Make R the subject of the fomula S = \(\sqrt{\frac{2R + T}{2RT}}\)

  • A R = \(\frac{T}{(TS^2 + 1)}\)
  • B R = \(\frac{T}{2(TS^2 - 2)}\)
  • C R = \(\frac{T}{2(TS^2 + 1)}\)
  • D R = \(\frac{R}{2(TS^2 + 1)}\)

The correct answer is B. R = \(\frac{T}{2(TS^2 - 2)}\)

We can make \(R\) the subject of the formula \(S = \sqrt{\frac{2R + T}{2RT}}\) by following these steps:

1. Square both sides of the equation to get rid of the square root: \(S^2 = \frac{2R + T}{2RT}\).
2. Multiply both sides by \(2RT\) to get rid of the fraction: \(2RTS^2 = 2R + T\).
3. Subtract \(T\) from both sides: \(2RTS^2 - T = 2R\).
4. Factor out \(R\) on the right side: \(2RTS^2 - T = 2R(1)\).
5. Divide both sides by \(2TS^2 - 1\): \(R = \frac{T}{2TS^2 - 1}\).

So, the correct answer is *\(R = \frac{T}{2TS^2 - 1}\)

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