Question

To one decimal place, square root of, 38, end square root
38
​
must lie between
and
because 6, squared, equals, start box, 36, end box6
2
=
36
​
and 7, squared, equals, start box, 49, end box7
2
=
49
​
, and 38 lies between these values.

Answer 1

To find the square root of 38 and identify the interval it lies in, we start with the following:

1. We know that \(6^2 = 36\) and \(7^2 = 49\).
2. Since \(38\) is between \(36\) and \(49\), it follows that the square root of \(38\) must lie between \(6\) and \(7\).

To express this mathematically:

\[
6 < \sqrt{38} < 7
\]

To find a more precise value for \(\sqrt{38}\):

Calculating the square root approximately, we can use the fact that:
\[
\sqrt{36} = 6 \quad \text{and} \quad \sqrt{49} = 7
\]

Using a calculator, we find:
\[
\sqrt{38} \approx 6.164
\]

Rounding this value to one decimal place gives:
\[
\sqrt{38} \approx 6.2
\]

So, to summarize:
\[
\sqrt{38} \text{ lies between } 6 \text{ and } 7 \text{ because } 6^2 = 36 \text{ and } 7^2 = 49.
\]

And the approximate value of \(\sqrt{38}\) rounded to one decimal place is \(6.2\).