answer:Let's call the particular number "x". According to the information given, we can set up the following equation: x - 29 + 64 = 76 Now, let's solve for x: x + (64 - 29) = 76 x + 35 = 76 Subtract 35 from both sides to isolate x: x = 76 - 35 x = 41 So, the particular number is boxed{41} .

answer:Let c denote the number of questions Lisa answered correctly, w denote the number of questions she answered incorrectly, and b denote the number of questions she did not answer. The conditions now give us: 1. The sum of questions answered and unanswered equals the total number of questions: [ c + w + b = 25 ] 2. The score calculation from correct and incorrect answers with a total score of 65: [ 4c - w = 65 ] We first solve for b: [ b = 25 - c - w ] We substitute from the second equation to find w: [ w = 4c - 65 ] Substituting w in the first equation: [ c + (4c - 65) + b = 25 Rightarrow 5c - 65 + b = 25 Rightarrow b = 90 - 5c ] Given that b geq 0, 90 - 5c geq 0 Rightarrow c leq 18. To ensure w is non-negative, 4c - 65 geq 0 Rightarrow c geq 16.25, thus c geq 17. Testing for maximum c, i.e., c = 18: [ w = 4(18) - 65 = 7 quad text{and} quad b = 90 - 5(18) = 0 ] This solution satisfies all conditions, so the maximum number of questions answered correctly is 18. The correct answer is C) boxed{18}.

answer:First simplify and rearrange the equation by factoring out common terms: [ 2(x^3 + 3x^2 sqrt{3} + 6x + 2 sqrt{3}) + (2x + 2 sqrt{3}) = 0. ] Factor out the common term 2(x + sqrt{3}): [ 2(x + sqrt{3})(x^2 + 2x sqrt{3} + 2) + 2(x + sqrt{3}) = 0. ] Combine like terms: [ 2(x + sqrt{3})(x^2 + 2x sqrt{3} + 3) = 0. ] Factoring further: [ 2(x + sqrt{3})((x + sqrt{3})^2 + 1) = 0. ] We then have: [ x + sqrt{3} = 0 quad text{or} quad (x + sqrt{3})^2 = -1. ] Solving these equations gives: [ x = -sqrt{3} quad text{or} quad x + sqrt{3} = pm i. ] Thus, the solutions to ( x ) are: [ x = -sqrt{3}, , x = -sqrt{3} + i, , x = -sqrt{3} - i. ] The final solutions are boxed{-sqrt{3}, -sqrt{3} + i, -sqrt{3} - i}.

answer:First, by using the triangle identity sqrt{1^2 + 3^2} = sqrt{10}, calculate: [cos angle AOC = frac{1}{sqrt{10}} quad text{and} quad sin angle AOC = frac{3}{sqrt{10}}.] Then, using angle addition formulas: begin{align*} cos angle AOB &= cos(angle AOC + angle BOC) &= cos angle AOC cos angle BOC - sin angle AOC sin angle BOC &= frac{1}{sqrt{10}} cdot frac{1}{sqrt{2}} - frac{3}{sqrt{10}} cdot frac{1}{sqrt{2}} &= -frac{1}{sqrt{5}}. end{align*} Using overrightarrow{OC} = m overrightarrow{OA} + n overrightarrow{OB}, and taking dot products: [frac{2}{sqrt{10}} = m - frac{1}{sqrt{5}} n] (with overrightarrow{OA}), [1 = -frac{1}{sqrt{5}} m + n] (with overrightarrow{OB}). Solving the system: begin{align*} m - frac{1}{sqrt{5}} n &= frac{2}{sqrt{10}}, -frac{1}{sqrt{5}} m + n &= 1. end{align*} Scaling and combining leads to: begin{align*} m &= frac{2 + sqrt{5}}{2}, n &= frac{1 + sqrt{5}}{2}. end{align*} Thus, (m, n) = boxed{left(frac{2 + sqrt{5}}{2}, frac{1 + sqrt{5}}{2}right)}.