answer:To multiply the scalar -frac{7}{32} by the matrix, we multiply each element of the matrix by the scalar: left( begin{array}{ccc} -4 cdot left(-frac{7}{32}right) & -4 cdot left(-frac{7}{32}right) & 0 cdot left(-frac{7}{32}right) -6 cdot left(-frac{7}{32}right) & -2 cdot left(-frac{7}{32}right) & 8 cdot left(-frac{7}{32}right) 0 cdot left(-frac{7}{32}right) & -10 cdot left(-frac{7}{32}right) & -2 cdot left(-frac{7}{32}right) end{array} right) Simplifying each element, we get: left( begin{array}{ccc} frac{7}{8} & frac{7}{8} & 0 frac{21}{16} & frac{7}{16} & -frac{7}{4} 0 & frac{35}{16} & frac{7}{16} end{array} right) So, the resulting matrix after multiplication is: left( begin{array}{ccc} frac{7}{8} & frac{7}{8} & 0 frac{21}{16} & frac{7}{16} & -frac{7}{4} 0 & frac{35}{16} & frac{7}{16} end{array} right)

answer:The balanced chemical equation is: PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) The equilibrium constant expression is: Kc = 83.3 = [PCl₅] / ([PCl₃] [Cl₂]) Initially, [PCl₃] = 0.301 M, [Cl₂] = 0.301 M, and [PCl₅] = 0. Let x be the change in molarity at equilibrium: 83.3 = x / ((0.301 - x)(0.301 - x)) Solving for x: 83.3 = x / (0.0906 - 0.602x + x²) 7.55 - 50.15x + 83.3x² = x 83.3x² - 51.15x + 7.55 = 0 x ≈ 0.247 M At equilibrium: [PCl₃] ≈ 0.301 - 0.247 = 0.054 M [Cl₂] ≈ 0.301 - 0.247 = 0.054 M [PCl₅] ≈ 0.247 M So, the equilibrium concentrations are: [PCl₃] = 0.054 M [Cl₂] = 0.054 M [PCl₅] = 0.247 M

answer:The complex number (3-i) e can be written in the form a + bi, where a = 3e and b = -1. The magnitude (norm) of a complex number a + bi is given by sqrt{a^2 + b^2}. Therefore, the magnitude of (3-i) e is: sqrt{(3e)^2 + (-1)^2} = sqrt{9e^2 + 1} = sqrt{10} e The argument (phase angle) of a complex number a + bi is given by tan^{-1} left(frac{b}{a}right). Therefore, the argument of (3-i) e is: tan^{-1} left(frac{-1}{3e}right) = -tan^{-1} left(frac{1}{3}right) Since the complex number lies in the first quadrant, we need to add pi to the argument to get the correct phase angle. Therefore, the argument of (3-i) e is: pi - tan^{-1} left(frac{1}{3}right) Magnitude (Norm): sqrt{10} e Argument (Phase Angle): pi - tan^{-1} left(frac{1}{3}right)

answer:The sequence (a_n) is midpoint-convex, as 2a_n leq a_{n-1} + a_{n+1}. Because (a_n) is bounded, it follows that it is also a weakly decreasing sequence (not strictly decreasing, but non-increasing). By the monotone convergence theorem, a bounded and weakly decreasing sequence converges. Now, consider the sequence (b_n) = (a_{n+1} - a_n). Since (a_n) converges, (a_{n+1}) also converges, and thus the difference sequence (b_n) converges to the difference of the limits: lim_{n to infty} b_n = lim_{n to infty} (a_{n+1} - a_n) = lim_{n to infty} a_{n+1} - lim_{n to infty} a_n However, we cannot directly conclude the existence of lim_{n to infty} a_n from the given information. Nonetheless, the convergence of (b_n) to zero indicates that the difference between consecutive terms of (a_n) approaches zero, which is a characteristic of a convergent sequence. Therefore, (b_n) converges to zero.