answer:To find the coefficient of x in the expansion of (1-3x)^8, we use the general term formula of the binomial expansion, which is T_{r+1} = C_{8}^{r} cdot (-1)^r cdot x^{frac{r}{3}}. Setting frac{r}{3} = 1 gives us r = 3. Therefore, the coefficient of x in the expansion is -56. Hence, the answer is boxed{-56}. In the general term formula of a binomial expansion, by setting the exponent of x to 1, we can find the value of r and thus determine the coefficient of x. This question primarily tests the application of the Binomial Theorem and the general term formula of a binomial expansion, and it is considered a basic question.

answer:Start by dividing the smallest four-digit number 1000 by the selected divisor 13. This division will give us the quotient and help determine if an additional multiplication step is needed. [ 1000 div 13 approx 76.92 ] This implies that 76 times 13 will still yield a three-digit number because: [ 76 times 13 = 988 ] However, 77 times 13 will give: [ 77 times 13 = 1001 ] Therefore, 1001 is the smallest four-digit multiple of 13. [ boxed{1001} ]

answer:We start by defining the problem clearly with the initial setup and the necessary constraints. Given: - n geq 2 is a positive integer. - lambda is a positive real number. - n fleas are on a horizontal line, initially not all at the same point. We define a move as choosing two fleas at points A (left) and B (right) such that the flea at A jumps over the flea at B to a new point C. The positions are chosen such that: [ frac{BC}{AB} = lambda ] Claim: We need to determine all values of lambda such that all fleas can be moved to positions strictly to the right of any given point M on the line. # Part 1: Sufficient Condition for lambda geq frac{1}{n-1} 1. **Strategy:** - We will prove that if lambda geq frac{1}{n-1}, it is possible to move all fleas arbitrarily far to the right. - To achieve this, select the leftmost flea and let it jump over the rightmost flea. 2. **Distance Notations:** - Let d denote the maximum distance between two fleas. - Let delta denote the minimum distance between any two fleas. Clearly, at any moment: [ d geq (n-1)delta ] 3. **Effect of a Move:** - When the leftmost flea, at point A, jumps over the rightmost flea, at point B, the new point C is such that: [ BC = lambda AB ] Since we need to ensure delta does not decrease: [ lambda d geq delta ] Given d geq (n-1)delta, we have: [ lambda (n-1) delta geq delta implies lambda (n-1) geq 1 implies lambda geq frac{1}{n-1} ] 4. **Conclusion for lambda geq frac{1}{n-1}:** - The position of the leftmost flea moves to the right by at least delta with each move. - Hence, we can move the fleas arbitrarily far to the right after a finite number of moves. # Part 2: Necessary Condition for lambda < frac{1}{n-1} 5. **Suppose lambda < frac{1}{n-1}:** - We will show that under this assumption there exists a point M that cannot be reached by any flea. 6. **Coordinate Notations and Sum:** - Assign each flea a coordinate on the real axis. - Let s_k be the sum of all flea coordinates at step k. - Let w_k be the coordinate of the rightmost flea at step k. Clearly: [ s_k leq n w_k ] 7. **Analyzing a Move:** - Suppose a flea at A (coordinate a) jumps over a flea at B (coordinate b) to C (coordinate c). - The change in sum due to this move: [ s_{k+1} - s_k = c - a ] Applying the given rule: [ lambda (b-a) = c - b ] Hence: [ c = b + lambda (b-a) ] Substituting, we get: [ s_{k+1} - s_k = (1+lambda)(b-a) = frac{1+lambda}{lambda}(c-b) ] Since b is the coordinate of the rightmost flea: [ w_{k+1} geq c geq b ] Hence: [ s_{k+1} - s_k geq frac{1+lambda}{lambda} (w_{k+1} - w_k) ] 8. **Summing Inequalities:** - Summing from k=0 to n-1: [ s_n - s_0 geq frac{1+lambda}{lambda} (w_n - w_0) ] Using s_n leq n w_n, we get: [ n w_n geq s_n geq s_0 + frac{1+lambda}{lambda} (w_n - w_0) ] Rearranging: [ left(frac{1+lambda}{lambda} - n right) w_n leq frac{1+lambda}{lambda} w_0 - s_0 ] Given lambda < frac{1}{n-1}: [ frac{1+lambda}{lambda} - n > 0 ] This implies w_n is bounded, as w_0 and s_0 are finite, ensuring that w_n must also remain finite within a bound. # Conclusion: Thus, the only values of lambda that ensure all fleas can be moved arbitrarily far to the right are: [ boxed{lambda geq frac{1}{n-1}} ]

answer:1. First, calculate the total resistance in the original circuit configuration. [ R_{text{total}} = frac{R}{2} + r ] Plugging in the given values: [ R = 50 text{ Ω} quad text{and} quad r = 20 text{ Ω} ] [ R_{text{total}} = frac{50}{2} + 20 = 25 + 20 = 45 text{ Ω} ] 2. Calculate the total current in the circuit. [ I_{text{total}} = frac{U_0}{R_{text{total}}} ] Given ( U_0 = 45 text{ V} ): [ I_{text{total}} = frac{45 text{ V}}{45 text{ Ω}} = 1 text{ A} ] 3. Determine the initial readings of the ammeter and the voltmeter. - The voltmeter measures the voltage across the internal resistance ( r ): [ U_1 = I_{text{total}} cdot r = 1 text{ A} cdot 20 text{ Ω} = 20 text{ V} ] - The ammeter measures the current through ( R/2 ): [ I_1 = frac{I_{text{total}}}{2} = frac{1 text{ A}}{2} = 0.5 text{ A} ] 4. Now consider the circuit with the ammeter and voltmeter positions swapped. 5. Calculate the new current passing through the circuit: [ I_2 = frac{U_0}{R} = frac{45 text{ V}}{50 text{ Ω}} = 0.9 text{ A} ] - This is the new reading of the ammeter. 6. Calculate the current flowing through the resistor ( r ): [ I = frac{U_0}{R + r} = frac{45 text{ V}}{50 text{ Ω} + 20 text{ Ω}} = frac{45 text{ V}}{70 text{ Ω}} approx 0.642857 text{ A} ] 7. Use this current to find the new reading of the voltmeter: [ U_2 = I cdot r = 0.642857 text{ A} cdot 20 text{ Ω} approx 12.86 text{ V} ] 8. Calculate the change in the ammeter reading: [ Delta I = I_2 - I_1 = 0.9 text{ A} - 0.5 text{ A} = 0.4 text{ A} ] 9. Calculate the change in the voltmeter reading: [ Delta U = U_1 - U_2 = 20 text{ V} - 12.86 text{ V} approx 7.14 text{ V} ] # Conclusion: The ammeter reading increased by ( 0.4 text{ A} ), and the voltmeter reading decreased by ( 7.14 text{ V} ). [ boxed{begin{aligned} &text{Ammeter increased by } 0.4 text{ A} &text{Voltmeter decreased by } 7.14 text{ V}end{aligned}} ]