answer:Let H be the foot of the perpendicular from E to overline{DC}. Since CD = AB = 7, FG = CD - DF - GC = 7 - 2 - 3 = 2, and triangle FEG is similar to triangle AEB, we have the proportion [ frac{EH}{EH+4} = frac{2}{7}. ] Solving this, we find: [ 7EH = 2EH + 8 quad Rightarrow quad 5EH = 8 quad Rightarrow quad EH = frac{8}{5}. ] Then the area of triangle AEB is given by: [ [triangle AEB] = frac{1}{2} times AB times EH = frac{1}{2} times 7 times frac{8}{5} = frac{28}{5}. ] Hence, the area of triangle AEB is boxed{frac{28}{5}}.

answer:To solve this problem, we need to express both sides of the equation with the same base. Since 16 is 4 squared (4^2), we can rewrite the left side of the equation in terms of base 4. 16^y = (4^2)^y Now, using the power of a power rule (where (a^b)^c = a^(b*c)), we can simplify the left side: (4^2)^y = 4^(2*y) Given that y = 8, we can substitute y with 8: 4^(2*8) = 4^16 So, the exponent on the right side of the equation is boxed{16} .

answer:To solve this problem, we consider the two different ways of folding a regular hexagon paper in half: 1. **Folding along the line passing through two opposite vertices:** - When we fold the hexagon in this manner, we align two opposite vertices and the sides adjacent to these vertices. This folding method results in a shape that has four sides, which is a quadrilateral. - The sum of the interior angles of a polygon can be calculated using the formula (n-2) times 180^{circ}, where n is the number of sides. For a quadrilateral (n=4), the sum of its interior angles is: [ (4-2) times 180^{circ} = 2 times 180^{circ} = 360^{circ} ] - Therefore, when folding along the line passing through two opposite vertices, the resulting shape is a quadrilateral with the sum of its interior angles (calculated per layer) being boxed{360^{circ}}. 2. **Folding along the line passing through the midpoints of the opposite sides:** - In this case, the fold aligns the midpoints of opposite sides. This method of folding reduces the hexagon into a shape that has five sides, which is a pentagon. - Using the same formula for the sum of the interior angles, (n-2) times 180^{circ}, for a pentagon (n=5), the sum of its interior angles is: [ (5-2) times 180^{circ} = 3 times 180^{circ} = 540^{circ} ] - Thus, when folding along the line passing through the midpoints of the opposite sides, the resulting shape is a pentagon with the sum of its interior angles (calculated per layer) being boxed{540^{circ}}. In summary, depending on the folding line chosen, the regular hexagon paper can be folded into either a quadrilateral with a sum of interior angles of 360^{circ} or a pentagon with a sum of interior angles of 540^{circ}.

answer:1. **Understanding the Problem:** We need to find all integers ( k ) such that ( 1 le k le n ) and for which the following property holds: If ( x_1, x_2, ldots, x_n ) are ( n ) real numbers such that ( x_i + x_{i+1} + cdots + x_{i+k-1} = 0 ) for all integers ( i ) (with indices taken modulo ( n )), then ( x_1 = x_2 = cdots = x_n = 0 ). 2. **Case 1: ( k ) is not relatively prime to ( n ):** - Suppose (gcd(k, n) = d > 1). - We can construct a counterexample where the sequence ( x_1, x_2, ldots, x_n ) is not all zeros. - Consider a sequence where we have blocks of length ( d ) of the form ( 1, 0, 0, ldots, 0, -1 ). This sequence satisfies the condition ( x_i + x_{i+1} + cdots + x_{i+k-1} = 0 ) but is not identically zero. - Therefore, if ( k ) is not relatively prime to ( n ), the property does not hold. 3. **Case 2: ( k ) is relatively prime to ( n ):** - Suppose (gcd(k, n) = 1). - By Bézout's identity, there exist integers ( x ) and ( y ) such that ( kx equiv 1 pmod{n} ). - This implies that for any ( i ), ( x_i = x_{i+k} = x_{i+2k} = cdots ). - Since ( k ) is relatively prime to ( n ), the sequence ( x_i, x_{i+k}, x_{i+2k}, ldots ) will eventually cover all indices modulo ( n ). - Therefore, ( x_i = x_{i+1} ) for all ( i ), implying that all ( x_i ) are equal. - Given that the sum of any ( k ) consecutive terms is zero, and all terms are equal, each term must be zero. - Hence, ( x_1 = x_2 = cdots = x_n = 0 ). Conclusion: - The property holds if and only if ( k ) is relatively prime to ( n ). The final answer is ( boxed{ k } ) such that (gcd(k, n) = 1).