Question Number 134185 by mnjuly1970 last updated on 28/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:{i}:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({tan}\left({x}\right)−{x}\right)}{{cos}\left({x}\right)}{dx}=\frac{\pi}{{e}} \\ $$$$\:\:\:\:{ii}:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}{e}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } =\frac{\pi}{{e}\sqrt{{e}}} \\ $$$$\:\:…
Question Number 3113 by Yozzi last updated on 05/Dec/15 $${Define},\:{for}\:{sets}\:{A}\:{and}\:{B}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}\ast{B}=\left({A}−{B}\right)\cup\left({B}−{A}\right). \\ $$$${Show}\:{that}\:{A}\ast{B}=\left({A}\cup{B}\right)−\left({A}\cap{B}\right). \\ $$$${Prove}\:{that}\:{A}\ast\left({B}\ast{C}\right)=\left({A}\ast{B}\right)\ast{C}.\: \\ $$ Commented by Yozzi last updated on 06/Dec/15…
Question Number 3112 by Filup last updated on 05/Dec/15 $$\mathrm{let}\:{S}=\left\{{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{n}} \right\}\::\:\mid{S}\mid={n} \\ $$$$ \\ $$$${S}\:\mathrm{contains}\:{n}\:\mathrm{random}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that} \\ $$$${a}\leqslant{a}_{{k}} \leqslant{b} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{I}\:\mathrm{wanted}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{average}\:\mathrm{value} \\…
Question Number 134182 by Algoritm last updated on 28/Feb/21 Answered by Ñï= last updated on 01/Mar/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)^{\mathrm{3}} \frac{\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 68642 by Maclaurin Stickker last updated on 14/Sep/19 $$\mathrm{Young}'\mathrm{s}\:\mathrm{modulus}\:\mathrm{of}\:\mathrm{a}\:\mathrm{material}\:\mathrm{measures} \\ $$$$\mathrm{its}\:\mathrm{resistance}\:\mathrm{caused}\:\mathrm{by}\:\mathrm{external}\:\mathrm{stresses}. \\ $$$$\mathrm{On}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{wall}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solid}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{specific} \\ $$$$\mathrm{mass}\:\rho\:\mathrm{and}\:\mathrm{Young}\:\varepsilon\:\mathrm{modulus}\:\mathrm{in}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{parallelepiped}\:\mathrm{shape},\:\mathrm{the}\:\mathrm{dimensions} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{which}\:\mathrm{are}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}.\: \\ $$$$\mathrm{Based}\:\mathrm{on}\:\mathrm{the}\:\mathrm{correlations}\:\mathrm{between}\:\mathrm{physical} \\ $$$$\mathrm{quantities},\:\mathrm{determine}\:\mathrm{the}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{that}…
Question Number 134179 by mohammad17 last updated on 28/Feb/21 Commented by mohammad17 last updated on 28/Feb/21 $${help}\:{me}\:{sir} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 3105 by prakash jain last updated on 04/Dec/15 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{next}\:\mathrm{term}\:\mathrm{in}\:\mathrm{the}\:\mathrm{below}\:\mathrm{sequence} \\ $$$$\mathrm{1},\mathrm{3},\mathrm{9},\mathrm{27},\mathrm{81},\mathrm{243},\mathrm{729},\:\mathrm{2123},\:\mathrm{5857},? \\ $$ Commented by Filup last updated on 04/Dec/15 $$\mathrm{3}^{\mathrm{10}} ? \\…
Question Number 134178 by mohammad17 last updated on 28/Feb/21 Commented by mohammad17 last updated on 28/Feb/21 $${help}\:{me}\:{sir} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 134173 by mathlove last updated on 28/Feb/21 $$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\frac{\sqrt[{\mathrm{3}}]{{x}}−\mathrm{2}}{\:\sqrt{{x}}−\mathrm{2}\sqrt{\mathrm{2}}}=? \\ $$ Answered by malwan last updated on 28/Feb/21 $$\underset{{x}\rightarrow\mathrm{8}} {{lim}}\:\frac{\:^{\mathrm{3}} \sqrt{{x}}−\mathrm{2}}{\:\sqrt{{x}}−\mathrm{2}\sqrt{\mathrm{2}}}×\frac{\:^{\mathrm{3}} \sqrt{{x}^{\mathrm{2}} }+\mathrm{2}\:^{\mathrm{3}}…
Question Number 68636 by ajfour last updated on 14/Sep/19 $$\frac{−{a}^{\mathrm{2}} +{a}+\mathrm{1}}{\:\:\:\:{a}^{\mathrm{2}} +{a}+\mathrm{1}}=\frac{−{b}^{\mathrm{2}} +{b}+\mathrm{1}}{\:\:\:\:{b}^{\mathrm{2}} +{b}+\mathrm{1}} \\ $$$$\:\:=\:\frac{\:\:\:\:\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{ab}+\left({b}−{a}\right)}{−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{ab}+\left({b}−{a}\right)} \\ $$$$\:\:=\frac{−\mathrm{2}{ab}+\left({a}+{b}\right)+\mathrm{2}}{\:\:\:\:\mathrm{2}{ab}+\left({a}+{b}\right)+\mathrm{2}} \\ $$$${Solve}\:{for}\:\boldsymbol{{a}}. \\ $$$$ \\…