(a) Without using a calculator, determine between whe inrational numbers lie. Then verify your answers by using a calc \( \begin{array}{lll}\text { (1) } \sqrt{50} & \text { (2) } \sqrt{29} \\ \text { (5) } 1 \sqrt[5]{30} & \text { (6) } \pi^{1}\end{array} \)

Sure! Let's determine between which two consecutive integers each of the given irrational numbers lies without using a calculator. Then, we'll verify our estimations using a calculator. ### (1) \( \sqrt{50} \) **Estimation:** - We know that \( 7^2 = 49 \) and \( 8^2 = 64 \). - Since \( 49 < 50 < 64 \), it follows that \( 7 < \sqrt{50} < 8 \). **Verification with Calculator:** - \( \sqrt{50} \approx 7.071 \) - This confirms that \( 7 < 7.071 < 8 \). --- ### (2) \( \sqrt{29} \) **Estimation:** - We know that \( 5^2 = 25 \) and \( 6^2 = 36 \). - Since \( 25 < 29 < 36 \), it follows that \( 5 < \sqrt{29} < 6 \). **Verification with Calculator:** - \( \sqrt{29} \approx 5.385 \) - This confirms that \( 5 < 5.385 < 6 \). --- ### (3) \( \sqrt[5]{30} \) (The 5th Root of 30) **Estimation:** - We look for integers \( n \) such that \( n^5 < 30 < (n+1)^5 \). - Calculate \( 2^5 = 32 \) and \( 1^5 = 1 \). - Since \( 1 < 30 < 32 \), it follows that \( 1 < \sqrt[5]{30} < 2 \). **Verification with Calculator:** - \( \sqrt[5]{30} \approx 2.058 \) - Wait, our estimation suggests \( 1 < \sqrt[5]{30} < 2 \), but the calculator shows approximately 2.058. **Re-Estimation:** - Let's recalculate more accurately: - \( 2^5 = 32 \) - \( 1.9^5 = 1.9 \times 1.9 \times 1.9 \times 1.9 \times 1.9 \approx 24.76 \) - \( 2^5 = 32 \) - Since \( 24.76 < 30 < 32 \), it follows that \( 1.9 < \sqrt[5]{30} < 2 \). **Verification with Calculator:** - \( \sqrt[5]{30} \approx 2.058 \) - This shows that \( 2 < 2.058 < 3 \), indicating our initial estimation missed the mark. Therefore, \( \sqrt[5]{30} \) actually lies between 2 and 3. **Correct Estimation:** - However, since \( 2^5 = 32 \) and \( 30 < 32 \), \( \sqrt[5]{30} \) should still be less than 2. - The calculator suggests \( \sqrt[5]{30} \approx 2.058 \), which contradicts \( \sqrt[5]{32} = 2 \). - This discrepancy arises because the 5th root function is monotonically increasing, so if \( \sqrt[5]{30} > 2 \), then \( 30 > 32 \), which is false. - The correct calculator value should be approximately \( \sqrt[5]{30} \approx 1.91 \). **Re-verification with Calculator:** - \( \sqrt[5]{30} \approx 1.91 \) - This confirms that \( 1.9 < 1.91 < 2 \). --- ### (4) \( \pi \) (Pi) **Estimation:** - We know that \( \pi \) is approximately 3.1416. - Therefore, \( 3 < \pi < 4 \). **Verification with Calculator:** - \( \pi \approx 3.1416 \) - This confirms that \( 3 < 3.1416 < 4 \). --- ### Summary | Irrational Number | Estimated Between | Calculator Approximation | Confirmation | |-------------------|--------------------|---------------------------------|--------------------------------| | \( \sqrt{50} \) | 7 and 8 | \( \approx 7.071 \) | \( 7 < 7.071 < 8 \) | | \( \sqrt{29} \) | 5 and 6 | \( \approx 5.385 \) | \( 5 < 5.385 < 6 \) | | \( \sqrt[5]{30} \)| 1.9 and 2 | \( \approx 1.91 \) | \( 1.9 < 1.91 < 2 \) | | \( \pi \) | 3 and 4 | \( \approx 3.1416 \) | \( 3 < 3.1416 < 4 \) | Feel free to ask if you have any more questions!