Number line

Distance equals the absolute difference of the two coordinates. In exams, treat the negative sign as direction, not distance.

Since (7^2=49) and (8^2=64), (\sqrt{52}) lies between them. In exams, locate square roots by comparing squares.

The coordinate of the midpoint is the average of the coordinates of the two endpoints. Thus, \(\frac{-1.25+2.75}{2}=\frac{1.50}{2}=0.75\). Therefore, option B is correct. Option A, 0.50, results from an arithmetic error in adding or averaging the coordinates. Exam tip: Add the endpoint coordinates and divide the sum by 2 to find the midpoint.

Since (\sqrt{10}) lies between (3) and (4), (-\sqrt{10}) lies between (-4) and (-3). In exams, place negative surds to the left.

Distance means (x=3\pm5), so the values are (-2) and (8). In exams, always consider both directions.

This decimal is non-recurring and non-terminating, so it is irrational. In exams, identify non-recurring infinite decimals as irrational.

(\frac{11}{4}=2.75), so it lies between (2) and (3). In exams, convert fractions to decimals or mixed numbers for placement.

(-\frac{5}{2}=-2.5), and (-2.45) is greater than (-2.5) but less than (-2.4). In exams, compare negative numbers carefully.

The direct answer is option C,
\(0.7\). On a number line, moving left means moving to a smaller number, so subtract the distance from the starting number:
\(a=1.6-0.9=0.7\). We can check it because the distance from
\(0.7\) to
\(1.6\) is
\(1.6-0.7=0.9\), and
\(0.7\) is indeed on the left. Option A,
\(2.5\), is obtained by adding and lies to the right, so it is wrong. Option B,
\(0.9\), is the distance moved, not the final coordinate. Option C is correct because it is the starting value minus the leftward distance. Option D,
\(-0.7\), subtracts too much and is not 0.9 units left of 1.6; its distance is
\(2.3\). Exam cue: right means add, left means subtract.

(-\frac{4}{5}=-0.8), which is the smallest among the given values. In exams, the farthest-left number is the least number.

From (x^2=\frac{1}{9}), (x=\pm\frac{1}{3}), and the given interval needs the negative value. In exams, use conditions to choose the correct sign.

By Pythagoras theorem, the hypotenuse is (\sqrt{2}). In exams, transfer the hypotenuse length to the number line for surd construction.

The middle number is the average, so (\frac{\frac{2}{7}+\frac{3}{7}}{2}=\frac{5}{14}). In exams, find the mean for the midway value.

The distance between two points on a number line is the absolute value of the difference of their coordinates. Thus, \(MN=|-1.8-(-6.2)|=|-1.8+6.2|=4.4\). Therefore, 4.4 is correct. A value such as 3.4 can result from handling the negative signs incorrectly. Exam tip: distance or length is always positive.

(\sqrt{18}\approx4.24), so the nearest integer is (4). In exams, first locate it between consecutive squares.

The number line represents real numbers, and each real number has a unique position. In exams, connect the number line with real numbers.

The equidistant point is the midpoint, so (\frac{-3+4}{2}=\frac{1}{2}). In exams, use the average for equidistant points.

(\sqrt{\frac{1}{2}}) lies between (0) and (1) and cannot be simplified to a rational form. In exams, identify square roots of non-perfect squares as irrational.

A positive number is greater than all negative numbers, so (0.003) is the greatest. In exams, the number to the right is greater.

In (\frac{5}{8}), the denominator is (8), so divide the unit into (8) equal parts and take the (5)th. In exams, the denominator gives the number of equal parts.

(\sqrt{80}\approx8.94), so it lies slightly left of (9). In exams, connect approximation with direction near the nearest integer.

The approximate values are (-2.25), (-2.2), (-2.236), and (-2.3); from the right, (-2.2) comes first and (-\sqrt{5}) second. In exams, order negative decimals using approximation.

On the number line, the greater number lies to the right, so (b) is to the right of (a). In exams, connect inequalities with direction.

\(\sqrt{7}\) is irrational because 7 is not a perfect square. Its decimal expansion is non-terminating and non-repeating, but it is a real number, so it has a unique point on the number line. Exam tip: the square root of a non-perfect square is irrational.

Since \(3^2=9\) and \(4^2=16\), and \(9<10<16\), we get \(3<\sqrt{10}<4\). Hence, \(\sqrt{10}\) is the correct option. \(\sqrt{8}<3\), while both \(\sqrt{17}\) and \(\sqrt{20}\) are greater than 4. Exam tip: locate a square root by comparing the number inside it with nearby perfect squares.