To approximate \(\sqrt{1935}\) using a sequence of values getting closer to \(\sqrt{1935}\), we can start by finding an initial approximation for \(\sqrt{1935}\) and then refine that approximation.
- Initial Approximation: First, look for two perfect squares between which 1935 lies:
- \(\sqrt{1936} = 44\) (since \(44^2 = 1936\))
- \(\sqrt{1935} \) is slightly less than 44.
So, we can start with an initial estimate of \(43.9\).
- Refining the Estimate: We can use the average method (or the method of "successive approximations") to refine the estimate. We use the formula:
\[ x_{n+1} = \frac{x_n + \frac{1935}{x_n}}{2} \]
Using \(x_0 = 43.9\):
Calculating \(x_1\): \[ x_1 = \frac{43.9 + \frac{1935}{43.9}}{2} \]
Calculating \(\frac{1935}{43.9}\): \[ \frac{1935}{43.9} \approx 44.1 \] Now, substitute this back into the equation for \(x_1\): \[ x_1 = \frac{43.9 + 44.1}{2} = \frac{88}{2} = 44.0 \]
Calculating \(x_2\): Now using \(x_1 = 44.0\): \[ x_2 = \frac{44.0 + \frac{1935}{44.0}}{2} \]
Calculating \(\frac{1935}{44.0}\): \[ \frac{1935}{44.0} \approx 43.9773 \] Substituting this back into the equation for \(x_2\): \[ x_2 = \frac{44.0 + 43.9773}{2} = \frac{87.9773}{2} \approx 43.98865 \]
Calculating \(x_3\): Using \(x_2 \approx 43.98865\): \[ x_3 = \frac{43.98865 + \frac{1935}{43.98865}}{2} \]
Calculating \(\frac{1935}{43.98865}\) gives a value close to 44.0, but for practical purposes, we can stop here.
After the sequence of approximations, we find that \(\sqrt{1935} \approx 44.0\).
Finally, to correct to one decimal place: \[ \sqrt{1935} \approx 44.0 \]
Therefore, the approximate value of \(\sqrt{1935}\) correct to one decimal place is 44.0.
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