\( \begin{array}{ll}\text { Question } 2 \\ 2.1 & \text { Simplify the following and give your answers as positive exponents. } \\ 2.1 .1 & -4\left(x^{3}\right)^{0} \\ 2.1 .2 & \left(-4 x^{3}\right)^{0} \\ 2.1 .3 & 3^{-1} a^{2} b^{-5} \\ 2.1 .4 & \left(-2 y^{2}\right)(-2 y)^{-2} \\ 2.1 .5 & \frac{x^{4} y^{3}}{x^{5} y^{3}} \\ 2.2 & \text { Calculate the following without the use of a calculator. Show all your work. } \\ & \end{array} \)

**2.1.1** Simplify \[ -4\left(x^3\right)^0 \] Any nonzero number raised to the 0 power equals 1, so \[ \left(x^3\right)^0 = 1. \] Thus, the expression simplifies to \[ -4 \cdot 1 = -4. \] --- **2.1.2** Simplify \[ \left(-4x^3\right)^0. \] Again, any nonzero expression raised to the 0 power equals 1, so \[ \left(-4x^3\right)^0 = 1. \] --- **2.1.3** Simplify \[ 3^{-1}a^2b^{-5}. \] Recall that a negative exponent indicates the reciprocal, i.e., \[ 3^{-1} = \frac{1}{3} \quad \text{and} \quad b^{-5} = \frac{1}{b^5}. \] Thus, the expression becomes \[ \frac{1}{3}a^2\frac{1}{b^5} = \frac{a^2}{3b^5}. \] --- **2.1.4** Simplify \[ \left(-2y^2\right)(-2y)^{-2}. \] First, rewrite the negative exponent: \[ (-2y)^{-2} = \frac{1}{(-2y)^2}. \] Compute the square in the denominator: \[ (-2y)^2 = 4y^2. \] Now, the expression becomes \[ (-2y^2) \cdot \frac{1}{4y^2}. \] Cancel \( y^2 \) in the numerator and denominator to obtain \[ \frac{-2}{4} = -\frac{1}{2}. \] --- **2.1.5** Simplify \[ \frac{x^4y^3}{x^5y^3}. \] Cancel the common factor \( y^3 \) in the numerator and denominator, and simplify the \( x \) terms using the quotient rule \( \frac{x^m}{x^n} = x^{m-n} \): \[ \frac{x^4}{x^5} = x^{4-5} = x^{-1}. \] Expressing the negative exponent as a positive exponent in the denominator gives \[ x^{-1} = \frac{1}{x}. \] --- **2.2** The problem statement for section 2.2 indicates that additional calculations are to be performed without a calculator, but no specific expressions or instructions are provided. Additional details are needed to proceed with section 2.2.