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question:Let the complex number z satisfy frac{z+2overline{z}}{1}=3i (where i is the imaginary unit). Find z.
answer:Let z=x+yi, thus overline{z}=x-yi. Substituting into z+2overline{z}, we get x+yi+2(x-yi)=3-i, which simplifies to 3x-yi=3-i. Therefore, x=1 and y=1. Hence, z=1+i. So, the answer is boxed{1+i}. By setting z=x+yi and then substituting overline{z}=x-yi into z+2overline{z}, and using the necessary and sufficient condition for the equality of complex numbers, we can find the values of x and y, and thus determine the answer. This problem tests the mixed operation of complex numbers in algebraic form and the necessary and sufficient condition for the equality of complex numbers, making it a basic question.
question:Ten distinct natural numbers are such that the product of any 5 of them is even, and the sum of all 10 numbers is odd. What is their smallest possible sum?
answer:1. Let's establish some basic properties using the information from the problem. Any set of 10 different natural numbers is given such that the product of any 5 of them is even, and their total sum is odd. 2. The product of any 5 numbers being even implies that at least one of these 5 numbers must be even; otherwise, the product would be odd. Hence, there cannot be 5 odd numbers within any subset of 10 numbers. Therefore, the total number of odd numbers must be at most 4. 3. Given the sum of these 10 numbers must be odd, an essential concept to consider is that the sum of an odd number of odd numbers and an even number of even numbers is odd. Notice that the sum of an even number of even numbers is even, and the sum of an even number of odd numbers is even. Thus, the sum of an odd number of odd numbers must offset this even sum to keep the total sum odd. 4. Therefore, the count of odd numbers must be an odd number. The maximum count of odd numbers can be 3, because having 4 odd numbers would result in an even total sum, contradicting the given condition of an odd total sum. Hence, there must be exactly 3 odd numbers among the 10. 5. Given the count of odd numbers is 3, the count of even numbers has to be (10 - 3 = 7). 6. To minimize the sum of these 10 numbers, we choose the smallest possible natural numbers. For the odd numbers, the smallest are: (1, 3, 5). For the even numbers, the smallest are: (2, 4, 6, 8, 10, 12, 14). 7. Now, we compute the sum of these chosen numbers: [ begin{align*} text{Sum of odd numbers} &= 1 + 3 + 5 = 9, text{Sum of even numbers} &= 2 + 4 + 6 + 8 + 10 + 12 + 14 = 56, text{Total sum} &= 9 + 56 = 65. end{align*} ] Hence, the minimum possible sum of the 10 numbers, adhering to the given conditions, is ( boxed{65} ).
question:A technician has to visit a total of 5 service centers (A, B, C, D, and E) along the same route, making a roundtrip by returning back to center A. The technician completes the drive to center B, then finishes 40% of the journey from center B to center C, and after that, completes 60% of the drive from center C to center D. What percent of the entire roundtrip, including the return trip to center A, has the technician completed?
answer:To calculate the percentage of the entire round-trip the technician has completed, we need to know the distances between the service centers or at least the relative distances. Since we don't have specific distances, we'll assume that the distances between each center are equal for the sake of calculation. Let's denote the distance between each center as "d". The round-trip consists of the following segments: - A to B - B to C - C to D - D to E - E to A (return trip) Since there are 5 centers, there are 4 segments in one direction and 4 segments in the return direction, making a total of 8 segments for the entire round-trip. The technician has completed: - The entire segment from A to B (1d) - 40% of the segment from B to C (0.4d) - 60% of the segment from C to D (0.6d) Adding these up, the technician has completed: 1d + 0.4d + 0.6d = 2d Since the entire round-trip consists of 8 segments (8d), the percentage of the round-trip completed is: (2d / 8d) * 100% = (1/4) * 100% = 25% Therefore, the technician has completed boxed{25%} of the entire round-trip.
question:The number of points on the circle ((x3)^{2}+(y3)^{2}=9) that are at a distance of 1 from the line (3x + 4y 11 = 0) is A. 1 B. 2 C. 3 D. 4
answer:1. **Identify the Circle and Line Equation:** The equation of the circle is ((x-3)^2 + (y-3)^2 = 9), and the equation of the line is (3x + 4y - 11 = 0). 2. **Calculate the Circle's Radius:** The given circle equation ((x-3)^2 + (y-3)^2 = 9) describes a circle centered at ((3, 3)), with radius (r = sqrt{9} = 3). 3. **Determine the Distance from Circle's Center to the Line:** We use the formula for the distance (d) from a point ((x_0, y_0)) to a line (Ax + By + C = 0): [ d = frac{|Ax_0 + By_0 + C|}{sqrt{A^2 + B^2}} ] Substituting ((x_0, y_0) = (3, 3)), and (A = 3), (B = 4), and (C = -11): [ d = frac{|3(3) + 4(3) - 11|}{sqrt{3^2 + 4^2}} = frac{|9 + 12 - 11|}{sqrt{9 + 16}} = frac{|10|}{sqrt{25}} = frac{10}{5} = 2 ] Thus, the distance from the circle's center to the line is (d = 2). 4. **Analyze the Relationship Between the Radius, Distance, and Given Condition:** Given the condition that points on the circle are at a distance of 1 from the line, examine: [ text{Radius} = text{Distance from center to line} + 1 ] Substituting the known values: [ r = d + 1 rightarrow 3 = 2 + 1 ] This confirms the condition holds. 5. **Determine the Number of Intersection Points:** Since the radius (r = 3) and the distance from the center to the line (d = 2), points that are at a distance (1) from the line and lie on the circle are determined geometrically. The configurations can be visualized as there are three such points on the circle due to the relative positions. **Conclusion:** Thus, the number of points on the circle that have a distance of 1 from the line is: [ boxed{text{C. 3}} ]