Representing real numbers on the number line

Since (-\sqrt{3}) is about (-1.73), it is left of (-1). In exams, order increases from left to right.

The governing concept is that the distance between two points on a number line is the absolute difference of their coordinates. Here Q is to the right of P because 5/6 > 2/3. Use a common denominator: 2/3 = 4/6. Therefore the distance is |5/6 − 4/6| = |1/6| = 1/6 unit. Option A is correct. Option B would result from an incorrect denominator operation, option C treats the numerators or denominators separately without equivalence, and option D adds the two coordinates instead of finding their separation. Because the points are ordered, ordinary subtraction gives a positive distance; absolute value would also give the same result.

The distance is \(\frac{1}{10}-\left(-\frac{4}{5}\right)=\frac{9}{10}\) unit. In exams, subtracting a negative fraction becomes addition.

The distance between two points on a number line is the absolute difference of their coordinates. Thus, the distance is \(3.75-1.20=2.55\) units. Writing 1.2 as 1.20 helps align the decimal places. The option 4.95 results from adding the coordinates, but distance here requires their difference. Exam tip: take the absolute value of the difference so that distance is positive.

Moving left decreases the number, so the point is (2-\frac{3}{4}). In exams, choose addition or subtraction according to direction.

Moving right gives (-3+\frac{5}{2}=-\frac{1}{2}). In exams, treat right movement as addition.

Moving left on the number line means subtracting. Therefore, \(x=1-2.5=-1.5\). Option C, \(1.5\), does not result from subtracting \(2.5\) from \(1\). Exam tip: move right by adding and move left by subtracting.

Since (5^2=25) and (6^2=36), (\sqrt{27}) lies between (5) and (6). In exams, remember nearby perfect squares.

Since (\sqrt{30}) lies between (5) and (6), (-\sqrt{30}) lies between (-6) and (-5). In exams, keep the negative direction in mind.

The governing concept is simplifying a square root by extracting perfect-square factors. Factor 50 as 25 × 2, where 25 is a perfect square. Using √(ab) = √a × √b, we get √50 = √(25 × 2) = √25 × √2 = 5√2. Therefore option A is correct. A numerical check gives √50 approximately 7.07, and 5√2 has the same approximate value. Although 2√5 is also approximately 4.47 and is not equal to √50, the other choices are much larger: 10√5 and 25√2 do not preserve the original value. The simplified radical should have no square factor remaining inside the root, which is why 5√2 is the preferred form.

(3\sqrt{2}) is about (4.24), so it lies between (4) and (5). In exams, you may estimate (\sqrt{2}) as about (1.414).

The governing concept is the order of real numbers: negative numbers lie to the left of zero and positive numbers lie to the right. Since √2 is positive, its negative, −√2, is negative and must lie to the left of 0. Thus option C is correct. Option A reverses the position of a negative number. Option B incorrectly places a positive number to the left of zero. Option D is false because √2 is irrational and approximately 1.414, not exactly 2; in fact, squaring 2 gives 4 rather than 2. The sign alone determines the location required in this question.

The governing concept is that the greater real number lies farther to the right on a number line. Compare the values exactly by squaring: 3.5²=12.25, and 12<12.25. Since both quantities are non-negative, taking square roots preserves the order, so √12<√12.25=3.5. Numerically, √12 is about 3.464, which confirms the result. Therefore b=3.5 is to the right and option B is correct. Option A is smaller, option C is false because the values are unequal, and option D is unnecessary because the comparison can be determined exactly.

(\frac{19}{6}=3+\frac{1}{6}), so it lies between (3) and (4). In exams, divide to identify the whole part.

(-\frac{22}{7}) is about (-3.14), so it lies between (-4) and (-3). In exams, note the order of negative decimals.

When the interval is divided into 10 equal parts, each part has a value of \\(\frac{1}{10}\\). Since \\(0.7=\frac{7}{10}\\), the seventh mark from 0 represents 0.7. The eighth mark represents \\(0.8\\), so it is incorrect. In exams, convert the decimal into a fraction with denominator 10, 100, and so on to identify the mark.

The exact midpoint of two numbers is their average: \(\frac{1.8+2.6}{2}=\frac{4.4}{2}=2.2\). Therefore, 2.2 is correct. The values 2.0 and 2.4 are not equally distant from both endpoints; for an exact midpoint, calculate the average of the two numbers.

There are two numbers at a distance of 1.5 units from 4 on the number line: \\(4-1.5=2.5\\) and \\(4+1.5=5.5\\). Since \\(x<4\\), the number to the left of 4 must be chosen; hence, \\(x=2.5\\). Exam tip: Always check the direction or inequality condition in distance questions.

Moving to the right on a number line means adding the given distance to the starting number. Thus, \\(y=-2+3.25=1.25\\). Option D results from subtracting the distance, which would represent moving to the left. Exam tip: add for a rightward move and subtract for a leftward move.

(-\frac{5}{4}=-1.25), which is to the left of (-1). In exams, convert negative fractions into decimals to check.