Question
V 76 square root is greater than the perfect square of what
Answers
GPT-4o mini
To determine the largest perfect square that is less than \( \sqrt{76} \), we first calculate \( \sqrt{76} \).
Calculating \( \sqrt{76} \):
1. \( \sqrt{76} \) can be simplified: \( \sqrt{76} = \sqrt{4 \times 19} = \sqrt{4} \times \sqrt{19} = 2\sqrt{19} \).
2. The approximate value of \( \sqrt{19} \) is about \( 4.36 \). Thus, \( 2\sqrt{19} \approx 2 \times 4.36 = 8.72 \).
Now, we can find the largest perfect square less than \( 8.72 \):
- The perfect squares less than \( 8.72 \) are:
- \( 0^2 = 0 \)
- \( 1^2 = 1 \)
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
Since \( 3^2 = 9 \) is greater than \( \sqrt{76} \), the largest perfect square less than \( \sqrt{76} \) would be \( 4 \) (which is \( 2^2 \)).
Thus, \( \sqrt{76} \) is greater than the perfect square of \( 2 \), and \( 4 \) is the largest perfect square less than \( 76 \).
Therefore, \( \sqrt{76} \) is greater than the perfect square of \( 4 \) (which is \( 2^2 \)).
Calculating \( \sqrt{76} \):
1. \( \sqrt{76} \) can be simplified: \( \sqrt{76} = \sqrt{4 \times 19} = \sqrt{4} \times \sqrt{19} = 2\sqrt{19} \).
2. The approximate value of \( \sqrt{19} \) is about \( 4.36 \). Thus, \( 2\sqrt{19} \approx 2 \times 4.36 = 8.72 \).
Now, we can find the largest perfect square less than \( 8.72 \):
- The perfect squares less than \( 8.72 \) are:
- \( 0^2 = 0 \)
- \( 1^2 = 1 \)
- \( 2^2 = 4 \)
- \( 3^2 = 9 \)
Since \( 3^2 = 9 \) is greater than \( \sqrt{76} \), the largest perfect square less than \( \sqrt{76} \) would be \( 4 \) (which is \( 2^2 \)).
Thus, \( \sqrt{76} \) is greater than the perfect square of \( 2 \), and \( 4 \) is the largest perfect square less than \( 76 \).
Therefore, \( \sqrt{76} \) is greater than the perfect square of \( 4 \) (which is \( 2^2 \)).