Rearrange to make s the subject for equation n=$$ \frac{2\pi}{s-3r} $$

To rearrange the equation $ n = \frac{2\pi}{s - 3r} $ to make $ s $ the subject, follow these steps:

### Step 1: Multiply both sides of the equation by $ (s - 3r) $
\[
n(s - 3r) = 2\pi
\]

### Step 2: Divide both sides by $ n $ to isolate $ s - 3r $
\[
s - 3r = \frac{2\pi}{n}
\]

### Step 3: Add $ 3r $ to both sides to isolate $ s $
\[
s = \frac{2\pi}{n} + 3r
\]

So, the equation rearranged to make $ s $ the subject is:
\[
s = \frac{2\pi}{n} + 3r
\]

Would you like further details on any of these steps?

Here are 5 related questions:
1. How would you rearrange this equation if $ s $ appeared more than once?
2. Can you apply this rearrangement method to a different formula, like $ a = \frac{b}{c - d} $?
3. How does this change if the equation involves an additional term in the denominator?
4. What happens if $ n $ depends on $ s $ in a nonlinear way (e.g., $ n = \frac{2\pi}{(s - 3r)^2} $)?
5. How do logarithms affect rearrangements involving exponential equations?

**Tip**: Always perform operations symmetrically on both sides of the equation to keep it balanced.

Rearrange to make s the subject for equation n = $ \frac{2\pi}{s - 3r} $

Learn how to rearrange the equation n = 2π / (s - 3r) to make s the subject with a step-by-step solution. This problem covers key algebraic concepts and is suitable for students in Grades 9-12.

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