Question Number 70167 by Kunal12588 last updated on 01/Oct/19 $${find}\:{minima}\:{of} \\ $$$$\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{5}+\sqrt{\mathrm{1}−\left({x}_{\mathrm{1}} \right)^{\mathrm{2}} }+\sqrt{\mathrm{4}{x}_{\mathrm{2}} }\:\:\forall\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} \in{R} \\ $$ Commented by MJS…
Question Number 4630 by sanusihammed last updated on 15/Feb/16 $${Solve}\:{for}\:{x}\:{in}\:{the}\:{equation} \\ $$$$\mathrm{4}^{{x}\:} =\:\mathrm{8}{x} \\ $$$$ \\ $$$${I}\:{know}\:{the}\:{answer}\:{is}\:\mathrm{2}.\:{I}\:{need}\:{the}\:{solution}.\:{Thanks} \\ $$$$ \\ $$ Commented by FilupSmith last…
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Question Number 70162 by TawaTawa last updated on 01/Oct/19 Answered by mind is power last updated on 01/Oct/19 $$\Rightarrow\mathrm{1}=\left({a}+{b}\right)\left(\frac{{sin}^{\mathrm{4}} \left(\theta\right)}{{a}}+\frac{{cos}^{\mathrm{4}} \left(\theta\right)}{{b}}\right)…{S} \\ $$$$\Rightarrow\frac{{b}}{{a}}{Sin}^{\mathrm{4}} \left(\theta\right)+\frac{{a}}{{b}}{cos}^{\mathrm{4}} \left(\theta\right)+{cos}^{\mathrm{4}}…
Question Number 70163 by A8;15: last updated on 01/Oct/19 Answered by ajfour last updated on 01/Oct/19 Commented by ajfour last updated on 02/Oct/19 $$\:\:\frac{{AB}}{{BC}}={Q}=\:\frac{{x}+{y}}{\:\sqrt{{y}^{\mathrm{2}} +{z}^{\mathrm{2}}…
Question Number 4625 by Yozzis last updated on 14/Feb/16 $${Solve}\:{the}\:{following}\:{system}\:{of}\:{differential}\: \\ $$$${equations}\:{for}\:{functions}\:{x}\left({t}\right)\:{and}\:{y}\left({t}\right). \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }=\frac{{kx}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }=\frac{{ky}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 70161 by TawaTawa last updated on 01/Oct/19 Answered by mind is power last updated on 01/Oct/19 $${let}\:{a}=\sqrt{\sqrt{\mathrm{5}}+\mathrm{2}}\:\:{b}=\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${ab}=\mathrm{1}…
Question Number 4624 by sanusihammed last updated on 14/Feb/16 Answered by Yozzii last updated on 15/Feb/16 $$\left(\mathrm{12}\right)\:{The}\:{equation}\:{of}\:{motion}\:{of}\:{the}\:{particle} \\ $$$${is}\:{given}\:{as} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{d}}{{ds}}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right)=−{g}−{kv}^{\mathrm{2}} . \\ $$$${v}\:{is}\:{a}\:{function}\:{of}\:{s}\:{as}\:{s}\:{and}\:{v}\:{are}\:…