Question Number 100404 by PengagumRahasiamu last updated on 26/Jun/20 Commented by PengagumRahasiamu last updated on 26/Jun/20 Thank you Commented by Rasheed.Sindhi last updated on 26/Jun/20 $$\left({i}\right)\Rightarrow\mathrm{x}=\frac{\mathrm{28}−\mathrm{y}}{\mathrm{y}+\mathrm{2}},\:\left({ii}\right)\Rightarrow\mathrm{x}=\frac{\mathrm{37}−\mathrm{z}}{\mathrm{z}+\mathrm{3}}…
Question Number 100385 by bobhans last updated on 26/Jun/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\:\mathrm{inequality} \\ $$$$\frac{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)\sqrt{\mathrm{x}+\mathrm{2}}}{\mathrm{x}+\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} }}\:\leqslant\:\mathrm{0} \\ $$ Commented by bobhans last updated on 26/Jun/20 $$\left(\mathrm{1}\right)\:{x}\:\geqslant−\mathrm{2}\:\:\:\:\left(\mathrm{2}\right)\:{x}\:\neq\:−\mathrm{1}\:\:\:\left(\mathrm{3}\right)\:\sqrt{{x}+\mathrm{2}}\:\geqslant\:\mathrm{0} \\…
Question Number 100370 by Rio Michael last updated on 26/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\:{f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2cosh}\:\left(\mathrm{ln}\:{x}\right)\:+\:\mathrm{3}} \\ $$ Commented by bemath last updated on 26/Jun/20 $$\mathrm{y}_{\mathrm{max}} =\:\mathrm{0}.\mathrm{6} \\ $$ Commented…
Question Number 100341 by peter frank last updated on 26/Jun/20 $$\mathrm{An}\:\mathrm{open}\:\mathrm{box}\:\mathrm{with}\:\mathrm{a}\:\mathrm{square} \\ $$$$\mathrm{base}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{made}\:\mathrm{out} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{given}\:\mathrm{quantity}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{cardboard}\:\mathrm{of}\:\mathrm{area}\:\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{square}\:\mathrm{units}.\mathrm{show}\:\mathrm{the} \\ $$$$\mathrm{maximum}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{box}\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{6}\sqrt{\mathrm{3}}}\:\:\mathrm{cubic}\:\mathrm{units} \\…
Question Number 165872 by pete last updated on 09/Feb/22 $$\mathrm{State}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{thefunction} \\ $$$$\mathrm{f}\left(×\right)=\sqrt{\mathrm{9}−\mathrm{x}^{\mathrm{2}} }+\mathrm{3} \\ $$ Answered by TheSupreme last updated on 09/Feb/22 $$−\mathrm{3}\leqslant{x}\leqslant\mathrm{3} \\ $$…
Question Number 165867 by Bagus1003 last updated on 09/Feb/22 $$\mathrm{2}\:×\:{x}!\:=\:\frac{\mathrm{96}}{\mathrm{2}\:+\:\mathrm{1}\:−\:\mathrm{1}} \\ $$$${How}\:{much}\:{the}\:{x}\:{is}? \\ $$ Commented by MJS_new last updated on 09/Feb/22 $$\mathrm{2}+\mathrm{1}−\mathrm{1}\:…\:\mathrm{this}\:\mathrm{is}\:{much}\:\mathrm{too}\:\mathrm{hard}!\:\mathrm{please},\:\mathrm{post} \\ $$$$\mathrm{some}\:\mathrm{easier}\:\mathrm{questions} \\…
Question Number 100330 by bemath last updated on 26/Jun/20 Commented by bobhans last updated on 26/Jun/20 $$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right)\:\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{xy}+\mathrm{xz}+\mathrm{yz}\right)=\mathrm{16} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)^{\mathrm{2}} =\mathrm{16}\:\Rightarrow\begin{cases}{\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{4}}\\{\mathrm{x}+\mathrm{y}+\mathrm{z}=−\mathrm{4}}\end{cases} \\ $$$$\mathrm{case}:\mathrm{1}\:…
Question Number 165851 by Bagus1003 last updated on 09/Feb/22 $$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\frac{\boldsymbol{\tau}}{{x}\:×\:\mathrm{2}}\:+\:\mathrm{1}\:−\:\frac{\mathrm{2}}{\mathrm{2}} \\ $$$${How}\:{much}\:{the}\:{x}\:{is}? \\ $$ Answered by MJS_new last updated on 09/Feb/22 $$\mathrm{2}\left(\mathrm{cos}\left(\mathrm{45}\right)\right)^{\mathrm{4}} \:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\tau}{{x}×\mathrm{2}}+\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}} \\…
Question Number 100302 by bemath last updated on 26/Jun/20 $$\sqrt{\mathrm{3}^{−\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{1}}\:=\:\frac{\sqrt{\mathrm{a}+\mathrm{1}}}{\mathrm{3}^{−\frac{\mathrm{1}}{\mathrm{4}}} }\:.\:\mathrm{find}\:\mathrm{a}\:? \\ $$ Commented by bobhans last updated on 26/Jun/20 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:+\mathrm{1}\:=\:\frac{\mathrm{a}+\mathrm{1}}{\mathrm{3}^{−\frac{\mathrm{1}}{\mathrm{2}}} }\:\Rightarrow\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:=\:\sqrt{\mathrm{3}}\:\left(\mathrm{a}+\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{a}+\mathrm{1}\:=\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 34760 by Tinkutara last updated on 10/May/18 Answered by Rasheed.Sindhi last updated on 11/May/18 $$\mathrm{Answer}\:\mathrm{in}\:\mathrm{more}\:\mathrm{technical}\:\mathrm{style} \\ $$$$\bullet\mathrm{At}\:\mathrm{least}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{and}\:\mathrm{each} \\ $$$$\mathrm{side}\:\mathrm{is}\:\mathrm{not}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{4}. \\ $$$$\bullet\mathrm{Sum}\:\mathrm{of}\:\mathrm{any}\:\mathrm{of}\:\mathrm{two}\:\mathrm{sides}>\mathrm{Third}\:\mathrm{side}. \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{integer}\:\mathrm{sides}\:\mathrm{are}…