Representing real numbers on the number line

( -\frac{43}{11}\approx-3.909 ), so it lies between (-4) and (-3). Convert negative fractions to decimals.

The coordinate of the midpoint is the average of the two coordinates: \(\frac{-4.125+1.875}{2}=\frac{-2.25}{2}=-1.125\). Therefore, option A is correct. Option C results from ignoring the negative sign of A. In an exam, add the coordinates algebraically and divide the result by 2.

\(0.0225=225/10000\), so \(\sqrt{0.0225}=15/100=0.15\), since \((0.15)^2=0.0225\). In option C, the decimal point is shifted one place too far to the left, making its square much smaller. Exam tip: For a decimal square root, convert the number into a fraction and take the square root of the numerator and denominator separately when both are perfect squares.

Since \(11^2=121<131<144=12^2\), we have \(11<\sqrt{131}<12\). Adding 2 to all parts gives \(13<\sqrt{131}+2<14\), so the expression lies between 13 and 14. Option A is incorrect because it corresponds to the bounds of \(\sqrt{131}\), not of the complete expression. Exam tip: compare the number under the square root with the nearest consecutive perfect squares.

Since \(9^2=81<99<100=10^2\), we have \(9<\sqrt{99}<10\). Therefore, \(-10< -\sqrt{99}< -9\), and adding 4 gives \(-6<p<-5\). Hence, \(p\) lies between −6 and −5. Exam tip: when multiplying an inequality by \(-1\), reverse its direction; this prevents choosing the nearby interval −5 to −4.

Moving left gives ( \frac{11}{12}-\frac{5}{18}=\frac{23}{36} ). Subtract the distance according to direction.

Estimate the square root by locating 199 between nearby perfect squares. Since 14² = 196 and 15² = 225, √199 lies just above 14. A more precise estimate is √199 ≈ 14.1067, because 14.1² = 198.81, which is very close to 199. The distance from 14.10 is about 0.0067, much smaller than the distances from 14.35, 13.90, or 14.90. Therefore option A is closest. The other choices are respectively about 0.24, 0.21, and 0.79 away from the root. The key concept is numerical approximation: compare the candidate’s distance from the irrational value rather than merely choosing a nearby-looking number.

For comparison, \(\sqrt{26}\approx 5.099\), while \(\frac{51}{10}=5.1\) and \(c=5.11\). Thus, \(5.11>5.1>5.099\), so \(c\) lies farthest to the right on the number line. The values of \(a\) and \(b\) are close, but they are not equal. Exam tip: Convert the numbers to comparable decimal forms before ordering them.

(9.2^2=84.64) and (9.3^2=86.49), so ( \sqrt{86} ) lies between them. Check squares for decimal bounds.

First evaluate the square root: \(\sqrt{225}=15\). Therefore, \(\frac{\sqrt{225}-6}{9}=\frac{15-6}{9}=\frac{9}{9}=1\). Hence, the point is \(1\). Exam tip: follow the order of operations carefully; \(\frac{9}{15}\) is incorrect because it reverses the intended numerator and denominator.

Moving ( \frac{13}{6} ) to the right of (-5) gives ( -5+\frac{13}{6}=-\frac{17}{6} ). Use the given interval to choose direction.

Since \(196=14^2\) and \(289=17^2\), \(\sqrt{\frac{196}{289}}=\frac{\sqrt{196}}{\sqrt{289}}=\frac{14}{17}\). The square-root symbol denotes the principal, non-negative square root, so \(-\frac{14}{17}\) is not the value. Exam tip: when the numerator and denominator are perfect squares, take their positive square roots separately.

( \sqrt{15}+\frac{1}{8}\approx3.998 ), so (3.95) is not greater than it. In this case no listed value satisfies the condition.

Since \(C=-\frac{68}{5}=-13.6\), points \(B\) and \(C\) coincide. Also, \(13.6^2=184.96<185\), so \(\sqrt{185}>13.6\), which gives \(-\sqrt{185}<-13.6\). On a number line, the smaller number lies farther left; hence point \(A\) is farthest left. Exam tip: For negative numbers, the one with the greater absolute value is the smaller number.

The governing concept is directed distance on the number line. A point whose distance from zero is √41 can have either coordinate √41 or −√41, since equal distances occur on opposite sides of zero. The additional condition “to the left of 0” selects the negative coordinate, so the answer is −√41. Option B is correct. Option A has the required distance but lies to the right of zero. Options C and D are not correct because the distance of either 41 or −41 from zero is 41, not √41. Thus both magnitude and direction must be used. This distinction is important: ordinary distance is nonnegative, while the coordinate records the side of zero through its sign.

To compare the two numbers, estimate the square root. Since \\(\\sqrt{11}\\) is about \\(3.316\\), the first expression is approximately \\(5-3.316=1.684\\). The second expression is \\(\\frac{17}{10}=1.7\\). These estimates are close, so careful calculation is useful, but they clearly show which value is smaller.

More precisely, \\(5-\\sqrt{11}\\approx1.6834\\), while \\(\\frac{17}{10}=1.7\\). Therefore \\(5-\\sqrt{11}<\\frac{17}{10}\\). Both values are positive and near 1.7, so the statement that both are less than \\(-1\\) is impossible. Hence option A is correct. A decimal approximation is sufficient here because the two values are not equal.

( \sqrt{3}\approx1.732 ), so ( \frac{a}{100} ) must be between (1.732) and (1.74). (a=173) gives (1.73), which is slightly smaller, so check the bound carefully.

Since \(245=49\times5\) and 49 is the largest perfect-square factor, \(\sqrt{245}=\sqrt{49\times5}=7\sqrt{5}\). Option A would represent \(\sqrt{175}\), not \(\sqrt{245}\), and \(\sqrt{49}+\sqrt{5}\) is not equal to \(\sqrt{49+5}\). Exam tip: factor the number using its largest perfect-square factor before simplifying a square root.

By checking carefully, ( -3.61<-\sqrt{13}\approx-3.606<-\frac{18}{5}=-3.6 ). For negative values, the smaller number comes first.

Since \(x\) is positive, squaring the inequality gives \(12.6^2<a<12.7^2\). Therefore, \(158.76<a<161.29\). Among the given options, only \(159\) lies in this interval. The value \(158\) is too small, while \(162\) and \(163\) are too large. Exam tip: For \(x=\sqrt a\), square the bounds on \(x\) to find the required interval for \(a\).