a) Step 1: Set up the differential equation and its solution.

Here are the solutions to the problems: a) Step 1: Set up the differential equation and its solution. Let M(t) be the amount of sugar still undissolved at time t. The rate of dissolution is proportional to the amount still undissolved, so we have the differential equation: (dM)/(dt) = -kM The solution to this equation is M(t) = M_0 e^-kt, where M_0 is the initial amount of sugar and k is the dissolution constant. Step 2: Use the given information to find the constant k. Given M_0 = 50 kg. At t=5 hours, M(5) = 20 kg. Substitute these values into the equation: 20 = 50 e^-k · 5 Divide by 50: (20)/(50) = e^-5k 0.4 = e^-5k Take the natural logarithm of both sides: (0.4) = -5k k = -((0.4))/(5) = ((1/0.4))/(5) = ((2.5))/(5) Step 3: Determine the amount of sugar remaining when 90% is dissolved. If 90% of the sugar is dissolved, then 10% remains. Remaining amount = 0.10 × M_0 = 0.10 × 50 kg = 5 kg. Step 4: Calculate the additional time required. We need to find how much longer it will take from t=5 hours until the amount of sugar remaining is 5 kg. Let this additional time be T_add. The amount of sugar at time t = 5 + T_add is M(5+T_add) = 5 kg. Using the formula M(t) = M_0 e^-kt: 5 = 50 e^-k(5+T_add) 5 = 50 e^-5k e^-kT_add We know from Step 2 that 50 e^-5k = 20. Substitute this into the equation: 5 = 20 e^-kT_add Divide by 20: (5)/(20) = e^-kT_add 0.25 = e^-kT_add Take the natural logarithm of both sides: (0.25) = -kT_add T_add = -((0.25))/(k) = ((1/0.25))/(k) = ((4))/(k) Now substitute the value of k = ((2.5))/(5): T_add = ((4))/((2.5))5 = (5 (4))/((2.5)) Using a calculator: (4) ≈ 1.38629 (2.5) ≈ 0.91629 T_add ≈ (5 × 1.38629)/(0.91629) ≈ (6.93145)/(0.91629) ≈ 7.564 hours It will take approximately 7.564 hours longer. b) Step 1: Identify restrictions for the function f(x) = (x)x. For the function to be defined, two conditions must be met: 1. The expression under the square root must be non-negative: x 0. 2. The denominator of the fraction cannot be zero: x 0. Step 2: Combine the restrictions to find the domain. Combining x 0 and x 0, we get x > 0. The cosine function is defined for all real numbers, so there are no further restrictions from the cosine itself. Thus, the domain of definition is all positive real numbers. The domain of definition is (0, ). c) Step 1: Define parity and evaluate f(-x). To study the parity of a function f(x), we evaluate f(-x). If f(-x) = f(x), the function is even. If f(-x) = -f(x), the function is odd. Otherwise, it is neither even nor odd. Given f(x) = x + 4x. Substitute -x for x: f(-x) = (-x) + (4(-x)) f(-x) = -x + (-4x) Step 2: Use trigonometric identities to simplify f(-x). Recall that the sine function is an odd function, meaning (-) = -(). Applying this identity: f(-x) = -x - (4x) Step 3: Compare f(-x) with f(x). Factor out -1 from f(-x): f(-x) = -(x + (4x)) Since f(x) = x + 4x, we can see that: f(-x) = -f(x) Therefore, the function f(x) = x + 4x is an odd function. The function is odd. Last free one today — make it count tomorrow, or type /upgrade for unlimited.