1. **State the problem:** We want to find the value of $\sqrt{10}$ using the binomial expansion of $(1 - x)^{-\frac{1}{2}}$ by substituting $x = \frac{1}{10}$. We will then approximate $\sqrt{10}$ to four significant figures. 2. **Recall the binomial expansion formula:** For any real number $n$, $$ (1 - x)^n = \sum_{k=0}^\infty \binom{n}{k} (-x)^k = 1 + n(-x) + \frac{n(n-1)}{2!}(-x)^2 + \frac{n(n-1)(n-2)}{3!}(-x)^3 + \cdots $$ 3. **Apply the formula for $n = -\frac{1}{2}$:** $$ (1 - x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-x) + \frac{-\frac{1}{2}(-\frac{3}{2})}{2!}(-x)^2 + \frac{-\frac{1}{2}(-\frac{3}{2})(-\frac{5}{2})}{3!}(-x)^3 + \cdots $$ 4. **Simplify the first few terms:** - First term: $1$ - Second term: $\frac{1}{2}x$ - Third term: $\frac{(-\frac{1}{2})(-\frac{3}{2})}{2} x^2 = \frac{3}{8} x^2$ - Fourth term: $\frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{6} (-x)^3 = -\frac{5}{16} x^3$ 5. **Substitute $x = \frac{1}{10}$:** $$ (1 - \frac{1}{10})^{-\frac{1}{2}} = 1 + \frac{1}{2} \times \frac{1}{10} + \frac{3}{8} \times \left(\frac{1}{10}\right)^2 - \frac{5}{16} \times \left(\frac{1}{10}\right)^3 + \cdots $$ 6. **Calculate each term:** - $1 = 1$ - $\frac{1}{2} \times \frac{1}{10} = 0.05$ - $\frac{3}{8} \times \frac{1}{100} = 0.00375$ - $-\frac{5}{16} \times \frac{1}{1000} = -0.0003125$ 7. **Sum the terms to approximate:** $$ 1 + 0.05 + 0.00375 - 0.0003125 = 1.0534375 $$ 8. **Interpretation:** Since $$ (1 - x)^{-\frac{1}{2}} = \frac{1}{\sqrt{1 - x}} \, \Rightarrow \, \sqrt{1 - x} = \frac{1}{(1 - x)^{-\frac{1}{2}}} $$ For $x = \frac{1}{10}$, $1 - x = \frac{9}{10}$, so $$ \frac{1}{\sqrt{\frac{9}{10}}} = \frac{1}{\frac{3}{\sqrt{10}}} = \frac{\sqrt{10}}{3} $$ Therefore, $$ (1 - \frac{1}{10})^{-\frac{1}{2}} = \frac{\sqrt{10}}{3} \approx 1.0534375 $$ 9. **Solve for $\sqrt{10}$:** $$ \sqrt{10} = 3 \times 1.0534375 = 3.1603125 $$ 10. **Round to four significant figures:** $$ \boxed{3.160} $$ This matches the known value of $\sqrt{10} \approx 3.1623$ closely, with a small error due to truncation of the series.