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) \( -\sqrt{54} \) \( = \) \( = \)
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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. \( -\sqrt{54} = -8 < -\sqrt{54} < -7 \)