1. Let's start by calculating each symbol value step-by-step. 2. For $$\Omega$$: $$\Omega = 28 - \frac{2}{6} = 28 - \frac{1}{3} = \frac{84}{3} - \frac{1}{3} = \frac{83}{3}$$ 3. For $$\varepsilon$$, the formula given is $$5 + \frac{4}{5} x$$ but no value for $$x$$ is provided, so we move to another expression with $$\varepsilon$$ later. 4. For $$\Delta$$: $$\Delta = -32 + 1 = -31$$ 5. For $$\bigcirc$$ (◎): $$\bigcirc = 14 - \left( \frac{3}{5} + \frac{11}{6} - \frac{2}{5} \right)$$ Calculate inside the bracket first: $$ \frac{3}{5} - \frac{2}{5} = \frac{1}{5}$$ So inside the bracket: $$ \frac{1}{5} + \frac{11}{6} = \frac{6}{30} + \frac{55}{30} = \frac{61}{30}$$ Therefore: $$ \bigcirc = 14 - \frac{61}{30} = \frac{420}{30} - \frac{61}{30} = \frac{359}{30}$$ 6. For $$\triangledown$$ (▼): $$ \triangledown = -1 \times \bigcirc + 2 = -1 \times \frac{359}{30} + 2 = - \frac{359}{30} + \frac{60}{30} = -\frac{299}{30}$$ 7. For $$\varepsilon$$ again: Given two expressions: $$ \varepsilon = 5 + \frac{4}{5} x $$ (unsolved without $$x$$) and $$ \varepsilon = 16 - (4 - 8) - 18 + (-2) $$ Calculate the second expression: $$ 4 - 8 = -4$$ So: $$ 16 - (-4) - 18 + (-2) = 16 + 4 - 18 - 2 = 0$$ Therefore, $$\varepsilon = 0$$ from the second expression. 8. For $$\bullet$$ (•): $$ \bullet = -6 \times \frac{3}{4} + \left(-\frac{1}{2}\right) = -\frac{18}{4} - \frac{1}{2} = -\frac{9}{2} - \frac{1}{2} = -5$$ 9. For $$\varepsilon$$ again, given: $$ \varepsilon = -1 - \blacksquare$$ We know $$\varepsilon = 0$$ from step 7, so: $$ 0 = -1 - \blacksquare \implies \blacksquare = -1$$ 10. For $$\blacksquare$$ (■): Given: $$ \blacksquare = (\bigcirc - \triangledown) + 2 $$ Calculate: $$ \bigcirc - \triangledown = \frac{359}{30} - \left(-\frac{299}{30}\right) = \frac{359}{30} + \frac{299}{30} = \frac{658}{30} = \frac{329}{15}$$ Add 2: $$ \blacksquare = \frac{329}{15} + 2 = \frac{329}{15} + \frac{30}{15} = \frac{359}{15}$$ But earlier we found $$\blacksquare = -1$$ indicating contradictory equations or different instances; the problem likely requires choosing the value from the previous equation (step 9). 11. Summary results: $$\Omega = \frac{83}{3}$$ $$\Delta = -31$$ $$\bigcirc = \frac{359}{30}$$ $$\triangledown = -\frac{299}{30}$$ $$\varepsilon = 0$$ $$\bullet = -5$$ $$\blacksquare = -1$$ 12. The tables provided list values with Arabic headers; they are data sets for verification but no operations requested. Final answers are as above.