answer:1. Let's denote the width of the original rectangle by (s) and the height by (h). 2. The area of the original rectangle is given by: [ text{Area}_{text{original}} = s cdot h ] 3. The problem states that if the width is increased by 3 cm and the height is decreased by 3 cm, the area remains unchanged. This new configuration gives us an equation: [ (s + 3)(h - 3) = s cdot h ] 4. Expanding the left-hand side of the equation yields: [ s cdot h - 3s + 3h - 9 = s cdot h ] 5. Subtracting ( s cdot h ) from both sides of the equation, we get: [ -3s + 3h - 9 = 0 ] 6. Solving for (h) in terms of (s), we obtain: [ 3(h - s) = 9 implies h - s = 3 implies h = s + 3 ] 7. Now, considering the second part of the problem, if we decrease the width by 4 cm and increase the height by 4 cm, the new dimensions are: [ text{New width} = s - 4 quad text{and} quad text{New height} = h + 4 = (s + 3) + 4 = s + 7 ] 8. The area of the new rectangle thus formed is: [ text{Area}_{text{new}} = (s - 4)(h + 4) ] 9. Substituting ( h = s + 3 ) into the equation, we get: [ text{Area}_{text{new}} = (s - 4)(s + 7) ] 10. Expanding the equation, we get: [ text{Area}_{text{new}} = s^2 + 7s - 4s - 28 = s^2 + 3s - 28 ] 11. Comparing it with the area of the original rectangle: [ text{Area}_{text{original}} = s cdot h = s cdot (s + 3) = s^2 + 3s ] 12. Hence, the change in the area is: [ Delta text{Area} = text{Area}_{text{new}} - text{Area}_{text{original}} = (s^2 + 3s - 28) - (s^2 + 3s) = -28 ] 13. Therefore, the area decreases by: [ 28 ; text{cm}^2 ] # Conclusion: [ boxed{28 ; text{cm}^2} ]

answer:1. Simplify left( frac{1}{3k} right)^{-2} using the power rule: left( frac{1}{3k} right)^{-2} = (3k)^2. 2. Simplify (-2k)^3 using the power rule: (-2k)^3 = -8k^3. 3. Multiply (3k)^2 by (-8k^3): (3k)^2 cdot (-8k^3) = 9k^2 cdot (-8k^3) = -72k^5. boxed{-72k^5}

answer:1. **Initial Assumption and Contradiction:** - Assume there exists a column with ( n ) grids, such that each grid is covered by exactly one ( 2 times 2 ) square. - Since ( n ) is odd, covering ( n ) grids in pairs (each pair being part of a ( 2 times 2 ) square) is impossible because ( n ) cannot be evenly divided into pairs. This leads to a contradiction. 2. **Upper Bound Calculation:** - Since each column can contain at most ( n-1 ) grids covered by exactly one ( 2 times 2 ) square, the total number of such grids on the board is at most ( n(n-1) ). 3. **Constructive Proof for Achievability:** - For ( n = 3 ), consider a ( 3 times 3 ) chessboard. It is easy to see that we can place ( 2 times 2 ) squares such that some grids are covered exactly once. - For a general odd ( n ), consider an L-shaped block composed of two rectangles of length ( n+2 ) and width ( 2 ). Combining this L-shaped block with an ( n times n ) block results in an ( (n+2) times (n+2) ) block. - Cover the grids in the block by successively placing the ( 2 times 2 ) squares from left to right and top to bottom, ensuring that only the two squares at the turn overlap at exactly one point. - This method covers ( 8 times frac{n+1}{2} - 2 ) grids, which simplifies to ( 4n + 2 ) grids covered exactly once. 4. **Inductive Step:** - By induction, if the desired configuration can be achieved for ( n ), it can also be achieved for ( n+2 ). The final answer is ( boxed{ n(n-1) } ).

answer:Let the sequence be denoted as [a, ar, ar^2, ar^3, ar^4,dots] with ( ar^2 = 3 ) and ( ar^4 = 27 ). From these terms, we find ( r^2 = frac{ar^4}{ar^2} = frac{27}{3} = 9 ), thus ( r = 3 ) or ( r = -3 ). Now, solving for ( a ): [ a = frac{ar^2}{r^2} = frac{3}{9} = frac{1}{3} ] Considering the possible values of ( r ), ( a ) could be: - ( frac{1}{3} ) if ( r = 3 ) - ( -frac{1}{3} ) if ( r = -3 ) The choices provided include: - A. -sqrt{9} = -3 - B. -frac{3sqrt{9}}{9} = -1 - C. -frac{sqrt{9}}{9} = -frac{1}{3} (Correct for r = -3) - D. sqrt{9} = 3 - E. 9 The correct option is (boxed{C}).