T varies directly as D and inversely as A. Given that T = 6, D = 3, and A = 2, find A when T = 3 and D = 6.
T = \(\frac{ \text{KD}}{\text{A}}\)
6 = \(\frac{3K}{2}\)
3K = 12
K = 4
T= \(\frac{4D}{A}\)
but T = 3, D = 6 and A =?
T = \(\frac{4D}{A}\)
A = \(\frac{4D}{T}\)
A = \(\frac{4 \times 6}{3}\) = 8
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