answer:1. **Examine the given equation**: Written as x^3(2x+2y+3) = y^3(2x+2y+3). 2. **Factor out common terms**: Factor (2x+2y+3) from both sides. - If 2x+2y+3 neq 0, divide both sides by 2x+2y+3. - Resulting in x^3 = y^3. 3. **Solve the simplified equation x^3 = y^3**: - x^3 - y^3 = 0. - Factor as (x - y)(x^2 + xy + y^2) = 0. - This results in two cases: - x = y. - x^2 + xy + y^2 = 0, which can be satisfied by complex numbers but not by real numbers unless x = y = 0. 4. **Analyze the line equation 2x+2y+3 = 0**: - Rearranging gives y = -x - frac{3}{2}. 5. **Graph description**: - From x = y, we have a line y = x. - The line y = -x - frac{3}{2} passes through different points and does not intersect y = x at any other point, except potentially at infinity or complex plane intersections which are not considered here. 6. **Conclusion**: The graph comprises of a line y = x, and another distinct line y = -x - frac{3}{2}. Therefore, the answer is two distinct lines in the plane. [ text{two intersecting lines} ] boxed{The correct answer is B) Two intersecting lines.}

answer:To find the highest point for each trajectory, set the derivative of y w.r.t. t equal to zero (maxima condition): [ v sin theta - g t = 0 implies t = frac{v sin theta}{g}. ] Plugging this t into the parametric equations, we get: [ x = frac{v^2}{g} sin theta cos theta, quad y = h + frac{v^2}{2g} sin^2 theta. ] Since x = frac{v^2}{2g} sin 2theta using the double angle formula, and substituting h = frac{v^2}{8g}, we have: [ y = frac{v^2}{8g} + frac{v^2}{2g} sin^2 theta = frac{v^2}{8g} + frac{v^2}{4g} left(frac{1 - cos 2 theta}{2}right) = frac{3v^2}{8g} - frac{v^2}{4g} cos 2 theta. ] Thus, the points (x,y) lie on an ellipse centered at left(0, frac{3v^2}{8g}right) with semi-major axis frac{3v^2}{8g} and semi-minor axis frac{v^2}{4g}. The area of this ellipse is: [ pi cdot frac{3v^2}{8g} cdot frac{v^2}{4g} = frac{3pi}{32} cdot frac{v^4}{g^2}. ] Thus, the value of c = boxed{frac{3pi}{32}}.

answer:From the equation (x-4)^2 + y^2 = r^2 (r > 0), we know that the center of the circle is at (4, 0), and the radius is r. Since the line sqrt{3}x - 2y = 0 is tangent to the circle (x-4)^2 + y^2 = r^2 (r > 0), The distance d from the center of the circle to the line is d = frac{4sqrt{3}}{sqrt{3 + 4}} = frac{4sqrt{21}}{7}, Thus, the radius of the circle is boxed{frac{4sqrt{21}}{7}}. Therefore, the correct answer is: C. This problem examines the relationship between a line and a circle, specifically the application of the formula for the distance from a point to a line, and is considered a basic question.

answer:(Ⅰ) Since we know the coordinates of the vertices, we can determine that P_1, P_2, and P_4 are all on the line y=1, so y_1=y_2=y_4=1, and P_3 is at the midpoint of OP_1 which lies on the line y=frac{1}{2}, so y_3=frac{1}{2}. For P_5, since it is the midpoint of P_1P_2, and P_2 has y-coordinate of 1, y_5=frac{3}{4}. Therefore, we have: a_1 = frac{1}{2}y_1 + y_2 + y_3 = frac{1}{2} cdot 1 + 1 + frac{1}{2} = 2 a_2 = frac{1}{2}y_2 + y_3 + y_4 = frac{1}{2} cdot 1 + frac{1}{2} + 1 = 2 a_3 = frac{1}{2}y_3 + y_4 + y_5 = frac{1}{2} cdot frac{1}{2} + 1 + frac{3}{4} = 2 Using the recursive nature of {P_n} as defined in the problem statement, for any n, P_{n-3} is the midpoint of P_n and P_{n+1}, which implies: y_{n-3} = frac{y_n + y_{n+1}}{2} Then we have the following for a_{n+1}: a_{n+1} = frac{1}{2}y_{n+1} + y_{n+2} + y_{n+3} = frac{1}{2}y_{n+1} + y_{n+2} + frac{y_n + y_{n+1}}{2} = frac{1}{2}y_n + y_{n+1} + y_{n+2} = a_n This shows that sequence {a_n} is a constant sequence, and therefore: a_n = a_1 = 2, forall n in mathbb{N}^* [boxed{ text{So, } a_n = 2 }] (Ⅱ) Dividing both sides of the equation frac{1}{2}y_n + y_{n+1} + y_{n+2} = 2 by 2 gives us: frac{1}{4}y_n + frac{1}{2}(y_{n+1} + y_{n+2}) = 1 Given that P_{n+4} is the midpoint of P_{n+1} and P_{n+2}, we have: y_{n+4} = frac{y_{n+1} + y_{n+2}}{2} Substituting the above relation into the equation, we get: y_{n+4} = 1 - frac{y_n}{4} [boxed{ y_{n+4} = 1 - frac{y_n}{4} }] (Ⅲ) Considering the definition of b_n, we have: b_{n-1} = y_{4n+3} - y_{4n} = left(1 - frac{y_{4n}}{4}right) - left(1 - frac{y_{4n+4}}{4}right) = -frac{1}{4}(y_{4n+4} - y_{4n}) = -frac{1}{4}b_n Since y_3 - y_4 = frac{1}{2} - 1 = -frac{1}{4} and it is non-zero, the sequence {b_n} is a geometric progression with the common ratio -frac{1}{4}. [boxed{ { b_n } text{ is a geometric progression with common ratio } -frac{1}{4} }]