1. Between which two integers do the following surds lie? \( \begin{array}{ll}\text { a) } \sqrt{56} & \text { b) }-\sqrt{12} \\ \text { c) }-\sqrt{78} & \text { d) } \sqrt{15} \\ \text { e) }-\sqrt{43} & \text { f) }-\sqrt{29} \\ \text { g) } \sqrt{99} & \text { h) } \sqrt{8}\end{array} \)
Solución de tutoría real
Responder
Solución
Revisado y aprobado por el equipo de tutoría de UpStudy
Beyond the Answer
To find between which two integers the given surds lie, we can estimate their values. For \( \sqrt{56} \), we know \( 7^2 = 49 \) and \( 8^2 = 64 \), so \( \sqrt{56} \) lies between 7 and 8. For \( -\sqrt{12} \), since \( \sqrt{12} \) is approximately 3.46, it lies between -4 and -3. For \( -\sqrt{78} \), knowing \( 8^2 = 64 \) and \( 9^2 = 81 \), \( \sqrt{78} \) is approximately 8.83, placing it between -9 and -8. For \( \sqrt{15} \), knowing \( 3^2 = 9 \) and \( 4^2 = 16 \), \( \sqrt{15} \) is about 3.87, so it lies between 3 and 4. For \( -\sqrt{43} \), with \( \sqrt{43} \) approximately 6.56, it falls between -7 and -6. For \( -\sqrt{29} \), since \( \sqrt{29} \) is approximately 5.39, it lies between -6 and -5. For \( \sqrt{99} \), with \( 9^2 = 81 \) and \( 10^2 = 100 \), \( \sqrt{99} \) is about 9.95, placing it between 9 and 10. Finally, for \( \sqrt{8} \), approximately 2.83, it lies between 2 and 3.