Question

1. apply the formula (t = \frac{sqrt{h}}{4}) represents the time (t) in seconds that it takes an object to fall from a height of (h) feet. if a rock falls from 125 feet, estimate how long it will take the rock to hit the ground. estimate the square - root to the nearest integer. locate the value on a number line. 13. the radius of a circle with area (a) can be approximated using the formula (r=\frac{1}{3}cdotsqrt{a}). estimate the radius of a wrestling mat circle with an area of 452 square feet. round to the nearest integer. locate the value on a number line. 14. find the error a classmate estimated (sqrt{397}) to be about 200. explain the mistake and correct it. 15. maintain accuracy explain how to write the exact value for the square root of a non - perfect square. give an example.

Explanation:

1. Question 12:

• Explanation:
• ## Step1: Given the formula \(t = \frac{\sqrt{h}}{4}\), and \(h = 125\).
• Substitute \(h = 125\) into the formula: \(t=\frac{\sqrt{125}}{4}\).
• ## Step2: Estimate \(\sqrt{125}\).
• We know that \(11^2=121\) and \(12^2 = 144\), so \(\sqrt{125}\approx11\).
• ## Step3: Calculate \(t\).
• \(t=\frac{\sqrt{125}}{4}\approx\frac{11}{4}=2.75\approx3\) (rounded to the nearest integer).
• Answer: \(t\approx3\) seconds.

1. Question 13:

• Explanation:
• ## Step1: Given the formula for the radius of a circle \(r = \frac{1}{3}\sqrt{A}\), and \(A = 452\).
• Substitute \(A = 452\) into the formula: \(r=\frac{1}{3}\sqrt{452}\).
• ## Step2: Estimate \(\sqrt{452}\).
• We know that \(21^2=441\) and \(22^2 = 484\), so \(\sqrt{452}\approx21\).
• ## Step3: Calculate \(r\).
• \(r=\frac{1}{3}\sqrt{452}\approx\frac{21}{3}=7\).
• Answer: \(r\approx7\) feet.

1. Question 14:

• Explanation:
• ## Step1: Calculate the actual value of \(\sqrt{397}\).
• We know that \(19^2 = 361\) and \(20^2=400\). Using a calculator, \(\sqrt{397}\approx19.92\).
• ## Step2: Analyze the error.
• The class - mate estimated \(\sqrt{397}\) to be about 200. The error is that the class - mate likely confused \(\sqrt{397}\) with \(397^2\). \(\sqrt{397}\) is a number between 19 and 20, not 200.
• Answer: The class - mate likely confused \(\sqrt{397}\) with \(397^2\). The actual value of \(\sqrt{397}\approx19.92\), which is much less than 200.

1. Question 15:

• Explanation:
• ## Step1: For a non - perfect square, say 17.
• We find the two closest perfect squares. \(4^2 = 16\) and \(5^2=25\).
• ## Step2: Estimate the square root.
• \(\sqrt{16}<\sqrt{17}<\sqrt{25}\), so \(4 < \sqrt{17}<5\). To get a more accurate estimate, we can use linear interpolation. Let \(x = 17\), \(a = 16\), \(b = 25\), \(y_1=4\), \(y_2 = 5\). The formula for linear interpolation is \(y=y_1+\frac{x - a}{b - a}(y_2 - y_1)\). Substituting the values: \(y = 4+\frac{17 - 16}{25 - 16}(5 - 4)=4+\frac{1}{9}\approx4.11\).
• Answer: For example, for the non - perfect square 17, \(\sqrt{17}\) is between 4 and 5. A more accurate estimate is about 4.11. We first find the two closest perfect squares (16 and 25), then use linear interpolation to get a better estimate.

Answer:

1. Question 12:

• Explanation:
• ## Step1: Given the formula \(t = \frac{\sqrt{h}}{4}\), and \(h = 125\).
• Substitute \(h = 125\) into the formula: \(t=\frac{\sqrt{125}}{4}\).
• ## Step2: Estimate \(\sqrt{125}\).
• We know that \(11^2=121\) and \(12^2 = 144\), so \(\sqrt{125}\approx11\).
• ## Step3: Calculate \(t\).
• \(t=\frac{\sqrt{125}}{4}\approx\frac{11}{4}=2.75\approx3\) (rounded to the nearest integer).
• Answer: \(t\approx3\) seconds.

1. Question 13:

• Explanation:
• ## Step1: Given the formula for the radius of a circle \(r = \frac{1}{3}\sqrt{A}\), and \(A = 452\).
• Substitute \(A = 452\) into the formula: \(r=\frac{1}{3}\sqrt{452}\).
• ## Step2: Estimate \(\sqrt{452}\).
• We know that \(21^2=441\) and \(22^2 = 484\), so \(\sqrt{452}\approx21\).
• ## Step3: Calculate \(r\).
• \(r=\frac{1}{3}\sqrt{452}\approx\frac{21}{3}=7\).
• Answer: \(r\approx7\) feet.

1. Question 14:

• Explanation:
• ## Step1: Calculate the actual value of \(\sqrt{397}\).
• We know that \(19^2 = 361\) and \(20^2=400\). Using a calculator, \(\sqrt{397}\approx19.92\).
• ## Step2: Analyze the error.
• The class - mate estimated \(\sqrt{397}\) to be about 200. The error is that the class - mate likely confused \(\sqrt{397}\) with \(397^2\). \(\sqrt{397}\) is a number between 19 and 20, not 200.
• Answer: The class - mate likely confused \(\sqrt{397}\) with \(397^2\). The actual value of \(\sqrt{397}\approx19.92\), which is much less than 200.

1. Question 15:

• Explanation:
• ## Step1: For a non - perfect square, say 17.
• We find the two closest perfect squares. \(4^2 = 16\) and \(5^2=25\).
• ## Step2: Estimate the square root.
• \(\sqrt{16}<\sqrt{17}<\sqrt{25}\), so \(4 < \sqrt{17}<5\). To get a more accurate estimate, we can use linear interpolation. Let \(x = 17\), \(a = 16\), \(b = 25\), \(y_1=4\), \(y_2 = 5\). The formula for linear interpolation is \(y=y_1+\frac{x - a}{b - a}(y_2 - y_1)\). Substituting the values: \(y = 4+\frac{17 - 16}{25 - 16}(5 - 4)=4+\frac{1}{9}\approx4.11\).
• Answer: For example, for the non - perfect square 17, \(\sqrt{17}\) is between 4 and 5. A more accurate estimate is about 4.11. We first find the two closest perfect squares (16 and 25), then use linear interpolation to get a better estimate.