Representing real numbers on the number line

The governing concept is that a zero of a polynomial or function is an x-value for which its output is exactly zero. Set r(x)=0: 3x+12=0. Subtracting 12 from both sides gives 3x=−12, and dividing by 3 gives x=−4. Thus the zero is located at the point −4 on the number line, so option B is correct. Substitution verifies the result: r(−4)=3(−4)+12=−12+12=0. Option A results from losing the negative sign while dividing, whereas options C and D select the constant term or its negative without solving the linear equation. The zero is an input value, not simply a coefficient or constant appearing in the expression.

\(x=0\) represents the origin on the number line. Zero is neither positive nor negative; positive numbers lie to the right of \(0\), while negative numbers lie to its left. Exam tip: Never classify \(0\) as positive or negative.

Both (-1) and (1) are (1) unit away from (0). In exams, recognize opposite signs with equal distance.

The opposite point of (4) is (-4), at the same distance from (0) on the other side. In exams, only the sign changes for the opposite number.

The opposite point of (-7) is (7) because both are equally distant from (0). In exams, change the sign to identify the opposite number.

(\frac{5}{4}=1+\frac{1}{4}), so it is one-fourth after (1). In exams, convert an improper fraction into mixed form.

Since \(\frac{3}{10}=0.3\), we get \(2+\frac{3}{10}=2+0.3=2.3\). Option A, \(2.03\), represents \(2+\frac{3}{100}\), so it is incorrect. Exam tip: When the denominator is 10, the decimal form has one digit after the decimal point.

Numbers increase to the right on the number line, and (-1>-2). In exams, connect the right direction with the greater number.

Numbers decrease to the left on the number line, and (2.9<3). In exams, connect the left direction with the smaller number.

The governing concept is distance on a number line, measured by the absolute difference from the reference point. For 1/10, the distance from 1 is |1−1/10|=|9/10|=9/10. For 9/10, the distance is |1−9/10|=1/10. Since 1/10 is less than 9/10, the point 9/10 is closer to 1. Therefore option B is correct. Option A reverses the two calculated distances. Option C would be valid only if the distances were equal, which they are not, and option D is impossible because both fractions denote definite real points on the number line. The comparison can also be seen geometrically: 9/10 lies only one tenth to the left of 1, whereas 1/10 lies nine tenths away.

(-0.2) is negative and greater than (-1). In exams, place small negative decimals between (-1) and (0).

The number 0.05 is positive, so it is greater than 0. It is also less than 1; therefore, it lies between 0 and 1. Option D is incorrect because 0.05 must not be confused with 5. Exam tip: First look at the integer part of a decimal to identify the interval between consecutive integers.

The distance between two points on a number line is the absolute value of the difference of their coordinates. Thus, distance = |1 − (−3)| = |4| = 4 units. The answer is not 2 units because the points lie four unit intervals apart. In exams, use brackets when subtracting a negative number.

The distance between two points on a number line is the absolute difference of their coordinates. Thus, the distance is \(|4-2.5|=1.5\) units. Option C is only the greater coordinate, while option D is the sum of the coordinates. In an exam, use the absolute difference for distance so that the answer is positive.

For (p(x)=x), putting (x=0) gives (p(x)=0). In exams, a zero is the value where the polynomial becomes (0).

The zero of a polynomial is the value of \(x\) for which the polynomial becomes zero. Thus, \(5x-10=0\), so \(5x=10\) and \(x=2\). Therefore, the correct point on the number line is 2. The value 5 is only the coefficient of \(x\), not the zero. In an exam, first set the polynomial equal to zero to find its zero.

An irrational number has a non-terminating, non-repeating decimal expansion, but it is still a real number and has one fixed point on the number line. For example, \(\sqrt{2}\) lies between 1 and 2. Exam tip: every real number corresponds to a point on the number line.

The direct answer is option B: \(2.5\). Starting at zero and moving 2.5 units to the right means adding 2.5: \(0+2.5=2.5\). On a number line, moving right increases the value, while moving left decreases it. Option B is correct because the point \(2.5\) is exactly the required distance and direction. Option A, \(-2.5\), is 2.5 units from zero but lies to the left, so it has the wrong direction. Option C, \(0.25\), is to the right but is only 0.25 unit away, not 2.5 units. Option D, \(5\), is to the right but is 5 units away. Exam cue: “right” means positive direction and the stated number gives the distance.

Moving \(1.75\) units to the left of \(0\) means subtract \(1.75\) from \(0\): \[0-1.75=-1.75\] So the point is \(-1.75\).

Since (8>2), (\sqrt{8}>\sqrt{2}). In exams, compare positive square roots using the numbers inside them.

Since (\sqrt{12}>\sqrt{3}), (-\sqrt{3}) is greater and lies to the right. In exams, negative signs reverse the comparison effect for square roots.

Since (3^2=9) and (4^2=16), (\sqrt{15}) lies between (3) and (4). In exams, use nearby perfect squares to locate square roots.

The direct answer is option B, the point 3. A zero of a polynomial is the value of x that makes its value equal to zero. Here solve step by step: \(u(x)=6x-18\). Put the expression equal to zero: \(6x-18=0\). Add 18 to both sides: \(6x=18\). Divide by 6: \(x=3\). Therefore the zero is located at 3 on the number line. Option A, 2, gives \(6(2)-18=-6\), not zero. Option B, 3, gives \(18-18=0\), so it is correct. Option C, 6, gives \(36-18=18\), not zero. Option D, 18, gives a still larger nonzero value. Remember: to find a zero, set the polynomial equal to zero, not just one of its terms.

The governing concept is that real-number values increase from left to right on a number line. Convert the fraction to a decimal for direct comparison: 1/2=0.5. The three values are therefore 0.4, 0.5, and 0.6, with 0.4<0.5<0.6. Replacing 0.5 by its equivalent fraction gives the left-to-right order 0.4, 1/2, 0.6, so option A is correct. Option B places 1/2 before the smaller value 0.4, option C reverses the order, and option D places 0.6 before 1/2 even though 0.6 is larger. Exact conversion avoids confusion caused by mixing decimal and fractional notation.

The governing concept is distance on a number line: the distance between a number x and 3 is |x − 3|. For 3.04, the distance is |3.04 − 3| = 0.04. For 3.4, it is |3.4 − 3| = 0.4. Because 0.04 is smaller than 0.4, the point 3.04 lies closer to 3. Equivalently, 3.04 is only four hundredths to the right of 3, whereas 3.4 is four tenths to the right. Therefore option A is correct. Option B confuses the larger decimal displacement with greater closeness, while option C is false because the distances are not equal. Option D is also false because both points have definite positions.