Representing real numbers on the number line

Moving right gives ( -\frac{3}{4}+\frac{5}{6}=\frac{1}{12}). Use a common denominator to add fractions.

For ( -\sqrt{10}<-\sqrt{m}), we need ( \sqrt{10}>\sqrt{m}), so (m<10). Inequality reverses with negative roots.

( \frac{5}{3}\approx1.667), (1.7=1.7), and ( \sqrt{3}\approx1.732). Convert mixed forms to decimals for ordering.

The point has coordinate \\( -\sqrt{48} \\). To locate it, compare 48 with nearby perfect squares: \\(6^2=36\\) and \\(7^2=49\\). Therefore \\(6<\sqrt{48}<7\\). Taking the negative changes the order on the number line, so \\(-7<-\sqrt{48}<-6\\). Thus the point lies between the consecutive integers \\(-7\\) and \\(-6\\), which is option A. A common mistake is to choose 6 and 7 because the square root itself lies there, without considering the negative sign.

The value is approximately \\(-6.93\\), confirming the interval. The negative of a number between 6 and 7 must lie between -7 and -6, not between -6 and -5. The endpoint values are not included because \\(48\\) is not a perfect square, so the inequalities are strict. Hence the supplied answer A follows directly from comparing perfect squares and reversing the order correctly for negative numbers.

( \sqrt{5}\approx2.236), so (2-\sqrt{5}\approx-0.236). Estimation is fastest for subtraction with roots.

( \sqrt{\frac{25}{16}}=\frac{5}{4}) because the principal square root is positive. Take the root of both numerator and denominator.

( \sqrt{6}\approx2.449) and ( \sqrt{7}\approx2.646), so (2.5) lies between them. Estimate carefully for close square roots.

The governing concept is that the distance between two points on a number line is the absolute value of the difference of their coordinates. Therefore AB=|1.2-(-3.8)|=|5.0|=5 units. Similarly, AC=|1.2-4.2|=|-3.0|=3 units. The question asks how much greater AB is than AC, so calculate 5-3=2 units. Hence option A is correct. The absolute value is essential because distance cannot be negative, even though the directed difference may be negative. Option B does not equal the difference of the two distances, while options C and D are not supported by the calculated values.

( \sqrt{1.44}=1.2) and ( \sqrt{1.69}=1.3), so (1.25) lies between them. Knowing decimal squares is useful.

( \pi\approx3.14159), ( \frac{19}{6}\approx3.1667), and ( \sqrt{10}\approx3.1623), so exact checking is needed. Estimate each value accurately while comparing.

Since ( \sqrt{5}\approx2.236), ( \sqrt{5}+\frac{1}{2}\approx2.736). Use estimation to identify the interval quickly.

( \sqrt{14}\approx3.742), so (3.8) lies between it and (4). Decimal estimation is important for nearby square roots.

( -\frac{11}{3}\approx-3.667) and ( -\sqrt{13}\approx-3.606), so ( -\frac{11}{3}) is smaller. On a number line, the smaller number lies farther left.

( \sqrt{31}\approx5.57 ) and ( \sqrt{12}\approx3.46 ) so the difference is about (2.11). Estimate both roots first.

( |x+3|=2.5 ) means the distance of (x) from (-3) is (2.5). Moving both ways gives (-0.5) and (-5.5).

The governing concept is comparing rational and irrational numbers by using exact values or sufficiently accurate decimal approximations. First, 13/4=3.25 exactly. Also, √11 is approximately 3.3166; this is reasonable because 3.3166² is close to 11 and √11 lies between √9=3 and √16=4. The three values are therefore 3.25, 3.28, and approximately 3.3166. Since 3.25<3.28<3.3166, the increasing order is 13/4, 3.28, √11. Hence option A is correct. Option B reverses the first two values, C makes √11 the smallest, and D places √11 before 3.28, both contrary to the approximations.

( -\sqrt{20}\approx-4.472 ) and (-4.4) is greater than it. The greater number lies to the right on a number line.

Since (1.7)^2=2.89), the principal square root is (\sqrt{2.89}=1.7). Therefore, it is marked at the point with coordinate 1.7. The squares of 1.6 and 1.8 are 2.56 and 3.24, respectively, so they are not correct. In an exam, check whether the decimal is a perfect square to find such roots quickly.

The distance is \( \left|\frac{5}{6}-\left(-\frac{7}{3}\right)\right|=\frac{19}{6} \). Always use absolute value for distance.

\(\sqrt{18}\approx 4.243\) and \(\sqrt{19}\approx 4.359\). Since \(4.243<4.3<4.359\), the value 4.3 lies between the two square roots. The value 4.1 is less than \(\sqrt{18}\), while 4.5 is greater than \(\sqrt{19}\). In the exam, estimate the decimal values of the relevant square roots to compare them quickly.