answer:Using Euler's formula, e^{itheta} = cos(theta) + isin(theta), substitute theta = frac{15pi}{2}: 1. Simplify theta: frac{15pi}{2} can be simplified by reducing the multiple of 2pi to find its equivalent simple angle. Considering 2pi is a full rotation: [ frac{15pi}{2} = 2pi times 3 + frac{3pi}{2} = 6pi + frac{3pi}{2} ] Here, angle frac{3pi}{2} corresponds to 270^circ, which is a three-quarter turn counterclockwise from the positive x-axis. 2. Determine cosine and sine values: [ cosleft(frac{15pi}{2}right) = cosleft(frac{3pi}{2}right) = 0 ] [ sinleft(frac{15pi}{2}right) = sinleft(frac{3pi}{2}right) = -1 ] 3. Substitute back into Euler's formula: [ e^{15pi i/2} = cosleft(frac{15pi}{2}right) + isinleft(frac{15pi}{2}right) = 0 - i = boxed{-i} ]

answer:1. **Identify the Set ( S ):** We are given the set ( S ) which consists of lattice points (points with integer coordinates) inside the circle ( x^2 + y^2 = 11 ). This means we are looking for points ((x, y)) such that ( x ) and ( y ) are integers and ( x^2 + y^2 leq 10.48 ) (since ( sqrt{11} approx 3.32 )). 2. **Identify Boundary Points:** Identify the points on the boundary by checking examples where ( x ) and ( y ) are integers: - Points with ( y = 0 ): ( (x, y) = (pm 3, 0) ) - Points with ( x = 0 ): ( (x, y) = (0, pm 3) ) - Other points: ( ( pm 1, pm 3) ) and ( ( pm 3, pm 1) ) - Diagonal points: ( ( pm 2, pm 2) ) Therefore, the relevant boundary points include ((pm 3, 0)), ((0, pm 3)), (( pm 1, pm 3)), (( pm 3, pm 1)), and (( pm 2, pm 2)). 3. **Determine Max Triangle Area:** To maximize the area of a triangle with vertices on the boundary of ( S ), we look for cases that maximize the determinant associated with the area formula for triangles. The determinant for the area of a triangle given vertices ((x_1, y_1)), ((x_2, y_2)), ((x_3, y_3)): [ A = frac{1}{2} left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) right| ] 4. **Compute Example Triangle Areas:** Compute the area for triangles formed from the vertices: - For the triangle with vertices ((2,2)), ( (1, -3) ), ((-3, 1)): [ A = frac{1}{2} left| 2(-3-1) + 1(1-2) + (-3)(2+3) right| = frac{1}{2} left| -8 - 1 - 15 right| = frac{1}{2} times 24 = 12 ] 5. **Identifying Other Maximum Area Triangles:** By symmetry and similar calculations: - Triangles such as ((3, -1)), ((-3, -1)), ((1, 3)) - Triangles such as ((3, -1)), ((-3, -1)), and ((0, 3)) Each of these configurations eventually calculates to the maximum area ( M = 12 ). 6. **Counting the Solutions:** We identify all possible sets of vertices that form triangles with maximum area: - Ensure each configuration matches one of the patterns and multiplicity by observing symmetry. 7. **Conclusion:** We conclude there are ( 16 ) distinct triangles with maximum area ( M = 12 ). [ boxed{16} ]

answer:Given the property ( f(x+y) = f(x) + f(y) ), we hypothesize that ( f(x) = cx ). We need to determine ( c ) using the given ( f(3) = 4 ). Therefore, [ f(3) = 3c = 4 ] [ c = frac{4}{3} ] Now, we want to find ( f(6) ). Based on our hypothesis, [ f(6) = 6c = 6 times frac{4}{3} = boxed{8} ]

answer:1. **Expressing in Binary:** The numerator (2^{225} + 1) in binary is (1 followed by 15 consecutive zeros repeated 14 times and ending with 1). The denominator (2^{15} + 1) in binary is (1 followed by 15 ones). 2. **Using Geometric Series:** [ frac{1}{2^{15}+1} = 2^0 - 2^{15} + 2^{30} - 2^{45} + cdots + 2^{210} - 2^{225} + ldots ] 3. **Multiplying and Simplifying:** When multiplying ((2^{15} + 1)) by ((2^0 - 2^{15} + 2^{30} - cdots + 2^{210} - 2^{225})), we get a telescoping series that simplifies to 1, meaning: [ 2^{225} + 1 = (2^{15} + 1)(2^0 + 2^{30} + 2^{60} + cdots + 2^{210}) ] 4. **Counting the Terms:** Each term (2^{n}) for (n = 0, 30, 60, ldots, 210) contributes to the sequence, resulting in a total of (1 + (210/30) = 8) terms. 5. **Conclusion:** The total number of terms, m, is then (8). (8) The final answer is boxed{textbf{(A)} 8}