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

Sure! Let’s simplify each expression step by step, ensuring we leave everything with positive exponents. 2.1.1) To simplify \(-4\left(x^{3}\right)^{0}\): Any non-zero base raised to the power of 0 equals 1. So, \((x^{3})^{0} = 1\). Therefore, \(-4\left(x^{3}\right)^{0} = -4 \cdot 1 = -4\). 2.1.2) For \(\left(-4 x^{3}\right)^{0}\): Similar to the previous one, any non-zero value raised to 0 is 1. Thus, \(\left(-4 x^{3}\right)^{0} = 1\). 2.1.3) When simplifying \(3^{-1} a^{2} b^{-5}\): The negative exponent means you take the reciprocal, so \(3^{-1} = \frac{1}{3}\) and \(b^{-5} = \frac{1}{b^{5}}\). Therefore, \(3^{-1} a^{2} b^{-5} = \frac{a^{2}}{3 b^{5}}\). 2.1.4) For \(\left(-2 y^{2}\right)(-2 y)^{-2}\): Simplifying \((-2 y)^{-2}\) means taking the reciprocal and squaring, thus \((-2 y)^{-2} = \frac{1}{(-2 y)^{2}} = \frac{1}{4 y^{2}}\). Now, \((-2 y^{2})(-2 y)^{-2} = (-2 y^{2}) \cdot \frac{1}{4 y^{2}} = \frac{-2}{4} = -\frac{1}{2}\). 2.1.5) In simplifying \(\frac{x^{4} y^{3}}{x^{5} y^{3}}\): When you divide like bases, you subtract the exponents: \(x^{4 - 5} = x^{-1}\) and \(y^{3 - 3} = y^{0} = 1\). Thus, \(\frac{x^{4} y^{3}}{x^{5} y^{3}} = \frac{1}{x} = x^{-1}\) (which is positive as it’s simply \(x^{-1}\)). 2.2) Since you mentioned to show all work without using a calculator, I assume you’re asking for something a bit more. For any calculations you have in mind, feel free to share, and I'll break them down for you!