Question Number 128179 by bounhome last updated on 05/Jan/21 $${show}\:{me}\:{that}\:{y}={x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{6}\:{is}\:{the}\:{solution}\:{of} \\ $$$${y}''−{y}'+\mathrm{2}\left({x}+\mathrm{1}\right)=\mathrm{0} \\ $$ Answered by bemath last updated on 05/Jan/21 $$\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{6}\:\rightarrow\begin{cases}{\mathrm{y}'=\mathrm{2x}+\mathrm{4}}\\{\mathrm{y}''=\mathrm{2}}\end{cases} \\…
Question Number 128172 by shaker last updated on 05/Jan/21 Commented by mr W last updated on 05/Jan/21 $${you}\:{can}\:{only}\:“{see}''\:{the}\:{solution}: \\ $$$${x}=\mathrm{0},\:{y}=\mathrm{1},\:{z}=\mathrm{2} \\ $$$$\left({x},\:{y},\:{z}\:{are}\:{interchangeable}\right) \\ $$$$ \\…
Question Number 128157 by naka3546 last updated on 04/Jan/21 $${Let}\:\:{P}\left({x}\right)\:\:{is}\:\:{monic}\:\:{polynomial}\:\:{function}\:\:{such}\:\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{P}\left(\mathrm{2}{x}\right)\right]^{\mathrm{2}} \:=\:{P}\left(\mathrm{4}{x}\right)\:+\:\mathrm{16}{P}\left({x}^{\mathrm{2}} \right) \\ $$$${P}\left({x}\right)\:=\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128150 by MJS_new last updated on 04/Jan/21 $$\mathrm{question}\:\mathrm{128091},\:\mathrm{trying}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{completely}\:\mathrm{in}\:\mathbb{R} \\ $$$${a}^{\mathrm{3}} =\mathrm{3}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${b}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\mathrm{25} \\ $$$${c}^{\mathrm{3}} =\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{25}…
Question Number 128135 by help last updated on 04/Jan/21 Commented by liberty last updated on 04/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{3sin}\:\mathrm{5x}}{\mathrm{x}}\right)^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{4x}}{\mathrm{x}^{\mathrm{2}} }\right)} =\:\mathrm{15}^{\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{16x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{x}^{\mathrm{2}} }\right)} \\…
Question Number 128124 by shaker last updated on 04/Jan/21 Answered by Dwaipayan Shikari last updated on 04/Jan/21 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt[{{n}}]{\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)…\left(\mathrm{1}+\frac{{n}}{{n}}\right)}={y} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{log}\left(\mathrm{1}+\frac{{k}}{{n}}\right)={logy} \\ $$$$\Rightarrow\int_{\mathrm{0}}…
Question Number 62576 by Sr@2004 last updated on 23/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 62556 by naka3546 last updated on 22/Jun/19 $$\frac{\mathrm{1}+\mathrm{3}}{\mathrm{3}}\:+\:\frac{\mathrm{1}+\mathrm{3}+\mathrm{5}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}+\mathrm{3}+\mathrm{5}+\mathrm{7}}{\mathrm{3}^{\mathrm{3}} }\:+\:…\:\:=\:\:\:\frac{{a}}{{b}}\:\:,\:\:{a},\:{b}\:\in\:\:\mathbb{Z}^{+} \: \\ $$ Commented by Prithwish sen last updated on 22/Jun/19 $$\mathrm{t}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{n}}…
Question Number 128091 by rydasss last updated on 04/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128080 by mohammad17 last updated on 04/Jan/21 Answered by mathmax by abdo last updated on 04/Jan/21 $$\mathrm{z}^{\mathrm{4}} −\mathrm{1}=\sqrt{\mathrm{3}}\mathrm{i}\:\Rightarrow\mathrm{z}^{\mathrm{4}} \:=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\:=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\mathrm{let}\:\mathrm{z}\:=\mathrm{re}^{\mathrm{i}\theta} \: \\ $$$$\mathrm{z}^{\mathrm{4}}…