answer:- Since angle EFJ = 90^{circ}, triangle EFJ is a right-angled triangle. By the Pythagorean theorem, [ FJ^2 = EJ^2 - EF^2 = 15^2 - 9^2 = 144 ] so, FJ = 12, as FJ > 0. - Since angle UKJ = 90^{circ}, triangle UKJ is a right-angled triangle with UJ = 12 (from FU = UJ = 12). By the Pythagorean theorem, [ UK^2 = JK^2 - UJ^2 = 25^2 - 12^2 = 529 - 144 = 385 ] Thus, UK = sqrt{385}, as UK > 0. Our final answer is boxed{12, sqrt{385}}.

answer:1. To begin, we need to understand the coloring condition provided by the problem: every 2018-cell block, regardless of its orientation or reflection, contains an equal number of black and white cells. 2. Let's assume the infinite plane is not colored in a checkerboard fashion. This assumption must be proven incorrect to conclude that the plane is indeed checkerboard-patterned. 3. If it is not colored in a checkerboard fashion, then there must exist at least one pair of neighboring cells of the same color (either both black or both white). 4. Let's consider two neighboring cells of the same color, say both white, in a row. For the coloring condition to hold, realizing that the two cells are white forces more cells into this pattern due to the equal black and white requirement in any 2018-cell sub-board arrangement. - As a critical result, if any two neighboring cells in a row are white, the entire row must be white. This comes from extending the observation that broken symmetry by two same-colored neighboring cells must imply a consecutive extension, thus leading to an infinite row being of the same color. 5. However, consider placing a 2018-cell block with its long dimension laid over this row which is entirely white. In such a block, constrained to the condition of exactly 1009 white and 1009 black cells, but containing one or more rows entirely white means there must be more than 1009 white cells. This is a contradiction because the condition enforces an exact half-split of black and white cells in any 2018-cell sub-board. 6. Reversing this contradiction proves that our initial wrong assumption of a non-checkerboard pattern necessarily leads to an inconsistency with the given problem's conditions. 7. Given no two adjacent cells can ever be the same color, we are forced to conclude that the only regular pattern satisfying the problem’s statement is the checkerboard pattern. # Conclusion: Thus, the infinite grid is definitively colored in a checkerboard pattern, ensuring every smaller 2018-cell section is evenly split between black and white cells, regardless of its position or orientation. boxed{text{Yes}}

answer:Let's consider the statements: - "x > 2" (Statement 1) - "x > 3" (Statement 2) To analyze the condition types, we proceed as follows: **Sufficiency**: We need to see if Statement 1 being true guarantees the truth of Statement 2. If there exists a case where x > 2 but not x > 3, then Statement 1 is not sufficient for Statement 2. For example, when x = frac{5}{2}, Statement 1 is satisfied as frac{5}{2} > 2, but Statement 2 is not, since frac{5}{2} leq 3. Thus, Statement 1 is not a sufficient condition for Statement 2. **Necessity**: We need to consider whether Statement 2 being true implies that Statement 1 must be true. If x > 3, it's clear that x must also be greater than 2. Therefore, Statement 1 is a necessary condition for Statement 2. Based on the reasoning above, we conclude that "x > 2" is a *necessary* but not a *sufficient* condition for "x > 3". Therefore, the correct answer is: [ boxed{B: text{Necessary but not sufficient}} ]

answer:1. **Identify Final Rectangle Perimeter**: The problem states that the perimeter of the final rectangle obtained after the cuts is 129 cm. 2. **Understanding the Cuts**: The cuts are made parallel to the sides of the original triangle and pass through the midpoints of those sides. This effectively halves the dimensions of the original sides. 3. **Perimeter Relation**: Let the original rectangle have sides (a) and (b). The final rectangle, after the cuts, will have sides (frac{a}{2}) and (frac{b}{2}). 4. **Calculate Final Perimeter**: The perimeter of a rectangle is given by: [ P = 2 text{ (length + width)} ] For the final rectangle: [ 129 = 2 left(frac{a}{2} + frac{b}{2}right) ] 5. **Simplify the Equation**: Simplifying, we get: [ 129 = frac{a}{1} + frac{b}{1} ] Therefore: [ frac{a + b}{1} = 129 ] 6. **Original Perimeter Relation**: Now, since (a + b) is the sum of the sides of the original rectangle, the perimeter of the original rectangle is: [ P = 2(a + b) ] 7. **Final Calculation**: Substitute (a + b) into the formula: [ P = 2 times 129 = 258 text{ cm} ] # Conclusion: The perimeter of the original rectangular sheet of cardboard before any cuts were made is: [ boxed{258 text{ cm}} ]