First, find the square root of \( 9.3 \). By approximation: \[ \sqrt{9.3} \approx 3.05 \] This is because \( \sqrt{9} = 3 \) and \( \sqrt{10} \approx 3.16 \). Since 9.3 is between 9 and 10, we estimate \( \sqrt{9.3} \) to be a little more than 3.
To represent \( \sqrt{9.3} \approx 3.05 \) on the number line: - First, mark the integer points 3 and 4. - Since \( 3.05 \) is slightly greater than 3, place a point just to the right of 3, closer to it than to 4.
The point \( 3.05 \) will be located a little bit to the right of 3, which can be visualized as:

The square root of \( 9.3 \), which is approximately \( 3.05 \), is represented slightly to the right of 3 on the number line.
For real number a, b (a > b > 0), let
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \leq a^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1 \right\} = 30\pi\)
and
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \geq b^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\} = 18\pi\)
Then the value of (a – b)2 is equal to _____.