Hint
Use the stated units and precision. Form the stated direct/inverse power model, determine its constant from the given pair and use the positive domain.
Worked solution
(a)
The given pair yields \(\displaystyle k=\)\(7\). Substituting this constant into the direct proportionality rule gives \(y=7 x^{3}\).
Answer: \(y=7 x^{3}\), x>0.
(b)
Insert the new positive input into \(7 x^{3}\); the corresponding power is exact.
Answer: \(189\)
(c)
After dividing by k and reversing the power, the positive solution is \(6\). Substitution recovers the stated y.
Answer: \(6\)
Maths4U indicative marking guidance
Maths4U indicative marking guidance, not an official mark scheme. Credit complete valid alternative methods and consequential reasoning. An unexplained result does not earn a method mark.
(a) · 3 marks
- 1 mark: Translate the stated proportional relationship into an equation with a constant.
- 1 mark: Use the given nonzero input-output pair to determine that constant.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
(b) · 2 marks
- 1 mark: Substitute into the determined model and evaluate the requested power accurately.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Check the answer against the original conditions.
(c) · 3 marks
- 1 mark: Rearrange the determined model and undo its power.
- 1 mark: Select the permitted positive input and check it in the original relationship.
- 1 mark: Give the complete requested result, using the required units, precision and boundary inclusion.
Alternative methods: Any complete mathematically valid alternative method with the same restrictions and equivalent exact result is acceptable.
Common error: Keep intermediate values exact until the final step.
Total: 8 marks.