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Question Number 171437 by solomonwells last updated on 15/Jun/22

$$\mathrm{make}\mathbf{r}\mathrm{the}\mathrm{subject}\mathrm{of}\mathrm{the}\mathrm{formula}$$$$$$$$\mathbf{y}={(\frac{\mathbf{pr}}{\mathbf{m}}-\frac{{\mathbf{p}}^3}{1})}^{-3/2}$$$$$$$$\mathrm{find}\mathbf{r}\mathrm{if}\mathrm{y}=-8,\mathrm{m}=-1,\mathrm{p}=3$$

Answered by alephzero last updated on 15/Jun/22

$$y={(\frac{pr}{m}-p^3)}^{-3/2}$$$$\Rightarrow y^{-2/3}=\frac{pr}{m}-\frac{{mp}^3}{m}$$$$\Rightarrow y^{-2/3}=\frac{pr-{mp}^3}{m}$$$$\Rightarrow {my}^{-2/3}=pr-{mp}^3$$$$\Rightarrow pr={my}^{-2/3}+{mp}^3$$$$\Rightarrow r=\frac{{my}^{-2/3}+{mp}^3}{p}$$

Commented by alephzero last updated on 17/Jun/22

$$y=-8,m=-1,p=3$$$$\Rightarrow r=\frac{-{(-8)}^{-2/3}-3^3}{3}=$$$$=-\frac{\sqrt[3]{{(-8)}^2}-27}{3}=-\frac{\sqrt[3]{64}-27}{3}=$$$$=-\frac{4-27}{3}=\frac{27-4}{3}=\frac{23}{3}=7\frac{2}{3}$$

Answered by elviselloin719A last updated on 18/Jun/22

$$Ans:7.667$$

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