Question Number 89010 by ajfour last updated on 14/Apr/20 Commented by ajfour last updated on 14/Apr/20 $${Find}\:{C}\left({h},{k}\right)\:{in}\:{terms}\:{of}\:{a},{b}. \\ $$ Commented by ajfour last updated on…
Question Number 154544 by amin96 last updated on 19/Sep/21 $$\begin{array}{|c|c|}{\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}}\\{\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} ^{\left(\mathrm{2}\right)} }{\boldsymbol{\mathrm{n}}^{\mathrm{3}} }+\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} \boldsymbol{\mathrm{H}}_{\boldsymbol{\mathrm{n}}} ^{\left(\mathrm{3}\right)} }{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }=\frac{\mathrm{21}}{\mathrm{8}}\boldsymbol{\zeta}\left(\mathrm{6}\right)+\boldsymbol{\zeta}^{\mathrm{2}} \left(\mathrm{3}\right)}\\\hline\end{array} \\ $$$${by}\:{Math}.{Amin}\:\:\mathrm{11}.{fb}.\mathrm{96}…
Question Number 154547 by peter frank last updated on 19/Sep/21 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{tunnel} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{4m}\:\mathrm{high}.\mathrm{if}\:\:\mathrm{the}\:\mathrm{initial}\:\mathrm{speed} \\ $$$$\mathrm{is}\:\mathrm{V}_{\mathrm{o}} .\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{maximum} \\ $$$$\mathrm{range}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{tunnel}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by}\:\:\:\:\:\:\mathrm{R}=\mathrm{4}\sqrt{\mathrm{2}}\:\sqrt{\frac{\mathrm{V}_{\mathrm{o}} ^{\mathrm{2}} }{\mathrm{g}}−\mathrm{8}} \\ $$ Answered…
Question Number 23472 by ajfour last updated on 31/Oct/17 Commented by ajfour last updated on 31/Oct/17 $${Q}.\mathrm{23390}\:\left({continued}..\right) \\ $$ Answered by ajfour last updated on…
Question Number 23471 by Tinkutara last updated on 31/Oct/17 $${Prove}\:{that} \\ $$$$\underset{\mathrm{0}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left(\frac{\mathrm{1}}{\:^{{n}} {C}_{{i}} }\:+\:\frac{\mathrm{1}}{\:^{{n}} {C}_{{j}} }\right)\:=\:\underset{{r}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{{n}\:−\:{r}}{\:^{{n}} {C}_{{r}} }\:+\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{r}}{\:^{{n}} {C}_{{r}} }…
Question Number 89007 by naka3546 last updated on 14/Apr/20 Answered by TANMAY PANACEA. last updated on 14/Apr/20 $$\frac{{a}^{\mathrm{2}} }{{b}+{c}+{d}}+{a} \\ $$$$\frac{{a}^{\mathrm{2}} +{ab}+{ac}+{ad}}{{b}+{c}+{d}} \\ $$$$\frac{{a}\left({a}+{b}+{c}+{d}\right)}{{b}+{c}+{d}} \\…
Question Number 154542 by SANOGO last updated on 19/Sep/21 Answered by qaz last updated on 19/Sep/21 $$\mathrm{Let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b} \\ $$$$\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}=\mathrm{x}^{\mathrm{2}} +\mathrm{a}\int_{\mathrm{0}} ^{\mathrm{x}} \left(\mathrm{x}−\mathrm{t}\right)\mathrm{dt}=\mathrm{x}^{\mathrm{2}} +\mathrm{a}\left(\mathrm{xt}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \right)_{\mathrm{0}}…
Question Number 88997 by jagoll last updated on 14/Apr/20 Commented by john santu last updated on 14/Apr/20 $$\mathrm{4}{R}+\mathrm{2}{R}\sqrt{\mathrm{2}}\:=\:\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${R}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\:=\:\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${R}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{2}+\sqrt{\mathrm{2}}}\:×\:\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}−\sqrt{\mathrm{2}}}\: \\ $$$${R}=\:\frac{\mathrm{8}\sqrt{\mathrm{2}}−\mathrm{8}}{\mathrm{4}−\mathrm{2}}\:=\:\mathrm{4}\sqrt{\mathrm{2}}\:−\mathrm{4}\:{dm} \\…
Question Number 23458 by ketto last updated on 31/Oct/17 Commented by mrW1 last updated on 31/Oct/17 $$\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{n}}} =\:^{\mathrm{n}} \sqrt{\mathrm{x}} \\ $$ Terms of Service Privacy…
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