answer:1. **Equation of the Circle and the Tangent Lines:** Given the equation of the circle: [ x^2 + y^2 = 25 ] We need to draw tangent lines from the point ( P(2, -8) ) to the circle. The equations representing the relationships between the coordinates of the tangency points ( A left( x_1, y_1 right) ) and ( B left( x_2, y_2 right) ) with the given point ( P left( x_0, y_0 right) = (2, -8) ) are: [ begin{cases} x^2 + y^2 = 25 x_0 x + y_0 y = 25 end{cases} ] 2. **Determine the Tangency Points ( A ) and ( B ):** By solving the system of equations, we find the coordinates of ( A ) and ( B ): [ begin{aligned} x_1 & = 4.58, & y_1 &= -1.98 x_2 & = -3.14, & y_2 &= -3.91 end{aligned} ] 3. **Volume of the Solid of Revolution:** The volume of the solid generated by rotating the triangle ( triangle ABP ) around the ( X )-axis is given by: [ V = 2 pi d T ] where ( d ) is the average of the ordinates (the distance from the ( X )-axis), which can be calculated as: [ d = frac{y_0 + y_1 + y_2}{3} ] 4. **Area of Triangle ( T ):** The area ( T ) of triangle ( triangle ABP ) can be calculated using the determinant formula: [ T = left| frac{1}{2} begin{vmatrix} x_0 & y_0 & 1 x_1 & y_1 & 1 x_2 & y_2 & 1 end{vmatrix} right| ] 5. **Substitute the Coordinates:** Substituting the values ( (x_0, y_0) = (2, -8) ), ( (x_1, y_1) = (4.58, -1.98) ), and ( (x_2, y_2) = (-3.14, -3.91) ) into the determinant: [ T = left| frac{1}{2} begin{vmatrix} 2 & -8 & 1 4.58 & -1.98 & 1 -3.14 & -3.91 & 1 end{vmatrix} right| ] Calculate the determinant: [ begin{vmatrix} 2 & -8 & 1 4.58 & -1.98 & 1 -3.14 & -3.91 & 1 end{vmatrix} = 2(-1.98 - (-3.91)) - (-8)(4.58 - (-3.14)) + 1(4.58(-3.91) - (-3.14)(-1.98)) = 2(1.93) + 8(7.72) + 1(approx -19.31) = 3.86 + 61.76 - 19.31 = 46.31 ] Therefore: [ T = left| frac{1}{2} times 46.31 right| = 23.155 ] 6. **Calculate ( d ):** [ d = frac{y_0 + y_1 + y_2}{3} = frac{-8 - 1.98 - 3.91}{3} = frac{-13.89}{3} approx -4.63 ] 7. **Calculate the Volume ( V ):** [ V = 2 pi left| -4.63 right| times 23.155 = 2 pi times 4.63 times 23.155 approx 2 pi times 107.15 approx 2 times 3.14 times 107.15 approx 672.1 ] Therefore, the volume of the solid formed by the rotation is: [ boxed{672.1} ]

answer:1. We start by understanding the problem conditions: - We are given a pentagon, each vertex of which corresponds to an integer. - The sum of these 5 integers is positive. - If three consecutive integers among these five integers are (x, y, z) and (y < 0), we perform the operation (x rightarrow x + y), (y rightarrow -y), and (z rightarrow z + y). - This operation continues as long as there is at least one negative integer among the five integers. 2. The question asks whether this operation will necessarily stop after a finite number of steps. 3. Consider a list of five integers: [ x, y, z, u, v, ] where (v) is adjacent to (x). 4. Given (y < 0), we need to assess what happens to certain sums after one operation. Let's investigate the sum of squares of each integer and their pairwise sums: [ begin{aligned} &x^2 + y^2 + z^2 + u^2 + v^2 & + (x+y)^2 + (-y)^2 + (z+y)^2 + u^2 + v^2. end{aligned} ] 5. Let’s calculate changes in the sum of squares after the transformation: [ (x+y)^2 + (-y)^2 + (z+y)^2 + u^2 + v^2 - left( x^2 + y^2 + z^2 + u^2 + v^2 right) = (x^2 + 2xy + y^2) + y^2 + (z^2 + 2zy + y^2) + u^2 + v^2 - left( x^2 + y^2 + z^2 + u^2 + v^2 + x^2 + y^2 + z^2 right). ] 6. Simplifying this expression results in: [ x^2 + 2xy + y^2 + y^2 + z^2 + 2zy + y^2 + u^2 + v^2 - x^2 - y^2 - z^2 - u^2 - v^2 = 2xy + 2zy + y^2. ] 7. Notice that (y < 0), so each transformation strictly reduces the squared sum of these elements: [ begin{aligned} & y(2x + 2z + y) quad < 0 end{aligned} ] 8. Since (x + z + y + u + v) is positive, if our transformed sum decreases, it eventually bounds us towards a solution where (y) cannot keep being negative. Consequently, each operation reduces the measure by a fixed amount. 9. We have proved that the sum of squares strictly decreases, ensuring there are no infinite steps and guaranteeing the process eventually terminates. # Conclusion: Therefore, the process must stop after a finite number of steps. [ boxed{text{Yes}} ]

answer:We first count the number of ways ants can correctly migrate without clashing on the same vertex and then calculate the probability. Let the cube be labeled with vertices A, B, C, D, E, F, G, H. Suppose, without loss of generality, that the ant from A moves to B. We must then consider different cases based on where ants from adjacent vertices of A (like D, E, and H) move. Case Analysis: Each ant on the cube has three choices for movement. The configurations in which no two ants end up in the same vertex are merely permutations where each vertex receives exactly one ant. Therefore, we are looking for derangements (permutations where no element appears in its original position) of the ants. The number of derangements D_n of n items is given by: D_n = n! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + ... + frac{(-1)^n}{n!}right) For n = 8 (the number of vertices/ants), D_8 = 8! left(1 - frac{1}{1!} + frac{1}{2!} - frac{1}{3!} + frac{1}{4!} - frac{1}{5!} + frac{1}{6!} - frac{1}{7!} + frac{1}{8!}right) D_8 = 40320 left(1 - 1 + frac{1}{2} - frac{1}{6} + frac{1}{24} - frac{1}{120} + frac{1}{720} - frac{1}{5040} + frac{1}{40320}right) D_8 = 40320 times 0.36788 approx 14833 Each ant moving independently has a probability of frac{1}{3} to choose a correct vertex. Therefore, the combined probability of all ants correctly dispersing is: frac{D_8}{3^8} = frac{14833}{6561} approx 0.00226 So, the probability that no two ants arrive at the same vertex is frac{14833{6561}}. The final answer is boxed{A}

answer:Let's assume the growth rate of the bacteria is constant and the bacteria doubles in size each day. This means that the bacteria's growth can be modeled by an exponential function. Let's call the growth rate "r" and the initial amount of bacteria "B0". The amount of bacteria after t days can be represented as B(t) = B0 * r^t. On day 26, the dish is 1/16 full, so we can write: B(26) = B0 * r^26 = 1/16 * B(30) On day 30, the dish is completely full, so we can write: B(30) = B0 * r^30 = 1 * B(30) Dividing the equation for day 26 by the equation for day 30 to eliminate B0, we get: (1/16 * B(30)) / (1 * B(30)) = (B0 * r^26) / (B0 * r^30) 1/16 = r^26 / r^30 1/16 = 1 / r^4 To find r, we take the fourth root of both sides: r^4 = 16 r = 16^(1/4) r = 2 So the growth rate of the bacteria each day is boxed{2,} or the bacteria doubles in size each day.