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Question Number 113816 by Study last updated on 15/Sep/20

0.095=h∙((h/(1+2h)))^(2/3)          h=?    & show the practice

$$\mathrm{0}.\mathrm{095}={h}\centerdot\left(\frac{{h}}{\mathrm{1}+\mathrm{2}{h}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} \:\:\:\:\:\:\:\:\:{h}=?\:\: \\ $$ $$\&\:{show}\:{the}\:{practice} \\ $$

Commented byStudy last updated on 15/Sep/20

help me i need to much

$${help}\:{me}\:{i}\:{need}\:{to}\:{much} \\ $$

Answered by MJS_new last updated on 15/Sep/20

((19)/(200))=h((h/(1+2h)))^(2/3)   (((19)/(200)))^3 =h^3 (h^2 /((1+2h)^2 ))  h^5 −4(((19)/(200)))^3 h^2 −4(((19)/(200)))^3 h−(((19)/(200)))^3 =0  we can only approximate  ⇒ h≈.292911

$$\frac{\mathrm{19}}{\mathrm{200}}={h}\left(\frac{{h}}{\mathrm{1}+\mathrm{2}{h}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$ $$\left(\frac{\mathrm{19}}{\mathrm{200}}\right)^{\mathrm{3}} ={h}^{\mathrm{3}} \frac{{h}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{h}\right)^{\mathrm{2}} } \\ $$ $${h}^{\mathrm{5}} −\mathrm{4}\left(\frac{\mathrm{19}}{\mathrm{200}}\right)^{\mathrm{3}} {h}^{\mathrm{2}} −\mathrm{4}\left(\frac{\mathrm{19}}{\mathrm{200}}\right)^{\mathrm{3}} {h}−\left(\frac{\mathrm{19}}{\mathrm{200}}\right)^{\mathrm{3}} =\mathrm{0} \\ $$ $$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$ $$\Rightarrow\:{h}\approx.\mathrm{292911} \\ $$

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