answer:Let int_{0}^{1} f(x)dx = t. Integrate both sides of f(x) = x^{2} + 2 int_{0}^{1}f(x)dx, we get: t = int_{0}^{1} x^2 dx + 2 int_{0}^{1} t dx = left[frac{x^3}{3}right]_0^1 + 2t int_{0}^{1} dx = frac{1}{3} + 2t Solving for t, we find int_{0}^{1}f(x)dx = t = -frac{1}{3}. So, the correct answer is boxed{B}. In this problem, treat the definite integral as a constant and integrate both sides. Simplify and solve for the desired result. This question tests understanding of definite integrals and the fundamental theorem of calculus.

answer:To solve this problem, we focus on the number of noncongruent triangles formed by any three of the points A, B, C, D, E, F. **Step 1: Total triangle combinations** There are binom{6}{3} = 20 combinations of three points from six. **Step 2: Identify collinear cases** - Points A, D, B are collinear. - Points B, E, C are collinear. - Whether A, F, C are collinear depends on alpha. If alpha = 1/2, they are collinear. Assume alpha neq 1/2 for a more general case. **Step 3: Count distinct triangle types** - Triangles using vertices of triangle ABC and one midpoint (DEF cases): - triangle ADE, triangle BDE (similar due to symmetry and midpoint D), and similarly for E and F (with adjustments based on alpha). - These triangles are not generally congruent if alpha neq 1/2. - Triangles using one vertex from triangle ABC and two midpoints: - triangle DEF is unique and its shape changes with alpha. - triangle AEF, triangle BFD, triangle CDE are also distinct based on alpha and other midpoint locations. **Step 4: Conclusion** Based on alpha not being frac{1}{2}, we generally have: - 3 non-congruent triangles from category (one vertex): triangle ADE, triangle BDE, triangle CDE - 4 non-congruent triangles from DEF-related (no original vertices): triangle DEF and variations with one original vertex and two defined points. Thus, the total non-congruent triangles are 7 if alpha neq 1/2. The correct answer is boxed{text{(B)} 7}.

answer:1. **Calculate radii of the frustum bases using their areas**: - Lower base area = 400pi implies a radius of r_L = sqrt{frac{400pi}{pi}} = 20 cm. - Upper base area = 100pi implies a radius of r_U = sqrt{frac{100pi}{pi}} = 10 cm. 2. **Establish proportional relationship**: - The radius of the upper base (r_U) is frac{1}{2} of the lower base radius (r_L). By the uniform slope property of a cone, this implies the small cut-off cone's height h is frac{1}{2} of the total height H of the original cone. 3. **Solve for the total height and the height of the smaller cone**: - Given the frustum height is frac{1}{2} of H, let the height of the frustum H_f = 30 cm. - Then, frac{1}{2}H = H_f implies H = 2H_f = 60 cm. - The small cone's height h = frac{1}{2}H = frac{1}{2} times 60 = 30 cm. boxed{30 text{ cm}} is the height of the smaller cone.

answer:To solve this modified problem, every row, every column, and the main diagonal must all have an odd sum. The configuration of odd and even numbers becomes crucial here. Step 1: Odd and Even Numbers From the numbers 1 to 9, there are 5 odd numbers (1, 3, 5, 7, 9) and 4 even numbers (2, 4, 6, 8). Step 2: Placement Requirement - In each row, column, and the main diagonal to be odd: - Each entity can have three odd numbers or one odd and two even numbers. Step 3: Placement of Even Numbers - Since there are only 4 even numbers, if we try to form the main diagonal odd, it must contain three odd numbers or one odd and two even numbers. But the limit of four evens across the entire grid doesn't fully support the latter mixed condition globally. Step 4: Focusing on Three Odds for the Main Diagonal - Place 3 odd numbers on main diagonal spots; possibilities for arranging 3 out of 5 odd numbers in these 3 spots: binom{5}{3} times 3! = 10 times 6 = 60 ways. - The remaining 2 odd numbers must be in the two squares that are not on the diagonal in any of these rows, with only one way to switch them. - Place 4 even numbers in the remaining 4 squares such that rows and columns add up to odd, which now defines distinct spots without a choice in arrangement. Step 5: Total Valid Placements - All remaining arrangements comply without inner row/column conflicts because we've forced diagonals and correct parity across structure. Step 6: Total Arrangements The total number of possible arrangements of the numbers from 1 to 9 is 9!. Step 7: Calculate Probability [ frac{60 cdot 2! cdot 4!}{9!} = frac{60 times 2 times 24}{362,880} = frac{2880}{362,880} = frac{1}{126} ] Conclusion The probability that the sum of the numbers in each row, each column, and the main diagonal is odd is frac{1{126}}. The final answer is C) boxed{frac{1}{126}}