answer:1. **Initial Shape**: The original shape is the letter ( mathrm{F} ). 2. **Reflecting in Line 1**: Reflect the letter ( mathrm{F} ) about Line 1, which is a vertical reflection. The reflection results in the letter ( mathrm{E} ). - Consider the shape ( mathrm{F} ) and its typical orientation with the vertical bar on the left and the three horizontal bars extending to the right. - Vertically reflecting ( mathrm{F} ) will reverse the direction of the horizontal bars to the left side, forming the letter ( mathrm{E} ). 3. **Reflecting in Line 2**: Reflect the obtained shape ( mathrm{E} ) about Line 2, which is a horizontal reflection. - The reflection flips the top and bottom parts of the letter ( mathrm{E} ). - A horizontal reflection of ( mathrm{E} ) transforms the position of its horizontal bars but such a reflection still retains the original configuration. All horizontal reflections of ( mathrm{E} ) retain the shape of the letter ( mathrm{E} ), effectively as the central vertical line and the 3 extending lines are symmetrical about a horizontal axis. - Since, numerically and typographically, horizontal reflection of "E" precisely brings in a letter "Э", however, looking at its visual invariance of "F" reflected vertically and then horizontally brings in the shape ( mathrm{H} ). 4. **Resulting Shape**: After both the vertical and horizontal reflections, the final shape that results is the letter ( boldsymbol{H} ). **Conclusion**: The shape that results is ( boldsymbol{H} ). [ boxed{D} ]

answer:Since A∩B={-3}, -3∈B. However, a²+3≠-3, When a-3=-3, a=0, A={0, 1, -3}, B={-3, 1, 3}, This leads to A∩B={-3, 1}, which contradicts A∩B={-3}; When 2a+1=-3, a=-2, which satisfies A∩B={-3} Therefore, boxed{a = -2}.

answer:1. Given that ( x = -2 ) is a root of the quadratic equation ( x^2 + ax + 2b = 0 ), we can substitute ( x = -2 ) into the equation to form a relationship between ( a ) and ( b ). 2. Substituting ( x = -2 ) into the equation: [ (-2)^2 + a(-2) + 2b = 0 ] Simplifying this, we get: [ 4 - 2a + 2b = 0 ] Rearranging the equation to isolate ( b ), we find: [ 2b = 2a - 4 implies b = a - 2 ] 3. Next, we want to find the minimum value of ( a^2 + b^2 ). Substituting ( b = a - 2 ): [ a^2 + b^2 = a^2 + (a-2)^2 ] Expanding ( (a-2)^2 ): [ a^2 + (a-2)^2 = a^2 + (a^2 - 4a + 4) ] Combining like terms, we obtain: [ a^2 + a^2 - 4a + 4 = 2a^2 - 4a + 4 ] 4. To find the minimum value of ( 2a^2 - 4a + 4 ), we can complete the square: [ 2a^2 - 4a + 4 = 2left(a^2 - 2aright) + 4 ] Completing the square inside the parentheses: [ a^2 - 2a = (a-1)^2 - 1 ] Substituting back: [ 2left((a-1)^2 - 1right) + 4 = 2(a-1)^2 - 2 + 4 = 2(a-1)^2 + 2 ] 5. The expression ( 2(a-1)^2 + 2 geq 2 ) since ( (a-1)^2 geq 0 ), and achieves its minimum value when ( (a-1)^2 = 0 ), i.e., ( a = 1 ). 6. Substituting ( a = 1 ) back into ( b = a - 2 ): [ b = 1 - 2 = -1 ] 7. Hence, ( a^2 + b^2 ) at this point is: [ 1^2 + (-1)^2 = 1 + 1 = 2 ] Conclusion: The minimum value of ( a^2 + b^2 ) is (boxed{2}).

answer:1. Problem Analysis We need to find how many rectangles on the grid contain exactly one gray cell. The given image has (2 times 20 = 40) gray cells. To solve this, we can color the gray cells into two distinct colors and then count the types of rectangles separately. 2. Coloring the Cells - Let's color some gray cells blue and others red. - Blue cells are the ones where they appear more frequently, and in the given image, there are 36 blue cells. - Red cells appear less frequently, and there are 4 red cells in the image. 3. Counting Rectangles Containing One Blue Cell Each blue cell is contained in exactly 4 rectangles made up only of gray cells. This can be visualized by selecting the gray cell and choosing positions to form a rectangle around it. Let's break down why each blue cell is contained in 4 such rectangles: - A rectangle can be formed by choosing vertical and horizontal boundaries around the gray cell such that it remains the only gray cell within that rectangle. Since there are 36 blue cells: [ text{Number of rectangles with one blue cell} = 36 times 4 = 144 ] 4. Counting Rectangles Containing One Red Cell Each red cell is contained in exactly 8 rectangles. This can be calculated similarly by selecting positions to form a rectangle around the red cell. Since there are 4 red cells: [ text{Number of rectangles with one red cell} = 4 times 8 = 32 ] 5. Total Rectangles Containing Exactly One Gray Cell Finally, we add both counts together: [ text{Total rectangles} = 144 + 32 = 176 ] # Conclusion Thus, the number of rectangles containing exactly one gray cell in the given grid is: [ boxed{176} ]