Question Number 184121 by Noorzai last updated on 03/Jan/23 Commented by Rasheed.Sindhi last updated on 03/Jan/23 $${See}\:{Q}#\mathrm{182423} \\ $$$${Why}\:{have}\:{you}\:{repeated}\:{your}\:\boldsymbol{{own}} \\ $$$$\boldsymbol{{answered}}\:{question}? \\ $$ Commented by…
Question Number 118513 by I want to learn more last updated on 18/Oct/20 $$\mathrm{If}\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:+\:\:\mathrm{f}\left(\frac{\mathrm{x}\:\:−\:\:\mathrm{1}}{\mathrm{x}}\right)\:\:\:=\:\:\:\mathrm{1}\:\:+\:\:\mathrm{x},\:\:\:\:\:\:\:\mathrm{find}\:\:\:\mathrm{f}\left(\mathrm{2}\right). \\ $$ Answered by Dwaipayan Shikari last updated on 18/Oct/20 $${f}\left(\mathrm{2}\right)+{f}\left(\frac{\mathrm{2}−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1}+\mathrm{2}\rightarrow\left({a}\right)\:\:\:\left({Takng}\:{x}=\mathrm{2}\right)…
Question Number 118488 by Lordose last updated on 18/Oct/20 $$\boldsymbol{\mathrm{Conjecture}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{formula}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{infinite}} \\ $$$$\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{series}}. \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{35}}+\:\centerdot\:\centerdot\:\centerdot\:\frac{\mathrm{1}}{\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}\right)\left(\mathrm{2}\boldsymbol{\mathrm{n}}+\mathrm{1}\right)} \\ $$$$\boldsymbol{\mathrm{And}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{formula}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{Induction}}. \\ $$ Answered by Olaf last updated on 18/Oct/20…
Question Number 118485 by bramlexs22 last updated on 18/Oct/20 $${If}\:{partial}\:{fraction}\: \\ $$$$\frac{\mathrm{10}{x}^{\mathrm{2}} +{px}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +{x}−\mathrm{2}}\:{can}\:{be}\:{written} \\ $$$${as}\:\frac{{q}}{\mathrm{2}{x}−\mathrm{1}}\:+\:\frac{\mathrm{4}}{{x}+\mathrm{2}}\:+\:\frac{{r}}{{x}+\mathrm{1}}.\:{Then}\:{find}\:{the} \\ $$$${value}\:{of}\:{p}−{q}+\mathrm{2}{r}\:. \\ $$ Answered by benjo_mathlover last…
Question Number 183989 by HeferH last updated on 01/Jan/23 $$\frac{\sqrt{{x}}\:+\:\sqrt{{x}\:−\:\mathrm{4}{a}}}{\:\sqrt{{x}}\:−\:\sqrt{{x}\:−\:\mathrm{4}{a}}}\:=\:{a}\:\neq\:\mathrm{0} \\ $$$$\:{find}\:“{x}''\:{in}\:{terms}\:{of}\:“{a}''.\: \\ $$ Commented by Frix last updated on 01/Jan/23 $$\mathrm{I}\:\mathrm{get}\:{x}=\left({a}+\mathrm{1}\right)^{\mathrm{2}} \\ $$ Answered…
Question Number 118448 by peter frank last updated on 17/Oct/20 Answered by Olaf last updated on 17/Oct/20 $$\mathrm{A}\:=\:\pi×\mathrm{7}^{\mathrm{2}} −\mathrm{2}\int_{−\mathrm{7}} ^{+\mathrm{7}} \left(\mathrm{7}−\sqrt{\mathrm{49}−{x}^{\mathrm{2}} }\right){dx} \\ $$$$\mathrm{A}\:=\:\mathrm{49}\pi−\mathrm{4}\int_{\mathrm{0}} ^{+\mathrm{7}}…
Question Number 183664 by Rasheed.Sindhi last updated on 28/Dec/22 $$ \\ $$$$\mathcal{H}{ow}\:{many}\:{artificially}\:{made}\:{weights}\:{are}\: \\ $$$$\boldsymbol{{required}}\:\left({minimally}\right)\:{to}\:{weigh} \\ $$$$\:\underline{\mathrm{ONLY}}\:\:\:\:\:\mathrm{1},\mathrm{3},\mathrm{5},…,\mathrm{79}\:{kg}\: \\ $$$${and}\:{what}\:{should}\:{be}\:{they}? \\ $$ Commented by Frix last updated…
Question Number 183646 by Rasheed.Sindhi last updated on 28/Dec/22 $$\mathcal{H}{ow}\:{many}\:{artificially}\:{made}\:{weights}\:{are}\: \\ $$$$\boldsymbol{{required}}\:\left({minimally}\right)\:{to}\:{weigh} \\ $$$$\:\mathrm{1},\mathrm{3},\mathrm{5},…,\mathrm{79}\:{kg}\: \\ $$$${and}\:{what}\:{should}\:{be}\:{they}? \\ $$ Commented by Rasheed.Sindhi last updated on 28/Dec/22…
Question Number 183617 by Rasheed.Sindhi last updated on 27/Dec/22 $$\mathcal{H}{ow}\:{many}\:{artificially}\:{made}\:{weights}\:{are}\: \\ $$$$\boldsymbol{{required}}\:\left({minimally}\right)\:{to}\:{weigh} \\ $$$$\:\mathrm{2},\mathrm{4},\mathrm{6},…,\mathrm{80}\:{kg}\: \\ $$$${and}\:{what}\:{should}\:{be}\:{they}? \\ $$ Commented by Rasheed.Sindhi last updated on 27/Dec/22…
Question Number 183535 by HeferH last updated on 26/Dec/22 $${Who}\:{is}\:{greater}?\:\mathrm{70}^{\mathrm{71}} \:{or}\:\:\mathrm{71}^{\mathrm{70}} \\ $$ Answered by Frix last updated on 26/Dec/22 $$\mathrm{1}^{\mathrm{2}} <\mathrm{2}^{\mathrm{1}} \\ $$$$\mathrm{2}^{\mathrm{3}} <\mathrm{3}^{\mathrm{2}}…