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Question-195799

Question Number 195799 by sonukgindia last updated on 10/Aug/23 Answered by som(math1967) last updated on 11/Aug/23 $$\:\frac{\mathrm{1}}{{a}+\frac{\mathrm{1}}{{a}}}\:+\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}}}=\mathrm{2}{a} \\ $$$$\Rightarrow\:\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}{a} \\ $$$$\Rightarrow{a}\left[\frac{\mathrm{1}}{{a}^{\mathrm{2}} +\mathrm{1}}\:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\right]=\mathrm{2}{a}…

Question-195742

Question Number 195742 by sonukgindia last updated on 09/Aug/23 Answered by MM42 last updated on 09/Aug/23 $$\angle\alpha_{{i}} =\mathrm{120}\Rightarrow\angle{BAC}=\angle{BDF}=\mathrm{90} \\ $$$${AB}={AC}\Rightarrow\angle{C}=\angle{B}_{\mathrm{1},\mathrm{2}} =\mathrm{45}\Rightarrow\angle{B}_{\mathrm{2}} =\angle{B}_{\mathrm{3}} =\mathrm{15} \\ $$$$\frac{{x}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{90}}\:\:\:\&\:\:\:\frac{{y}}{{sin}\mathrm{15}}=\frac{{BE}}{{sin}\mathrm{60}}…

500-499-0-002-plz-soon-

Question Number 195732 by bbbbbbbb last updated on 09/Aug/23 $$\frac{\mathrm{500}!+\mathrm{499}!}{\mathrm{0}.\mathrm{002}}=? \\ $$$$\boldsymbol{\mathrm{plz}}\:\boldsymbol{\mathrm{soon}}\: \\ $$ Answered by HeferH last updated on 09/Aug/23 $$\frac{\mathrm{500}\centerdot\mathrm{499}!\:+\:\mathrm{499}!}{\frac{\mathrm{2}}{\mathrm{1000}}}\:=\:\frac{\mathrm{499}!\centerdot\mathrm{501}\centerdot\mathrm{1000}}{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{499}!\:\centerdot\mathrm{501}\centerdot\mathrm{500}}{\mathrm{1}}\:=\:\mathrm{501}!\: \\…

Question-195746

Question Number 195746 by Humble last updated on 09/Aug/23 Answered by mr W last updated on 09/Aug/23 $$\rho={x}\rho_{\mathrm{2}} +\left(\mathrm{1}−{x}\right)\rho_{\mathrm{1}} \\ $$$$\Rightarrow{x}=\frac{\rho−\rho_{\mathrm{1}} }{\rho_{\mathrm{2}} −\rho_{\mathrm{1}} }=\frac{\mathrm{0}.\mathrm{917}−\mathrm{0}.\mathrm{876}}{\mathrm{1}.\mathrm{049}−\mathrm{0}.\mathrm{876}}\approx\mathrm{23}.\mathrm{7\%} \\…

Prove-that-log-a-b-a-b-1-

Question Number 195708 by mokys last updated on 08/Aug/23 $$\boldsymbol{{Prove}}\:\boldsymbol{{that}}\::\:\boldsymbol{{log}}_{\left(\sqrt{\boldsymbol{{a}}}\:−\:\boldsymbol{{b}}\right)} \left(\sqrt{\boldsymbol{{a}}}\:+\boldsymbol{{b}}\right)\:=\:−\mathrm{1} \\ $$ Commented by mr W last updated on 08/Aug/23 $${you}\:{can}\:{not}\:{prove}\:{things}\:{like}\:{x}+{y}=\mathrm{2}. \\ $$ Commented…

Question-195704

Question Number 195704 by sonukgindia last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $${A}=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\frac{\left({a}+{b}\right){b}}{\mathrm{2}} \\ $$$${A}_{\mathrm{2}} =\left(\frac{{b}}{\:\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}}…