Расшифруйте название прибора, применяемого для измерения углов.
| О | $-\frac{1}{6} + (-\frac{3}{5})$ |
|---|---|
| Р | $-\frac{7}{8} + \frac{2}{9}$ |
| А | $\frac{1}{4} + (-\frac{11}{12})$ |
| Т | $-\frac{17}{27} + (-\frac{1}{3})$ |
| И | $-\frac{5}{14} + \frac{3}{4}$ |
| С | $-\frac{5}{6} + (-\frac{2}{9})$ |
| Б | $-\frac{13}{24} + \frac{5}{16}$ |
| Я | $\frac{7}{8} + (-\frac{3}{10})$ |
| Л | $\frac{2}{7} + (-\frac{5}{14})$ |
| $-\frac{2}{3}$ | $-1\frac{1}{18}$ | $-\frac{26}{27}$ | $-\frac{47}{72}$ | $-\frac{23}{30}$ | $-\frac{1}{14}$ | $\frac{23}{40}$ | $-\frac{11}{48}$ | $\frac{11}{28}$ | $\frac{23}{40}$ |
|---|---|---|---|---|---|---|---|---|---|
О = $-\frac{1}{6} + (-\frac{3}{5}) = -(\frac{5}{30} + \frac{18}{30}) = -\frac{23}{30}$
Р = $-\frac{7}{8} + \frac{2}{9} = -\frac{63}{72} + \frac{16}{72} = -(\frac{63}{72} - \frac{16}{72}) = -\frac{47}{72}$
А = $\frac{1}{4} + (-\frac{11}{12}) = \frac{3}{12} + (-\frac{11}{12}) = -(\frac{11}{12} - \frac{3}{12}) = -\frac{8}{12} = -\frac{2}{3}$
Т = $-\frac{17}{27} + (-\frac{1}{3}) = -(\frac{17}{27} + \frac{9}{27}) = -\frac{26}{27}$
И = $-\frac{5}{14} + \frac{3}{4} = -\frac{10}{28} + \frac{21}{28} = \frac{21}{28} - \frac{10}{28} = \frac{11}{28}$
С = $-\frac{5}{6} + (-\frac{2}{9}) = -(\frac{15}{18} + \frac{4}{18}) = -\frac{19}{18} = -1\frac{1}{18}$
Б = $-\frac{13}{24} + \frac{5}{16} = -\frac{26}{48} + \frac{15}{48} = -(\frac{26}{48} - \frac{15}{48}) = -\frac{11}{48}$
Я = $\frac{7}{8} + (-\frac{3}{10}) = \frac{35}{40} + (-\frac{12}{40}) = \frac{35}{40} - \frac{12}{40} = \frac{23}{40}$
Л = $\frac{2}{7} + (-\frac{5}{14}) = \frac{4}{14} + (-\frac{5}{14}) = -(\frac{5}{14} - \frac{4}{14}) = -\frac{1}{14}$
| $-\frac{2}{3}$ | $-1\frac{1}{18}$ | $-\frac{26}{27}$ | $-\frac{47}{72}$ | $-\frac{23}{30}$ | $-\frac{1}{14}$ | $\frac{23}{40}$ | $-\frac{11}{48}$ | $\frac{11}{28}$ | $\frac{23}{40}$ |
|---|---|---|---|---|---|---|---|---|---|
| А | С | Т | Р | О | Л | Я | Б | И | Я |
Ответ: АСТРОЛЯБИЯ