Representing real numbers on the number line

The governing property is that for a non-negative number a, if 0<a<1, then its principal square root also satisfies 0<\sqrt{a}<1. Here 0<0.49<1, so taking the non-negative square root gives 0<\sqrt{0.49}<1. In fact, \sqrt{0.49}=0.7, which directly confirms the statement. Therefore option A is correct. Option B contradicts the given value, option C confuses the square root with the number 49, and option D is impossible because a principal square root is never negative. The interval property, rather than an unsupported approximation, provides the justification.

From (5<\sqrt{n}<6), we get (25<n<36), so (30) is possible. Square positive sides in square-root inequalities.

The governing concept is bounding an irrational expression between consecutive integers. Because 1² < 2 < 2², taking positive square roots gives 1 < √2 < 2. Adding 3 to every part preserves the order, so 4 < 3 + √2 < 5. Thus the expression lies strictly between 4 and 5, making option A correct. A decimal check gives √2 ≈ 1.414 and therefore 3+√2 ≈ 4.414, which confirms the interval. Option B is incorrect because the value is greater than 4. Option C is incorrect because the value is less than 5, and option D places it even farther below the actual value. The exact inequality is sufficient; decimal approximation is only a confirmation.

Since (2<\sqrt{5}<3), (1<-1+\sqrt{5}<2). When adding or subtracting a constant, adjust the whole inequality.

The governing concept is simplification of a square root before performing division. Because 16 is a perfect square, its principal square root is √16 = 4, not ±4 in this context. Hence √16/2 = 4/2 = 2. The corresponding point on the number line is therefore 2, so option A is correct. Option B results from stopping after evaluating the square root and forgetting the division by 2. Option C incorrectly multiplies 4 by 2, while option D reverses the division and treats the expression as 2/4. The expression is positive, so no negative interpretation is involved.

(\sqrt{18}=3\sqrt{2}\approx4.242), so (4.24) is closest. Simplification makes estimation easier.

With common denominator (30), (\frac{18}{30}<\frac{20}{30}<\frac{25}{30}). Use a common denominator to order fractions.

(\sqrt{2}\approx1.414) and (\sqrt{3}\approx1.732), so the sum is about (3.146). Add approximate values for sums.

The length is \(PQ=\left|\frac{11}{2}-\frac{3}{2}\right|=4\). For distance, always take absolute difference.

Because (11<13<15), (\sqrt{11}<\sqrt{13}<\sqrt{15}). For positive square roots, the order of radicands is preserved.

(|x-2|=3) means (x) is (3) units away from (2), so (x=-1,5). Place the distance on both sides of the center.

Since (3<\sqrt{12}<4), (-4<-\sqrt{12}<-3). Multiplying by a negative reverses the inequality.

This decimal is non-terminating and non-repeating, so it is irrational and lies between (0) and (1). Identify rationality by the decimal pattern.

The principal square root is positive. Since \(3.1^2=3.1\times3.1=9.61\), it follows that \(\sqrt{9.61}=3.1\). Option B is incorrect because \(3.01^2=9.0601\), not \(9.61\). In the exam, verify a decimal square by multiplying the number by itself.

Both numbers are positive, so their squares can be compared. \(x^2=\left(\frac{17}{10}\right)^2=\frac{289}{100}=2.89\), whereas \(y^2=(\sqrt{3})^2=3\). Thus, \(x^2<y^2\) implies \(x<y\), so \(x\) lies to the left on the number line. The distractor \(y\) is incorrect because \(y\approx1.732\), while \(x=1.7\). Exam tip: For positive numbers, comparing their squares is often easier than comparing the numbers directly.

A negative number lies to the left of (0), and its distance is its absolute value (\frac{7}{4}). Identify direction and distance separately.

(\sqrt{25}=5) and (\sqrt{4}=2), so (\sqrt{25}-\sqrt{4}=3). Simplify perfect squares immediately.

(\sqrt{8}=2\sqrt{2}), so the midpoint is (\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}). Simplify first, then average.

The governing concept is order preservation when squaring positive numbers. Since both 2.6 and 2.7 are positive, the statement 2.6 < √7 < 2.7 is equivalent to 2.6² < 7 < 2.7². Calculate the endpoint squares: 2.6² = 6.76 and 2.7² = 7.29. Because 6.76 < 7 < 7.29, the required test is true, and √7 does lie in that interval; numerically it is about 2.646. Therefore option A is correct. Option B reverses the natural order of the endpoint squares. Option C says 7 is below the lower square, and option D says it is above the upper square; both contradict the actual inequalities and fail to establish that √7 lies between the given bounds.

For negative numbers, the one with larger magnitude is farther left, so (-\frac{5}{6}<-\frac{3}{4}<-\frac{2}{3}). Compare positive values first, then reverse the order.