A household's monthly electricity bill, \(\pounds C\), is made up of a fixed charge and a charge per unit of electricity used, \(u\), shown in the diagram, ...
Question 1 Report
A household's monthly electricity bill, \(\pounds C\), is made up of a fixed charge and a charge per unit of electricity used, \(u\), shown in the diagram, so \(C=15+0.2u\).
Rearrange the formula to make \(u\) the subject. (3)
Answer Details
Two different laws of indices are being compared here: raising a power to a power multiplies the indices, while multiplying two powers of the same base adds the indices; applying each correctly, then comparing the resulting coefficients, decides which model has the greater factor.
(a) Raising a power to a power: \((2v^{2})^{2}=2^{2}\times v^{2\times2}=4v^{4}\). [1 mark]
(b) Multiplying powers of the same base, \(v^{3}\times v=v^{3+1}=v^{4}\), so \(8v^{3}\times v=8v^{4}\). [1 mark]
(c) Both expressions share the factor \(v^{4}\), which is positive for every \(v\gt0\), so comparing them reduces to comparing their coefficients: since \(8\gt4\), \(8v^{4}\gt4v^{4}\) for every \(v\gt0\), meaning Model B always has the greater factor. [1 mark]
Once both expressions are written with the same power of \(v\), the comparison in part (c) becomes a simple comparison of the numerical coefficients, which only works because \(v^{4}\) cannot be negative or zero for \(v\gt0\).
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