answer:1. **Understanding the Problem:** Peter wants to assemble a large square using smaller squares such that no rectangle formed by the small squares has all four vertices of the same color. We need to determine the largest possible size of such a square. 2. **Valid Coloring for (4 times 4) Square:** The solution claims that the largest possible square is (4 times 4). Let's verify this by providing a valid coloring: [ begin{array}{cccc} 1 & 1 & 0 & 0 1 & 0 & 1 & 0 0 & 1 & 0 & 1 0 & 0 & 1 & 1 end{array} ] Here, '1' represents black and '0' represents white. We need to check that no rectangle formed by these squares has all four vertices of the same color. 3. **Verification for (4 times 4) Square:** - Consider any rectangle formed by selecting two rows and two columns. - For example, selecting rows 1 and 2, and columns 1 and 2, we get: [ begin{array}{cc} 1 & 1 1 & 0 end{array} ] The vertices are not all the same color. - Checking all possible rectangles, we find that no rectangle has all four vertices of the same color. 4. **Attempting a (5 times 5) Square:** - Assume we have a (5 times 5) square with 13 black cells and 12 white cells. - Define a "bp" (black pair) as a pair of black cells in the same column. - By convexity, the minimum number of black pairs is calculated as follows: [ 3binom{3}{2} + 2binom{2}{2} = 3 cdot 3 + 2 cdot 1 = 9 + 2 = 11 ] - There are only 10 pairs of rows in a (5 times 5) square. - By the pigeonhole principle, there must be at least one pair of rows that contains at least two black pairs in the same columns. - This configuration forms a rectangle with all four vertices being black, which contradicts the requirement. 5. **Conclusion:** - Since a (5 times 5) square cannot be colored without forming a rectangle with all four vertices of the same color, the largest possible square Peter can assemble is (4 times 4). The final answer is ( boxed{ 4 times 4 } )

answer:Since tan alpha = 2, we can express this in terms of sine and cosine as frac{sin alpha}{cos alpha} = 2. Our goal is to find the value of 7sin^2alpha + 3cos^2alpha, and we start by recognizing the Pythagorean identity, which states that for any angle alpha, sin^2alpha + cos^2alpha = 1. Let's rewrite the given expression by incorporating this identity: 7sin^2alpha + 3cos^2alpha = frac{7sin^2alpha + 3cos^2alpha}{sin^2alpha + cos^2alpha} As the denominator simplifies to 1 due to the Pythagorean identity, we can disregard it: frac{7sin^2alpha + 3cos^2alpha}{sin^2alpha + cos^2alpha} = frac{7sin^2alpha + 3cos^2alpha}{1} Now, we divide both terms in the numerator by cos^2alpha: = frac{7frac{sin^2alpha}{cos^2alpha} + 3}{frac{sin^2alpha}{cos^2alpha} + 1} Using the fact that tan alpha = frac{sin alpha}{cos alpha}, we substitute tan^2alpha for frac{sin^2alpha}{cos^2alpha}: = frac{7tan^2alpha + 3}{tan^2alpha + 1} We then substitute the given value of tan alpha = 2 into the expression. Recall that substituting tan alpha requires squaring the value because the expression has tan^2alpha: = frac{7 cdot 2^2 + 3}{2^2 + 1} = frac{7 cdot 4 + 3}{4 + 1} = frac{28 + 3}{5} = frac{31}{5} Therefore, the value of 7sin^2alpha + 3cos^2alpha is equal to frac{31}{5}, which is simplified, and we can box the final result: boxed{frac{31}{5}}

answer:Let's calculate the work done by John and Jane together in one day. John can complete the task in 20 days, so his work rate is 1/20 of the task per day. Jane can complete the task in 12 days, so her work rate is 1/12 of the task per day. When they work together, their combined work rate is: (1/20) + (1/12) = (3/60) + (5/60) = 8/60 = 2/15 of the task per day. Let's assume they worked together for x days before Jane was indisposed. In those x days, they would have completed (2/15) * x of the task. After Jane was indisposed, John had to complete the remaining task alone. Since Jane was indisposed 4 days before the work got over, John worked alone for those 4 days. In those 4 days, John would have completed (1/20) * 4 = 4/20 = 1/5 of the task. The total work done when they worked together and when John worked alone should add up to 1 (the whole task). So we have: (2/15) * x + 1/5 = 1 Now, let's solve for x: (2/15) * x = 1 - 1/5 (2/15) * x = 5/5 - 1/5 (2/15) * x = 4/5 Now, multiply both sides by the reciprocal of (2/15) to solve for x: x = (4/5) * (15/2) x = (4 * 15) / (5 * 2) x = 60 / 10 x = 6 So, John and Jane worked together for 6 days before Jane was indisposed. Since Jane was indisposed 4 days before the work got over, the total time from when they started working together until the task was completed is: 6 days (together) + 4 days (John alone) = 10 days Therefore, the work got over in boxed{10} days from the time John and Jane started to work on it together.

answer:The equation sqrt {(x-2)^{2}+y^{2}}+ sqrt {(x+2)^{2}+y^{2}}=10 represents the sum of the distances from the moving point M(x,y) to two fixed points (±2,0) is a constant value 10=2a, and 10 > 2+2. According to the definition of the problem, it can be concluded that the trajectory of the moving point M is an ellipse, and b^{2}=a^{2}-c^{2}=5^{2}-2^{2}=21. Thus, the equation of the ellipse is: dfrac {x^{2}}{25}+ dfrac {y^{2}}{21}=1. Therefore, the answer is: boxed{B}. This can be obtained by using the definition of an ellipse. This problem tests the understanding of the definition of an ellipse and is a basic problem.