Representing real numbers on the number line

(\sqrt{3}) lies between (1) and (2), so (2+\sqrt{3}) lies between (3) and (4). In exams, adding a number shifts the whole interval forward.

(\sqrt{2}) lies between (1) and (2), so (5-\sqrt{2}) lies between (3) and (4). In exams, change limits carefully while subtracting.

(\frac{13}{4}=3+\frac{1}{4}), so it lies one-fourth after (3). In exams, convert an improper fraction into a mixed number.

(-\frac{17}{5}=-3-\frac{2}{5}), so it lies between (-4) and (-3). In exams, keep the sign of a negative mixed number correct.

The denominator of (\frac{5}{6}) is (6), so the segment from (0) to (1) has (6) equal parts. In exams, the denominator tells the number of parts.

(1+\frac{3}{8}) is the third of (8) equal parts after (1). In exams, count the fractional parts after the integer.

The distance is \(\frac{5}{2}-\left(-\frac{3}{2}\right)=4\) units. In exams, always take distance as positive.

The distance is \(1.25-\left(-2.75\right)=4\) units. In exams, do not miss signs while subtracting negative decimals.

The midpoint is (\frac{-\frac{7}{3}+\frac{2}{3}}{2}=-\frac{5}{6}). In exams, first add and then divide by (2).

The midpoint of two numbers is their average. Thus, \(\frac{2.4+5.6}{2}=\frac{8}{2}=4\), so option B is correct. Options A and C do not give the correct average, while 8 is the sum of the two numbers, not their midpoint. In an exam, add the two numbers and divide the result by 2 to find the midpoint.

From (2x+5=0), we get (x=-\frac{5}{2}=-2.5). In exams, find the zero and place it like a real number.

From (3x-7=0), (x=\frac{7}{3}), which lies between (2) and (3). In exams, first find the zero and then locate it.

From (4x+6=0), (x=-\frac{3}{2}), which lies between (-2) and (-1). In exams, identify the interval of a negative fraction carefully.

To find the zeros, set \(s(x)=0\): \(x^2-16=0\). Factoring gives \((x-4)(x+4)=0\), so \(x=4\) or \(x=-4\). Therefore, the corresponding points on the number line are \(-4\) and \(4\). Exam tip: for an equation of the form \(x^2=a^2\), check both values \(x=\pm a\).

From (x^2-7=0), (x=\pm\sqrt{7}). In exams, take both positive and negative square roots in a square equation.

(x^2-2x=x(x-2)), so the zeros are (0) and (2). In exams, taking a common factor is an easy method.

(\sqrt{6}) is about (2.45), so (2.4) is closest. In exams, think of (2.4^2) and (2.5^2) for estimation.

(\sqrt{11}) is about (3.32), so (3.4) is to its right. In exams, a point to the right is greater.

(\sqrt{14}) is about (3.74), so (3.5) is greater than (3) and less than (\sqrt{14}). In exams, make a rough estimate of the square root.

The governing concept is that a point lies to the right of another point when its numerical value is greater. Since √5 is between 2 and 2.25 (because 2² = 4 and 2.25² = 5.0625), −√5 is approximately −2.236. Comparing the choices with −2.236, −2 is greater and therefore lies to its right. Options −3 and −4 are smaller, so they lie to the left. The value −2.5 is also smaller than approximately −2.236 and is therefore to the left. Thus option C is correct. With negative numbers, the number closer to zero is greater, which is the key point that prevents reversing the order.