Question Number 69778 by Rio Michael last updated on 27/Sep/19 $${prove}\:{that}\:{the}\:{equation}\: \\ $$$$\:\:\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right){x}^{\mathrm{2}} \:+\:\mathrm{4}\left({a}\:+\:{c}\right){x}\:−\mathrm{4}\:=\:\mathrm{0}\:{is}\:{always}\:{real}. \\ $$ Commented by prakash jain last updated on 28/Sep/19…
Question Number 4242 by prakash jain last updated on 05/Jan/16 $$\mathrm{Find}\:\mathrm{that}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{2}^{\mathrm{2}^{\mathrm{2}\centerdot\centerdot\centerdot} } \:\:\left(\mathrm{continued}\:\mathrm{power}\:\mathrm{of}\:\mathrm{2}\right) \\ $$$$\mathrm{using}\:\mathrm{analytical}\:\mathrm{continuation}. \\ $$ Commented by RasheedSindhi last updated on…
Question Number 135309 by mnjuly1970 last updated on 12/Mar/21 $$\:\:\:\:\:\:\:\:\:\:….\:\:\mathscr{N}{ice}\:\:\:\:\mathscr{C}{alculus}\:…. \\ $$$$\:\:\:\:\:\:\:\:{prove}\:\:{that}\:::: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left({ksin}\alpha\right){x}}{\:\sqrt{{x}}}{dx}=\sqrt{\frac{\pi}{{k}}}\:{sin}\left(\frac{\alpha}{\mathrm{2}}\right)… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\: \\ $$ Answered by mathmax…
Question Number 135308 by bobhans last updated on 12/Mar/21 Commented by mr W last updated on 12/Mar/21 $${at}\:{t}=\mathrm{3}\:{s},\:{the}\:{height}\:{is}\:\mathrm{71}.\mathrm{65}{m} \\ $$$${at}\:{t}=\mathrm{7}.\mathrm{8009}{s},\:{the}\:{ball}\:{hits}\:{the}\:{ground} \\ $$$${max}.\:{height}\:{is}\:\mathrm{75}.\mathrm{4235}{m} \\ $$$${in}\:\mathrm{6}.\mathrm{1554}−\mathrm{1}.\mathrm{5997}=\mathrm{4}.\mathrm{5557}{s}\:{the}\:{ball} \\…
Question Number 135305 by JulioCesar last updated on 12/Mar/21 Answered by MJS_new last updated on 12/Mar/21 $$\int\frac{{dx}}{{a}^{{x}} +{b}}= \\ $$$$\:\:\:\:\:\left[{t}={a}^{{x}} +{b}\:\rightarrow\:{dx}=\frac{{dt}}{{a}^{{x}} \mathrm{ln}\:{a}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:{a}}\int\frac{{dt}}{{t}\left({t}−{b}\right)}=\frac{\mathrm{1}}{{b}\mathrm{ln}\:{a}}\int\left(\frac{\mathrm{1}}{{t}−{b}}−\frac{\mathrm{1}}{{t}}\right){dt}= \\…
Question Number 4234 by Filup last updated on 04/Jan/16 $${n},\:{m}\in\mathbb{Z} \\ $$$${n}!={n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)… \\ $$$$ \\ $$$$\mathrm{if}\:{n}!={m}\:\mathrm{where}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{value} \\ $$$${m}\:\mathrm{but}\:\mathrm{not}\:{n},\:\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{for}\:{n}? \\ $$ Commented by prakash jain last…
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Question Number 4232 by Yozzii last updated on 03/Jan/16 Commented by Yozzii last updated on 04/Jan/16 $${I}'{m}\:{experiencing}\:{difficulty}\: \\ $$$${to}\:{solve}\:{the}\:\mathrm{2}{nd}\:{part}\:{without}\:{use}\:{of}\:{a}\:{theorem} \\ $$$${I}\:{posted}\:{sometime}\:{ago},\:{and}\: \\ $$$${generating}\:{functions}. \\ $$…
Question Number 69766 by Rio Michael last updated on 27/Sep/19 $${find}\:\:\frac{{dy}}{{dx}}\:\:{at}\:{the}\:{point}\:\:\left(\mathrm{0},\mathrm{3}\right)\:\:{when}\:\:\mathrm{2}{x}^{\mathrm{2}} {y}\:+\:{y}\:+\:\mathrm{4}{xy}^{\mathrm{2}} \:=\:\mathrm{2}{x}\:+\:\mathrm{3}\: \\ $$ Commented by kaivan.ahmadi last updated on 27/Sep/19 $${f}\left({x},{y}\right)=\mathrm{2}{x}^{\mathrm{2}} {y}+{y}+\mathrm{4}{xy}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}…
Question Number 135300 by bobhans last updated on 12/Mar/21 $$\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{3}}\:+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)=? \\ $$ Answered by Olaf last updated on 12/Mar/21 $$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\:=\:−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\…