answer:To solve the problem, we analyze each set A and B separately and then find their intersection.Step 1: Analyzing Set AGiven A={x| lg x leqslant 0}, we can rewrite the inequality in terms of x by comparing it to lg 1, since lg 1 = 0. This gives us:[lg x leqslant lg 1 implies x leqslant 1]However, we must remember that the domain of lg x is x > 0. Thus, combining these two conditions, we get:[0 < x leqslant 1]Therefore, the set A can be expressed as:[A = (0,1]]Step 2: Analyzing Set BGiven B={x| |x^{2}-1| leqslant 1}, we can solve the inequality by considering the absolute value:[-1 leqslant x^{2}-1 leqslant 1]Adding 1 to all sides of the inequality gives us:[0 leqslant x^{2} leqslant 2]Solving this inequality for x, we find:[-sqrt{2} leq x leq sqrt{2}]Thus, the set B can be expressed as:[B = {x|-sqrt{2} leq x leq sqrt{2}}]Step 3: Finding A cap BNow, we need to find the intersection of sets A and B. Since A = (0,1] and B = {x|-sqrt{2} leq x leq sqrt{2}}, their intersection is the set of elements common to both, which is the set A itself, because all elements in (0,1] are also in [-sqrt{2}, sqrt{2}].Therefore, the intersection A cap B is:[A cap B = (0,1] = A]Hence, the correct answer is:[boxed{text{A}}]

answer:Let overline{alpha beta gamma} be the given three-digit number, then the new three-digit number is overline{alpha gamma beta}. Therefore, we havebegin{array}{l}overline{alpha beta gamma}+overline{alpha gamma beta}=100 alpha+10 beta+gamma+100 alpha+10 gamma+beta =200 alpha+11(beta+gamma) .end{array}Since this is a four-digit number starting with 173, we have alpha=8.Let x be its last digit, then we have1600+11(beta+gamma)=1730+x,which simplifies to 11(beta+gamma)=130+x.From this relationship and 111132, we get x=2.Thus, the four-digit number is 1732.Therefore, beta+gamma=12 and overline{8 beta gamma}+overline{8 gamma beta}=1732.Hence, the given three-digit positive integer A is one of 839,848,857,866,875,884,893 text { . }

answer:(I) Since (2a-c)cos B = bcos C, we have (2sin A - sin C)cos B = sin Bcos C which implies 2sin Acos B = sin Bcos C + sin Ccos B = sin(B + C) Since A + B + C = pi, we have 2sin Acos B = sin A. Given 0 1, the maximum value of overrightarrow{m} cdot overrightarrow{n} occurs when t = 1. According to the problem, -2 + 4k + 1 = 5, therefore k = boxed{frac{3}{2}}.

answer:At 90 steps per minute and 75 cm per step, Dave walks at a rate of 90times75 cm per minute, and with 16 minutes, the distance Dave walks to school is 90times75times16. Also, at 100 steps per minute and 60 cm per step, Jack walks at a rate of 100times60 cm per minute. Jack must walk 90times75times16 cm, so it takes him frac{90times75times16}{100times60} minutes. Canceling some of the factors, this comes out to 18 text{minutes}, boxed{text{C}}.

answer:To solve the problem, we need to find all integers ( n ) for which there exists a table with ( n ) rows and ( 2022 ) columns, such that subtracting any two rows entry-wise leaves every remainder modulo ( 2022 ).1. Assume the existence of such a table: Let the rows of the table be denoted by ( mathbf{a}_1, mathbf{a}_2, ldots, mathbf{a}_n ), where each ( mathbf{a}_i ) is a vector of length ( 2022 ) with integer entries.2. Condition on the differences: For any two rows ( mathbf{a}_i ) and ( mathbf{a}_j ), the difference ( mathbf{a}_i - mathbf{a}_j ) must have all entries congruent modulo ( 2022 ). This means that for any ( k ) and ( l ), the difference ( a_{ik} - a_{jk} equiv a_{il} - a_{jl} pmod{2022} ).3. Consider three rows: Suppose we have three rows ( mathbf{a}_1, mathbf{a}_2, mathbf{a}_3 ). Define the sums of the differences: [ x = sum_{k=1}^{2022} (a_{3k} - a_{1k}), ] [ y = sum_{k=1}^{2022} (a_{1k} - a_{2k}), ] [ z = sum_{k=1}^{2022} (a_{2k} - a_{3k}). ]4. Modulo condition: By the problem's condition, each of these sums ( x, y, z ) must be congruent to the same value modulo ( 2022 ). Let ( m = 1011 ). Then: [ x equiv y equiv z equiv m pmod{2022}. ]5. Sum of differences: Notice that: [ x + y + z = sum_{k=1}^{2022} (a_{3k} - a_{1k}) + sum_{k=1}^{2022} (a_{1k} - a_{2k}) + sum_{k=1}^{2022} (a_{2k} - a_{3k}) = 0. ] This is because each term ( a_{ik} ) cancels out.6. Contradiction: However, if ( x equiv y equiv z equiv m pmod{2022} ), then: [ x + y + z equiv m + m + m equiv 3m pmod{2022}. ] Since ( m = 1011 ), we have: [ 3m = 3 times 1011 = 3033. ] But ( 3033 notequiv 0 pmod{2022} ). This is a contradiction.7. Conclusion: Therefore, it is not possible to have three rows satisfying the given condition. Hence, the maximum number of rows ( n ) for which such a table can exist is ( n leq 2 ).The final answer is ( boxed{ n leq 2 } ).

answer:Solving, we knowbegin{array}{l} a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c =(a+b-c)^{2}+2 c(a+c-b) . text { Let } I=frac{(a+b-c)^{2}+2 c(a+c-b)}{2 c(a+b-c)}=frac{a+b-c}{2 c}+frac{a+c-b}{a+b-c} .end{array}Since a geqslant frac{1}{3}(b+c), we have a geqslant frac{1}{4}(a+b-c)+frac{c}{2}, thus a+c-b=2 a-(a+b-c) geqslant-frac{1}{2}(a+b-c)+c. Therefore, I geqslant frac{a+b-c}{2 c}-frac{frac{1}{2}(a+b-c)}{a+b-c}+frac{c}{a+b-c}=-frac{1}{2}+frac{a+b-c}{2 c}+frac{c}{a+b-c} geqslant-frac{1}{2}+2 sqrt{frac{1}{2}}=sqrt{2}-frac{1}{2} text {, }which means frac{a c+b c-c^{2}}{a^{2}+b^{2}+3 c^{2}+2 a b-4 b c}. Thus, the required lambda leqslant frac{2 sqrt{2}+1}{7}.On the other hand, when a=frac{sqrt{2}}{4}+frac{1}{2}, b=frac{3 sqrt{2}}{4}+frac{1}{2}, c=1, thena c+b c-c^{2}=sqrt{2}, a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=4-sqrt{2} text {, }so frac{1}{2 I}=frac{sqrt{2}}{4-sqrt{2}}=frac{sqrt{2}(4+sqrt{2})}{14}=frac{2 sqrt{2}+1}{7}.Thus, lambda geqslant frac{2 sqrt{2}+1}{7}.In summary, lambda=frac{2 sqrt{2}+1}{7}.