Solving for \(s\), we have \(s = \frac{10}{\sqrt{2}} = \frac{10 \times \sqrt{2}}{2} = 5\sqrt{2}\) cm.

["Solving for ( s ): A Clear Guide to Simplifying ( s = \frac{10}{\sqrt{2}} )", "When working with irrational expressions in algebra, simplifying square roots is essential for clarity, accuracy, and further problem-solving. One common expression students encounter is:", "[\ns = \frac{10}{\sqrt{2}}\n]", "At first glance, this form may appear complex, but with a few algebraic steps, it can be simplified to a more elegant and usable form:\n[\ns = 5\sqrt{2} \ \ ext{cm}\n]", "In this article, we’ll explore how to solve for ( s ) step by step, why simplification matters, and how this cleansed form enhances understanding and applications.", "---", "### Step 1: Understanding the Original Expression", "We begin with:", "[\ns = \frac{10}{\sqrt{2}}\n]", "This expression contains a square root in the denominator, which is not ideal for most mathematical contexts because irrational denominators complicate calculations and interpretations. To improve simplicity and standardize the form, we apply rationalizing the denominator.", "---", "### Step 2: Rationalizing the Denominator", "To rationalize the denominator, multiply both the numerator and denominator by ( \sqrt{2} ):", "[\ns = \frac{10}{\sqrt{2}} \ imes \frac{\sqrt{2}}{\sqrt{2}} = \frac{10\sqrt{2}}{2}\n]", "This multiplication uses the identity ( \sqrt{2} \cdot \sqrt{2} = 2 ), eliminating the radical from the denominator.", "---", "### Step 3: Simplifying the Fraction", "Now simplify the fraction:", "[\n\frac{10\sqrt{2}}{2} = 5\sqrt{2}\n]", "Thus,\n[\ns = 5\sqrt{2} \ \ ext{cm}\n]", "---", "### Why Simplifying ( s = \frac{10}{\sqrt{2}} ) to ( 5\sqrt{2} ) Matters", "1. Enhanced Clarity:\n The simplified form ( 5\sqrt{2} ) is cleaner and easier to interpret, especially when dimensions or measurements are involved.", "2. Easier Calculations:\n Using exact radical values simplifies further algebra, integration, or geometric applications compared to retaining an irrational denominator.", "3. Standard Mathematical Communication:\n In academic, scientific, and real-world contexts, simplifying radicals is considered best practice, ensuring consistency and precision.", "4. Foundation for Further Problems:\n When solving equations involving ( s ), a simplified expression allows faster substitution and manipulation in subsequent steps.", "---", "### Expressing ( s ) in Centimeters with Proper Units", "Given the original value stems from a measurement, retaining centimeters as the unit and expressing the result in standard form ensures practicality:", "[\ns = 5\sqrt{2} ,\ ext{cm} \approx 7.07,\ ext{cm} \quad (\ ext{since } \sqrt{2} \approx 1.414)\n]", "This standardized form is valuable in fields such as engineering, physics, and geometry.", "---", "### Conclusion", "Transforming ( s = \frac{10}{\sqrt{2}} ) into ( s = 5\sqrt{2} ) cm through rationalization and simplification is more than just algebraic manipulation—it’s a key step in precision and clarity. By mastering this process, you develop stronger algebra skills essential for advanced mathematics and real-world problem-solving.", "Remember, simplifying expressions like ( s = \frac{10}{\sqrt{2}} ) unlocks deeper understanding and broader application possibilities. Whether solving geometry problems, computing distances, or analyzing functions, clean, rational expressions empower confident, efficient work.", "---", "Keywords: solve for ( s ), simplify ( \frac{10}{\sqrt{2}} ), rationalize denominator, ( 5\sqrt{2} ) cm, algebraic simplification, exact values in measurements, algebra reference, mathematical precision", "---", "Understanding and practicing such simplifications equips learners and professionals alike with tools crucial for success in STEM disciplines. Keep simplifying—because clarity starts with a well-rationalized expression."]