a) \( 2^{-2}+2^{-3} \quad \) b) \( 4^{0}-4^{-1} \)

Below is the step-by-step solution. ## Part a) \( 2^{-2}+2^{-3} \) 1. Rewrite the negative exponents as fractions: - \( 2^{-2} = \frac{1}{2^2} = \frac{1}{4} \) - \( 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \) 2. Add the fractions: \[ \frac{1}{4} + \frac{1}{8} \] 3. Find a common denominator. The common denominator of 4 and 8 is 8: \[ \frac{1}{4} = \frac{2}{8} \] Therefore: \[ \frac{2}{8} + \frac{1}{8} = \frac{3}{8} \] Thus, the answer for part a) is: \[ \frac{3}{8} \] ## Part b) \( 4^{0}-4^{-1} \) 1. Evaluate \( 4^{0} \) and \( 4^{-1} \): - \( 4^{0} = 1 \) - \( 4^{-1} = \frac{1}{4} \) 2. Subtract: \[ 1-\frac{1}{4} \] 3. Express 1 as \(\frac{4}{4}\) and subtract: \[ \frac{4}{4}-\frac{1}{4} = \frac{3}{4} \] Thus, the answer for part b) is: \[ \frac{3}{4} \]