T varies inversely as the square root of F when T = 7, F = 2\(\frac{1}{4}\). Find T when F = \(\frac{27}{9}\)
T ∝ \(\frac{1}{\sqrt{F}}\)
T = \(\frac{K}{\sqrt{F}}\)
K = T x \(\sqrt{F}\)
K = 7 x \(\frac{3}{2}\) = 10.5
T = \(\frac{10.5}{\sqrt{F}}\)
T = T = \(\frac{10.5}{\sqrt{3}}\) ≈ 6.1
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