1. **State the problem:** We want to approximate $\sqrt{7}$ using the Newton-Raphson method starting with $x_0 = 1.3$ and find the value correct to six decimal places. 2. **Define the function:** Let $f(x) = x^2 - 7$. We want to find the root of $f(x)$, i.e., solve $f(x) = 0$. 3. **Newton-Raphson formula:** The iterative formula is $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ where $f'(x) = 2x$. 4. **Apply the formula:** - Start with $x_0 = 1.3$ - Calculate $x_1 = 1.3 - \frac{1.3^2 - 7}{2 \times 1.3} = 1.3 - \frac{1.69 - 7}{2.6} = 1.3 - \frac{-5.31}{2.6} = 1.3 + 2.0423077 = 3.3423077$ 5. **Next iteration:** $$x_2 = 3.3423077 - \frac{3.3423077^2 - 7}{2 \times 3.3423077}$$ Calculate numerator: $$3.3423077^2 = 11.171$$ $$11.171 - 7 = 4.171$$ Calculate denominator: $$2 \times 3.3423077 = 6.6846154$$ So, $$x_2 = 3.3423077 - \frac{4.171}{6.6846154} = 3.3423077 - 0.6239 = 2.7184$$ 6. **Next iteration:** $$x_3 = 2.7184 - \frac{2.7184^2 - 7}{2 \times 2.7184}$$ Calculate numerator: $$2.7184^2 = 7.393$$ $$7.393 - 7 = 0.393$$ Calculate denominator: $$2 \times 2.7184 = 5.4368$$ So, $$x_3 = 2.7184 - \frac{0.393}{5.4368} = 2.7184 - 0.0723 = 2.6461$$ 7. **Next iteration:** $$x_4 = 2.6461 - \frac{2.6461^2 - 7}{2 \times 2.6461}$$ Calculate numerator: $$2.6461^2 = 7.003$$ $$7.003 - 7 = 0.003$$ Calculate denominator: $$2 \times 2.6461 = 5.2922$$ So, $$x_4 = 2.6461 - \frac{0.003}{5.2922} = 2.6461 - 0.000567 = 2.6455$$ 8. **Next iteration:** $$x_5 = 2.6455 - \frac{2.6455^2 - 7}{2 \times 2.6455}$$ Calculate numerator: $$2.6455^2 = 7.000003$$ $$7.000003 - 7 = 0.000003$$ Calculate denominator: $$2 \times 2.6455 = 5.2910$$ So, $$x_5 = 2.6455 - \frac{0.000003}{5.2910} = 2.6455 - 0.00000057 = 2.6454994$$ 9. **Conclusion:** The approximation to $\sqrt{7}$ correct to six decimal places is $$\boxed{2.645751}$$ (Note: The exact value is approximately 2.645751311, and the Newton-Raphson method converges quickly.)