Representing real numbers on the number line

(5) is positive, so it lies (5) units to the right of (0). In exams, remember right direction for positive numbers.

(-6) is negative, so it lies (6) units to the left of (0). In exams, choose the left direction for negative numbers.

(\frac{2}{5}) is greater than (0) and less than (1). In exams, place a proper fraction between (0) and (1).

(-\frac{3}{4}) is negative and greater than (-1). In exams, show such fractions between (-1) and (0).

(1.25) is greater than (1) and less than (2). In exams, compare a decimal with nearby integers.

(-2.5) is greater than (-3) and less than (-2). In exams, be careful because order with negative decimals can feel reversed.

(0) is treated as the origin of the number line. In exams, place positive numbers to the right and negative numbers to the left.

(\frac{1}{3}) has the smallest distance from (0). In exams, check distance for closeness and not only the sign.

The direct answer is option B: 3 units. Distance on a number line means the positive separation between two points, so use the absolute difference: \(|10-7|=|3|=3\). Since 10 is to the right of 7, simply subtract the smaller number from the larger number. Option B is correct. Option A, 2 units, would be the distance between numbers differing by 2, not these two points. Option C, 7 units, is just one coordinate and does not measure the gap between the points. Option D, 10 units, is the other coordinate, not the separation. Distance cannot be negative, so the absolute-value idea is important if the order is reversed: \(|7-10|\) is also 3. On the number line, count the three unit intervals from 7 to 10. Memory cue: distance between two numbers is \(|a-b|\), or larger minus smaller.

The distance between two numbers is the absolute value of their difference: |−1 − (−5)| = |4| = 4 units. Therefore, option C is correct. Choosing 6 units incorrectly adds the magnitudes directly instead of finding the difference between the positions. In an exam, remember that distance is always positive.

The midpoint of two numbers is their average: \(\frac{2+6}{2}=\frac{8}{2}=4\). Therefore, 4 is correct. The numbers 3 and 5 are not equidistant from both given points, while 8 does not lie between them. In an exam, use the midpoint formula \(\frac{x_1+x_2}{2}\).

The direct answer is option B: 3. The symbol \(\sqrt{9}\) means the principal, non-negative square root of 9. Since \(3\times3=9\), we have \(\sqrt9=3\), so the point is located at 3 on the number line. Option B is correct. Option A, 2, is wrong because \(2^2=4\), not 9. Option C, 9, is the number inside the root, not its square root; \(9^2=81\). Option D, -3, is also a square root of the equation \(x^2=9\), but it is not the value of the principal symbol \(\sqrt9\). This distinction matters: solving \(x^2=9\) gives \(x=\pm3\), whereas evaluating \(\sqrt9\) gives only 3. On the number line, mark the point three units to the right of zero. Memory cue: the radical sign gives the non-negative root unless a separate equation asks for all roots.

(\sqrt{16}=4), so (-\sqrt{16}=-4). In exams, apply the outside negative sign after finding the square root.

Since (4^2=16) and (5^2=25), (\sqrt{20}) lies between (4) and (5). In exams, find nearby perfect squares.

Since (\sqrt{7}) lies between (2) and (3), (-\sqrt{7}) lies between (-3) and (-2). In exams, note the direction of a negative square root.

(\frac{7}{2}=3.5), so it lies between (3) and (4). In exams, convert an improper fraction into a decimal or mixed number.

(-\frac{9}{2}=-4.5), so it lies between (-5) and (-4). In exams, place a negative mixed number in the correct interval.

(1+\frac{2}{5}=1.4), so it is two parts after (1). In exams, take each equal part as (\frac{1}{5}).

Since \(0.6\) has one digit after the decimal point, it can be written as \(\frac{6}{10}\). Reducing numerator and denominator by 2 gives \(\frac{3}{5}\), so option A is correct. Options B, C and D represent approximately 1.67, 6 and 0.167 respectively. Exam tip: use 10, 100 or 1000 as the denominator according to the number of decimal places, then simplify the fraction.

(\sqrt{10}) is greater than (3) because (3^2=9) and (10) is larger. In exams, check square root positions using squares.

Since (1^2=1) and (2^2=4), (\sqrt{3}) lies between (1) and (2). In exams, bracket square roots using perfect squares.

The direct answer is option B: at -3. A zero of a polynomial is an input value that makes its value equal to zero. Set \(p(x)=x+3\) equal to zero: \(x+3=0\). Subtract 3 from both sides, giving \(x=-3\). Therefore the zero is represented by the point -3 on the number line. Option B is correct. Option A, 3, is not the zero because \(p(3)=3+3=6\), not 0. Option C, 0, gives \(p(0)=3\), so it is not a zero. Option D, 1, gives \(p(1)=4\), also not zero. The number -3 works because \(p(-3)=-3+3=0\). On a horizontal number line, -3 lies three units to the left of zero. Memory cue: to find a linear polynomial's zero, put the polynomial equal to zero and solve; the zero is the x-coordinate, not the output value.

The zero of a polynomial is the value of \(x\) that makes the polynomial equal to zero. Thus, putting \(q(x)=0\), we get \(x-8=0\), so \(x=8\). Therefore, the zero is represented by the point 8 on the number line. Exam tip: the zero of a polynomial of the form \(x-a\) is \(a\), not \(-a\).

The zero (-5) is negative, so it lies to the left of (0). In exams, the sign of the zero tells its direction.

The zero of a polynomial is the value of \(x\) for which \(p(x)=0\). Thus, \(2x-6=0\Rightarrow 2x=6\Rightarrow x=3\). Therefore, the correct point on the number line is 3. The value 6 is the constant term, not the zero. Exam tip: Set a linear polynomial equal to zero and solve for \(x\) to find its zero.