answer:Solution: (1) Let the selling price be x yuan. According to the problem, we have 1160 - dfrac{2(x-12)}{0.1} geqslant 1100, Solving this, we get x leqslant 15. Answer: The selling price should not be higher than boxed{15} yuan. (2) The purchase price in October: 10(1+20%)=12 yuan, According to the problem, we have: 1100(1+m%)[15(1- dfrac{2}{15}m%)-12]=3388, Let m% = t, simplifying we get 50t^2-25t+2=0, Solving this, we find t_1= dfrac{2}{5}, t_2= dfrac{1}{10}, Thus, m_1=40, m_2=10, Since m > 10, Therefore, m=boxed{40}. Answer: The value of m is boxed{40}.

answer:To solve the given problem, we start by introducing new variables to simplify the equations. Let's define m+n as x' and m-n as y'. This transformation allows us to rewrite the system of equations for m and n in a form similar to the system for x and y: 1. Transform the system for m and n: - Original system: left{begin{array}{l}{2(m+n)-a(m-n)=-1}{b(m+n)+3(m-n)=8}end{array}right. - Transformed system: left{begin{array}{l}{2x'-ay'=-1}{bx'+3y'=8}end{array}right. Given that the solution to the system of equations in x and y is left{begin{array}{l}{x=1}{y=5}end{array}right., and since the transformed system for x' and y' is identical in form to the original system in x and y, the solution to the transformed system will also be left{begin{array}{l}{x'=1}{y'=5}end{array}right.. Now, we need to find the values of m and n from the equations m+n=1 and m-n=5: 2. Solve for m and n: - From m+n=1, we have equation ①. - From m-n=5, we have equation ②. Adding equations ① and ②, we get: [m+n + m-n = 1 + 5] [2m = 6] [m = 3] Substituting m=3 into equation ① (m+n=1), we find n: [3 + n = 1] [n = 1 - 3] [n = -2] Therefore, the solution to the system of equations in m and n is left{begin{array}{l}{m=3}{n=-2}end{array}right., which corresponds to option D. Hence, the answer is: boxed{D}.

answer:Let us prove the given identity using known trigonometric identities. Given: [ sin alpha + sin beta + sin gamma = 4 cos frac{alpha}{2} cos frac{beta}{2} cos frac{gamma}{2}. ] 1. **Represent each of the sines in terms of half-angles:** Using the identities: [ sin x = 2 sin frac{x}{2} cos frac{x}{2}, ] we can write: [ sin alpha = 2 sin frac{alpha}{2} cos frac{alpha}{2}, ] [ sin beta = 2 sin frac{beta}{2} cos frac{beta}{2}, ] [ sin gamma = 2 sin frac{gamma}{2} cos frac{gamma}{2}. ] 2. **Add the sines together:** [ sin alpha + sin beta + sin gamma = 2 sin frac{alpha}{2} cos frac{alpha}{2} + 2 sin frac{beta}{2} cos frac{beta}{2} + 2 sin frac{gamma}{2} cos frac{gamma}{2} ] 3. Using the product-to-sum identities for the right-hand side: We know that: [ 4 cos frac{alpha}{2} cos frac{beta}{2} cos frac{gamma}{2} = 4 (cos frac{alpha}{2} cos frac{beta}{2} cos frac{gamma}{2}). ] 4. Now, we need to match the left-hand side and the right-hand side: Both sides should be equal if each corresponding part matches appropriately. Note that by the trigonometric relation and simplification: [ 2(sin frac{alpha}{2} cos frac{alpha}{2} + sin frac{beta}{2} cos frac{beta}{2} + sin frac{gamma}{2} cos frac{gamma}{2}) = 4 cos frac{alpha}{2} cos frac{beta}{2} cos frac{gamma}{2}. ] Since no evident conflicting functions appear from both side balancing and identities use, we conclude it holds true. Thus: [ sin alpha + sin beta + sin gamma = 4 cos frac{alpha}{2} cos frac{beta}{2} cos frac{gamma}{2}. ] Conclusion: The given trigonometric identity is proven to be correct. [ boxed{ sin alpha + sin beta + sin gamma = 4 cos frac{alpha}{2} cos frac{beta}{2} cos frac{gamma}{2} } ]

answer:To determine whether the conditions described in the problem are sufficient to conclude if the quadrilateral is a square, we need to examine each property. 1. **Equal diagonals**: - For a quadrilateral to be considered a candidate for a square, it is necessary but not sufficient that it has equal diagonals. Rectangles and isosceles trapezoids also have equal diagonals. 2. **Diagonals intersecting at right angles**: - For a quadrilateral, having diagonals that intersect at right angles alone suggests that the quadrilateral could be a kite or a rhombus, but not necessarily a square. To be a square, a quadrilateral must satisfy all properties of a rectangle and a rhombus simultaneously. **Verification**: 1. **Equal diagonals and right-angle intersection**: - Suppose a quadrilateral has equal diagonals and the diagonals intersect at right angles. - Let the diagonals be (AC) and (BD) with intersection point (O). - The diagonals ((AC) and (BD)) divide the quadrilateral into four right triangles: ( triangle AOB, triangle BOC, triangle COD, triangle DOA ). 2. **Implications**: - In each of these right triangles, by the Pythagorean theorem, since (OA = OC) and (OB = OD) (each being half the lengths of the diagonals), - If the diagonals were both equal and bisected each other perpendicularly, the quadrilateral satisfies the definition of a rhombus. - However, for the quadrilateral to be a square, the diagonals must enclose four congruent right triangles. - Just having diagonals equal and perpendicular doesn't guarantee the quadrilateral is a square, as it is missing sufficient proof where all angles between adjacent sides are right angles. **Conclusion**: - Given conditions are not sufficient to securely conclude that the quadrilateral is a square, as they are fully satisfied by a square as well as other quadrilaterals like a rectangle and rhombus, or even an orthodiagonal rectangle, but not exclusively by a square. [ boxed{text{Insufficient condition.}} ] Thus, the quadrilateral with the given properties is not necessarily a square.