answer:To solve this problem, we need to analyze it in two parts: the condition for the linear function y=(a-1)x+2 to pass through the second and fourth quadrants, and the condition for the system of inequalities left{{begin{array}{l}{x+1leq 2a}{a-xleq 2}end{array}}right. to have a solution. Let's break it down step by step. # Part 1: Condition for the Linear Function to Pass Through the Second and Fourth Quadrants For the linear function y=(a-1)x+2 to pass through the second and fourth quadrants, the slope of the line, which is a-1, must be negative. This is because lines with a negative slope descend from left to right, which is necessary for the line to cross the second (top-left) and fourth (bottom-right) quadrants. Therefore, we have: [a-1 < 0] [a < 1] # Part 2: Condition for the System of Inequalities to Have a Solution The system of inequalities is given by: [left{{begin{array}{l}{x+1leq 2a}{a-xleq 2}end{array}}right.] Solving each inequality for x gives us: [left{{begin{array}{l}{xleq 2a-1}{xgeq a-2}end{array}}right.] For these inequalities to have a solution, the set of x values satisfying xleq 2a-1 must intersect with the set of x values satisfying xgeq a-2. This means the upper bound of the first inequality must be greater than or equal to the lower bound of the second inequality: [2a-1 geq a-2] Solving this inequality for a gives: [a geq -1] # Combining the Conditions Combining the conditions from parts 1 and 2, we find that a must satisfy both a < 1 and a geq -1. Therefore, the range of a that satisfies both conditions is: [-1 leq a < 1] From the given cards, the numbers that fall within this range are -1 and 0. Since there are 5 cards in total, the probability that a card drawn at random satisfies both conditions is: [frac{2}{5}] Therefore, the probability that the linear function y=(a-1)x+2 passes through the second and fourth quadrants, and the system of inequalities about x left{{begin{array}{l}{x+1leq 2a}{a-xleq 2}end{array}}right. has a solution is boxed{frac{2}{5}}.

answer:Since a < b < c < d, we have a-b < 0, a-c < 0, a-d < 0, b-c < 0, b-d < 0, c-d < 0, Since x = (a+b)(c+d), y = (a+c)(b+d), z = (a+d)(b+c), thus x-y = (a+b)(c+d) - (a+c)(b+d), = ac + ad + bc + bd - ab - ad - bc - cd, = ac + bd - ab - cd, = (ac-cd) + (bd-ab), = c(a-d) - b(a-d), = (a-d)(c-b) < 0, y-z = (a+c)(b+d) - (a+d)(b+c), = ab + ad + bc + cd - ab - ac - bd - cd, = ad + bc - ac - bd, = (ad-bd) + (bc-ac), = (a-b)(d-c) < 0, Therefore, x-y < 0, y-z < 0, which means x < y, y < z, Thus, x < y < z. Hence, the correct option is boxed{A}.

answer:To determine the last two digits of ( 9^{2008} ), we need to compute ( 9^{2008} mod 100 ). 1. **Step 1: Identify the pattern in exponents of 9 modulo 100.** We begin by finding powers of 9 modulo 100: [ 9^2 equiv 81 pmod{100} ] [ 9^4 equiv (9^2)^2 equiv 81^2 equiv 6561 equiv 61 pmod{100} ] [ 9^6 equiv 9^4 cdot 9^2 equiv 61 cdot 81 equiv 4941 equiv 41 pmod{100} ] [ 9^8 equiv 9^4 cdot 9^4 equiv 61 cdot 61 equiv 3721 equiv 21 pmod{100} ] [ 9^{10} equiv 9^8 cdot 9^2 equiv 21 cdot 81 equiv 1701 equiv 01 pmod{100} ] 2. **Step 2: Further confirmation and numerical pattern.** Observing the results up to ( 9^{10} ), the exponents’ pattern suggests a repeat at ( 9^{10} equiv 01 pmod{100} ). 3. **Step 3: Utilize pattern for large exponent.** Now, since ( 9^{10} equiv 01 pmod{100} ), we can use this periodicity to simplify: [ 9^{2008} = (9^{10})^{200} times 9^8 pmod{100} ] [ (9^{10})^{200} equiv 01^{200} equiv 1 pmod{100} ] 4. **Step 4: Multiply by ( 9^8 ) to find the last two digits.** Since ( 9^{8} equiv 21 pmod{100} ): [ 9^{2008} equiv 1 times 21 equiv 21 pmod{100} ] **Conclusion:** The last two digits of ( 9^{2008} ) are ( boxed{21} ).

answer:To solve this problem, we need to analyze the expression [ frac{a}{|a|} + frac{b}{|b|} + frac{c}{|c|} + frac{abc}{|abc|} ] for all possible combinations of signs of (a), (b), and (c). 1. **Understanding the terms:** - (frac{a}{|a|}) is the sign of (a). It is (1) if (a) is positive and (-1) if (a) is negative. - Similarly, (frac{b}{|b|}) is the sign of (b), and (frac{c}{|c|}) is the sign of (c). - (frac{abc}{|abc|}) is the sign of the product (abc). It is (1) if (abc) is positive and (-1) if (abc) is negative. 2. **Case Analysis:** - **Case 1: All positive ((a > 0), (b > 0), (c > 0))** [ frac{a}{|a|} = 1, quad frac{b}{|b|} = 1, quad frac{c}{|c|} = 1, quad frac{abc}{|abc|} = 1 ] [ 1 + 1 + 1 + 1 = 4 ] - **Case 2: Two positive, one negative ((a > 0), (b > 0), (c < 0))** [ frac{a}{|a|} = 1, quad frac{b}{|b|} = 1, quad frac{c}{|c|} = -1, quad frac{abc}{|abc|} = -1 ] [ 1 + 1 - 1 - 1 = 0 ] This result is the same for any permutation of two positive and one negative. - **Case 3: One positive, two negative ((a > 0), (b < 0), (c < 0))** [ frac{a}{|a|} = 1, quad frac{b}{|b|} = -1, quad frac{c}{|c|} = -1, quad frac{abc}{|abc|} = 1 ] [ 1 - 1 - 1 + 1 = 0 ] This result is the same for any permutation of one positive and two negative. - **Case 4: All negative ((a < 0), (b < 0), (c < 0))** [ frac{a}{|a|} = -1, quad frac{b}{|b|} = -1, quad frac{c}{|c|} = -1, quad frac{abc}{|abc|} = -1 ] [ -1 - 1 - 1 - 1 = -4 ] 3. **Conclusion:** From the above cases, the possible values of the expression are (4), (0), and (-4). Therefore, the set of all numbers formed is ({-4, 0, 4}). The final answer is ( boxed{ {-4, 0, 4} } )