To determine which of the given options can be a possible irrational number \( d \) such that \( 6 < d < 7 \), let's evaluate each option.
A. \( \sqrt{6.2} \)
To approximate \( \sqrt{6.2} \), consider the squares:
- \( \sqrt{6} \approx 2.449 \) (since \( 2.449^2 \approx 6 \))
- \( \sqrt{7} \approx 2.646 \) (since \( 2.646^2 \approx 7 \))
Calculating \( \sqrt{6.2} \):
Using a calculator, \( \sqrt{6.2} \approx 2.489 \).
Thus, \( \sqrt{6.2} \approx \) 2.489 is not relevant since it is not in the range \( 6 < d < 7 \).
B. \( 2\pi \)
Approximating \( 2\pi \):
\[ 2\pi \approx 2 \times 3.14 \approx 6.28 \]
This value \( 6.28 \) lies in the interval \( (6, 7) \), so \( 2\pi \) is a possible value.
C. \( \sqrt{13} \)
To approximate \( \sqrt{13} \):
Calculating \( 3.6^2 = 12.96 \) and \( 3.7^2 = 13.69 \).
Thus, \( \sqrt{13} \) is approximately \( 3.605 \), which does not lie within \( (6, 7) \).
D. \( 6\pi \)
Calculating \( 6\pi \):
\[ 6\pi \approx 6 \times 3.14 \approx 18.84 \]
This value is far outside the range \( (6, 7) \).
From the options, only option B \( 2\pi \) satisfies the condition \( 6 < d < 7 \).
Thus, the possible value for \( d \) is:
Answer: B. \( 2\pi \)