^{2} b^{-5} \) To express this with positive exponents, we can rewrite \( 3^{-1} \) and \( b^{-5} \): \[ 3^{-1} = \frac{1}{3} \] \[ b^{-5} = \frac{1}{b^{5}} \] Thus, we have: \[ 3^{-1} a^{2} b^{-5} = \frac{a^{2}}{3 b^{5}} \] ### 2.1.4: \( \left(-2 y^{2}\right)(-2 y)^{-2} \) First, simplify \( (-2 y)^{-2} \): \[ (-2 y)^{-2} = \frac{1}{(-2 y)^{2}} = \frac{1}{4 y^{2}} \] Now, substituting back into the expression: \[ \left(-2 y^{2}\right)(-2 y)^{-2} = (-2 y^{2}) \cdot \frac{1}{4 y^{2}} = \frac{-2 y^{2}}{4 y^{2}} = \frac{-2}{4} = -\frac{1}{2} \] ### 2.1.5: \( x^{4} y^{3} \) and \( x^{5} y^{3} \) This part seems to be a comparison or simplification of two expressions. We can express both with positive exponents: 1. \( x^{4} y^{3} \) is already in the correct form. 2. \( x^{5} y^{3} \) is also in the correct form. ### Summary of Answers 1. \( -4\left(x^{3}\right)^{0} = -4 \) 2. \( \left(-4 x^{3}\right)^{0} = 1 \) 3. \( 3^{-1} a^{2} b^{-5} = \frac{a^{2}}{3 b^{5}} \) 4. \( \left(-2 y^{2}\right)(-2 y)^{-2} = -\frac{1}{2} \) 5. \( x^{4} y^{3} \) and \( x^{5} y^{3} \) are already simplified.