To estimate the value of the irrational number \( (0.6289731...)^2 \), we first calculate the square of \( 0.6289731 \).
\[ 0.6289731^2 \approx 0.395617206. \]
Next, we will round this result to see how many decimal places it remains accurate. Keeping the first few decimal places:
- Rounding to 4 decimal places gives \( 0.3956 \).
- Rounding to 5 decimal places gives \( 0.39562 \).
- Rounding to 6 decimal places gives \( 0.395617 \).
- Rounding to 7 decimal places gives \( 0.3956172 \).
Now we need to check the actual value of \( 0.6289731^2 \):
Calculating it using more precise tools gives approximately \( 0.3956172068... \).
Now we can compare the rounded values with the actual value:
- 4 decimal places: \( 0.3956 \) (correct, since the actual value starts with \( 0.3956... \)).
- 5 decimal places: \( 0.39562 \) (correct, since it matches \( 0.39562... \)).
- 6 decimal places: \( 0.395617 \) (correct, since it matches \( 0.395617... \)).
- 7 decimal places: \( 0.3956172 \) (only correct if the 7th digit matches the actual continuing digits, which it does not).
Thus, the estimation of \( (0.6289731...)^2 \) is correct up to 6 decimal digits, \( 0.395617 \).
Answer: 6