answer:To calculate the total number of accidents across all three highways, we need to find the accident rate for each highway and then apply it to the total traffic for that highway. For Highway A: The accident rate is 75 accidents per 100 million vehicles. Total traffic is 2.5 billion vehicles, which is 2,500 million vehicles. So, the number of accidents on Highway A = (75 accidents / 100 million vehicles) * 2,500 million vehicles = (75/100) * 2,500 = 0.75 * 2,500 = 1,875 accidents For Highway B: The accident rate is 50 accidents per 80 million vehicles. Total traffic is 1.6 billion vehicles, which is 1,600 million vehicles. So, the number of accidents on Highway B = (50 accidents / 80 million vehicles) * 1,600 million vehicles = (50/80) * 1,600 = 0.625 * 1,600 = 1,000 accidents For Highway C: The accident rate is 90 accidents per 200 million vehicles. Total traffic is 1.9 billion vehicles, which is 1,900 million vehicles. So, the number of accidents on Highway C = (90 accidents / 200 million vehicles) * 1,900 million vehicles = (90/200) * 1,900 = 0.45 * 1,900 = 855 accidents Now, we add the number of accidents from all three highways to get the total number of accidents: Total accidents = Accidents on Highway A + Accidents on Highway B + Accidents on Highway C = 1,875 + 1,000 + 855 = 3,730 accidents Therefore, the total number of accidents across all three highways last year was boxed{3,730} .

answer:In the context of the new problem, the distance given is the space diagonal of the cube. In terms of geometry, the space diagonal d of a cube with side length s is given by the formula: [ d = ssqrt{3} ] Given that the space diagonal is 10sqrt{3}, we can solve for the side length s: [ 10sqrt{3} = ssqrt{3} ] [ s = 10 ] Thus, the volume V of the cube can be calculated as: [ V = s^3 = 10^3 = 1000 ] So, the volume of the cube is boxed{1000} cubic units.

answer:1. **Initial Setup and Definitions**: - Let the initial rectangle be denoted as Pi with vertices A, B, C, D. - After a series of moves, the tokens form another rectangle Pi' which is similar to Pi but not equal to Pi. - We need to prove that Pi is a square. 2. **Properties of Rectangles and Similarity**: - Since Pi and Pi' are similar, the ratio of corresponding sides of Pi and Pi' is constant. - Let the side lengths of Pi be a and b with a neq b (assuming Pi is not a square initially). 3. **Invariance Under Moves**: - The allowed move is to shift one token in the direction perpendicular to the diagonal of Pi not containing this token. - This move preserves the perpendicularity of the sides and the angles of the quadrilateral. 4. **Diagonal Lengths**: - For a rectangle Pi with side lengths a and b, the lengths of the diagonals are given by: [ d_1 = sqrt{a^2 + b^2} ] [ d_2 = sqrt{a^2 + b^2} ] - Since Pi' is similar to Pi, the diagonals of Pi' will also be proportional to the diagonals of Pi. 5. **Invariance of the Expression**: - Consider the expression AB^2 + CD^2 - AC^2 - BD^2 for the quadrilateral ABCD. - For a rectangle, AB = CD = a and AC = BD = sqrt{a^2 + b^2}. - Thus, the expression simplifies to: [ AB^2 + CD^2 - AC^2 - BD^2 = a^2 + a^2 - (a^2 + b^2) - (a^2 + b^2) = 2a^2 - 2(a^2 + b^2) = -2b^2 ] - This expression is invariant under the allowed moves. 6. **Implication of Similarity**: - Since Pi' is similar to Pi but not equal, the side lengths of Pi' are ka and kb for some k neq 1. - The invariance of the expression implies that the ratio of the sides must remain the same, i.e., frac{a}{b} = frac{ka}{kb}. 7. **Conclusion**: - The only way for the rectangles to remain similar under the given moves and for the expression to remain invariant is if a = b. - Therefore, Pi must be a square. [ blacksquare ]

answer:To find the greatest number n that divides 1305, 4665, and 6905 leaving the same remainder in each case, we need to find the greatest common divisor (GCD) of the differences between these numbers. Let's calculate the differences between the pairs of numbers: 4665 - 1305 = 3360 6905 - 4665 = 2240 6905 - 1305 = 5600 Now, we need to find the GCD of these three differences: GCD(3360, 2240, 5600) We can simplify this by first finding the GCD of the first two numbers and then using that result to find the GCD with the third number. Let's start with the GCD of 3360 and 2240: GCD(3360, 2240) We can use the Euclidean algorithm to find the GCD: 3360 = 2240 * 1 + 1120 2240 = 1120 * 2 + 0 So the GCD of 3360 and 2240 is 1120. Now we find the GCD of 1120 and 5600: GCD(1120, 5600) Again, using the Euclidean algorithm: 5600 = 1120 * 5 + 0 So the GCD of 1120 and 5600 is 1120. Therefore, the greatest number n that divides 1305, 4665, and 6905 leaving the same remainder is 1120. Now, we need to find the sum of the digits in n: 1 + 1 + 2 + 0 = 4 The sum of the digits in n is boxed{4} .