answer:To find the first order Taylor expansion of the composition f(g(x)), we can use the chain rule. The chain rule states that if f is a function of g, then the derivative of f(g(x)) with respect to x is equal to f'(g(x))g'(x). In this case, f(x) = x^2 and g(x) = cos (x^2). So, f'(x) = 2x and g'(x) = -2x sin (x^2). Evaluating these derivatives at x = 1, we get f'(1) = 2 and g'(1) = -2 sin (1). Therefore, the first order Taylor expansion of f(g(x)) about x = 1 is: f(g(x)) approx f(g(1)) + f'(g(1))(x-1) = f(cos (1)) + f'(cos (1))(-2 sin (1))(x-1) = 1+cos (1) + 2(1-cos (1))(x-1) = (x-1) (2-2 sin (1))+1+cos (1) The answer is (x-1) (2-2 sin (1))+1+cos (1)

answer:Centroid: left(frac{1}{3} left(1+sin left(frac{11 pi }{45}right) cos left(frac{7 pi }{36}right) csc left(frac{pi }{20}right)right), frac{1}{3} cos left(frac{7 pi }{36}right) cos left(frac{11 pi }{45}right) csc left(frac{pi }{20}right)right) Incenter: left(frac{2 cos left(frac{7 pi }{36}right) left(sin left(frac{11 pi }{90}right)+cos left(frac{11 pi }{90}right)right)^2}{sqrt{2 left(1+sin left(frac{pi }{90}right)right)}+2 sin left(frac{pi }{20}right)+2 cos left(frac{7 pi }{36}right)}, frac{2 cos left(frac{7 pi }{36}right) cos left(frac{11 pi }{45}right)}{sqrt{2 left(1+sin left(frac{pi }{90}right)right)}+2 sin left(frac{pi }{20}right)+2 cos left(frac{7 pi }{36}right)}right) Orthocenter: left(sin left(frac{11 pi }{45}right) cos left(frac{7 pi }{36}right) csc left(frac{pi }{20}right), tan left(frac{11 pi }{45}right) left(1-sin left(frac{11 pi }{45}right) cos left(frac{7 pi }{36}right) csc left(frac{pi }{20}right)right)right) Circumcenter: left(frac{1}{2}, frac{1}{4} csc left(frac{pi }{20}right) sec left(frac{11 pi }{45}right) left(sin left(frac{37 pi }{180}right)+cos left(frac{7 pi }{36}right)right)right) Nine-point center: left(frac{1}{4} left(2+cos left(frac{11 pi }{180}right) csc left(frac{pi }{20}right)right), frac{1}{8} csc left(frac{pi }{20}right) sec left(frac{11 pi }{45}right) left(cos left(frac{7 pi }{36}right)-sin left(frac{11 pi }{60}right)right)right) Symmedian point: left(-frac{4 cos left(frac{7 pi }{36}right) left(3 cos left(frac{7 pi }{36}right)-sin left(frac{37 pi }{180}right)right)}{sqrt{2 left(5+sqrt{5}right)}-4 left(3+sin left(frac{pi }{90}right)+sin left(frac{pi }{9}right)right)}, -frac{8 sin left(frac{pi }{20}right) cos left(frac{7 pi }{36}right) cos left(frac{11 pi }{45}right)}{sqrt{2 left(5+sqrt{5}right)}-4 left(3+sin left(frac{pi }{90}right)+sin left(frac{pi }{9}right)right)}right)

answer:Given: - Mass of CO2, m = 2 kg - Initial pressure, P1 = 110 kPa - Initial temperature, T1 = 24°C = 297 K - Final pressure, P2 = 1200 kPa - Final temperature, T2 = 308°C = 581 K - Gas constant for CO2, R = 0.1889 kJ/(kg·K) - Heat capacity at constant volume, CV = 0.657 kJ/(kg·K) The initial volume, V1, can be calculated using the ideal gas law: 110 kPa = (2 kg) * (0.1889 kJ/(kg·K)) * (297 K) / V1 V1 = 1.02 m³ The final volume, V2, can be found similarly: 1200 kPa = (2 kg) * (0.1889 kJ/(kg·K)) * (581 K) / V2 V2 = 0.182 m³ The work done, W, during the process is given by: W = (P1 + P2) / 2 * (V2 - V1) W = (110 kPa + 1200 kPa) / 2 * (0.182 m³ - 1.02 m³) = -548.28 kJ The negative sign indicates work is done on the system. The change in internal energy, ΔU, is: ΔU = m * CV * (T2 - T1) ΔU = (2 kg) * (0.657 kJ/(kg·K)) * (581 K - 297 K) = 1153.69 kJ Using the first law of thermodynamics (Q - W = ΔU), the heat transfer, Q, is: Q = ΔU + W = 1153.69 kJ + 548.28 kJ = 1601.97 kJ Hence, the heat transfer to the system is 1601.97 kJ, and the work done by the system is 548.28 kJ.

answer:To find the real solutions, we need to consider the two possible cases for the absolute value: 1. 9x - 10 = 14 (when the expression inside the absolute value is positive or zero) 2. 9x - 10 = -14 (when the expression inside the absolute value is negative) Solving each case separately: 1. 9x - 10 = 14 Adding 10 to both sides gives us 9x = 24. Dividing by 9, we find x = frac{24}{9} = frac{8}{3}. 2. 9x - 10 = -14 Adding 10 to both sides gives us 9x = -4. Dividing by 9, we find x = frac{-4}{9}. So, the real solutions are x = frac{8}{3} and x = -frac{4}{9}. In LaTeX format: x = left{frac{8}{3}, -frac{4}{9}right}