1. Problem 65: A miller takes 1/10 of the flour he grinds as his fee. A customer has exactly one bushel left after the fee. How much flour did the miller grind? 2. Let the total amount of flour ground be $x$ bushels. 3. The miller takes $\frac{1}{10}x$ as his fee. 4. The customer is left with $x - \frac{1}{10}x = \frac{9}{10}x$ bushels. 5. We know the customer has exactly 1 bushel left, so: $$\frac{9}{10}x = 1$$ 6. Solve for $x$: $$x = \frac{1}{\frac{9}{10}} = \frac{10}{9}$$ 7. Therefore, the miller ground $\frac{10}{9}$ bushels of flour. 8. Problem 66: In rectangle ABCD, find the measure of $\angle a + \angle b$ in terms of $x$. 9. Since ABCD is a rectangle, all interior angles are right angles, i.e., $90^\circ$. 10. Angles $a$ and $b$ are adjacent angles at one corner inside the rectangle. 11. The sum of angles around a point is $360^\circ$, but inside the rectangle, the sum of angles at a corner is $90^\circ$. 12. Given the problem context, $\angle a + \angle b$ must equal the right angle at that corner, so: $$\angle a + \angle b = 90^\circ$$ 13. If the problem provides expressions for $a$ and $b$ in terms of $x$, sum them accordingly. Without explicit expressions, the sum remains $90^\circ$. Final answers: - Problem 65: The miller ground $\frac{10}{9}$ bushels of flour. - Problem 66: $\angle a + \angle b = 90^\circ$.