answer:The first twelve prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. To satisfy both conditions (sum being prime and greater than 20), one number must still be 2 (since sum of two odd primes would be even, thus not prime), and the other number must be such that their sum is a prime number and greater than 20. Checking each prime added to 2 for being greater than 20 and prime: - 2+19 = 21 (not prime) - 2+23 = 25 (not prime) - 2+29 = 31 (prime) - 2+31 = 33 (not prime) - 2+37 = 39 (not prime) Only 2+29 = 31 meets the criteria. Thus, there is only 1 valid pair (2, 29). The total number of ways to choose 2 primes from 12 is given by binom{12}{2} = 66. Therefore, the probability that their sum is a prime number greater than 20 is frac{1}{66}. boxed{frac{1}{66}}

answer:Let's analyze the given polynomial: [P(X) = X^4 + square X^3 + square X^2 + square X + square ] where each square represents a placeholder for a non-zero integer, selected by two players alternately. The second player wins if the resulting polynomial has at least two integer roots; otherwise, the first player wins. Assumptions: We must distinguish between two cases: 1. **Distinct Integer Roots**: The polynomial must have two different integer roots. 2. **Repeated Integer Roots**: The polynomial must have a double (repeated) integer root. We will show the first player can always win in both scenarios by choosing an appropriate strategy. Player 1 Strategy: 1. **First Move**: The first player places (-1) in the constant term of the polynomial, leaving the polynomial as: [ P(X) = X^4 + a_3 X^3 + a_2 X^2 + a_1 X - 1 ] This choice ensures that the potential integer roots are only (pm 1) (positive or negative factors of (-1)). 2. **Second Player's Move**: The second player places a non-zero integer ((a_i neq 0)) in one of the remaining empty positions. 3. **First Player's Move**: The first player places another non-zero integer ((a_j neq 0)) in one of the remaining empty positions. 4. **Second Player's Move**: The second player places a non-zero integer ((a_k neq 0)) in the only remaining position. After all moves, we need to verify that there cannot be two integer roots for the polynomial: Verifying Potential Roots: Observe that after all placements, the polynomial takes the form: [ P(X) = X^4 + a_3 X^3 + a_2 X^2 + a_1 X - 1 ] We examine the polynomial to ensure it does not have (pm 1) as roots. Suppose (X=1) and (X=-1) are roots: 1. For (X = 1): [ 1^4 + a_3 cdot 1^3 + a_2 cdot 1^2 + a_1 cdot 1 - 1 = 0 implies 1 + a_3 + a_2 + a_1 - 1 = 0 implies a_3 + a_2 + a_1 = 0 ] 2. For (X = -1): [ (-1)^4 + a_3 cdot (-1)^3 + a_2 cdot (-1)^2 + a_1 cdot (-1) - 1 = 0 implies 1 - a_3 + a_2 - a_1 - 1 = 0 implies -a_3 + a_2 - a_1 = 0 ] We now have two simultaneous equations: [ 1. quad a_3 + a_2 + a_1 = 0 ] [ 2. quad -a_3 + a_2 - a_1 = 0 ] Adding these equations, we get: [ a_3 + a_2 + a_1 - a_3 + a_2 - a_1 = 0 implies 2a_2 = 0 implies a_2 = 0 ] However, if (a_2 = 0), this would violate the rule that coefficients must be non-zero. Hence, having (X = 1) and (X = -1) as roots is impossible. Scenario of Root Multiplicity: Now consider the possibility of double roots: 1. **Double Root (X=1):** If (X=1) is a double root, the conditions are: [ a_3 + a_2 + a_1 = 0 ] [ 4 + 3a_3 + 2a_2 + a_1 = 0 ] This can be written as: [ a_3 + a_2 + a_1 = 0 ] [ 4 + 3a_1 + 2a_2 + a_3 = 0 ] Subtracting these equations gives: [ (4 + 3a_3 + 2a_2 + a_1) - (a_3 + a_2 + a_1) = 0 implies 4 + 2a_3 + a_2 = 0 implies a_2 = -4 - 2a_3 ] Similarly: 2. **Double Root (X=-1):** If (X=-1) is a double root: [ -a_3 + a_2 - a_1 = 0 ] [ 4 - 3a_3 + 2a_2 - a_1 = 0 ] This gives: [ 2a_2 = -4 - a_1 ] If (a_2 = 0), this would again contradict the non-zero requirement for the coefficients. In each case, regardless of the coefficients ((a_1, a_3)) chosen by the second player, they must place values such that the polynomial results in configurations violating the rules, hence leaving only the scenario with no two integer roots possible. Thus, the first player can always strategize to avoid polynomials with two integer roots, securing a win. boxed{The first player always wins.}

answer:Since cos (α+ dfrac {π}{4})= dfrac {7 sqrt {2}}{10}, We can derive that cos α-sin α= dfrac {7}{5}, Given that cos 2α= dfrac {7}{25}=cos ^{2}α-sin ^{2}α=(cos α-sin α)(cos α+sin α)= dfrac {7}{5}×(cos α+sin α), We can solve for cos α+sin α= dfrac {1}{5}. Therefore, the answer is boxed{D}. By using the cosine formula for the sum of two angles, we can find cos α-sin α. Then, using the given information and the cosine formula for double angles, we can find the solution. This problem primarily tests the understanding of the cosine formula for the sum of two angles and the cosine formula for double angles in trigonometric function simplification and value finding. It also examines the ability to transform thinking, and it is a basic problem.

answer:1. **Find the radii of the circles**: The first circle has a center at (5,5), so its radius is 5 (distance from the y-coordinate of the center to the x-axis). The second circle has a center at (22,13), so its radius is 13. 2. **Calculate the distance between the centers of the circles**: [ text{Distance} = sqrt{(22 - 5)^2 + (13 - 5)^2} = sqrt{17^2 + 8^2} = sqrt{289 + 64} = sqrt{353} ] 3. **Calculate the distance between the closest points of the circles**: This is the distance between the centers minus the sum of the radii. [ text{Distance between closest points} = sqrt{353} - (5 + 13) = sqrt{353} - 18 ] Conclusion: The distance between the closest points of the two circles is boxed{sqrt{353} - 18}.