Question Number 135413 by sdfg last updated on 13/Mar/21 Commented by sdfg last updated on 13/Mar/21 $${Q}\mathrm{24} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 135415 by JulioCesar last updated on 13/Mar/21 Commented by liberty last updated on 13/Mar/21 $$\left({nx}+\mathrm{1}\right)^{{p}} \:=\:\underset{{i}=\mathrm{0}} {\overset{{p}} {\sum}}\:{C}_{{i}} ^{\:{p}} \:\left({nx}\right)^{{p}−{i}} \: \\ $$$$\:=\:\left({nx}\right)^{{p}}…
Question Number 135364 by liberty last updated on 12/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135354 by liberty last updated on 12/Mar/21 $${Algebra} \\ $$A & B together can finish a work in 24 days, B & C…
Question Number 4266 by Rasheed Soomro last updated on 06/Jan/16 $$\mathrm{Analyze}\:\mathrm{for}\:\:\mathrm{integer}\:\mathrm{solution}\:: \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{are}\:\mathrm{fixed}\:\mathrm{positive}\:\mathrm{integers}. \\ $$$$\mathrm{ax}^{\mathrm{a}} +\mathrm{by}^{\mathrm{b}} =\mathrm{cz}^{\mathrm{c}} \\ $$ Commented by Yozzii last updated on…
Question Number 4250 by Yozzii last updated on 06/Jan/16 $${Let}\:{u}=\frac{{ln}\left(\mathrm{3}−\left\{\frac{{ln}\left(\mathrm{3}−\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left(\mathrm{2}+\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}{{ln}\left(\mathrm{2}+\left\{\frac{{ln}\left(\mathrm{3}−\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left(\mathrm{2}+\left[\frac{{ln}\left(\mathrm{3}−\left(\ldots\right.\right.}{{ln}\left(\mathrm{2}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}. \\ $$$${What}\:{is}\:{the}\:{value}\:{of}\:{u}?\: \\ $$$$ \\ $$$${Let}\:{k}=\frac{{ln}\left({x}−\left\{\frac{{ln}\left({x}−\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left({x}−\mathrm{1}+\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}{{ln}\left({x}−\mathrm{1}+\left\{\frac{{ln}\left({x}−\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}{{ln}\left({x}−\mathrm{1}+\left[\frac{{ln}\left({x}−\left(\ldots\right.\right.}{{ln}\left({x}−\mathrm{1}+\left(\ldots\right.\right.}\right.\right.}\right\}\right)}. \\ $$$${For}\:{what}\:{values}\:{of}\:{x}\:{does} \\ $$$$\left({i}\right)\:{k}\:{converge}\:\left({ii}\right)\:{k}\:{diverge}? \\ $$ Commented by Yozzii…
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…
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 $$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$=\:−\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}\:=\:−\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}\right)\:=\:\mathrm{2}−\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 4232 by Yozzii last updated on 03/Jan/16 Commented by Yozzii last updated on 04/Jan/16 $${I}'{m}\:{experiencing}\:{difficulty}\: \\ $$$${to}\:{solve}\:{the}\:\mathrm{2}{nd}\:{part}\:{without}\:{use}\:{of}\:{a}\:{theorem} \\ $$$${I}\:{posted}\:{sometime}\:{ago},\:{and}\: \\ $$$${generating}\:{functions}. \\ $$…