answer:Since the two sets are equal, {2, -1} = {2, a^2 - 2a}, we can equate their corresponding elements. From the sets, we notice that 2 is common in both, thus we only need to equate the other elements. Therefore, we have the following equation to solve for a: a^2 - 2a = -1 To solve this quadratic equation, we first bring all the terms to one side: a^2 - 2a + 1 = 0 Now, we attempt to factorize the quadratic expression: (a - 1)(a - 1) = 0 Setting each factor to zero gives us: a - 1 = 0 Solving for a: a = 1 Hence, the real number a is boxed{1}.

answer:1. **Given the circle equation and points on the circle:** The circle is given by the equation (x^2 + y^2 = r^2). The points where the circle meets the coordinate axes are (A = (r, 0)), (B = (-r, 0)), (C = (0, r)), and (D = (0, -r)). 2. **Points (P) and (Q) on the circle:** Let (P = (u, v)) and (Q = (-u, v)) be two points on the circumference of the circle. Since these points lie on the circle, they satisfy the circle's equation: [ u^2 + v^2 = r^2 ] 3. **Finding point (N):** The point (N) is the intersection of the line (PQ) with the (y)-axis. The line (PQ) is horizontal because both points have the same (y)-coordinate (v). Therefore, the line (PQ) is given by (y = v). The intersection of this line with the (y)-axis occurs at (x = 0), so (N = (0, v)). 4. **Finding point (M):** The point (M) is the foot of the perpendicular from (P) to the (x)-axis. Since (P = (u, v)), the perpendicular from (P) to the (x)-axis will be at (M = (u, 0)). 5. **Calculating distances:** - (|AM|): [ |AM| = |r - u| ] - (|BM|): [ |BM| = |-r - u| = r + u ] - (|DN|): [ |DN| = |v - (-r)| = v + r ] - (|PQ|): [ |PQ| = 2u ] 6. **Given conditions and solving for (u) and (v):** Given (r^2) is odd, (u = p^m > q^n = v), where (p) and (q) are prime numbers and (m) and (n) are natural numbers. We need to show that: [ |AM| = 1, |BM| = 9, |DN| = 8, |PQ| = 8 ] 7. **Solving the equation (x^2 + y^2 = r^2) with integer solutions:** Since (r) is odd, one of (x) or (y) must be even. Suppose (x = 2^n) for some positive integer (n). Then: [ y^2 = (r - 2^n)(r + 2^n) ] Since (r - 2^n) and (r + 2^n) are powers of the same prime and their greatest common divisor is 1, we conclude that: [ r - 2^n = 1 ] Therefore: [ y^2 = (r - 2^n)(r + 2^n) = 2^{n+1} + 1 ] Also: [ 2^{n+1} = y^2 - 1 = (y - 1)(y + 1) ] From this, (y - 1) and (y + 1) are two powers of two that are two units apart, so: [ y - 1 = 2 quad text{and} quad y + 1 = 4 quad Rightarrow quad y = 3 ] Thus: [ n = 2, quad x = 4, quad r = 5 ] 8. **Verifying the distances:** - (|AM| = |r - u| = |5 - 4| = 1) - (|BM| = r + u = 5 + 4 = 9) - (|DN| = v + r = 3 + 5 = 8) - (|PQ| = 2u = 2 times 4 = 8) Thus, the given conditions are satisfied. (blacksquare)

answer:To find out how many minutes Mark spends kneading the bread, we need to subtract the time spent on rising and baking from the total time. The bread needs to rise for 120 minutes twice, so that's 120 minutes + 120 minutes = 240 minutes. He also spends 30 minutes baking it. So, the total time spent on rising and baking is 240 minutes + 30 minutes = 270 minutes. The total time to finish making the bread is 280 minutes. Therefore, the time spent kneading the bread is 280 minutes - 270 minutes = boxed{10} minutes.

answer:To find the angle between the hands of a clock at 3:30, calculate the positions of the hour and minute hands separately and then find the difference between these positions. 1. **Position of the Hour Hand**: - The hour hand moves at (30^circ) per hour. - At 3:00, the hour hand is at (3 times 30^circ = 90^circ) from the top (12:00 position). - By 3:30, the hour hand moves further. In 30 minutes, the hour hand moves: [ left(frac{30^circ}{60}right) times 30 = 15^circ ] - Therefore, the position of the hour hand at 3:30 is: [ 90^circ + 15^circ = 105^circ ] 2. **Position of the Minute Hand**: - The minute hand moves at (6^circ) per minute. - At 30 minutes, the minute hand is: [ 30 times 6^circ = 180^circ ] - This is directly at the 6:00 position. 3. **Calculating the Angle Between the Hands**: - The angle between the hands is the absolute difference between their positions: [ |180^circ - 105^circ| = 75^circ ] The final angle between the clock hands at 3:30 is calculated to be (75^circ). Conclusion: [ 75^circ ] The final answer is boxed{textbf{(B)} 75^circ}