answer:The cell membrane plays a crucial role in maintaining homeostasis by regulating the transport of materials into and out of the cell. It acts as a selective barrier, distinguishing the cell's internal environment from the external surroundings. This membrane allows certain substances to pass through while controlling the passage of others. Transport mechanisms across the cell membrane include active transport, which requires energy, and passive transport, which occurs due to the natural diffusion of molecules. By controlling these processes, the cell membrane ensures a stable internal environment, maintaining homeostasis.

answer:We can rearrange the terms of the given expression as follows: x_{1}.x_{2}.x_{3}+x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014} =(x_1 + x_4 + dots+x_{2014})(x_2 + x_5 + dots+x_{2012})(x_3 + x_6 + dots+x_{2013}) =(x_1 + x_2 + x_3 + dots + x_{2014} - x_2 - x_5 - dots - x_{2012})(x_2 + x_3 + x_4 + dots + x_{2014} - x_1 - x_4 - dots - x_{2013})(x_3 + x_4 + x_5 + dots + x_{2014} - x_1 - x_2 - dots - x_{2012}) =(2014 - x_2 - x_5 - dots - x_{2012})(2014 - x_1 - x_4 - dots - x_{2013})(2014 - x_1 - x_2 - dots - x_{2012}) Now, we can apply the AM-GM inequality to each of the three factors in the above expression: 2014 - x_2 - x_5 - dots - x_{2012} le frac{2014 - x_2 - x_5 - dots - x_{2012}}{3} 2014 - x_1 - x_4 - dots - x_{2013} le frac{2014 - x_1 - x_4 - dots - x_{2013}}{3} 2014 - x_1 - x_2 - dots - x_{2012} le frac{2014 - x_1 - x_2 - dots - x_{2012}}{3} Multiplying these three inequalities, we get: (2014 - x_2 - x_5 - dots - x_{2012})(2014 - x_1 - x_4 - dots - x_{2013})(2014 - x_1 - x_2 - dots - x_{2012}) le left(frac{2014 - x_2 - x_5 - dots - x_{2012}}{3}right)^3 x_{1}.x_{2}.x_{3}+x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014} le left(frac{2014 - x_2 - x_5 - dots - x_{2012}}{3}right)^3 x_{1}.x_{2}.x_{3}+x_{2}.x_{3}.x_{4}+x_{3}.x_{4}.x_{5}+...+x_{2012}.x_{2013}.x_{2014} le left(frac{2014}{3}right)^3 Since x_{1}+x_{2}+...+x_{2014}=2014, we have x_2 + x_5 + dots + x_{2012} = x_1 + x_4 + dots + x_{2013} = x_1 + x_2 + dots + x_{2012} = frac{2014}{3}. Therefore, the maximum value of the given expression is left( frac{2014}3 right)^3. The maximum value of the given expression is left( frac{2014}3 right)^3.

answer:To find the third point C, let's denote the sides of the triangle as a, b, and c, with corresponding angles A, B, and C, respectively. You know c, gamma, and the ratio a/b. Your goal is to determine a, b, A, and B. 1. Using the law of cosines, we can relate the sides and angles: c^2 = a^2 + b^2 - 2abcos C Since cos C is known, we can express the product ab in terms of c and the given ratio: ab = frac{c^2}{frac{a}{b} + frac{b}{a} - 2cos C} Then, a and b can be found separately: a = sqrt{frac{c^2}{frac{a}{b} + frac{b}{a} - 2cos C} cdot frac{b}{a}} quad text{and} quad b = sqrt{frac{c^2}{frac{a}{b} + frac{b}{a} - 2cos C} cdot frac{a}{b}} 2. Next, use the law of sines to find the angles A and B: frac{sin A}{a} = frac{sin B}{b} = frac{sin C}{c} Thus, the angles are: A = arcsinleft(frac{a cdot sin C}{c}right) quad text{and} quad B = arcsinleft(frac{b cdot sin C}{c}right) Now, with a, b, A, and B determined, you can calculate the coordinates of point C using basic trigonometry. Note that you may need to adjust the signs of A and B based on the orientation of the triangle.

answer:The inverse of the given matrix is left( begin{array}{ccc} -frac{1}{3} & frac{1}{2} & 0 frac{1}{2} & -frac{1}{2} & frac{1}{2} frac{1}{6} & -frac{1}{2} & frac{1}{2} end{array} right)