answer:12.5(sqrt{2}+1).From the condition, we have|x-1|+|y-3|=|x-6|+|y-9| text {. }When y geqslant 9,|x-1|+y-3=|x-6|+y-9 text {, }which simplifies to |x-1|+6=|x-6|, with no solution;When 3 leqslant y leqslant 9,|x-1|+y-3=|x-6|+9-y text {, }which simplifies to 2 y-12=|x-6|-|x-1|.When x leqslant 1, y=8.5, the length of the line segment is 1.When 1 leqslant x leqslant 6, x+y=9.5, the length of the line segment is 5 sqrt{2}.When x geqslant 6, y=3.5, the length of the line segment is 4.When y leqslant 3,|x-1|+3-y=|x-6|+9-y,which simplifies to |x-1|=6+|x-6|, with no solution.In summary, the total length of the line segments formed by the trajectory of point C is 1+5 sqrt{2}+4=5(sqrt{2}+1).

answer:a) The first two cards can "encode" the suit of the second card, the next two cards can encode the suit of the fourth card, and so on. When only two cards are left in the deck, it is sufficient to encode only their order, which can be done using the back of the 35th card. Thus, the clairvoyant will guess the suits of 19 cards.b) We will call the 1, 3, 5, ldots, 35-th cards odd. Consider the following 17 cards: all odd cards except the first and the second-to-last, and the second card. Among them, there will be five cards of the same suit. We will call this suit the main suit. The position of the first two cards in the deck can encode the main suit. The position of the (2k-1)-th and 2k-th cards (for 2 leq k leq 17) can encode the suit of the 2k-th card. The position of the back of the second-to-last card can encode the suits of the last two cards (see the solution to part a). The clairvoyant should name the main suit for each of the selected 17 cards. Then he will guess the suits of at least five of these 17 cards. In addition, the clairvoyant will guess the suits of all even cards except the second and the last, as well as the suits of the last two cards. In total, he will guess the suits of at least 23 cards.## Answera), b) Yes.

answer:Let the number of ways to travel from city A to city B (neq A) by taking k buses be alpha(k), and the number of ways to travel from city A to city A by taking k buses be beta(k).It is easy to see that, starting from city A and taking a bus k times, there are (n-1)^{k} ways. At the same time, if returning to city A has beta(k) ways, and if arriving at a city different from city A has (n-1) alpha(k) ways, thenbeta(k)+(n-1) alpha(k)=(n-1)^{k} text {. }Assume k geqslant 2. It is easy to see that,beta(k)=(n-1) alpha(k-1) text {. }Substituting beta(k)=(n-1) alpha(k-1) into equation (1), we know that when k geqslant 2,begin{array}{l}(n-1) alpha(k-1)+(n-1) alpha(k)=(n-1)^{k} Rightarrow alpha(k)=(n-1)^{k-1}-alpha(k-1) .end{array}Using mathematical induction on k can prove the conclusion.

answer:Since set A={x|x^2+2x-a=0, xin mathbb{R}}, and A is a non-empty set, then x^2+2x-a=0 has solutions, thus, Delta=4-4(-a)geq0, solving this gives ageq-1, therefore, the range of real number a is [-1, +infty). Hence, the answer is boxed{[-1, +infty)}.

answer:6. frac{41}{81}.Let the probability that the player makes the (n-1)-th shot be a_{n-1}, and the probability of missing be 1-a_{n-1}. Then the probability that he makes the n-th shot isbegin{array}{l}a_{n}=frac{2}{3} a_{n-1}+frac{1}{3}left(1-a_{n-1}right)=frac{1}{3}+frac{1}{3} a_{n-1} Rightarrow a_{n}-frac{1}{2}=frac{1}{3}left(a_{n-1}-frac{1}{2}right) . text { Also, } a_{n}-frac{1}{2} =frac{a_{n}-frac{1}{2}}{a_{n-1}-frac{1}{2}} cdot frac{a_{n-1}-frac{1}{2}}{a_{n-2}-frac{1}{2}} cdots cdot frac{a_{2}-frac{1}{2}}{a_{1}-frac{1}{2}}left(a_{1}-frac{1}{2}right) =left(frac{1}{3}right)^{n-1} times frac{1}{6}=frac{1}{2}left(frac{1}{3}right)^{n} Rightarrow a_{n}=frac{1}{2}+frac{1}{2} times frac{1}{3^{n}} .end{array}Given a_{1}=frac{2}{3}, we know a_{n}=frac{1}{2}left(1+frac{1}{3^{n}}right).Thus, a_{4}=frac{1}{2}left(1+frac{1}{3^{4}}right)=frac{41}{81}.

answer:Let's calculate: 28 times 7 + 11 = 196 + 11 = 207 Therefore, the dividend is 207. Hence, the answer is boxed{207}.