The depression in freezing point $\Delta T_f = i \cdot K_f \cdot m$, where $i$ = van't Hoff factor (number of particles from ionization).

The compound with the fewest ions in solution gives the smallest $\Delta T_f$ and therefore the highest freezing point.

• $[\text{Co(H}_2\text{O)}_4\text{Cl}_2]\text{Cl}$: ionizes to give 2 particles → $i = 2$
• $[\text{Co(H}_2\text{O)}_3\text{Cl}_3]$: non-electrolyte → $i = 1$ (no ions) — smallest depression
• $[\text{Co(H}_2\text{O)}_6]\text{Cl}_3$: gives 4 particles → $i = 4$
• $[\text{Co(H}_2\text{O)}_5\text{Cl}]\text{Cl}_2$: gives 3 particles → $i = 3$

Highest freezing point → smallest $i$ → Option B ($i=1$).

Split into right and left limits.

Right limit ($x\to0^+$): $[x]=0$, so $\sin(x[x])=0$, $|x|=x$.

$\dfrac{\tan(\pi\sin^2x)+x^2}{x^2}$. Using $\tan(\pi\sin^2x)\approx\pi\sin^2x\approx\pi x^2$:

$=\pi+1$

Left limit ($x\to0^-$): $[x]=-1$, $\sin(x[x])=\sin(-x)=-\sin x$, $|x|=-x$.

$=\dfrac{\tan(\pi\sin^2x)+(-x-(-\sin x))^2}{x^2}=\dfrac{\pi x^2+(\sin x-x)^2}{x^2}\to\pi+0=\pi$

Left $\neq$ Right, so the limit does not exist.

With 80 employees, the first quartile (Q1) covers the bottom 25%, which is the bottom 20 salaries (positions 1–20). Mark is the 2nd-highest in Q1, meaning he is at position 19 out of 80.

After hiring 8 new employees at the lowest salaries, there are 88 employees. The new employees occupy positions 1–8. Everyone else shifts up by 8 positions.

Mark's new position: $19 + 8 = 27$ out of 88.

New Q1 covers positions 1–22 (bottom 25% of 88). Mark at position 27 is above Q1, now in Q2.

Mark is the 5th from the bottom of Q2 (positions 23, 24, 25, 26, 27 → 5th lowest in Q2).

Answer: The fifth-lowest salary in the second quartile.

To travel straight north, the swimmer must aim upstream (westward) to cancel river drift.

If $\phi$ is the angle west of north:

$v_{\text{swim}}\sin\phi = v_{\text{river}} \Rightarrow 20\sin\phi = 10 \Rightarrow \sin\phi = \dfrac{1}{2} \Rightarrow \phi = \mathbf{30°}$ west of north