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Class work Without using a Ealculator, determine between whieh tho integers the following irrational numbers lie D) \( \sqrt{50} \) \( = \) \( = \) \( \therefore \) (2) \( \sqrt{29} \) \( = \) \( = \) \( \therefore \) \( 3 \sqrt[3]{45} \) \( = \) \( = \) - \( \quad \) - (1) \( -\sqrt{54} \) \( = \) \( = \)

Ask by Frank Mccoy. in South Africa
Jan 23,2025

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\[ -\sqrt{54} \quad \text{lies between} \quad -8 \quad \text{and} \quad -7 \] \[ \sqrt{29} \quad \text{lies between} \quad 5 \quad \text{and} \quad 6 \] \[ \sqrt{50} \quad \text{lies between} \quad 7 \quad \text{and} \quad 8 \] \[ 3\sqrt[3]{45} \quad \text{lies between} \quad 10 \quad \text{and} \quad 11 \]

Solution

Sure, let's determine between which two integers each of the given irrational numbers lie without using a calculator. --- ### (1) \( -\sqrt{54} \) **Step 1:** Find the nearest perfect squares around 54. - \( 7^2 = 49 \) - \( 8^2 = 64 \) **Step 2:** Determine where 54 falls. - \( 49 < 54 < 64 \) **Step 3:** Apply square roots. - \( 7 < \sqrt{54} < 8 \) **Step 4:** Since we have a negative sign: - \( -8 < -\sqrt{54} < -7 \) **Conclusion:** \[ -\sqrt{54} \quad \text{lies between} \quad -8 \quad \text{and} \quad -7 \] --- ### (2) \( \sqrt{29} \) **Step 1:** Find the nearest perfect squares around 29. - \( 5^2 = 25 \) - \( 6^2 = 36 \) **Step 2:** Determine where 29 falls. - \( 25 < 29 < 36 \) **Step 3:** Apply square roots. - \( 5 < \sqrt{29} < 6 \) **Conclusion:** \[ \sqrt{29} \quad \text{lies between} \quad 5 \quad \text{and} \quad 6 \] --- ### (D) \( \sqrt{50} \) **Step 1:** Find the nearest perfect squares around 50. - \( 7^2 = 49 \) - \( 8^2 = 64 \) **Step 2:** Determine where 50 falls. - \( 49 < 50 < 64 \) **Step 3:** Apply square roots. - \( 7 < \sqrt{50} < 8 \) **Conclusion:** \[ \sqrt{50} \quad \text{lies between} \quad 7 \quad \text{and} \quad 8 \] --- ### (3) \( 3\sqrt[3]{45} \) **Step 1:** Find the nearest perfect cubes around 45. - \( 3^3 = 27 \) - \( 4^3 = 64 \) **Step 2:** Determine where 45 falls. - \( 27 < 45 < 64 \) **Step 3:** Apply cube roots. - \( 3 < \sqrt[3]{45} < 4 \) **Step 4:** Multiply by 3. - \( 3 \times 3 = 9 \) - \( 3 \times 4 = 12 \) - So, \( 9 < 3\sqrt[3]{45} < 12 \) **Refined Estimate:** For a more precise interval: - \( \sqrt[3]{45} \) is slightly greater than \( 3.5 \) (since \( 3.5^3 = 42.875 \) and \( 3.6^3 = 46.656 \)) - Therefore, \( 3.5 < \sqrt[3]{45} < 3.6 \) - Multiplying by 3: \( 10.5 < 3\sqrt[3]{45} < 10.8 \) **Conclusion:** \[ 3\sqrt[3]{45} \quad \text{lies between} \quad 10 \quad \text{and} \quad 11 \] --- ### Summary 1. **\( -\sqrt{54} \)** lies between **-8** and **-7**. 2. **\( \sqrt{29} \)** lies between **5** and **6**. 3. **\( \sqrt{50} \)** lies between **7** and **8**. 4. **\( 3\sqrt[3]{45} \)** lies between **10** and **11**. --- Feel free to ask if you need further explanations or assistance!

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To determine between which two integers the irrational numbers lie, let’s start with \( \sqrt{50} \). We know that \( 7^2 = 49 \) and \( 8^2 = 64 \). Since \( 50 \) is between \( 49 \) and \( 64 \), we conclude that \( \sqrt{50} \) lies between \( 7 \) and \( 8 \). Hence, \( 7 < \sqrt{50} < 8 \). Next, for \( \sqrt{29} \), we compare it to \( 5^2 = 25 \) and \( 6^2 = 36 \). Since \( 29 \) is between these two perfect squares, we find \( 5 < \sqrt{29} < 6 \). For \( 3 \sqrt[3]{45} \), first find \( \sqrt[3]{45} \). We know \( 3^3 = 27 \) and \( 4^3 = 64 \). So \( \sqrt[3]{45} \) lies between \( 3 \) and \( 4 \). Multiplying through by \( 3 \) gives \( 9 < 3\sqrt[3]{45} < 12 \). Finally, considering \( -\sqrt{54} \), we must keep in mind the negative sign. \( \sqrt{54} \) lies between \( 7 \) (since \( 7^2 = 49 \)) and \( 8 \) (since \( 8^2 = 64 \)). Thus, \( -8 < -\sqrt{54} < -7 \). Summarizing: 1. \( \sqrt{50} = 7 < \sqrt{50} < 8 \) 2. \( \sqrt{29} = 5 < \sqrt{29} < 6 \) 3. \( 3 \sqrt[3]{45} = 9 < 3 \sqrt[3]{45} < 12 \) 4. \( -\sqrt{54} = -8 < -\sqrt{54} < -7 \)

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