Question Number 132301 by mnjuly1970 last updated on 13/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\:{nice}…….{calculus}…. \\ $$$$\:\:\:\:\:\:\:\:{prove}\:\:\:{that}\::: \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}{arctan}\left(\frac{{x}}{\mathrm{2}}\right)\right)}{\left({x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} \right){sinh}\left(\pi{x}\right)}{dx}=\frac{\mathrm{7}}{\mathrm{8}}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$ \\ $$ Terms of…
Question Number 1231 by Rasheed Soomro last updated on 16/Jul/15 $${i}^{\mathrm{2}} ={i}.{i}=\sqrt{−\mathrm{1}}.\sqrt{−\mathrm{1}}=\sqrt{−\mathrm{1}×−\mathrm{1}}=\sqrt{\mathrm{1}}=\mathrm{1}??? \\ $$$${i}^{\mathrm{2}} =\mathrm{1}\Rightarrow−\mathrm{1}=\mathrm{1}???\: \\ $$$${Resolve}\:{the}\:{contradiction}. \\ $$ Commented by prakash jain last updated…
Question Number 66767 by John Kaloki Musau last updated on 19/Aug/19 $${In}\:{a}\:{school}\:{there}\:{are}\:\mathrm{30}\:{more}\:{boys} \\ $$$${than}\:{girls}.\:{One}-{quarter}\:{of}\:{the}\:{boys} \\ $$$${and}\:{two}-{thirds}\:{of}\:{the}\:{girls}\:{are}\:{boarders}. \\ $$$${If}\:{there}\:{are}\:\mathrm{255}\:{boarders},\:{find}\:{the} \\ $$$${number}\:{of}\:{students}\:{in}\:{the}\:{school}. \\ $$ Commented by Prithwish…
Question Number 132303 by mohammad17 last updated on 13/Feb/21 Answered by EDWIN88 last updated on 13/Feb/21 $$\:\overset{\rightarrow} {\mathrm{a}}=\left(−\mathrm{1},\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:,\:\mathrm{1}\right)\: \\ $$$$\:\overset{\rightarrow} {\mathrm{b}}=\left(−\mathrm{3},\:−\mathrm{5ln}\:\mathrm{2},\:−\mathrm{4}\right) \\ $$$$\:\overset{\rightarrow} {\mathrm{c}}=\:\left(\mathrm{4},\:−\mathrm{5},\:\mathrm{5}\:\right)\: \\…
Question Number 1229 by Rasheed Soomro last updated on 16/Jul/15 $${f}\left(\:{f}\left({x}\right)\:\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({Modification}\:{of}\:{Q}\:\mathrm{1147}\right) \\ $$ Commented by 123456 last updated on 17/Jul/15…
Question Number 132302 by Lordose last updated on 13/Feb/21 $$\boldsymbol{\mathrm{Simplify}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\boldsymbol{\Gamma}\left(\frac{\boldsymbol{\mathrm{p}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$ Answered by Ar Brandon last updated on 13/Feb/21 $$\Gamma\left(\mathrm{m}\right)\Gamma\left(\mathrm{m}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2m}−\mathrm{1}} }\Gamma\left(\mathrm{2m}\right) \\…
Question Number 66765 by John Kaloki Musau last updated on 19/Aug/19 $${Simba}\:{had}\:\mathrm{57}\:{denomination}\:{notes} \\ $$$${which}\:{he}\:{deposited}\:{in}\:{his}\:{account}. \\ $$$${He}\:{had}\:{six}\:{times}\:{as}\:{many}\:{two}-{hundred} \\ $$$${shilling}\:{notes}\:{as}\:{one}-{thousand}\: \\ $$$${shilling}\:{notes}\:{and}\:{twice}\:{as}\:{many}\: \\ $$$${one}-{hundred}\:{shilling}\:{notes}\:{as}\:{two}- \\ $$$${hundred}\:{shilling}\:{notes}.\:{The}\:{rest}\:{were} \\…
Question Number 66760 by John Kaloki Musau last updated on 19/Aug/19 $${By}\:{writing}\:{your}\:{answer}\:{in}\:{the} \\ $$$${form}\:{a}^{{y}} \:{simplify} \\ $$$$\left(\mathrm{3}^{\mathrm{5}{x}} ×\mathrm{5}^{\mathrm{2}{x}} ×\mathrm{3}^{−{x}} \boldsymbol{\div}\mathrm{5}^{−\mathrm{2}{x}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$ Commented by…
Question Number 132293 by Algoritm last updated on 13/Feb/21 Commented by mr W last updated on 13/Feb/21 $${x}=\mathrm{1},\:{y}=\mathrm{0} \\ $$$${x}=\mathrm{0},\:{y}=\mathrm{1} \\ $$$${the}\:{proof}\:{you}\:{may}\:{ask}\:{for}\:{is}\:{thinking}. \\ $$ Answered…
Question Number 132292 by Algoritm last updated on 13/Feb/21 Answered by mathmax by abdo last updated on 13/Feb/21 $$\frac{\mathrm{1}}{\left[\mathrm{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2x}\right]}=\left[\mathrm{x}\right]+\frac{\mathrm{1}}{\mathrm{3}}\:\:\mathrm{let}\:\left[\mathrm{x}\right]=\mathrm{n}\:\Rightarrow\mathrm{n}\leqslant\mathrm{x}<\mathrm{n}+\mathrm{1}\:\Rightarrow\mathrm{2n}\leqslant\mathrm{2x}<\mathrm{2n}+\mathrm{2} \\ $$$$\mathrm{if}\:\mathrm{2x}\in\left[\mathrm{2n},\mathrm{2n}+\mathrm{1}\left[\:\Rightarrow\mathrm{x}\in\left[\mathrm{n},\mathrm{n}+\frac{\mathrm{1}}{\mathrm{2}}\left[\:\Rightarrow\left[\mathrm{2x}\right]=\mathrm{2n}\right.\right.\right.\right. \\ $$$$\mathrm{e}\Rightarrow\frac{\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{1}}{\mathrm{2n}}=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{2n}}=\mathrm{n}+\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\frac{\mathrm{9}}{\mathrm{2n}}=\mathrm{3n}+\mathrm{1}\:\Rightarrow\mathrm{9}=\mathrm{2n}\left(\mathrm{3n}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{6n}^{\mathrm{2}}…