answer:Through the vertex of the trapezoid, draw a line parallel to the diagonal.## SolutionLet the bases AD and BC of the isosceles trapezoid ABCD be 40 and 24, respectively, and the diagonals AC and BD are perpendicular. Draw a line through vertex C parallel to diagonal BD. Let this line intersect the extension of base AD at point K. Then triangle ACK is an isosceles right triangle with base AK = AD + DK = 64 and height equal to half of AK. Therefore,S_{mathrm{ABCD}}=S_{triangle mathrm{ACK}}=frac{1}{2} cdot 64 cdot 32=32^{2}=1024 .## Answer1024.

answer:To find the positions of three points ( A, B, C ) on the boundary of a unit cube such that ( min{AB, AC, BC} ) is the greatest possible, we need to consider the distances between these points. 1. Identify the possible distances between vertices of a unit cube: - The edge length of the unit cube is (1). - The face diagonal (distance between two vertices on the same face but not on the same edge) is ( sqrt{2} ). - The space diagonal (distance between two opposite vertices of the cube) is ( sqrt{3} ).2. Consider the minimum distance ( min{AB, AC, BC} ): - For the minimum distance to be maximized, we need to ensure that the smallest distance among ( AB, AC, ) and ( BC ) is as large as possible.3. Analyze the possible configurations: - If ( A, B, ) and ( C ) are all vertices of the cube, the minimum distance between any two points can be (1), ( sqrt{2} ), or ( sqrt{3} ). - If ( A, B, ) and ( C ) are on the same face, the minimum distance can be (1) or ( sqrt{2} ). - If ( A, B, ) and ( C ) are on different faces but not all vertices, the minimum distance can be (1) or ( sqrt{2} ).4. Maximize the minimum distance: - To maximize ( min{AB, AC, BC} ), we need to ensure that the smallest distance among ( AB, AC, ) and ( BC ) is as large as possible. - The largest possible minimum distance is ( sqrt{2} ), which occurs when ( A, B, ) and ( C ) are vertices of the cube such that no two points are on the same edge.5. Verify the configuration: - Consider the vertices ( (0,0,0) ), ( (1,1,0) ), and ( (0,1,1) ). - The distances are: [ AB = sqrt{(1-0)^2 + (1-0)^2 + (0-0)^2} = sqrt{2} ] [ AC = sqrt{(0-0)^2 + (1-0)^2 + (1-0)^2} = sqrt{2} ] [ BC = sqrt{(1-0)^2 + (1-1)^2 + (0-1)^2} = sqrt{2} ] - Thus, ( min{AB, AC, BC} = sqrt{2} ).Conclusion:The positions of the points ( A, B, ) and ( C ) on the boundary of the unit cube that maximize ( min{AB, AC, BC} ) are such that each pair of points is separated by a distance of ( sqrt{2} ). One such configuration is ( A = (0,0,0) ), ( B = (1,1,0) ), and ( C = (0,1,1) ).The final answer is ( boxed{ sqrt{2} } ).

answer:B2. 4 Dat 2216 bij deling door a rest 29 geeft, betekent precies dat 2216-29=2187 deelbaar is door a en dat a groter is dan 29 (de rest is altijd kleiner dan de deler a). The divisors of 2187=3^{7} that are greater than 29 are 81, 243, 729, and 2187. There are thus 4 possibilities in total.

answer:If 3x^3-9x^2+kx-12 is divisible by x-3, then by the [Remainder Theorem](https://artofproblemsolving.com/wiki/index.php/Remainder_Theorem), plugging in 3 in the cubic results in 0.[3 cdot 3^3 - 9 cdot 3^2 + 3k - 12 = 0]Combine like terms to get[3k - 12 = 0]Thus, k=4. The cubic is 3x^3-9x^2+4x-12, and it can be factored (by grouping or synthetic division) into[(3x^2+4)(x-3)]Thus, the answer is boxed{textbf{(C)}}.

answer:The total ways the textbooks can be arranged in the 3 boxes is 12textbf{C}3cdot 9textbf{C}4, which is equivalent to frac{12cdot 11cdot 10cdot 9cdot 8cdot 7cdot 6}{144}=12cdot11cdot10cdot7cdot3. If all of the math textbooks are put into the box that can hold 3 textbooks, there are 9!/(4!cdot 5!)=9textbf{C}4 ways for the other textbooks to be arranged. If all of the math textbooks are put into the box that can hold 4 textbooks, there are 9 ways to choose the other book in that box, times 8textbf{C}3 ways for the other books to be arranged. If all of the math textbooks are put into the box with the capability of holding 5 textbooks, there are 9textbf{C}2 ways to choose the other 2 textbooks in that box, times 7textbf{C}3 ways to arrange the other 7 textbooks. 9textbf{C}4=9cdot7cdot2=126, 9cdot 8textbf{C}3=9cdot8cdot7=504, and 9textbf{C}2cdot 7textbf{C}3=9cdot7cdot5cdot4=1260, so the total number of ways the math textbooks can all be placed into the same box is 126+504+1260=1890. So, the probability of this occurring is frac{(9cdot7)(2+8+(4cdot5))}{12cdot11cdot10cdot7cdot3}=frac{1890}{27720}. If the numerator and denominator are both divided by 9cdot7, we have frac{(2+8+(4cdot5))}{4cdot11cdot10}=frac{30}{440}. Simplifying the numerator yields frac{30}{10cdot4cdot11}, and dividing both numerator and denominator by 10 results in frac{3}{44}. This fraction cannot be simplified any further, so m=3 and n=44. Therefore, m+n=3+44=boxed{047}.

answer:Analysis Let the three prime numbers be represented by p, q, r, and using the properties of prime numbers, it is easy to solve for p, q, r.Solution Let the three prime numbers be p, q, r,then p q r=5(p+q+r),thus one of p, q, r must be equal to 5. Without loss of generality, let r=5,then 5 p q=5(p+q+5),which simplifies to p q=p+q+5,(p-1)(q-1)=6.Given p>1, q>1,we haveleft{begin{array}{l}p-1=2 q-1=3end{array}right. or left{begin{array}{l}p-1=3 q-1=2end{array}right. or left{begin{array}{l}p-1=1 q-1=6end{array}right. or left{begin{array}{l}p-1=6 q-1=1end{array}right.Since p=4 or q=4 is not a prime number,the only solutions are p=2, q=7 or p=7, q=2.Therefore, the prime numbers are 2,5,7.