Question Number 4579 by Rasheed Soomro last updated on 08/Feb/16 $${In}\:{a}\:{right}\:{triangle},\:{the}\:{mid}-{point}\:{of}\:{the} \\ $$$${hypotenuse}\:{is}\:{equidistant}\:{from}\:{all}\:{the} \\ $$$${three}\:{vertices}\:{of}\:{the}\:{triangle}. \\ $$ Commented by Yozzii last updated on 08/Feb/16 $${Prove}\:{this}\:{theorem}?…
Question Number 4578 by Rasheed Soomro last updated on 08/Feb/16 $${The}\:{segment}\:{between}\:{the}\:{mid}-{points} \\ $$$${of}\:{two}\:{sides}\:{of}\:{a}\:{triangle}\:{is}\:{parallel} \\ $$$${to}\:{the}\:{third}\:{side}\:{and}\:{half}\:{as}\:{long}.\: \\ $$ Commented by Yozzii last updated on 08/Feb/16 $${Prove}\:{this}\:{theorem}?…
Question Number 4575 by FilupSmith last updated on 08/Feb/16 $$\int_{{a}} ^{\:\infty} \frac{\mathrm{1}}{{x}}{dx}=\infty \\ $$$$ \\ $$$$\int_{{a}} ^{\:\infty} \lfloor\frac{\mathrm{1}}{{x}}\rfloor{dx}={S} \\ $$$${a}>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Is}\:{S}\:\mathrm{finite}?\:\mathrm{Can}\:\mathrm{you}\:\mathrm{solve}\:\mathrm{for}\:{S}? \\…
Question Number 135647 by Dwaipayan Shikari last updated on 14/Mar/21 $$…+\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{2}\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}−\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{\left(\mathrm{1}+\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{2}\pi^{\mathrm{2}} \right)^{\mathrm{2}} }+…=\frac{{csc}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\pi}\right)}{\pi^{\mathrm{2}} } \\ $$ Terms of…
Question Number 70108 by naka3546 last updated on 01/Oct/19 Answered by MJS last updated on 01/Oct/19 $$\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{x}} >\mathrm{3}^{\mathrm{36}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{18}{x}} \\ $$$$\left(−\mathrm{2ln}\:\mathrm{2}\right){x}>\mathrm{36ln}\:\mathrm{3}\:+\mathrm{18}\left(\mathrm{ln}\:\mathrm{2}\:−\mathrm{ln}\:\mathrm{3}\right){x} \\ $$$$\left(\mathrm{18ln}\:\mathrm{3}\:−\mathrm{20}\:\mathrm{ln}\:\mathrm{2}\right){x}>\mathrm{36ln}\:\mathrm{3} \\ $$$${x}>\frac{\mathrm{18ln}\:\mathrm{3}}{\mathrm{9ln}\:\mathrm{3}\:−\mathrm{10ln}\:\mathrm{2}}\approx\mathrm{6}.\mathrm{68970287}…
Question Number 135646 by metamorfose last updated on 14/Mar/21 $$\int\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\:{dx}=…? \\ $$ Answered by Ñï= last updated on 15/Mar/21 $$\int\left[\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)−{x}\right]{dx} \\ $$$$=\int\left[\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({ln}\left({x}+\mathrm{1}\right)−{lnx}\right)−{x}\right]{dx} \\ $$$$=\int\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){ln}\left({x}+\mathrm{1}\right){dx}−\int\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right){lnxdx}−\int{xdx} \\…
Question Number 135642 by Raxreedoroid last updated on 14/Mar/21 $$\mathrm{Write}\:\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\mathrm{3}^{{k}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{n}.\:\mathrm{Where}\:{n}\in\mathbb{N}. \\ $$ Answered by Ñï= last updated on 14/Mar/21 $$\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{2}} {\sum}}\mathrm{3}^{{k}}…
Question Number 70103 by Shamim last updated on 01/Oct/19 $$\mathrm{if}\:\mathrm{m}^{\mathrm{3}} +\mathrm{2p}^{\mathrm{3}} =\mathrm{3mn},\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} =\mathrm{p}^{\mathrm{3}} \:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{n}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}+\mathrm{b}=\mathrm{m}. \\ $$ Answered by mind is…
Question Number 135638 by Dwaipayan Shikari last updated on 14/Mar/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\frac{\mathrm{1}}{\mathrm{12}}{log}^{\mathrm{2}} \left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 135633 by metamorfose last updated on 14/Mar/21 $$\int\frac{\mathrm{1}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}{dx}=…? \\ $$ Answered by Ñï= last updated on 14/Mar/21 $$\int\frac{{dx}}{{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{1}}\overset{{t}={x}^{\frac{\mathrm{1}}{\mathrm{3}}} } {=}\int\frac{\mathrm{3}{t}^{\mathrm{2}} {dt}}{{t}+\mathrm{1}}=\mathrm{3}\int\frac{\left({t}+\mathrm{1}\right)\left({t}−\mathrm{1}\right)+\mathrm{1}}{{t}+\mathrm{1}}{dt}=\mathrm{3}\int\left\{\left({t}−\mathrm{1}\right)+\frac{\mathrm{1}}{{t}+\mathrm{1}}\right\}{dt}…