Representing real numbers on the number line

( \sqrt{10}\approx3.162 ), so (2-\sqrt{10}\approx-1.162). Estimation is important in root subtraction.

Since \(7^2=49<57<64=8^2\), we have \(7<\sqrt{57}<8\). Comparing this with \(m<\sqrt{57}<m+1\) gives \(m=7\). Option 6 is too small because \(6^2=36<57\), while 8 is an upper bound, not the required integer. Exam tip: locate a square root between two consecutive perfect squares to find its integer part.

\(\sqrt{0.36}=0.6\) because \(0.6^2=0.36\). The negative sign outside the radical makes the entire value negative, so \(-\sqrt{0.36}=-0.6\). Option B ignores the outside negative sign, while C and D have incorrect decimal values. Exam tip: first evaluate the principal square root, then apply any sign outside the radical.

Since \(\sqrt{81}=9\), we get \(x=\frac{9}{4}\). Option B incorrectly uses 81 instead of its square root, while options C and D do not result from the given expression. In the exam, simplify the square root first and then perform the division.

( \sqrt{5}\approx2.236 ), which is slightly less than (2.24). Compare close decimals to more places.

Moving left on the number line means subtracting the given distance from the starting coordinate. Hence, \(C=-1-\sqrt{6}\). Option B adds the distance and therefore represents a point to the right of \(-1\). Exam tip: subtract distances when moving left and add them when moving right.

The point exactly midway between two numbers is their arithmetic mean. For \(\frac25\) and \(\frac35\), add them and divide by 2: \(\frac{\frac25+\frac35}{2}=\frac{\frac55}{2}=\frac12\). Therefore the midpoint is \(\frac12\), so option A is correct.

This also makes sense geometrically. The two endpoints are 0.4 and 0.6 on the number line, and 0.5 is equally distant from each: the distance from 0.4 to 0.5 is 0.1, as is the distance from 0.5 to 0.6. The values \(\frac52\), \(\frac15\), and \(\frac45\) do not have equal distances from both endpoints.

The key concept is comparing negative real numbers in their actual number-line order. Since √3≈1.732, we have -√3≈-1.732. Also, -5/3=-1.6666... . A number between these endpoints must be greater than -1.732 and less than -1.6666..., because the more negative number is farther to the left. Among the choices, -1.70 satisfies -1.732<-1.70<-1.6666..., so it lies between the two given values. Therefore option A is correct. The value -1.60 is to the right of the interval, while -1.80 and -2.00 are more negative and lie to its left. Treating the negative values as positive magnitudes would reverse the comparison incorrectly.

( \sqrt{27}=3\sqrt{3} ) and ( \sqrt{12}=2\sqrt{3} ), so the difference is ( \sqrt{3} ). Subtract like radicals.

The decimal expansion is non-terminating and non-repeating because the number of 7s after each 2 keeps increasing. Therefore, it cannot be expressed in the form p/q and is irrational. A rational number has a terminating or recurring decimal expansion, so option B is incorrect. Exam tip: a non-terminating, non-repeating decimal is irrational.

Since \(6.4<\sqrt{n}<6.5\), squaring both sides gives \(6.4^2<n<6.5^2\). Therefore, \(40.96<n<42.25\). Among the given choices, only 42 lies in this interval. The values 40, 45 and 49 fall outside it. Exam tip: for positive quantities, squaring preserves the direction of the inequality.

\(-\frac{17}{5}=-3.4\). On the number line, \(-3.4\) lies to the right of \(-4\) and to the left of \(-3\), so it lies between \(-4\) and \(-3\). Exam tip: for negative numbers, the number with the greater magnitude is farther to the left on the number line.

The midpoint of two points \(x_1\) and \(x_2\) is \(\frac{x_1+x_2}{2}\). Therefore, \(\frac{-2.75+1.25}{2}=\frac{-1.50}{2}=-0.75\), so option A is correct. Option B is only the sum of the two coordinates, not their midpoint. Exam tip: always divide the sum of the coordinates by 2 to find a midpoint.

Since \(0.09\times0.09=0.0081\), we get \(\sqrt{0.0081}=0.09\). The principal square root is taken as non-negative, so the point is represented at \(0.09\) on the number line. The closest distractor, \(0.9\), gives \(0.9^2=0.81\), not \(0.0081\). In the exam, verify the decimal places by squaring the chosen value.

Since \(6^2=36<37<49=7^2\), we get \(6<\sqrt{37}<7\). Adding 1 to all parts gives \(7<\sqrt{37}+1<8\), so the expression lies between 7 and 8. Exam tip: To bound a square root, compare the number with the consecutive perfect squares around it. Option B is incorrect because it represents the interval for \(\sqrt{37}\), not for \(\sqrt{37}+1\).

Since \(6<\sqrt{45}<7\), multiplying by \(-1\) reverses the inequalities: \(-7< -\sqrt{45}< -6\). Adding 2 throughout gives \(-5<p<-4\). Therefore, \(p\) lies between \(-5\) and \(-4\). Option B is incorrect because \(p\) is less than \(-4\). Exam tip: Compare a square root with neighbouring perfect squares instead of calculating its decimal value.

Movement to the left on a number line means subtracting the stated distance from the starting coordinate. The starting point is 7/10 and the movement is 3/20 unit, so calculate 7/10 − 3/20. Convert 7/10 to twentieths: 7/10 = 14/20. Then 14/20 − 3/20 = 11/20. Therefore the required point is 11/20, making option A correct. Option B adds the distance and therefore represents movement to the right. Option C is 10/20 and does not result from the required subtraction, while option D uses an incorrect denominator conversion. The direction word “left” determines the operation.

The governing concept is estimating the position of an irrational square root by comparing nearby perfect squares and testing decimal approximations. Since 7² = 49 and 8² = 64, √63 lies between 7 and 8 and is very close to 8. Squaring the most plausible option gives 7.94² = 63.0436, whose difference from 63 is only 0.0436. In fact, √63 is approximately 7.937, so 7.94 is the closest listed decimal and therefore option A is correct. Option B is much farther below the root, option C is greater than 8 and hence cannot be close, and option D is well below the interval from 7 to 8.

\(\sqrt{8}=2\sqrt{2}\approx2.828\), \(\frac{14}{5}=2.8\), and \(c=2.81\). Thus, \(2.8<2.81<2.828\), so \(a\) is the greatest number. The closest distractor is \(c\), but \(2.828>2.81\). In an exam, converting radicals and fractions to comparable decimal forms is a quick way to compare them.

Since \(4.8^2=23.04\) and \(4.9^2=24.01\), and \(23.04<24<24.01\), it follows that \(4.8<\sqrt{24}<4.9\). Options B and D give intervals that are too small, while both endpoints in option C have squares greater than 24. Exam tip: To verify a decimal bound for a square root, compare the squares of its endpoints with the given number.