To approximate \(\sqrt{1131}\) and demonstrate a method that gets closer to \(\sqrt{31}\), we can use the fact that \(\sqrt{1131}\) is approximately equal to \(\sqrt{31} \times \sqrt{37}\) (since \(1131 = 31 \times 37\)).
First, we need to calculate \(\sqrt{31}\) to a one-decimal place:
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Estimate \(\sqrt{31}\): We know that \(5.5^2 = 30.25\) and \(5.6^2 = 31.36\). Hence, \(\sqrt{31}\) is between \(5.5\) and \(5.6\).
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Refining the estimate: Let's try \(5.57\): \[ 5.57^2 = 31.0649 \quad (\text{too high}) \]
Now try \(5.56\): \[ 5.56^2 = 30.9136 \quad (\text{too low}) \]
Since \(5.56 < \sqrt{31} < 5.57\), we can say that \(\sqrt{31} \approx 5.57\) to one decimal place.
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Finding \(\sqrt{1131}\): Since \(\sqrt{1131} = \sqrt{31 \times 37} = \sqrt{31} \cdot \sqrt{37}\) and we have \(\sqrt{31} \approx 5.57\), we will also need to estimate \(\sqrt{37}\).
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Estimate \(\sqrt{37}\): Similarly, we know that \(6^2 = 36\) and \(7^2 = 49\). Thus, \(\sqrt{37}\) is between \(6\) and \(7\).
Let's try \(6.1\): \[ 6.1^2 = 37.21 \quad (\text{too high}) \] And then \(6.05\): \[ 6.05^2 = 36.6025 \quad (\text{too low}) \]
Therefore, \(\sqrt{37} \approx 6.1\) to one decimal place.
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Multiply the estimates: \[ \sqrt{1131} \approx 5.57 \cdot 6.1 \approx 33.977 \]
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Round to one decimal place: \[ \sqrt{1131} \approx 34.0 \]
Thus, the approximation of \(\sqrt{1131}\) correct to one decimal place is \(\boxed{34.0}\).
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