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: $$\frac{dM}{dt} = -kM$$ 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: $$20 = 50 e^{-k \cdot 5}$$ Divide by 50: $$\frac{20}{50} = e^{-5k}$$ $$0.4 = e^{-5k}$$ Take the natural logarithm of both sides: $$\ln(0.4) = -5k$$ $$k = -\frac{\ln(0.4)}{5} = \frac{\ln(1/0.4)}{5} = \frac{\ln(2.5)}{5}$$ Step 3: Determine the amount of sugar remaining when 90% is dissolved. If 90% of the sugar is dissolved, then 10% remains. Remaining amount $= 0.10 \times M_0 = 0.10 \times 50 \text{ kg} = 5 \text{ kg}$. 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: $$\frac{5}{20} = e^{-kT_{add}}$$ $$0.25 = e^{-kT_{add}}$$ Take the natural logarithm of both sides: $$\ln(0.25) = -kT_{add}$$ $$T_{add} = -\frac{\ln(0.25)}{k} = \frac{\ln(1/0.25)}{k} = \frac{\ln(4)}{k}$$ Now substitute the value of $k = \frac{\ln(2.5)}{5}$: $$T_{add} = \frac{\ln(4)}{\frac{\ln(2.5)}{5}} = \frac{5 \ln(4)}{\ln(2.5)}$$ Using a calculator: $$\ln(4) \approx 1.38629$$ $$\ln(2.5) \approx 0.91629$$ $$T_{add} \approx \frac{5 \times 1.38629}{0.91629} \approx \frac{6.93145}{0.91629} \approx 7.564 \text{ hours}$$ It will take approximately $\boxed{\text{7.564 hours}}$ longer. b) Step 1: Identify restrictions for the function $f(x) = \cos\frac{\sqrt{x}}{x}$. For the function to be defined, two conditions must be met: 1. The expression under the square root must be non-negative: $x \ge 0$. 2. The denominator of the fraction cannot be zero: $x \ne 0$. Step 2: Combine the restrictions to find the domain. Combining $x \ge 0$ and $x \ne 0$, 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 $\boxed{(0, \infty)}$. 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 $f(x) = x + \sin 4x$. Substitute $-x$ for $x$: $$f(-x) = (-x) + \sin(4(-x))$$ $$f(-x) = -x + \sin(-4x)$$ Step 2: Use trigonometric identities to simplify $f(-x)$. Recall that the sine function is an odd function, meaning $\sin(-\theta) = -\sin(\theta)$. Applying this identity: $$f(-x) = -x - \sin(4x)$$ Step 3: Compare $f(-x)$ with $f(x)$. Factor out $-1$ from $f(-x)$: $$f(-x) = -(x + \sin(4x))$$ Since $f(x) = x + \sin 4x$, we can see that: $$f(-x) = -f(x)$$ Therefore, the function $f(x) = x + \sin 4x$ is an odd function. The function is $\boxed{\text{odd}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.