answer:# Problem 1: Calculation We start with the expression: -|-1|+sqrt{9}-sqrt[3]{-8}+left(pi +1right)^{0}-(frac{1}{2})^{-2}-|sqrt{3}-2| Breaking it down step by step: 1. -|-1| simplifies to -1 because the absolute value of -1 is 1, and the negative sign outside makes it -1. 2. sqrt{9} simplifies to 3 because 3^2 = 9. 3. -sqrt[3]{-8} simplifies to 2 because -2^3 = -8. 4. left(pi +1right)^{0} simplifies to 1 because anything to the power of 0 is 1. 5. -(frac{1}{2})^{-2} simplifies to -4 because (frac{1}{2})^{-2} = 4. 6. |sqrt{3}-2| simplifies to 2-sqrt{3} because sqrt{3} < 2, so the absolute value changes the sign of the expression inside. Putting it all together: -1 + 3 + 2 + 1 - 4 - (2 - sqrt{3}) = sqrt{3} - 1 Therefore, the final answer is boxed{sqrt{3} - 1}. # Problem 2: Solve the Equation Given the equation: frac{2}{x^{2}-4}+frac{x}{x-2}=1 1. Combine the fractions: frac{2}{x^{2}-4}+frac{x(x-2)}{x^{2}-4}=1 2. Simplify the equation: 2 + x(x+2) = x^2 - 4 3. This simplifies to: 2 + x^2 + 2x = x^2 - 4 4. Rearranging gives: 2x = -6 5. Solving for x: x = -3 Checking the solution in the original equation: When x = -3, x^2 - 4 = 9 - 4 = 5, which is not equal to 0, so the denominator is valid. Therefore, the solution to the equation is boxed{x = -3}.

answer:Let's designate the events in a simpler notation. Let: - A square B: Event "Alyosha wins against Borya" - B square V: Event "Borya wins against Vasya" - V square A: Event "Vasya wins against Alyosha" We are given: - mathrm{P}(A square B) = 0.6 - mathrm{P}(B square V) = 0.4 We need to find the probability of the event C that "There is no absolute winner." This will happen if and only if all three players have exactly one point, which is possible in two scenarios: 1. Scenario 1: - Alyosha wins against Borya (A square B) - Borya wins against Vasya (B square V) - Vasya wins against Alyosha (V square A) The probabilities for this scenario are: [ mathrm{P}(A square B) times mathrm{P}(B square V) times mathrm{P}(V square A) ] Since mathrm{P}(V square A) is an unknown, let it be x. Thus, the probability for Scenario 1: [ mathrm{P}(A square B) times mathrm{P}(B square V) times x = 0.6 times 0.4 times x = 0.24x ] 2. Scenario 2: - Alyosha wins against Borya (A square B), with probability 0.6 - Vasya wins against Borya (V square B), with probability 1 - mathrm{P}(B square V) = 1 - 0.4 = 0.6 - Alyosha loses to Vasya (A square V), with probability 1 - mathrm{P}(V square A) = 1 - x Thus, the probability for Scenario 2: [ mathrm{P}(A square B) times mathrm{P}(V square B) times mathrm{P}(A square V) = 0.6 times 0.6 times (1 - x) = 0.36 times (1 - x) ] The total probability of event C happening is the sum of the probabilities of these two scenarios: [ mathrm{P}(C) = 0.24x + 0.36(1 - x) ] Next, simplify the expression: [ 0.24x + 0.36 - 0.36x = 0.36 - 0.12x ] For event C, since everyone has to win one match among the three, let's assume: [ mathrm{P}(V square A) = x = 1 ] which does not change our calculation: [ mathrm{P}(C) = 0.36 ] Hence, [ boxed{0.24} ]<<|vq_6797|>

answer:1. Given a heptagon ( A_1, A_2, ldots, A_7 ) inscribed in a circle, we need to prove that if the center of the circle lies inside the heptagon, then the sum of the angles at vertices ( A_1, A_3, ) and ( A_5 ) is less than 450°. 2. We start by noting each angle's measure using the inscribed angle theorem: [ angle A_1 = 180^circ - frac{text{arc } A_2 A_7}{2}, quad angle A_3 = 180^circ - frac{text{arc } A_4 A_2}{2}, quad text{and} quad angle A_5 = 180^circ - frac{text{arc } A_6 A_4}{2}. ] 3. Summing these angles: [ angle A_1 + angle A_3 + angle A_5 = left(180^circ - frac{text{arc } A_2 A_7}{2}right) + left(180^circ - frac{text{arc } A_4 A_2}{2}right) + left(180^circ - frac{text{arc } A_6 A_4}{2}right). ] 4. Simplifying the sum: [ angle A_1 + angle A_3 + angle A_5 = 3 cdot 180^circ - frac{text{arc } A_2 A_7}{2} - frac{text{arc } A_4 A_2}{2} - frac{text{arc } A_6 A_4}{2}. ] 5. Recognizing that the angles make up partial arcs around the circle, and since the full circle is 360°, we can combine the arcs: [ text{arc } A_2 A_7 + text{arc } A_4 A_2 + text{arc } A_6 A_4 = 360^circ - text{arc } A_7 A_6. ] 6. Substituting back into the equation: [ angle A_1 + angle A_3 + angle A_5 = 3 cdot 180^circ - frac{360^circ - text{arc } A_7 A_6}{2}. ] 7. This reduces to: [ angle A_1 + angle A_3 + angle A_5 = 360^circ + frac{text{arc } A_7 A_6}{2}. ] 8. Since the center of the circle lies inside the heptagon, the arc ( A_7 A_6 ) subtended by these angles must be less than 180°: [ text{arc } A_7 A_6 < 180^circ. ] 9. Thus, [ angle A_1 + angle A_3 + angle A_5 < 360^circ + frac{180^circ}{2} = 360^circ + 90^circ = 450^circ. ] Conclusion: [ boxed{angle A_1 + angle A_3 + angle A_5 < 450^circ} ]

answer:Given the parabola y=a(x-b)(x-1), where a neq 0, and it passes through the point P(3,0), we aim to find the coordinates of another intersection point of the parabola with the x-axis. First, let's express the parabola in its standard quadratic form: begin{align*} y &= a(x-b)(x-1) &= ax^2 - a(b+1)x + ab. end{align*} The axis of symmetry for a parabola in the form y = ax^2 + bx + c is given by x = -frac{b}{2a}. Applying this to our equation, we find the axis of symmetry as: begin{align*} x &= -frac{-a(b+1)}{2a} &= frac{b+1}{2}. end{align*} Given that the parabola passes through (3,0), we substitute x = 3 and y = 0 into the equation to find the value of b: begin{align*} 0 &= a(3-b)(3-1) &= 2a(3-b). end{align*} Since a neq 0, for the equation to hold true, we must have 3-b = 0, which implies b = 3. With b = 3, the axis of symmetry is: begin{align*} x &= frac{3+1}{2} &= 2. end{align*} The parabola is symmetric about the line x = 2. Given that one intersection with the x-axis is at x = 3, the other intersection, due to symmetry, must be at: begin{align*} x &= 2 - (3 - 2) &= 1. end{align*} Therefore, the coordinates of the other intersection point of the parabola with the x-axis are (1,0). Thus, the correct answer is boxed{C}.