Representing real numbers on the number line

Correct answer: A. (5) units to the right

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

Correct answer: B. (6) units to the left of (0)

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

Correct answer: A. (0) and (1)

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

Correct answer: B. Between (-1) and (0)

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

Correct answer: B. (1) and (2)

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

Correct answer: B. (-3) and (-2)

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

Correct answer: A. Origin

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

Correct answer: B. (\frac{1}{3})

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

Correct answer: B. (3) units

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.

Correct answer: C. 4 units

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.

Correct answer: B. 4

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 \(\frac{x_1+x_2}{2}\).

Correct answer: B. (3)

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.

Correct answer: C. (-4)

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

Correct answer: C. (4) and (5)

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

Correct answer: B. (-3) and (-2)

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.

Correct answer: B. (3) and (4)

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

Correct answer: B. (-5) and (-4)

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

Correct answer: B. (1.4)

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

Correct answer: A. \(\frac{3}{5}\)

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.

Correct answer: D. (\sqrt{10})

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

Correct answer: A. (\sqrt{3})

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

Correct answer: B. At (-3)

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.

Correct answer: C. 8

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 \(-a\).

Correct answer: B. To the left of (0)

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

Correct answer: B. 3

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.