Use linear approximation, i.e. the tangent line, to approximate $\frac{1}{0.203}$ as follows: Let $f(x)=\frac{1}{x}$ and find the equation of the tangent line to $f(x)$ at a "nice" point near 0.203 . Then use this to approximate $\frac{1}{0.203}$

Step 1 ：Let's use linear approximation to solve the problem. We start by defining the function \(f(x)=\frac{1}{x}\).

Step 2 ：We need to find a 'nice' point near 0.203. A 'nice' point is a value that is easy to work with. In this case, we choose 0.2.

Step 3 ：Next, we find the derivative of the function \(f(x)=\frac{1}{x}\), which is \(f'(x)=-\frac{1}{x^2}\).

Step 4 ：Evaluating this derivative at the point x=0.2 gives us the slope of the tangent line at that point. The slope is -25.

Step 5 ：We can then use the point-slope form of a line to find the equation of the tangent line. The equation is \(y=-25x+10\).

Step 6 ：Finally, we substitute x=0.203 into this equation to approximate the value of \(\frac{1}{0.203}\). The approximation is 4.925.

Step 7 ：Final Answer: \(\boxed{4.925}\)