What’s the square root of 41 to the nearest hundredth?
The Correct Answer and Explanation is :
The square root of 41, rounded to the nearest hundredth, is approximately 6.40.
Explanation
To understand how to find the square root of 41, we begin with the definition of a square root: it is a number that, when multiplied by itself, gives the original number. In mathematical terms, if ( x = \sqrt{41} ), then ( x^2 = 41 ).
Finding the Square Root
- Estimating the Value: Since 41 is not a perfect square, we can start by estimating. The perfect squares near 41 are ( 36 ) (which is ( 6^2 )) and ( 49 ) (which is ( 7^2 )). Therefore, we know that ( 6 < \sqrt{41} < 7 ).
- Refining the Estimate: To narrow it down further, we can test values between 6 and 7:
- For ( 6.4 ):
[
6.4^2 = 40.96
] - For ( 6.5 ):
[
6.5^2 = 42.25
]
From this, we can conclude that ( 6.4 < \sqrt{41} < 6.5 ).
- Using a Calculator: For a precise calculation, we can use a calculator or a mathematical software. Inputting ( \sqrt{41} ) yields approximately ( 6.403124237 ).
- Rounding: To round to the nearest hundredth, we consider the third decimal place:
- The number ( 6.403124237 ) has a ‘3’ in the hundredths place and a ‘1’ in the thousandths place. Since ‘1’ is less than 5, we round down, which gives us ( 6.40 ).
Conclusion
The square root of 41 is approximately 6.40 when rounded to the nearest hundredth. This process involves estimating the value, refining the range, using precise calculations, and finally rounding to the desired accuracy. Understanding square roots is essential in various fields, including mathematics, engineering, and physics, as they often relate to geometric calculations and real-world applications.