jelyk hulle dan deur \( <,> \) of \( = \) in te vul: \( \begin{array}{lll}\frac{4}{5} \square \frac{9}{10} & \text { b) } \frac{3}{8} \square \frac{5}{16} & \text { c) } \frac{5}{6} \square \frac{11}{12} \\ \frac{13}{18} \square \frac{2}{3} & \text { e) } \frac{9}{21} \square \frac{3}{7} & \text { f) } \frac{6}{11} \square \frac{12}{33} \\ \text { yskik hierdie breuke in dalende volgorde deur almal } \\ \text { semer van } 48 \text { neer te skrvf: }\end{array} \)

Om die ongelykhede te bepaal, sal ons elke breuk vergelyk deur hulle na 'n gemeenskaplike noemer om te skakel. In hierdie geval is die gemeenskaplike noemer 48, wat ons sal gebruik om die breuke te herskryf. ### Stap 1: Breuke herskryf na 'n gemeenskaplike noemer 1. **a)** \(\frac{4}{5}\) en \(\frac{9}{10}\) - \(\frac{4}{5} = \frac{4 \times 48/5}{48} = \frac{38.4}{48}\) - \(\frac{9}{10} = \frac{9 \times 48/10}{48} = \frac{43.2}{48}\) 2. **b)** \(\frac{3}{8}\) en \(\frac{5}{16}\) - \(\frac{3}{8} = \frac{3 \times 48/8}{48} = \frac{18}{48}\) - \(\frac{5}{16} = \frac{5 \times 48/16}{48} = \frac{15}{48}\) 3. **c)** \(\frac{5}{6}\) en \(\frac{11}{12}\) - \(\frac{5}{6} = \frac{5 \times 48/6}{48} = \frac{40}{48}\) - \(\frac{11}{12} = \frac{11 \times 48/12}{48} = \frac{44}{48}\) 4. **d)** \(\frac{13}{18}\) en \(\frac{2}{3}\) - \(\frac{13}{18} = \frac{13 \times 48/18}{48} = \frac{36}{48}\) - \(\frac{2}{3} = \frac{2 \times 48/3}{48} = \frac{32}{48}\) 5. **e)** \(\frac{9}{21}\) en \(\frac{3}{7}\) - \(\frac{9}{21} = \frac{9 \times 48/21}{48} = \frac{20.57}{48}\) - \(\frac{3}{7} = \frac{3 \times 48/7}{48} = \frac{20.57}{48}\) 6. **f)** \(\frac{6}{11}\) en \(\frac{12}{33}\) - \(\frac{6}{11} = \frac{6 \times 48/11}{48} = \frac{26.18}{48}\) - \(\frac{12}{33} = \frac{12 \times 48/33}{48} = \frac{17.45}{48}\) ### Stap 2: Vergelyk die breuke Nou kan ons die breuke vergelyk: 1. **a)** \(\frac{4}{5} < \frac{9}{10}\) 2. **b)** \(\frac{3}{8} > \frac{5}{16}\) 3. **c)** \(\frac{5}{6} < \frac{11}{12}\) 4. **d)** \(\frac{13}{18} > \frac{2}{3}\) 5. **e)** \(\frac{9}{21} = \frac{3}{7}\) 6. **f)** \(\frac{6}{11} > \frac{12}{33}\) ### Stap 3: Vul die ongelykhede in Die finale ongelykhede is: \[ \begin{array}{lll} \frac{4}{5} < \frac{9}{10} & \text { b) } \frac{3}{8} > \frac{5}{16} & \text { c) } \frac{5}{6} < \frac{11}{12} \\ \frac{13}{18} > \frac{2}{3} & \text { e) } \frac{9}{21} = \frac{3}{7} & \text { f) } \frac{6}{11} > \frac{12}{33} \\ \end{array} \] ### Stap 4: Rangskik die breuke in dalende volgorde Die breuke in dalende volgorde, gebaseer op die waarde wat ons bereken het, is: 1. \(\frac{11}{12}\) 2. \(\frac{9}{10}\) 3. \(\frac{5}{6}\) 4. \(\frac{13}{18}\) 5. \(\frac{3}{8}\) 6. \(\frac{6}{11}\) 7. \(\frac{3}{7}\) 8. \(\frac{9}{21}\) 9. \(\frac{2}{3}\) 10. \(\frac{12}{33}\) 11. \(\frac{5}{16}\) Die finale rangskikking is: \[ \frac{11}{12}, \frac{9}{10}, \frac{5}{6}, \frac{13}{18}, \frac{3}{8}, \frac{6}{11}, \frac{3}{7}, \frac{9}{21}, \frac{2}{3}, \frac{12}{33}, \frac{5}{16} \]