answer:To solve this problem, first, address the exponential decay of volume: V = V0 * e^(-At). Then, rewrite the original differential equation for substrate concentration as: (1/(V_m * C)) * (K_m + C) * (dC/dt) = -1/(V0) * e^(At) Integrate both sides. On the left, you obtain: K_m * ln|C| + C / V_m On the right: -1/(V0 * A) * e^(At) + C1, where C1 is the constant of integration. Using the initial condition (t=0, C=C0), we can express C1 in terms of the given parameters and C0. This leads to an implicit equation linking C and t. While you can solve explicitly for t in terms of C, solving for C in terms of t is not possible using elementary functions. Therefore, numerical methods would be required for practical solutions. For completeness, the implicit equation is: K_m * ln|C| + C / V_m = -1/(V0 * A) * e^(At) + C1 However, it's important to note that finding an explicit solution for C in terms of t is not feasible due to the complexity introduced by the volume change.

answer:The result of multiplying the scalar frac{1}{2} with the vector left( begin{array}{c} -4 end{array} right) is left( begin{array}{c} -4 cdot frac{1}{2} end{array} right) = left( begin{array}{c} -2 end{array} right)

answer:Given the hydronium ion concentration ([H₃O⁺]) is 1.0 × 10⁻⁶ M, we can determine the solution's pH and hydroxide ion concentration ([OH⁻]). First, let's calculate the pH: pH = -log[H₃O⁺] pH = -log(1.0 × 10⁻⁶ M) pH = 6.0 Next, we find the [OH⁻] using the ionization constant of water (Kw) at 25°C: Kw = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ [OH⁻] = Kw / [H₃O⁺] [OH⁻] = (1.0 × 10⁻¹⁴) / (1.0 × 10⁻⁶ M) [OH⁻] = 1.0 × 10⁻⁸ M Since [H₃O⁺] > [OH⁻] and the pH is less than 7, we can classify the solution as acidic at this temperature. Therefore, the solution has a pH of 6.0 and an [OH⁻] of 1.0 × 10⁻⁸ M, and it is acidic.

answer:The factored form of the quadratic expression is 2 (-x - 3) (x + 18).