In an isosceles triangle, two angles are equal. The sum of all angles in a triangle is $180°$.

Case 1: If $\angle RST = 40°$ is one of the two equal angles, then the two equal angles sum to $2 \times 40° = 80°$, and the third angle is $180° - 80° = 100°$.

Case 2: If $\angle RST = 40°$ is the unique angle (not one of the equal pair), then the two equal angles each measure $\frac{180° - 40°}{2} = 70°$, and their sum is $2 \times 70° = 140°$.

In Case 1, Column A = $80° < 120°$, so Column B is greater.

In Case 2, Column A = $140° > 120°$, so Column A is greater.

Since both cases are possible, the relationship cannot be determined.

The subject-centered curriculum organizes content around academic disciplines and often prioritizes content coverage over meaningful understanding. This can encourage memorization (rote learning) rather than deep comprehension. Learner-centered and activity-centered curricula are less associated with this problem.

Correct Answer: A — Subject-centered curriculum.

Technique is the best answer. A technique refers to a specific skillful way of doing something, especially in a professional or medical context. Technology refers to tools, machinery, or systems — too broad for a surgical approach. Method is more general and less precise than technique in the context of a specific surgical procedure.

Correct Answer: B — technique.

Using $\vec{s} = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2$:

• $x = 2 + 5(2) + \frac{1}{2}(4)(2^2) = 2 + 10 + 8 = 20$
• $y = 4 + 4(2) + \frac{1}{2}(4)(2^2) = 4 + 8 + 8 = 20$
• Distance from origin $D = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m}$.

Let the price of the book $= p$. Total paid $= 1.04p$.

She paid $10 and received change less than $3.00, meaning total paid was more than $7.00:

$1.04p > 7.00 \Rightarrow p > 6.73$

She also received some change, so total paid $< \$10.00$:

$1.04p < 10.00 \Rightarrow p < 9.615$

So $6.73 < p < 9.615$. Now check each statement:

• Statement A: $p < \$9.50$ — Not necessarily, since $p$ can be up to $9.615$. ✗
• Statement B: $p > \$6.90$ — Not necessarily, $p$ can be $6.80$ (giving change $\approx \$2.93 < \$3.00$). ✗
• Statement C: Sales tax $= 0.04p < 0.04 \times 9.615 \approx \$0.385 < \$0.45$. This is always true within the valid range. ✓
• Statement D: Tax $> \$0.50$ would require $p > 12.50$, which is impossible here. ✗

Only Statement C must be true.

Using Gay-Lussac's Law at constant volume:

$\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$

$P_1 = 300\ \text{kPa} = 3 \times 10^5\ \text{Pa},\ T_1 = 300\ \text{K}$

$P_2 = 1.2 \times 10^6\ \text{Pa}$

$T_2 = T_1 \times \dfrac{P_2}{P_1} = 300 \times \dfrac{1.2\times10^6}{3\times10^5} = 300 \times 4 = 1200\ \text{K}$

$T_2\ (°C) = 1200 - 273 = \mathbf{927}°\text{C}$