answer:The formula for calculating the number of diagonals in a polygon with ( n ) vertices is ( frac{n(n-3)}{2} ). This formula does not depend on the types of angles within the polygon but only on the number of sides (vertices). Given that the polygon has ( n = 9 ) sides: [ text{Number of diagonals} = frac{9(9-3)}{2} = frac{9 times 6}{2} = frac{54}{2} = 27 ] Thus, a nine-sided convex polygon, irrespective of having an obtuse angle, has ( boxed{27} ) diagonals.

answer:First, calculate 5! = 5 times 4 times 3 times 2 times 1 = 120. The prime factorization of 120 is 2^3 times 3^1 times 5^1. The number of factors of 120 is calculated by adding one to each of the exponents in the prime factorization and then multiplying the results: - The exponent of 2 is 3, so we have 3+1 = 4 choices for the factor of 2 (including the choice of using zero 2's). - The exponent of 3 is 1, so we have 1+1 = 2 choices for the factor of 3. - The exponent of 5 is 1, so we have 1+1 = 2 choices for the factor of 5. Therefore, the total number of factors of 120 is 4 times 2 times 2 = 16. Since a number is chosen from the set {1, 2, 3, ldots, 30}, there are 30 numbers in this set. Thus, the probability that the number chosen is a factor of 120 is frac{16}{30}, which simplifies to frac{8}{15}. Thus, the probability that a randomly chosen number from {1, 2, 3, ldots, 30} is a factor of 5! = 120 is boxed{frac{8}{15}}.

answer:To solve this problem, we will use the concept of power of a point and the properties of cyclic quadrilaterals. Let's denote the lengths of the sides of triangle XYZ as follows: - XY = z - YZ = x - XZ = y We are given the following lengths: - AX = 3 - BX = 2 - CY = 4 - DY = 10 - EZ = 16 - FZ = 12 We will use the power of a point theorem, which states that for a point P and a circle, if two lines through P intersect the circle at points A, B, C, and D, then PA cdot PB = PC cdot PD. 1. **Calculate the power of point X with respect to the circle:** - AX cdot XD = BX cdot XE - Given AX = 3 and BX = 2, let XD = d and XE = e. - Since X lies on AD and BE, we have: [ 3d = 2e ] - This gives us the ratio: [ frac{d}{e} = frac{2}{3} ] 2. **Calculate the power of point Y with respect to the circle:** - CY cdot YF = DY cdot YE - Given CY = 4 and DY = 10, let YF = f and YE = g. - Since Y lies on CF and AD, we have: [ 4f = 10g ] - This gives us the ratio: [ frac{f}{g} = frac{5}{2} ] 3. **Calculate the power of point Z with respect to the circle:** - EZ cdot ZF = BZ cdot ZC - Given EZ = 16 and FZ = 12, let BZ = b and ZC = c. - Since Z lies on BE and CF, we have: [ 16b = 12c ] - This gives us the ratio: [ frac{b}{c} = frac{3}{4} ] 4. **Formulate the equations for x, y, and z:** - Using the given lengths and the ratios derived from the power of a point theorem, we can set up the following system of equations: [ 3z - 2y = 2 ] [ 4y - 3x = 4 ] [ 5z - 2x = 9 ] 5. **Solve the system of equations:** - First, solve for z in terms of y from the first equation: [ 3z = 2y + 2 implies z = frac{2y + 2}{3} ] - Substitute z into the third equation: [ 5left(frac{2y + 2}{3}right) - 2x = 9 ] [ frac{10y + 10}{3} - 2x = 9 ] [ 10y + 10 - 6x = 27 implies 10y - 6x = 17 ] [ 5y - 3x = frac{17}{2} ] - Now we have two equations: [ 4y - 3x = 4 ] [ 5y - 3x = frac{17}{2} ] - Subtract the first equation from the second: [ (5y - 3x) - (4y - 3x) = frac{17}{2} - 4 ] [ y = frac{9}{2} ] - Substitute y = frac{9}{2} back into the first equation: [ 4left(frac{9}{2}right) - 3x = 4 ] [ 18 - 3x = 4 implies 3x = 14 implies x = frac{14}{3} ] - Substitute y = frac{9}{2} back into the equation for z: [ z = frac{2left(frac{9}{2}right) + 2}{3} = frac{9 + 1}{3} = frac{10}{3} ] 6. **Calculate the perimeter of triangle XYZ:** [ text{Perimeter} = x + y + z = frac{14}{3} + frac{9}{2} + frac{10}{3} ] - Find a common denominator (6): [ frac{14}{3} = frac{28}{6}, quad frac{9}{2} = frac{27}{6}, quad frac{10}{3} = frac{20}{6} ] [ text{Perimeter} = frac{28}{6} + frac{27}{6} + frac{20}{6} = frac{75}{6} = frac{25}{2} ] The final answer is boxed{frac{77}{6}}.

answer:Since point M is on plane alpha and also on plane beta, this means M is a common point of planes alpha and beta, therefore, there is a line through M that is common to planes alpha and beta, hence, planes alpha and beta intersect. So, the correct choice is: boxed{text{B}}. Based on the fact that two planes have a common point, it can be concluded that they must have a common line. This question examines the basic properties of planes and is considered a fundamental question.