Question Number 202116 by Calculusboy last updated on 21/Dec/23 $$\sqrt{\mathrm{3}^{\boldsymbol{{x}}} }\:+\mathrm{1}=\mathrm{2}^{\boldsymbol{{x}}} \:\:\:\boldsymbol{{find}}\:\boldsymbol{{x}} \\ $$ Answered by Frix last updated on 21/Dec/23 $$\mathrm{Obviously}\:{x}=\mathrm{2} \\ $$$$\sqrt{\mathrm{3}^{\mathrm{2}} }+\mathrm{1}=\sqrt{\mathrm{9}}+\mathrm{1}=\mathrm{3}+\mathrm{1}=\mathrm{4}=\mathrm{2}^{\mathrm{2}}…
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Question Number 202146 by hardmath last updated on 21/Dec/23 $$ \\ $$There are three children in a family. Two of the children have blood group…
Question Number 202114 by MATHEMATICSAM last updated on 21/Dec/23 $$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{and}\:\alpha\:+\:{k}\:\mathrm{and}\:\beta\:+\:{k}\:\:\mathrm{are}\: \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{lx}^{\mathrm{2}} \:+\:{mx}\:+\:{n}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{k}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{b}}{{a}}\:−\:\frac{{m}}{{l}}\right). \\ $$ Answered by aleks041103 last updated…
Question Number 202109 by hardmath last updated on 20/Dec/23
Question Number 202104 by necx122 last updated on 20/Dec/23 Commented by necx122 last updated on 20/Dec/23 $${please}\:{help}\:{with}\:{this} \\ $$ Commented by cortano12 last updated on…
Question Number 202100 by sonukgindia last updated on 20/Dec/23 Answered by mr W last updated on 21/Dec/23 $${f}'={g} \\ $$$${f}''={g}'={f} \\ $$$${f}''−{f}=\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{r}=\pm\mathrm{1}…
Question Number 202102 by MATHEMATICSAM last updated on 20/Dec/23 $$\mathrm{If}\:{x}\:=\:\mathrm{3}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{then}\:\frac{{x}^{\mathrm{6}} \:+\:{x}^{\mathrm{4}} \:+\:{x}^{\mathrm{2}} \:+\:\mathrm{1}}{{x}^{\mathrm{3}} }\:=\:? \\ $$ Answered by AST last updated on 20/Dec/23 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{6} \\…
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Question Number 202093 by sonukgindia last updated on 20/Dec/23 Commented by a.lgnaoui last updated on 22/Dec/23 Commented by a.lgnaoui last updated on 22/Dec/23 $$\Delta\mathrm{AMN}\:\:\measuredangle\:\mathrm{MNA}=\frac{\mathrm{3}\pi}{\mathrm{4}}−\alpha \\…