Free Math Worksheets

- Compare the magnitudes by squaring the positive values. - Recall how the order changes when positive numbers are replaced by their opposites. - Compute \(3.5^2\) exactly.

1. Compare the positive magnitudes. Since \((\sqrt{13})^2=13\) and \(3.5^2=12.25\), \(\sqrt{13}>3.5\). 2. For negative numbers, the number with the smaller magnitude is greater. Therefore, \(-3.5>-\sqrt{13}\), so \(b\) lies farther to the right.

\(b=-3.5\) lies farther to the right because \(-3.5>-\sqrt{13}\).

- How does comparing radicands to perfect squares help compare their square roots? - If a square root lies between two numbers, where must its radicand lie relative to their squares? - Calculate the two boundary squares first.

1. The boundary squares are \(13^2 = 169\) and \(14^2 = 196\). 2. Since \(150 < 169\), \(\sqrt{150} < 13\). 3. Since \(169 < 180 < 196\), \(13 < \sqrt{180} < 14\). 4. Since \(210 > 196\), \(\sqrt{210} > 14\).

Only \(\sqrt{180}\) lies between \(13\) and \(14\), because \(169 < 180 < 196\).

- A lower decimal bound is found by truncating at the requested place value. - The upper bound is the next decimal at that place value. - Use systematic testing or a calculator to determine digits of a square root when permitted.

1. At increments of \(0.1\), \(3.8 < \sqrt{15} < 3.9\). 2. At increments of \(0.01\), \(3.87 < \sqrt{15} < 3.88\). 3. At increments of \(0.0001\), \(3.8729 < \sqrt{15} < 3.8730\). 4. Since \(\sqrt{0.8} \approx 0.89442719\ldots\), the first five digits after the decimal point are \(89442\).

1. Increment \(0.1\): lower \(3.8\), upper \(3.9\) Increment \(0.01\): lower \(3.87\), upper \(3.88\) Increment \(0.0001\): lower \(3.8729\), upper \(3.8730\) 2. \(89442\)

- Estimate each square root to several decimal places before multiplying. - Multiply each square-root estimate by its coefficient. - Round both results to the same place value before comparing.

1. Using \(\sqrt{13} \approx 3.6056\), \(A \approx 2 \cdot 3.6056 = 7.2112\), so \(A \approx 7.2\). 2. Using \(\sqrt{2} \approx 1.4142\), \(B \approx 5 \cdot 1.4142 = 7.0710\), so \(B \approx 7.1\). 3. Since \(7.2 > 7.1\), \(A\) is greater than \(B\).

\(A \approx 7.2\) and \(B \approx 7.1\). Therefore, \(A > B\).

- Substitute the given square-root approximations first. - Follow the order of operations carefully. - Round only after completing the calculation. - Use the thousandths digit to round to the nearest hundredth.

1. For a), \(1.5 \cdot 1.414 + 2 = 2.121 + 2 = 4.121\), so the result is approximately \(4.12\). 2. For b), \(2 \cdot 1.732 - 1.414 = 3.464 - 1.414 = 2.050\), so the result is approximately \(2.05\). 3. For c), \(\frac{1.732 + 1.414}{2} = \frac{3.146}{2} = 1.573\), so the result is approximately \(1.57\).

a) \(\approx 4.12\) b) \(\approx 2.05\) c) \(\approx 1.57\)

- Keep enough decimal places in each square-root estimate to round the final answer reliably. - Can any radical, such as \(\sqrt{8}\), be simplified before estimating? - Use the thousandths digit to round to the nearest hundredth. - An accuracy of \(0.01\) means rounding to the nearest hundredth.

1. Using \(\sqrt{3} \approx 1.7321\), \(2.45 + \sqrt{3} \approx 4.1821\), which rounds to \(4.18\). 2. Using \(\sqrt{10} \approx 3.1623\), \(7.18 - \sqrt{10} \approx 4.0177\), which rounds to \(4.02\). 3. Using \(\sqrt{8} \approx 2.8284\) and \(\sqrt{2} \approx 1.4142\), the sum is approximately \(4.2426\), which rounds to \(4.24\). 4. Using \(\sqrt{5} \approx 2.2361\), \(3\sqrt{5} - 4 \approx 6.7083 - 4 = 2.7083\), which rounds to \(2.71\).

1) \(\approx 4.18\) 2) \(\approx 4.02\) 3) \(\approx 4.24\) 4) \(\approx 2.71\)

- Estimate each square root before multiplying. - Keep more decimal places in intermediate values than the final answer requires. - Use the thousandths digit to round to the nearest hundredth.

1. \(4\sqrt{7} \approx 4 \cdot 2.64575 = 10.583\), so the result is approximately \(10.58\). 2. \(9\sqrt{2} \approx 9 \cdot 1.41421 = 12.72789\), so the result is approximately \(12.73\). 3. \(2\sqrt{15} \approx 2 \cdot 3.87298 = 7.74596\), so the result is approximately \(7.75\).

1) \(\approx 10.58\) 2) \(\approx 12.73\) 3) \(\approx 7.75\)

- Evaluate any square roots of perfect squares first. - To compare a positive decimal with a square root, compare their squares. - Use the order of the radicands to compare positive square roots.

1. Since \(\sqrt{16}=4\), those two values are equal. 2. Because \(4.1^2=16.81<18\), \(4.1<\sqrt{18}\). 3. Since \(18<20\), \(\sqrt{18}<\sqrt{20}\). 4. Therefore, \(4=\sqrt{16}<4.1<\sqrt{18}<\sqrt{20}\).

\(4=\sqrt{16}<4.1<\sqrt{18}<\sqrt{20}\)

- Evaluate square roots of perfect-square decimals when possible. - When both values are nonnegative, compare their squares. - Remember how squaring affects numbers between \(0\) and \(1\).

1. In a), \(\sqrt{0.09}=0.3\), and \(0.3>0.09\). 2. In b), \(1.1^2=1.21\), so \(\sqrt{1.21}=1.1\). 3. In c), \(4.5^2=20.25>20\), so \(\sqrt{20}<4.5\). 4. In d), both quantities are positive, and \(\left(\frac{1}{2}\right)^2=\frac{1}{4}<\frac{1}{2}\). Therefore, \(\sqrt{\frac{1}{2}}>\frac{1}{2}\).

a) \(\sqrt{0.09}>0.09\) b) \(\sqrt{1.21}=1.1\) c) \(\sqrt{20}<4.5\) d) \(\sqrt{\frac{1}{2}}>\frac{1}{2}\)

- Rewrite each radicand as \(5\) times a power of \(10\) that is a perfect square. - Which property lets you separate the square root of a product? - How does multiplying by \(10\), \(100\), or \(0.1\) move the decimal point? - Look for a perfect-square factor inside each radical.

1. \(\sqrt{500} = \sqrt{5 \cdot 100} = \sqrt{5} \cdot 10 \approx 2.236 \cdot 10 = 22.36\). 2. \(\sqrt{0.05} = \sqrt{5 \cdot 0.01} = \sqrt{5} \cdot 0.1 \approx 2.236 \cdot 0.1 = 0.2236\). 3. \(\sqrt{50{,}000} = \sqrt{5 \cdot 10{,}000} = \sqrt{5} \cdot 100 \approx 2.236 \cdot 100 = 223.6\).

a) \(\sqrt{500} \approx 22.36\) b) \(\sqrt{0.05} \approx 0.2236\) c) \(\sqrt{50{,}000} \approx 223.6\)

- Is \(32\) a perfect square? - Which familiar perfect squares are closest to \(32\)? - How does a square’s area change when its side length is multiplied by a factor?

1. The side length is \(s = \sqrt{32}\,\text{m} \approx 5.66\,\text{m}\). 2. Since \(5^2 = 25\) and \(6^2 = 36\), and \(25 < 32 < 36\), \(\sqrt{32}\) lies between \(5\) and \(6\). 3. Doubling the side length multiplies the area by \(2^2 = 4\). The new area is \(4 \cdot 32\,\text{m}^2 = 128\,\text{m}^2\).

a) \(s = \sqrt{32}\,\text{m} \approx 5.66\,\text{m}\) b) The side length lies between \(5\,\text{m}\) and \(6\,\text{m}\). c) The new floor area is \(128\,\text{m}^2\).

- Square nearby hundredths to determine whether they are below or above \(45\). - What does the upper endpoint of the interval tell you when comparing \(x\) with \(6.71\)?

1. Since \(6.70^2 = 44.89\) and \(6.71^2 = 45.0241\), \(44.89 < 45 < 45.0241\). 2. Therefore, \(6.70 < \sqrt{45} < 6.71\), so an interval of width \(0.01\) containing \(x\) is \([6.70, 6.71]\). 3. Because \(x < 6.71 = y\), \(y\) is greater than \(x\).

a) \(x \in [6.70, 6.71]\) b) \(y > x\), because \(6.71^2 = 45.0241 > 45\), so \(\sqrt{45} < 6.71\).

- A number is in \([a, b]\) when it is at least \(a\) and at most \(b\). - Compare a square root with positive bounds by squaring the bounds. - Convert the fraction to a decimal for a direct comparison.

1. For a), square the positive interval endpoints: \(5.38^2 = 28.9444\) and \(5.39^2 = 29.0521\). Since \(28.9444 < 29 < 29.0521\), it follows that \(5.38 < \sqrt{29} < 5.39\). Therefore, \(\sqrt{29}\) is in the interval. 2. For b), \(\frac{43}{8} = 5.375\). Since \(5.375 < 5.38\), the number is not in the interval. 3. For c), \(5.38 < 5.382 < 5.39\), so \(5.382\) is in the interval.

a) Yes, because \(5.38^2 < 29 < 5.39^2\). b) No, because \(\frac{43}{8} = 5.375 < 5.38\). c) Yes, because \(5.38 < 5.382 < 5.39\).

- Relate the side lengths of a rectangle to its area. - How do you find the mean of two numbers? - If one side changes, how can the other side preserve the area? - Compare the square of the approximation with the target area.

1. Since \(A=a_0b_0\), \(b_0=18\div6=3\,\text{cm}\). 2. The mean is \(a_1=\frac{6+3}{2}=4.5\,\text{cm}\). 3. To preserve the area, \(b_1=18\div4.5=4\,\text{cm}\). 4. The square is \(a_1^2=(4.5\,\text{cm})^2=20.25\,\text{cm}^2\). 5. The absolute error is \(\left|20.25-18\right|\,\text{cm}^2=2.25\,\text{cm}^2\).

a) \(b_0=3\,\text{cm}\) b) \(a_1=4.5\,\text{cm}\) c) \(b_1=4\,\text{cm}\) d) \(a_1^2=20.25\,\text{cm}^2\), and the absolute error is \(2.25\,\text{cm}^2\).

- Use \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{A}{x_n}\right)\) with \(A=10\). - Keep the first two iterations as fractions to avoid rounding error. - Dividing by a fraction means multiplying by its reciprocal. - Compare the square of the second approximation with \(10\).

1. Using \(x_{n+1}=\frac{1}{2}\left(x_n+\frac{10}{x_n}\right)\), the first approximation is \(x_1=\frac{1}{2}\left(3+\frac{10}{3}\right)=\frac{19}{6}\). 2. The second approximation is \(x_2=\frac{1}{2}\left(\frac{19}{6}+\frac{10}{19/6}\right)=\frac{1}{2}\left(\frac{19}{6}+\frac{60}{19}\right)=\frac{721}{228}\). 3. As a decimal, \(x_2=\frac{721}{228}\approx3.162281\). 4. The squared value is \(x_2^2=\frac{519841}{51984}\). Therefore, the absolute difference is \(\left|x_2^2-10\right|=\frac{1}{51984}\approx0.00001924\).

a) \(x_1=\frac{19}{6}\) and \(x_2=\frac{721}{228}\) b) \(x_2\approx3.162281\) c) \(\left|x_2^2-10\right|=\frac{1}{51984}\approx0.00001924\)

- Apply the iteration formula separately to each starting value. - Compare the absolute errors of the two approximations. - Which starting value is already closer to \(\sqrt{2}\)?

1. For Student A, \(x_1=\frac{1}{2}\left(1.5+\frac{2}{1.5}\right)=\frac{1}{2}\left(\frac{3}{2}+\frac{4}{3}\right)=\frac{17}{12}\approx1.4166667\). 2. For Student B, \(x_1=\frac{1}{2}\left(2+\frac{2}{2}\right)=\frac{3}{2}=1.5\). 3. Student A's absolute error is approximately \(0.0024531\), while Student B's absolute error is approximately \(0.0857864\). Therefore, the starting value \(1.5\) gives the more accurate result after one iteration. 4. A starting value closer to the actual square root generally requires fewer iterations to reach a given level of accuracy. Here, \(1.5\) is much closer to \(\sqrt{2}\) than \(2\) is.

a) Student A: \(x_1=\frac{17}{12}\approx1.4166667\) Student B: \(x_1=\frac{3}{2}=1.5\) b) The starting value \(1.5\) gives the more accurate result. c) A starting value closer to the square root begins with a smaller error, so fewer iterations are usually needed.

- Substitute the current approximation into the iteration formula. - Use common denominators when adding fractions. - Square the third approximation before subtracting \(2\).

1. The second approximation is \(x_2=\frac{1}{2}\left(x_1+\frac{2}{x_1}\right)=\frac{1}{2}(1+2)=\frac{3}{2}\). 2. The third approximation is \(x_3=\frac{1}{2}\left(\frac{3}{2}+\frac{2}{3/2}\right)=\frac{1}{2}\left(\frac{3}{2}+\frac{4}{3}\right)=\frac{17}{12}\). 3. Squaring gives \(x_3^2=\left(\frac{17}{12}\right)^2=\frac{289}{144}\). 4. Therefore, \(x_3^2-2=\frac{289}{144}-\frac{288}{144}=\frac{1}{144}\).

\(x_2=\frac{3}{2}\), \(x_3=\frac{17}{12}\), and \(x_3^2-2=\frac{1}{144}\).

- Find the mean of \(3\) and \(\frac{10}{3}\). - Compare \(x_2^2\) with \(10\). - What happens to an inequality between positive numbers when reciprocals are taken? - Relate multiplication and division when checking the square-root bounds.

1. The second approximation is \(x_2=\frac{1}{2}\left(3+\frac{10}{3}\right)=\frac{19}{6}\). 2. Squaring gives \(x_2^2=\left(\frac{19}{6}\right)^2=\frac{361}{36}>10\). Since \(x_2\) is positive, \(x_2>\sqrt{10}\). 3. If \(x_n>\sqrt{10}>0\), taking reciprocals reverses the inequality: \(\frac{1}{x_n}<\frac{1}{\sqrt{10}}\). Multiplying by \(10\) gives \(\frac{10}{x_n}<\frac{10}{\sqrt{10}}=\sqrt{10}\).

a) \(x_2=\frac{19}{6}\) b) \(x_2^2=\frac{361}{36}>10\), so \(x_2>\sqrt{10}\). c) Since reciprocals reverse inequalities for positive numbers, \(x_n>\sqrt{10}\) implies \(\frac{10}{x_n}<\sqrt{10}\).

- Which perfect squares are closest to \(11\)? - Square consecutive tenths to narrow the interval. - Continue by testing consecutive hundredths.

1. Since \(3^2 = 9\) and \(4^2 = 16\), \(3 < \sqrt{11} < 4\). The first interval is \([3, 4]\). 2. Since \(3.3^2 = 10.89\) and \(3.4^2 = 11.56\), \(3.3 < \sqrt{11} < 3.4\). The second interval is \([3.3, 3.4]\). 3. Since \(3.31^2 = 10.9561\) and \(3.32^2 = 11.0224\), \(3.31 < \sqrt{11} < 3.32\). The third interval is \([3.31, 3.32]\).

1. \([3, 4]\) 2. \([3.3, 3.4]\) 3. \([3.31, 3.32]\)

- What happens to an interval’s width when it is divided into \(10\) equal parts? - Square the proposed endpoints to check whether \(\sqrt{2}\) lies between them. - A closed interval includes both endpoints.

1. The width of \(I_1\) is \(1\). Each subdivision divides the width by \(10\), so the widths of \(I_2\), \(I_3\), and \(I_4\) are \(0.1\), \(0.01\), and \(0.001\), respectively. 2. Since \(1.4^2 = 1.96\) and \(1.5^2 = 2.25\), \(I_2 = [1.4, 1.5]\). 3. Since \(1.41^2 = 1.9881\) and \(1.42^2 = 2.0164\), \(I_3 = [1.41, 1.42]\). 4. Since \(1.414^2 = 1.999396\) and \(1.415^2 = 2.002225\), \(I_4 = [1.414, 1.415]\). The number \(1.414\) is the lower endpoint, so it belongs to this closed interval.

a) The width of \(I_4\) is \(0.001\). b) \(I_4 = [1.414, 1.415]\), and \(1.414\) belongs to the interval.

- Begin with neighboring perfect squares. - The midpoint determines which of two approximations is closer. - Find the midpoint of \(4.2\) and \(4.3\), then compare its square with \(18\).

1. Since \(4^2 = 16 < 18 < 25 = 5^2\), \(4 < \sqrt{18} < 5\). Thus, \(\sqrt{18}\) lies in \([4, 5]\). 2. The midpoint of \(4.2\) and \(4.3\) is \(4.25\). Since \(4.2^2 = 17.64 < 18\) and \(4.25^2 = 18.0625 > 18\), \(4.2 < \sqrt{18} < 4.25\). 3. Because \(\sqrt{18}\) is below the midpoint, it is closer to \(4.2\) than to \(4.3\).

a) \(\sqrt{18} \in [4, 5]\) b) \(\sqrt{18}\) is closer to \(4.2\), because \(4.2 < \sqrt{18} < 4.25\), and \(4.25\) is the midpoint of \(4.2\) and \(4.3\).

- Substitute the latest approximation into the iteration formula. - Keep fractions until the final decimal conversion. - The absolute error is the absolute value of the difference. - Increase the number of decimal places one at a time until the rounded values first differ.

1. The first approximation is \(x_1=\frac{1}{2}\left(2+\frac{5}{2}\right)=\frac{9}{4}=2.25\). 2. The second approximation is \(x_2=\frac{1}{2}\left(\frac{9}{4}+\frac{5}{9/4}\right)=\frac{1}{2}\left(\frac{9}{4}+\frac{20}{9}\right)=\frac{161}{72}\approx2.236111\). 3. Since \(\sqrt{5}\approx2.236068\), the absolute error is \(\left|\frac{161}{72}-\sqrt{5}\right|\approx0.00004313\). 4. Rounded to four digits after the decimal point, both values are \(2.2361\). Rounded to five digits, they are \(2.23611\) and \(2.23607\), so four digits is the greatest number of digits after the decimal point for which the two values round identically.

a) \(x_1=\frac{9}{4}=2.25\) and \(x_2=\frac{161}{72}\approx2.236111\) b) The absolute error is approximately \(0.00004313\). The values round identically to four digits after the decimal point.

- Use neighboring perfect squares. - To compare \(3.6\) and \(3.61\), square their midpoint. - A positive number is greater than \(\sqrt{13}\) when its square is greater than \(13\).

1. Since \(3^2 = 9 < 13 < 16 = 4^2\), \(3 < \sqrt{13} < 4\). 2. The midpoint of \(3.6\) and \(3.61\) is \(3.605\). Since \(3.605^2 = 12.996025 < 13\), \(\sqrt{13}\) lies above the midpoint. Therefore, it is closer to \(3.61\). 3. \(\left(\frac{11}{3}\right)^2 = \frac{121}{9} > 13\). Since \(\frac{11}{3}\) is positive, \(\frac{11}{3} > \sqrt{13}\).

a) \(3 < \sqrt{13} < 4\) b) \(\sqrt{13}\) is closer to \(3.61\). c) Since \(\left(\frac{11}{3}\right)^2 > 13\), \(\frac{11}{3} > \sqrt{13}\).

- Square both proposed approximations first. - Then find the midpoint of the two values. - Determine which side of that midpoint contains \(\sqrt{5}\).

1. \(2.236^2 = 4.999696 < 5\), while \(2.25^2 = 5.0625 > 5\). 2. The midpoint of the two approximations is \(m = \frac{2.236 + 2.25}{2} = 2.243\). Since \(2.243^2 = 5.031049 > 5\), \(\sqrt{5} < 2.243\). 3. Because \(2.236 < \sqrt{5} < 2.243\), \(\sqrt{5}\) is closer to \(2.236\) than to \(2.25\).

The approximation \(2.236\) is more precise. Since \(2.236 < \sqrt{5} < 2.243\), \(\sqrt{5}\) lies closer to \(2.236\).

- Which familiar perfect squares are nearest to \(18\)? - Compare how the square changes as the decimal input increases by tenths. - Recall the difference between rational and irrational numbers.

1. Since \(4^2 = 16\) and \(5^2 = 25\), \(4 < \sqrt{18} < 5\). 2. \(4.1^2 = 16.81\), \(4.2^2 = 17.64\), and \(4.3^2 = 18.49\). Therefore, \(4.2 < \sqrt{18} < 4.3\), so the tenths digit is \(2\). 3. The integer \(18\) is not a perfect square, so \(\sqrt{18}\) is irrational. Its decimal expansion is nonterminating and nonrepeating, so increasingly narrow decimal intervals never end at an exact terminating decimal.

a) \(4\,\text{m} < s < 5\,\text{m}\) b) \(4.2 < \sqrt{18} < 4.3\), so the tenths digit is \(2\). c) \(\sqrt{18}\) is irrational, so its decimal expansion is nonterminating and nonrepeating.

- Find the exact value of \(A\). - Combine the numbers inside \(B\). - Compare two positive values by comparing their squares.

1. Evaluate \(A\): \(A=0.6+0.5=1.1\). 2. Combine the numbers under the radical in \(B\): \(B=\sqrt{0.61}\). 3. Both values are positive. Since \(1.1^2=1.21>0.61\), \(1.1>\sqrt{0.61}\). Therefore, \(A>B\).

\(A>B\)

- Write each number as a decimal to enough places to distinguish it from nearby values. - Recall the meaning of a bar over a decimal digit. - Approximate \(\sqrt{5}\) beyond the thousandths place. - Compare enough decimal places; two places may not distinguish nearby values.

1. Write comparable decimal forms: \(2.\overline{2}=2.2222\ldots\), \(2.23=2.2300\ldots\), \(2.236=2.2360\ldots\), \(\sqrt{5}\approx2.2360679\ldots\), and \(\frac{9}{4}=2.25\). 2. Comparing place values gives \(2.\overline{2}<2.23<2.236<\sqrt{5}<\frac{9}{4}\).

\(2.\overline{2}<2.23<2.236<\sqrt{5}<\frac{9}{4}\)

- Which formula relates a circle’s area and radius? - Rearrange the formula so the squared radius is isolated before taking a square root. - Use the third decimal place to round to the nearest hundredth. - Include the correct length unit in each answer.

1. From \(A = \pi r^2\), solve for the radius: \(r = \sqrt{\frac{A}{\pi}}\). 2. For a), \(r = \sqrt{\frac{10}{\pi}}\,\text{cm} \approx 1.78\,\text{cm}\). 3. For b), \(r = \sqrt{\frac{35.5}{\pi}}\,\text{m} \approx 3.36\,\text{m}\). 4. For c), \(r = \sqrt{\frac{0.8}{\pi}}\,\text{dm} \approx 0.50\,\text{dm}\).

a) \(r \approx 1.78\,\text{cm}\) b) \(r \approx 3.36\,\text{m}\) c) \(r \approx 0.50\,\text{dm}\)

- How does multiplying the radicand by \(4\) affect a square root? - Substitute \(2r\) for the radius in the area formula. - Decide whether area and radius have a linear or quadratic relationship.

1. For a), \(r = \sqrt{\frac{50}{\pi}}\,\text{cm} \approx 3.99\,\text{cm}\). 2. If the radius doubles, \(A_{\text{new}} = \pi(2r)^2 = 4\pi r^2 = 4A\). Therefore, \(A_{\text{new}} = 4 \cdot 50\,\text{cm}^2 = 200\,\text{cm}^2\). 3. Equivalently, \(r_{\text{new}} = \sqrt{\frac{4A}{\pi}} = 2\sqrt{\frac{A}{\pi}} = 2r\). Multiplying the area by \(4\) doubles the radius.

a) \(r \approx 3.99\,\text{cm}\) b) \(A_{\text{new}} = 200\,\text{cm}^2\) c) The radius doubles when the area is multiplied by \(4\).

- What happens to a small approximation error when the value is later multiplied by a large factor? - At which stage does each method discard information by rounding? - For method b, begin by squaring \(15\).

1. Method a: \(\sqrt{2} \approx 1.4\), so \(15 \cdot 1.4 = 21.0\). 2. Method b: \(15\sqrt{2} = \sqrt{15^2 \cdot 2} = \sqrt{450}\). Since \(21.2^2 = 449.44\) and \(21.25^2 = 451.5625\), \(21.2 < \sqrt{450} < 21.25\). Therefore, \(\sqrt{450}\) rounds to \(21.2\) to the nearest tenth. 3. Method b is more accurate. Method a rounds before multiplying, so the initial rounding error is multiplied by \(15\). Method b rounds only once, at the end.

a) \(21.0\) b) \(21.2\) Method b is more accurate because it delays rounding until the final result.

- How can a coefficient be moved inside a square root? - Which perfect squares are closest to the new radicand? - Use enough decimal places in the square-root estimate before rounding the final product.

1. Move the coefficient inside the radical by squaring it: \(4\sqrt{5} = \sqrt{4^2 \cdot 5} = \sqrt{80}\). 2. Since \(8^2 = 64\) and \(9^2 = 81\), \(8 < \sqrt{80} < 9\). 3. Using \(\sqrt{5} \approx 2.236\), \(4\sqrt{5} \approx 4 \cdot 2.236 = 8.944\), which rounds to \(8.9\).

a) The value lies between \(8\) and \(9\). b) \(4\sqrt{5} \approx 8.9\)

- How are a square’s area and side length related? - Which formula gives the perimeter of a square? - Use the given diagonal formula or the Pythagorean theorem. - Keep extra decimal places in intermediate calculations.

1. The side length is \(s = \sqrt{7}\,\text{cm} \approx 2.65\,\text{cm}\). 2. The perimeter is \(P = 4s = 4\sqrt{7}\,\text{cm} \approx 10.58\,\text{cm}\). 3. The diagonal is \(d = \sqrt{7}\sqrt{2}\,\text{cm} = \sqrt{14}\,\text{cm} \approx 3.74\,\text{cm}\).

a) \(s \approx 2.65\,\text{cm}\); \(P \approx 10.58\,\text{cm}\) b) \(d \approx 3.74\,\text{cm}\)

- Isolate \(x^2\) before taking a square root. - Include both the positive and negative solutions. - Use the thousandths digit to round to the nearest hundredth.

1. \(x^2 = 13\), so \(x = \pm\sqrt{13} \approx \pm 3.61\). 2. Multiply by \(2\): \(x^2 = 22\). Therefore, \(x = \pm\sqrt{22} \approx \pm 4.69\). 3. Add \(30\) and divide by \(4\): \(x^2 = 20\). Therefore, \(x = \pm\sqrt{20} \approx \pm 4.47\).

1) \(x \approx 3.61\) or \(x \approx -3.61\) 2) \(x \approx 4.69\) or \(x \approx -4.69\) 3) \(x \approx 4.47\) or \(x \approx -4.47\)

- Which familiar perfect squares are nearest to \(40\)? - Compare \(40\) with those squares. - If a positive number has square \(40\), what can you say about its opposite?

1. Since \(6^2 = 36\) and \(7^2 = 49\), and \(36 < 40 < 49\), the positive solution \(\sqrt{40}\) lies between \(6\) and \(7\). 2. \(\sqrt{40} \approx 6.32455\ldots\), so the positive solution is approximately \(6.32\). 3. The negative solution is its opposite, approximately \(-6.32\).

a) The positive solution lies between \(6\) and \(7\). b) \(x \approx 6.32\) or \(x \approx -6.32\)

- How are the side length and diagonal of the two squares related to the circle's radius? - How can you find a square's area from its diagonal? - What does it imply about \(\pi\) when the circle's area lies between two polygon areas? - As a regular polygon gains more sides, how do its bounds on the circle's area change?

1. For the circumscribed square, the side length equals the circle's diameter, so \(s = 2r\). Therefore, \(A_c = s^2 = (2r)^2 = 4r^2\). 2. For the inscribed square, the diagonal equals the diameter, so \(d = 2r\). Using \(A = \frac{1}{2}d^2\), \(A_i = \frac{1}{2}(2r)^2 = 2r^2\). 3. Because \(A_i < A_{\text{circle}} < A_c\), \(2r^2 < \pi r^2 < 4r^2\). Dividing by \(r^2\) gives \(2 < \pi < 4\). 4. The dodecagon areas give \(3r^2 < \pi r^2 < 3.215r^2\), so \(3 < \pi < 3.215\). 5. The interval width is \(3.215 - 3 = 0.215\).

a) \(A_i = 2r^2\) and \(A_c = 4r^2\) b) \(2 < \pi < 4\) c) The interval is \(3 < \pi < 3.215\), and its width is \(0.215\).

- Begin with the inner square root and bound it between nearby whole numbers. - Use those bounds to estimate the entire radicand of the outer square root. - Which perfect squares surround that resulting interval?

1. Since \(11^2 = 121\) and \(12^2 = 144\), \(11 < \sqrt{125} < 12\). 2. Add \(40\) to each part: \(51 < 40 + \sqrt{125} < 52\). 3. Because \(7^2 = 49\) and \(8^2 = 64\), \(49 < 40 + \sqrt{125} < 64\). 4. Taking square roots gives \(7 < \sqrt{40 + \sqrt{125}} < 8\). Therefore, \(T\) lies between \(7\) and \(8\).

The value of \(T\) lies between \(7\) and \(8\).

- Keep extra digits during intermediate calculations. - Compare the digits after the decimal point from left to right. - Stop counting at the first digit that differs.

1. The second approximation is \(x_2=\frac{1}{2}\left(3.5+\frac{13}{3.5}\right)=\frac{101}{28}\approx3.6071\). 2. The third approximation is \(x_3=\frac{1}{2}\left(\frac{101}{28}+\frac{13}{101/28}\right)=\frac{20393}{5656}\approx3.605551627\). 3. Comparing \(3.605551627\ldots\) with \(3.605551275\ldots\), the first six digits after the decimal point, \(605551\), agree. The seventh digits differ.

a) \(x_2=\frac{101}{28}\approx3.6071\) and \(x_3=\frac{20393}{5656}\approx3.605551627\) b) The first six digits after the decimal point agree.

- First square each fraction to determine which side of \(\sqrt{10}\) it lies on. - For two values on opposite sides, compare \(\sqrt{10}\) with their midpoint. - Square the midpoint rather than directly evaluating \(\sqrt{10}\).

1. The approximations lie on opposite sides of \(\sqrt{10}\): \(\left(\frac{22}{7}\right)^2 = \frac{484}{49} < 10\), while \(\left(\frac{19}{6}\right)^2 = \frac{361}{36} > 10\). 2. Their midpoint is \(m = \frac{1}{2}\left(\frac{22}{7} + \frac{19}{6}\right) = \frac{265}{84}\). Since \(m^2 = \frac{70225}{7056} < 10\), the midpoint is below \(\sqrt{10}\). Therefore, \(\sqrt{10}\) is closer to the larger approximation, \(\frac{19}{6}\). 3. Using \(\sqrt{10} \approx 3.162278\), the errors are \(\left|\frac{19}{6} - \sqrt{10}\right| \approx 0.004389\) and \(\left|\frac{22}{7} - \sqrt{10}\right| \approx 0.019421\). Lucas is correct.

a) \(\frac{19}{6}\) is closer to \(\sqrt{10}\). b) Lucas’s absolute error is approximately \(0.004389\), and Sofia’s is approximately \(0.019421\). Lucas is correct.