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Exercise 6 g Each question starts with a formula. A letter appears in heavy type after it. Make that letter the subject of the formula. If more than one lette is given, make each letter the subject in turn. (1) \( c=2 \pi r \) (2) \( P=a W+b \) (3) \( P=\frac{N+2}{\mathrm{D}} \) \( \begin{array}{ll}\text { (4) } A=P+\frac{P R T}{100} & \boldsymbol{W} \\ \text { (5) } v^{2}=u^{2}+2 a s & \mathbf{N}, \boldsymbol{D} \\ \text { (6) } s=\frac{n}{2}(a+l) & \mathbf{S}, \boldsymbol{u}\end{array} \)

Ask by Joseph Morrison. in Nigeria
Mar 15,2025

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(1) \( r = \frac{c}{2\pi} \) (2) \( W = \frac{P - b}{a} \) (3) \( D = \frac{N + 2}{P} \) (4) Cannot solve for \( W \) as it does not appear in the equation. (5) Cannot solve for \( N \) or \( D \) as they do not appear in the equation. (6) \( S = \frac{n}{2}(a + l) \) Cannot solve for \( u \) as it does not appear in the equation.

Solución

(1) Solve for \(r\) in \[ c=2\pi r. \] Divide both sides by \(2\pi\): \[ r=\frac{c}{2\pi}. \] (2) Solve for \(W\) in \[ P=aW+b. \] Subtract \(b\) from both sides: \[ P-b=aW. \] Then divide by \(a\): \[ W=\frac{P-b}{a}. \] (3) Solve for \(D\) in \[ P=\frac{N+2}{D}. \] Multiply both sides by \(D\): \[ PD=N+2. \] Then divide by \(P\): \[ D=\frac{N+2}{P}. \] (4) The equation is \[ A=P+\frac{PRT}{100}, \] but the heavy‐type letter given is \(W\), which does not appear in the equation. No rearrangement can make \(W\) the subject. (5) The equation is \[ v^2=u^2+2as, \] while the heavy‐type letters given are \(N\) and \(D\). Since neither \(N\) nor \(D\) occurs in the equation, neither can be made the subject. (6) The equation is \[ s=\frac{n}{2}(a+l). \] The heavy type shows \(S\) and \(u\). Interpreting \(S\) as the same as \(s\), we have \[ S=\frac{n}{2}(a+l). \] The variable \(u\) does not appear in the equation, so it cannot be made the subject.

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Bonus Knowledge

To make each letter the subject of the formulas, we can rearrange the equations accordingly. (1) For \( c=2 \pi r \) and making \( r \) the subject, we have: \[ r = \frac{c}{2 \pi} \] (2) For \( P=a W+b \) and making \( W \) the subject, we can rearrange it to: \[ W = \frac{P-b}{a} \] (3) For \( P=\frac{N+2}{D} \) and making \( N \) the subject, we rearrange it to: \[ N = PD - 2 \] And for \( D \): \[ D = \frac{N+2}{P} \] (4) For \( A=P+\frac{P R T}{100} \) and making \( W \) the subject is a bit tricky as \( W \) isn't in that equation, but if we assume we need a different letter, we can rearrange for \( P \): \[ P = \frac{100A}{100 + RT} \] (5) For \( v^{2}=u^{2}+2as \) and making \( N \) the subject doesn't apply as \( N \) isn't in the equation, but if \( a \) is of interest: \[ a = \frac{v^2 - u^2}{2s} \] And for \( s \): \[ s = \frac{v^2 - u^2}{2a} \] (6) For \( s=\frac{n}{2}(a+l) \) and making \( S \) the subject is a bit complicated, as we would typically focus on \( n \) or \( l \) instead since \( S \) doesn't appear on that list. However, if we're focusing on \( n \): \[ n = \frac{2s}{a + l} \] And for \( u \): In this formula, \( u \) isn't part of it but if we were to involve the surroundings (for instance, if \( u \) relates to a velocity in a similar context) we would require an alternate formula or define a new equation. Each of these equations has its specific purpose and can be rearranged depending on which variable you wish to make the subject. Happy rearranging!

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