answer:1. **Initial Setup**: The Tigers and Sharks have played 3 games, with the Tigers winning 2 games and the Sharks winning 1 game. Let N be the number of additional games played, all of which are won by the Sharks (to maximize the Sharks' winning percentage). 2. **Setting Up the Equation**: The total number of games played after N additional games is 3 + N. The total number of games won by the Sharks is 1 + N. We need the fraction of games won by the Sharks to be at least 95%, or frac{95}{100} = frac{19}{20}. Therefore, we set up the equation: [ frac{1 + N}{3 + N} = frac{19}{20} ] 3. **Solving the Equation**: - Cross-multiplying to eliminate the fractions, we get: [ 20(1 + N) = 19(3 + N) ] - Expanding both sides: [ 20 + 20N = 57 + 19N ] - Simplifying by moving all terms involving N to one side and constants to the other: [ 20N - 19N = 57 - 20 ] [ N = 37 ] 4. **Conclusion**: The minimum number of additional games N that the Sharks need to win to ensure their winning percentage is at least 95% is 37. Thus, the minimum possible value for N is boxed{textbf{(B)}; 37}.

answer:**Analysis** This problem mainly tests the simplification and evaluation of trigonometric functions, examining the student's computational ability. By using the auxiliary angle formula to simplify the function, we can find A. By using the induction formula to simplify the function, we can find B. **Solution** Given the information, we have: sqrt{2}sin left(A+ frac{pi}{4}right) = sqrt{2}, because 0 < A < pi, therefore A= frac{pi}{4}. From the given information, we also have sqrt{3}cos A = sqrt{2}sin B. Substituting A= frac{pi}{4} into this, we get sin B= frac{sqrt{3}}{2}, because 0 < B < frac{3pi}{4}, therefore B= frac{pi}{3} or frac{2pi}{3}. Therefore, the answer is boxed{frac{pi}{3} text{ or } frac{2pi}{3}}.

answer:1. **Introduce the variables and given data:** Let ( Q ) be the center of circle ( K ). Given the proportions: [ AB = 5x, quad AM = 2x ] Also, let ( AN ) be the height of triangle ( triangle ABC ). 2. **Calculate the height ( AN ) and length ( BN ):** Since ( angle B = arcsinleft(frac{3}{5}right) ), where sine of the angle is the ratio of the opposite side to the hypotenuse in a right-angled triangle, we have: [ AB sin(angle B) = 5x cdot frac{3}{5} = 3x ] The length ( BN ) can be calculated by the Pythagorean Theorem: [ BN = AB cos(angle B) = 5x cdot frac{4}{5} = 4x ] 3. **Find the relationships involving the circle ( K ):** Since ( Q ) lies on the circumcircle of ( triangle ABC ), it implies the circle ( Q ) equally bisects ( angle B ). Using this, the additional angles around point ( Q ) and triangle relationships can be inferred. [ angle AMC = frac{1}{2} angle AQC = frac{1}{2} left(180^circ - angle Bright) = 90^circ - frac{1}{2} angle B ] Given ( QA = QC ), the segment ( BQ ) is the bisector of ( angle B ). 4. **Determine perpendicular relationships:** Since ( BQ ) bisects ( angle B ), we have: [ angle QBM = frac{1}{2} angle B ] And thus, [ MC perp BQ ] 5. **Identify isosceles triangle properties:** Using the perpendicularity, triangles involving ( M ) and ( C ) lead us to deduce that: [ triangle MBC text{ is isosceles, with } BM = BC = 7x ] Therefore, [ NC = BC - BN = 7x - 4x = 3x = NA ] 6. **Identifying ( triangle ANC ):** [ NC = NA ] From this, we conclude ( triangle ANC text{ is isosceles, implying } triangle ANCtext{ is }angle NC = angle NCA = 45^circ ). # Conclusion: [ boxed{45^circ} ]

answer:Since the element a is in the set {a, b, c}, we have a in {a, b, c}. Since the element 0 is in the set {x | x^2 neq 0}, we have 0 in {x | x^2 neq 0}. Therefore, the answer is: boxed{in, in}.