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Question Number 209991 by efronzo1 last updated on 28/Jul/24 $$\:\:\:\:\:\frac{\mathrm{10}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{6}\right)} .\:\mathrm{15}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)} }{\mathrm{6}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)} .\:\mathrm{5}^{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{4}}{\mathrm{3}}\right)} }\:=?\: \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 210017 by klipto last updated on 28/Jul/24 $$ \\ $$$$\boldsymbol{\mathrm{MATH}}−\boldsymbol{\mathrm{WHIZZKID}} \\ $$$$\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{kamke}}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{genral}} \\ $$$$\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}} \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}''+\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}'−\mathrm{2}\boldsymbol{\mathrm{y}}=\mathrm{0} \\ $$$$−−−−−−−−− \\ $$$$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{using}}\:\boldsymbol{\mathrm{forbenius}}\:\boldsymbol{\mathrm{mtd}} \\…
Question Number 209986 by a.lgnaoui last updated on 28/Jul/24 $$\mathrm{Solve}\: \\ $$$$\:\boldsymbol{\mathrm{ax}}^{\mathrm{3}} −\boldsymbol{\mathrm{bx}}\sqrt{\boldsymbol{\mathrm{x}}}\:+\boldsymbol{\mathrm{c}}=\mathrm{0}\:\:\:\:\: \\ $$$$\:\left(\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}}\right)\in\mathbb{R}^{\mathrm{3}} \:\:\:\:\mathrm{and}\:\boldsymbol{\mathrm{x}}\in\mathbb{R} \\ $$$$\left(\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{x}}\:\boldsymbol{{for}}\:\boldsymbol{{a}}=\mathrm{1},\:\:\boldsymbol{{b}}=\mathrm{9},\boldsymbol{{c}}=\mathrm{8}\right) \\ $$ Answered by mr W last…
Question Number 209980 by a.lgnaoui last updated on 27/Jul/24 $$\mathrm{determiner}\:\mathrm{h}\:? \\ $$$$\boldsymbol{\mathrm{CD}}=\mathrm{20}\:\:\:\:\boldsymbol{\mathrm{AB}}=\mathrm{30} \\ $$$$\boldsymbol{\mathrm{h}}\mathrm{1}=\mathrm{25} \\ $$$$ \\ $$ Commented by a.lgnaoui last updated on 27/Jul/24…
Question Number 209965 by mnjuly1970 last updated on 27/Jul/24 Answered by mr W last updated on 27/Jul/24 Commented by mr W last updated on 27/Jul/24…
Question Number 209976 by Abdullahrussell last updated on 27/Jul/24 Commented by mr W last updated on 27/Jul/24 $$\Rightarrow{Q}\mathrm{208555} \\ $$ Terms of Service Privacy Policy…
Question Number 209960 by a.lgnaoui last updated on 27/Jul/24 $$\mathrm{Determiner}\:\:\mathrm{Aire}\:\left(\boldsymbol{\mathrm{ABH}}\right) \\ $$$$\:\:\:\mathrm{AH}\bot\mathrm{CE} \\ $$ Commented by a.lgnaoui last updated on 27/Jul/24 Answered by mr W…
Question Number 209972 by Abdullahrussell last updated on 27/Jul/24 Answered by efronzo1 last updated on 27/Jul/24 $$\:\:\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{9x}^{\mathrm{2}} }{\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} }\:=\:\mathrm{16}\: \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} \left(\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{9}\right)=\:\mathrm{16}\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} \\…
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