Representing real numbers on the number line

Correct answer: A. ( \frac{7}{11} )

For a nonnegative fraction, the principal square root can be found by taking the positive square root of the numerator and denominator separately. Since \(49=7^2\) and \(121=11^2\), we have \(\sqrt{49/121}=\sqrt{49}/\sqrt{121}=7/11\). Thus the point is represented by \(7/11\), which is option A.

The positive sign matters because the symbol \(\sqrt{\,}\) denotes the principal, nonnegative square root. The value \(49/11\) forgets to take the square root of the denominator, while \(11/7\) reverses the fraction. Also, \(7/121\) is not the result of taking either required square root. Therefore option A follows.

Correct answer: A. 1.99

Since \(\sqrt{3}\approx1.732\), we get \(\sqrt{3}+\frac{1}{4}\approx1.982\). Thus, \(1.99>1.982\) and \(1.99<2\), so 1.99 is correct. The values 1.95 and 1.90 are smaller than the expression, while 2.10 is greater than 2. Exam tip: First find a suitable decimal approximation of the irrational expression, then compare it with the given values.

Correct answer: A. C

Since \(\sqrt{28}\approx 5.292\), we get \(A\approx -5.292\). Also, \(B=-5.3\) and \(C=-\frac{16}{3}\approx -5.333\). Among negative numbers, the smaller number lies farther to the left on the number line. Thus, \(-5.333<-5.3<-5.292\), so \(C\) is farthest left. Exam tip: When comparing negative numbers, the one with the greater magnitude is the smaller number.

Correct answer: A. \(3-\sqrt{2}<\frac{8}{5}\)

To compare the two numbers, rewrite \(3-\sqrt{2}<\frac{8}{5}\) as \(\sqrt{2}>\frac{7}{5}\). Both sides are positive, and \(2=\frac{50}{25}>\frac{49}{25}=\left(\frac{7}{5}\right)^2\), so \(\sqrt{2}>\frac{7}{5}\). Hence \(3-\sqrt{2}<\frac{8}{5}\). Their approximate values, \(1.586\) and \(1.6\), confirm this; therefore option B is incorrect. Exam tip: check that both sides are non-negative before squaring an inequality.

Correct answer: A. (14.5)

( \sqrt{2}\approx1.414 ), so ( \frac{a}{10} ) must be between (1.414) and (1.5). (a=14.5) gives (1.45).

Correct answer: A. 4√5

Since 80 = 16 × 5, we have √80 = √16 × √5 = 4√5. This is the simplest form because 5 has no perfect-square factor greater than 1. Its approximate value is 4 × 2.236 ≈ 8.94, so it lies between 8 and 9 on the number line.

Correct answer: A. ( -1.42,-\sqrt{2},-\frac{7}{5} )

Since ( -1.42<-1.414<-1.4 ), the order is ( -1.42,-\sqrt{2},-\frac{7}{5} ). For negative decimals the smaller value comes first.

Correct answer: A. 28

Since \\(5.2<x<5.3\\) and both bounds are positive, squaring preserves the inequality: \\(5.2^2<a<5.3^2\\). Thus, \\(27.04<a<28.09\\). Among the given options, only 28 lies in this interval; 26 and 25 are too small, while 30 is too large. Exam tip: when both sides of an inequality are positive, squaring does not reverse its order.

Correct answer: A. (0.41)

( \sqrt{7}\approx2.646 ) and ( \sqrt{5}\approx2.236 ), so the difference is about (0.41). The difference of nearby roots is small.

Correct answer: A. ( -1.8 )

( -\sqrt{3}\approx-1.732 ), and (-1.8) lies between it and (-2). Check left-to-right order in negative intervals.

Correct answer: A. 11

Since \(\sqrt{49}=7\) and \(\sqrt{16}=4\), we get \(\sqrt{49}+\sqrt{16}=7+4=11\). Therefore, the coordinate of point \(P\) on the number line is \(11\). The value 65 is the product \(7\times 4\), not the required sum. Exam tip: evaluate each square root first and then perform the indicated operation.

Correct answer: A. \(A\) and \(C\)

\(C=-\frac{25}{4}=-(25\div 4)=-6.25\), so \(A=C\) and both points represent the same position on the number line. \(B=-\sqrt{39}\) is different because \(\sqrt{39}\neq 6.25\); in fact, \(6.25^2=39.0625\), not \(39\). Exam tip: Convert a fraction to decimal form when comparing it with a terminating decimal.

Correct answer: B. (10)

( \sqrt{90}\approx9.49 ), which is closer to (10) than to (9). Check distance for the nearest integer.

Correct answer: A. (2\sqrt{13})

Adding like radicals gives ( \sqrt{13}+\sqrt{13}=2\sqrt{13} ). Do not add the numbers inside the radicals directly.

Correct answer: A. 15.21

Given \(\sqrt{x}=3.9\), squaring both sides gives \(x=(3.9)^2=3.9\times3.9=15.21\). Hence, \(x\) represents the point 15.21 on the number line. The values 7.8, 12.21, and 3.09 do not result from correctly squaring 3.9. Exam tip: when squaring a decimal, multiply the numbers as whole numbers first and then place the decimal point according to the total number of decimal places.

Correct answer: A. 2.4

\(\frac{9}{4}=2.25\) and \(\sqrt{6}\approx 2.449\). Thus, 2.4 lies between 2.25 and 2.449. Option B, 2.2, is less than \(\frac{9}{4}\), while 2.5 and 3.0 are greater than \(\sqrt{6}\). Exam tip: Convert the fraction and square root to suitable decimal approximations before comparing them.

Correct answer: A. Between ( -3 ) and ( -2 )

( -\sqrt{2}-1\approx-2.414 ), so it lies between (-3) and (-2). Estimate negative sums carefully.

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

( \sqrt{48}=4\sqrt{3} ) and ( \sqrt{27}=3\sqrt{3} ), so the difference is ( \sqrt{3} ). Simplify the radicals first.

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

( \sqrt{91}\approx9.54 ) and ( \sqrt{55}\approx7.42 ), so the difference is about (2.12). Estimate both roots first.

Correct answer: A. (1.25) and (-6.25)

( |x+2.5|=3.75 ) means the distance of (x) from (-2.5) is (3.75). Moving both ways gives (1.25) and (-6.25).

Correct answer: A. ( \sqrt{13},3.61,\frac{29}{8} )

( \sqrt{13}\approx3.606 ), (3.61), and ( \frac{29}{8}=3.625 ). Compare close values to more decimal places.

Correct answer: B. (Q)

( -\sqrt{74}\approx-8.602 ) and (-8.55) is greater. The greater number lies to the right on a number line.

Correct answer: C. 2.4

The principal square root is non-negative, and 2.4 × 2.4 = 5.76. Therefore, √5.76 = 2.4, so mark the point 2.4 on the number line.

Correct answer: A. \( \frac{47}{12} \)

The distance is \( \left|\frac{7}{6}-\left(-\frac{11}{4}\right)\right|=\frac{47}{12} \). Use absolute value while finding distance.

Correct answer: B. (5.75)

( \sqrt{33}\approx5.745 ) and ( \sqrt{34}\approx5.831 ), so (5.75) lies between them. Keep estimates accurate for close roots.