Transpose each of the following formulae to make the given variable the subject: (a) \(x=\frac{c}{y}\), for \(y\) (b) \(x=\frac{c}{y}\), for \(c\) (c) \(k=\frac{2 n+5}{n+3}\), for \(n\) (d) \(T=2 \pi \sqrt{\frac{R-L}{g}}\), for \(R\)

Short Answer

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Question: Rewrite each of the following equations to make the given variable the subject: a) Make y the subject in the equation $$xy = c$$. b) Make c the subject in the equation $$xy = c$$. c) Make n the subject in the equation $$\frac{k}{n+3} = 2n + 5$$. d) Make R the subject in the equation $$2 \pi \sqrt{\frac{R-L}{g}} = T.$$

Step by step solution

01

(a) Make y the subject

To make y the subject, we need to isolate y. To do this, we start by multiplying both sides of the equation by y. This will give us: $$xy = c$$ Now, we divide both sides by x to isolate y: $$y = \frac{c}{x}$$
02

(b) Make c the subject

To make c the subject, we need to isolate c. Since c is already in the numerator, we simply multiply both sides of the equation by y to obtain: $$c = xy$$
03

(c) Make n the subject

To make n the subject, we first need to clear the fractions. To do this, we multiply both sides of the equation by \((n+3)\), which gives: $$k(n+3) = 2n + 5$$ Next, we distribute k on the left side of the equation: $$kn + 3k = 2n + 5$$ Now, we need to isolate n. To do this, we first subtract 2n from both sides to have all n terms at one side: $$kn - 2n + 3k = 5$$ Now, we factor out n: $$n(k - 2) = 5 - 3k$$ Finally, we divide both sides by \((k-2)\) to isolate n: $$n = \frac{5 - 3k}{k-2}$$
04

(d) Make R the subject

To make R the subject, we start by isolating the square root on one side. We do this by dividing both sides by \(2\pi\): $$\sqrt{\frac{R-L}{g}} = \frac{T}{2 \pi}$$ Now, we square both sides of the equation to eliminate the square root: $$\frac{R-L}{g} = \left(\frac{T}{2 \pi}\right)^2$$ Next, we multiply both sides by g to clear the denominator: $$R-L = g\left(\frac{T}{2 \pi}\right)^2$$ Finally, we add L to both sides to isolate R: $$R = g\left(\frac{T}{2 \pi}\right)^2 + L$$

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