Mastery checklist

Work. Distribute to get 10x − 5 = 3x + 16. Then 7x = 21 and x = 3. An independent check substitutes into the original equation: both sides equal 25.

Answer: x = 3; original-equation substitution is a suitable check.

Check. 5(6 − 1) = 25 and 9 + 16 = 25.

Key takeaway. Passing this probe requires the equation-solving steps and a valid check. Recognizing the final number from memory is not enough.

Work. The slope is (0 − 9)/(5 − 2) = −3. Thus f(x) = −3x + 15, so f(0) = 15. Solving −3x + 15 = 3 gives x = 4.

Answer: Slope −3; f(0) = 15; input x = 4.

Check. The rule reproduces both original outputs.

Key takeaway. A single skill can be checked through a rate, an intercept, and a reverse lookup. These are separate pieces of evidence.

Work. No. At x = −5, the left side is 5 and the right side is −5. The principal square root is nonnegative, so the general identity is $\sqrt{{\text{x}}^{\text{2}}}$ = |x|. It equals x only when x ≥ 0.

Answer: $\sqrt{{\text{x}}^{\text{2}}}$ = |x|

Check. For negative x, |x| = −x > 0; for nonnegative x, |x| = x.

Key takeaway. One counterexample disproves a universal statement. Mastery includes the conditions attached to an identity.

Work. Factor the numerator as (x − 3)(x + 2). Thus f(x) = x + 2 for x ≠ 3. The original denominator is zero at 3, so f(3) does not exist.

Answer: f(x) = x + 2 for x ≠ 3; f(3) is undefined.

Check. The point (3, 5) would lie on the simplified line but is missing from the original function.

Key takeaway. Require a domain statement whenever practice involves cancellation.

Work. Write f(x) = a(x − 2)(x − 8). Then f(0) = 16a = 32, so a = 2. Thus f(x) = 2(x − 2)(x − 8). The two zeros alone would not determine the vertical scale.

Answer: f(x) = 2(x − 2)(x − 8)

Check. The zeros are 2 and 8, and the intercept is 2(−2)(−8) = 32.

Key takeaway. Mastery includes knowing which facts fix a model and which leave a parameter free.

Work. The constant ratio is 3, and the output at 0 is 6. The model is f(x) = 6 · 3^x. At 4, the output is 6(81) = 486. A linear model would require constant output differences, but the differences here are 12 and 36.

Answer: f(x) = 6 · 3^x; f(4) = 486.

Check. The model gives all three tabulated outputs.

Key takeaway. The stated model family matters. A finite table alone does not rule out every other imaginable function.

Work. Write A = 1.4B. The difference is 0.4B, but the comparison less than A uses A as its base. The fraction is 0.4B/(1.4B) = 2/7, or 200/7% ≈ 28.57%.

Answer: About 28.57%

Check. For B = 100, A = 140 and the decrease back to 100 is 40/140.

Key takeaway. Reverse percent comparisons change the base. A memorized 40% is not transferable reasoning.

Work. The interval is 48% to 56%. It includes values at or below 50%, so it does not establish that the true proportion exceeds one-half. The estimate itself is above 50%, but the interval expresses uncertainty around it.

Answer: No; the interval is [48%, 56%].

Check. 49% is inside the interval and is not a majority.

Key takeaway. For a threshold claim, compare the whole interval with the threshold rather than only the point estimate.

Work. The overlap is 20 in both calculations. Conditioning on art gives 20/32 = 5/8; conditioning on music gives 20/50 = 2/5.

Answer: $\frac{\text{5}}{\text{8}}$ and $\frac{\text{2}}{\text{5}}$, respectively.

Check. The numerator group belongs to both conditioning groups, but the denominators differ.

Key takeaway. Ask for a reversed version of a skill to test whether the student understands its structure rather than its wording.

Work. No. A single matching angle is insufficient. The remaining angle sets differ, so angle-angle similarity does not apply. No common side scale factor is justified by these facts.

Answer: No; the triangles are not similar.

Check. If they were similar, all corresponding angles would match, which they do not.

Key takeaway. A correct proportion starts with a justified correspondence, not a visually convenient pairing.

Work. For complementary angles, the side opposite A is adjacent to B, while the hypotenuse is the same. Thus cos B = sin A = 5/13.

Answer: $\frac{\text{5}}{\text{13}}$

Check. For a 5–12–13 right triangle, the same side of length 5 appears in both requested ratios.

Key takeaway. Derive a remembered identity from side roles when checking whether it has been understood.

Work. The condition is l + w = 20, which leaves many positive pairs. A 5 × 15 rectangle has area 75; an 8 × 12 rectangle has area 96. Both have perimeter 40, so the area is not uniquely determined.

Answer: No; areas 75 and 96 are both possible.

Check. 2(5 + 15) = 2(8 + 12) = 40.

Key takeaway. Producing two valid outcomes is a decisive way to show that the given information is insufficient.

Work. The fresh set is more informative about transfer under those conditions. Review its five misses by skill and earliest error. The repeated set still shows that the student can reproduce familiar work, but it is not interchangeable with unseen performance.

Answer: Use the fresh-set errors to choose the next repair tasks.

Check. The two percentages, 100% and 50%, describe different conditions; averaging them would obscure that distinction.

Key takeaway. Track whether an item is fresh, repeated, or solved with help. These labels improve the meaning of every accuracy figure.

Work. Begin with the radical equations, especially any repeated restriction or extraneous-root error. The observed rates are 87.5% and about 33.3%, but small samples are noisy; use the scratch work, not percentages alone, to diagnose the cause. Retain some later maintenance practice for linear skills.

Answer: Prioritize the radical-equation error pattern, then retest on fresh items.

Check. A consistent error across several questions is more actionable than a single overall score.

Key takeaway. These are instructional observations, not stable ability estimates or an official mastery threshold.

Work. Use a fresh system on another day, change the representation or requested expression, and work without the old solution. Require a check in both original equations. If the same error returns, revisit the underlying step rather than rereading the answer.

Answer: A delayed, fresh, unassisted transfer check with original-equation verification.

Check. The check should not merely reuse the same numbers or the same visible method cue.

Key takeaway. The timing of a retest is an adjustable study choice. Its purpose here is to distinguish remembered steps from independently selected reasoning.