answer:**Analysis** This question examines the operation of geometric probability models. The approach is to first determine the total length of the experiment and the length of the region constituting the event, and then calculate the ratio, which is a basic question. Let Delta =4m^2-2 > 0, solving for m in the interval (0,1) gives the range (frac{sqrt{2}}{2},1), then the probability sought is the ratio of the length of the interval that meets the condition to the length of the interval (0,1). **Solution** Given: The equation x^2+2mx+frac{1}{2}=0 has two distinct real roots, Therefore, Delta =4m^2-2 > 0, solving gives m > frac{sqrt{2}}{2} or m < -frac{sqrt{2}}{2} (discard), Therefore, frac{sqrt{2}}{2} < m < 1, Therefore, the probability that the equation x^2+2mx+frac{1}{2}=0 has two distinct real roots is P= frac{1- frac{sqrt{2}}{2}}{1-0}= frac{2- sqrt{2}}{2}, Hence, the answer is boxed{frac{2- sqrt{2}}{2}}.

answer:Given a sphere O with a radius of 9, and the center of the sphere is O. The section of the sphere O cut by a certain plane is circle M. We are tasked with finding the maximum volume of a cone with circle M as the base and O as the vertex. Let's denote the radius of circle M as r and the height of the cone as h. According to the Pythagorean theorem, in the right triangle formed by the radius of the sphere, the radius of the circle M, and the height of the cone, we have: [r^{2} + h^{2} = 9^2 = 81.] The volume V of a cone is given by: [V = frac{1}{3}pi r^2 h.] Substituting r^2 = 81 - h^2 into the formula for volume, we get: [V = frac{1}{3}pi (81 - h^2) h.] So, we define a function f(h) = frac{1}{3}pi (81 - h^2) h where the height h ranges from 0 to 9. To find the maximum volume, we differentiate f(h) with respect to h: [f'(h) = frac{1}{3}pi [(-2h)h + (81 - h^2)] = frac{1}{3}pi (81 - 3h^2).] Now, analyzing f'(h), we note that for h in (0,3sqrt{3}), f'(h) > 0, indicating that f(h) is increasing. For h in [3sqrt{3},9), f'(h) < 0, indicating that f(h) is decreasing. Therefore, f(h) reaches its maximum when h = 3sqrt{3}. Substituting h = 3sqrt{3} into f(h), we find the maximum volume: [f(3sqrt{3}) = frac{1}{3}pi (81 - (3sqrt{3})^2) (3sqrt{3}) = frac{1}{3}pi (81 - 27) (3sqrt{3}) = frac{1}{3}pi cdot 54 cdot 3sqrt{3} = 54sqrt{3}pi.] Hence, the maximum volume of the cone is boxed{54sqrt{3}pi}.

answer:1. **Determine the number of multiples of 5 between 1 and 2020**: - A number ( k ) is a multiple of 5 if ( k = 5 times n ) for some integer ( n ). - We are to find the largest ( n ) such that ( 5 times n leq 2020 ). [ n = leftlfloor frac{2020}{5} rightrfloor = leftlfloor 404 rightrfloor = 404 ] Hence, there are 404 multiples of 5 in the range, i.e., ( M = 404 ). 2. **Determine the number of multiples of 20 between 1 and 2020**: - A number ( k ) is a multiple of 20 if ( k = 20 times n ) for some integer ( n ). - We are to find the largest ( n ) such that ( 20 times n leq 2020 ). [ n = leftlfloor frac{2020}{20} rightrfloor = leftlfloor 101 rightrfloor = 101 ] Hence, there are 101 multiples of 20 in the range, i.e., ( N = 101 ). 3. **Calculate ( 10M div N )**: - Substitute the values of ( M ) and ( N ) into the expression ( 10 M div N ). [ 10 M = 10 times 404 = 4040 ] [ frac{4040}{101} = 40 ] 4. **Conclusion**: The value of ( 10 M div N ) is [ boxed{40} ]

answer:To solve this problem, we start by clarifying the given conditions and constructing a potential counterexample: 1. The problem deals with a 4028 times 4028 board where the cells are painted either black or white. 2. The board's coloring is such that any 2018 times 2018 sub-grid, regardless of its orientation or flipping (rotations and reflections), contains an equal number of black and white cells. Given this, we need to determine if the whole board must necessarily be colored in a chessboard pattern. Let's examine the reference solution step by step. 1. **Construct a Hypothetical Coloring Pattern:** - Paint the entire 4028 times 4028 board in a chessboard pattern (alternating black and white cells). 2. **Modify the Center:** - Now, modify the coloring of a central 4 times 4 block, as illustrated (this specific pattern isn't included but we'll assume a possible modification where the color alternation is broken). 3. **Verification:** - Test to see if any 2018 times 2018 sub-grid still maintains the property of having equal numbers of black and white cells. 4. **Verification in Non-Modified Zones:** - Clearly, if a 2018 times 2018 grid does not include the modified 4 times 4 block, it will have an equal number of black and white cells due to the initial chessboard pattern. This is intuitive because the chessboard inherently balances the black and white cells across any sufficiently large region. 5. **Verification Including the Modified Block:** - Consider a 2018 times 2018 sub-grid intersecting the modified 4 times 4 block. Given 2017 = (4028 / 2) + 3, this sub-grid will span across the boundary of the central modification by at most 1 cell row or column, ensuring that the deviation introduced by the modification is minimal: - When this sub-grid intersects the 4 times 4 block, it will either include exactly 2 black and 2 white cells from this 4 times 4 pattern's rows/columns: - Fundamentally, any additions or removals within the overall pattern still satisfy the black-white balance as long as symmetrical conceptual eliminations are done (2 white, 2 black). This reasoning shows that the balance persists even with the modification. Thus, it's clear that: **Conclusion:** Not all cells need to be painted strictly in a chessboard pattern. The example modification shows that certain alterations still allow the grid to fulfill the required conditions. [ boxed{text{No, it is not necessarily colored in a chessboard pattern}} ]