Step 1: Understanding the Concept
We can find the approximate value of a function near a known point using linear approximation (or differentials). The formula is $f(x+\Delta x) \approx f(x) + f'(x)\Delta x$. For this problem, $f(x) = \sqrt{x}$.
Step 2: Key Formula or Approach
1. Let $f(x) = \sqrt{x}$.
2. Choose a perfect square $x$ close to 6560.
3. Calculate the small change $\Delta x$ such that $x+\Delta x = 6560$.
4. Find the derivative $f'(x) = \frac{1}{2\sqrt{x}}$.
5. Apply the linear approximation formula: $\sqrt{x+\Delta x} \approx \sqrt{x} + \frac{1}{2\sqrt{x}} \Delta x$.
Step 3: Detailed Explanation
We need to approximate $\sqrt{6560}$.
1. Choose a nearby perfect square:
We know that $80^2 = 6400$ and $81^2 = 6561$. The number 6561 is very close to 6560.
So we choose $x=6561$.
2. Define the function and variables:
Let $f(x) = \sqrt{x}$.
Let $x = 6561$ and $\Delta x = 6560 - 6561 = -1$.
3. Calculate $f(x)$ and $f'(x)$ at $x=6561$:
$f(6561) = \sqrt{6561} = 81$.
The derivative is $f'(x) = \frac{1}{2\sqrt{x}}$.
$f'(6561) = \frac{1}{2\sqrt{6561}} = \frac{1}{2 \times 81} = \frac{1}{162}$.
4. Apply the approximation formula:
\[ \sqrt{6560} = f(6561 - 1) \approx f(6561) + f'(6561) \cdot (-1) \]
\[ \sqrt{6560} \approx 81 - \frac{1}{162} \]
Now we calculate the decimal value of $\frac{1}{162}$:
\[ \frac{1}{162} \approx 0.0061728... \]
So, the approximate value is:
\[ \sqrt{6560} \approx 81 - 0.0061728 = 80.9938272... \]
Rounding to four decimal places, we get 80.9938.
Wait, my calculation gives 80.9938, which is option B. The checkmark is on option A. Let me recheck the calculation. $1/162 = 0.0061728...$ is correct. $81 - 0.00617... = 80.9938...$.
The calculation is correct. The answer should be option B. The provided key marking option A seems incorrect. Let's provide the solution based on the calculation.
Step 4: Final Answer
The calculated approximate value is $80.9938...$. This matches option (B). (Note: There might be an error in the provided answer key which indicates option A.)