1. **Convert decimal 0.6402 to binary (5 steps):** To convert a decimal fraction to binary, multiply by 2 repeatedly and record the integer part each time. Step 1: $0.6402 \times 2 = 1.2804$ (integer part = 1) Step 2: $0.2804 \times 2 = 0.5608$ (integer part = 0) Step 3: $0.5608 \times 2 = 1.1216$ (integer part = 1) Step 4: $0.1216 \times 2 = 0.2432$ (integer part = 0) Step 5: $0.2432 \times 2 = 0.4864$ (integer part = 0) Binary after 5 steps: $(0.6402)_{10} \approx (0.10100)_2$ 2. **False Position method for $f(x) = x^3 - 3x + 1$ with $x_0=0$, $x_1=1$:** Formula: $$x_{n+1} = \frac{x_{n-1}f(x_n) - x_n f(x_{n-1})}{f(x_n) - f(x_{n-1})}$$ Calculate $f(0) = 1$, $f(1) = -1$ Calculate $x_2$: $$x_2 = \frac{0 \times (-1) - 1 \times 1}{-1 - 1} = \frac{-1}{-2} = 0.5$$ Calculate $f(0.5) = (0.5)^3 - 3(0.5) + 1 = -0.375$ Calculate $x_3$: $$x_3 = \frac{0 \times (-0.375) - 0.5 \times 1}{-0.375 - 1} = \frac{-0.5}{-1.375} \approx 0.363636$$ 3. **Newton-Raphson method for $f(x) = x - 0.8 - 0.2 \sin(x)$ starting at $x_0 = \frac{\pi}{4}$:** Formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ Where $f'(x) = 1 - 0.2 \cos(x)$ Iteration 1: $f(x_0) = 0.785398 - 0.8 - 0.2 \times 0.707107 = -0.156023$ $f'(x_0) = 1 - 0.2 \times 0.707107 = 0.858579$ $x_1 = 0.785398 - \frac{-0.156023}{0.858579} = 0.967121$ Iteration 2: $f(x_1) = 0.967121 - 0.8 - 0.2 \times 0.823159 = 0.002489$ $f'(x_1) = 1 - 0.2 \times 0.567801 = 0.886440$ $x_2 = 0.967121 - \frac{0.002489}{0.886440} = 0.964313$ 4. **One iteration of Muller's method for $f(x) = x^3 + x$ starting with $x_0 = -1$, $x_1 = 0$, $x_2 = 1$:** Formula: $$x_3 = x_2 - \frac{2c}{b \pm \sqrt{b^2 - 4ac}}$$ Calculate $f(-1) = -2$, $f(0) = 0$, $f(1) = 2$ Calculate $h_0 = 1$, $h_1 = 1$ Calculate $\delta_0 = 2$, $\delta_1 = 2$ Calculate $a = 0$, $b = 2$, $c = 2$ Calculate $x_3$: $$x_3 = 1 - \frac{2 \times 2}{2 + \sqrt{4 - 0}} = 1 - \frac{4}{4} = 0$$