Representing real numbers on the number line

sqrt{50} = sqrt{25times2} = 5sqrt{2}. Since sqrt{2} is approximately 1.414, sqrt{50} approx 5times1.414 = 7.07. Therefore, the correct value is 7.07. The distractor 6.07 is incorrect because sqrt{50} is slightly greater than sqrt{49}=7. Exam tip: First check the bounds using 7^2<50<8^2.

(\sqrt{9}=3), and moving left means subtracting, so (P=4-3=1). In direction questions, simplify the distance first.

(\pi\approx3.14) and (\sqrt{10}\approx3.16), so (\pi<\sqrt{10}). Good approximations help compare close irrationals.

Because (4^2+1^2=17), the hypotenuse will be (\sqrt{17}). Pythagoras theorem is useful for square-root construction on the number line.

(\sqrt{5}\approx2.236) and (\sqrt{2}\approx1.414), so the difference is about (0.822). Use short approximations to locate differences of irrationals.

Since \(4^2=16<18<25=5^2\), it follows that \(4<\sqrt{18}<5\). Therefore, \(\sqrt{18}\) lies between 4 and 5 on the number line. Option B is incorrect because \(\sqrt{18}>4\). Exam tip: compare the number with the nearest consecutive perfect squares to locate its square root.

The distance from (2.357) to (2.36) is (0.003), and to (2.35) is (0.007). The smaller distance gives the closer point.

Since (2<\sqrt{7}<3), we have (-3<-\sqrt{7}<-2). For negative numbers, remember the order reverses.

( \frac{17}{5}=3.4 ), so the point lies between (3) and (4). Convert the fraction to a decimal to locate it.

Moving right means addition, so the coordinate is (1+\sqrt{6}). Choose the sign by checking the direction.

To locate \(\frac{-13}{4}\), divide 13 by 4 and retain the negative sign. Since \(13\div4=3.25\), the number is \(-3.25\). On the number line, \(-3.25\) lies to the left of \(-3\) but to the right of \(-4\). Therefore it lies between \(-4\) and \(-3\), making option A correct.

The negative sign is important: negative numbers appear on the left side of zero, and a more negative number is farther left. We can also compare using fourths: \(-4=-16/4\) and \(-3=-12/4\), while \(-13/4\) is between them. It is not between \(-3\) and \(-2\), nor in the positive intervals.

Since (2<3<5), ( \sqrt{2}<\sqrt{3}<\sqrt{5}). For positive square roots, order follows the radicand.

Since (6^2=36) and (7^2=49), ( \sqrt{45}) is about (6.7). Estimate using nearby perfect squares.

The smallest number is farthest left, and (-1.8) is the smallest. For negative decimals, larger magnitude means farther left.

Since \(3^2<12<4^2\), we have \(3<\sqrt{12}<4\). Subtracting 1 from all parts gives \(2<\sqrt{12}-1<3\), so the expression lies between 2 and 3. Option B describes the interval for \(\sqrt{12}\), not for the complete expression. Exam tip: first bound the square root between consecutive integers, then apply the remaining operation to all parts of the inequality.

The point exactly midway between two numbers is their average. This works because the midpoint must be equally far from both endpoints. For the numbers 0 and 1, add them and divide by 2. The result is \\(\\frac{0+1}{2}=\\frac12\\).

On the number line, \\(\\frac12\\) is half a unit from 0 and half a unit from 1, so the two distances are equal. The value \\(\\frac13\\) is closer to 0, while \\(\\frac23\\) and \\(\\frac34\\) are closer to 1. Hence the point exactly in the middle is \\(\\frac12\\), making option A correct. The average formula is a quick and general way to locate the midpoint of any two real numbers.

The distance between two points on a number line is the absolute value of the difference of their coordinates: \(|3.6-(-2.4)|=|6|=6\). Therefore, the length of AB is \(6\). Remember that distance is always non-negative; \(1.2\) results from incorrectly subtracting the decimal parts only.

( \frac{7}{8}) means (7) parts out of (8) equal parts from (0) to (1). The denominator gives the number of equal parts.

(99) is very close to (100), so ( \sqrt{99}) is about (10). The nearest perfect square gives a quick answer.

( |x-2|=3) means the distance of (x) from (2) is (3), so (x=-1) or (x=5). Check both directions in distance questions.