Question Number 191859 by mathlove last updated on 02/May/23 Commented by mr W last updated on 03/May/23 $${all}\:{are}\:{of}\:{type}\:{y}'+{P}\left({x}\right){y}={Q}\left({x}\right). \\ $$ Answered by qaz last updated…
Question Number 191796 by mathlove last updated on 30/Apr/23 Commented by mathlove last updated on 01/May/23 $${pleas}\:{solve}\:{this} \\ $$ Answered by AST last updated on…
Question Number 60695 by maxmathsup by imad last updated on 24/May/19 $${solve}\:\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }{y}^{''} \:\:\:\:−\left(\mathrm{2}{x}+\mathrm{1}\right){y}^{'} \:={x}^{\mathrm{2}} \:{e}^{−{x}^{\mathrm{2}} \:\:\:} \\ $$ Commented by maxmathsup by imad last…
Question Number 126070 by benjo_mathlover last updated on 17/Dec/20 Commented by liberty last updated on 17/Dec/20 $${h}\left({x}\right)=\int_{\mathrm{1}} ^{\:{x}} {f}\left({t}\right)\:{dt}\:\Leftrightarrow\:{h}'\left({x}\right)=\:{f}\left({x}\right) \\ $$$${h}'\left(\mathrm{4}\right)={f}\left(\mathrm{4}\right)=\:\mathrm{2} \\ $$$$ \\ $$…
Question Number 126039 by snipers237 last updated on 16/Dec/20 $${Let}\:{A}=\left[\mathrm{2};\infty\left[^{\mathrm{2}} \:\:{and}\:{f}\:\in\left(\mathbb{R}×\mathbb{R}\right)^{\mathbb{R}} \:{such}\:{as}\:\right.\right. \\ $$$${f}\left({x},{y}\right)=\frac{\chi_{{A}} \left({x},{y}\right)}{{E}\left({x}\right)^{{E}\left({y}\right)} }\:\:{where}\:\chi_{{A}} \:{is}\:{the}\:{caracteristic}\:{function}\:{of}\:{A} \\ $$$$\:{Prove}\:{that}\:{f}\:{is}\:{a}\:{density}\:{of}\:{a}\:{probability}\:{P} \\ $$ Answered by mindispower last…
Question Number 126010 by bramlexs22 last updated on 16/Dec/20 Answered by liberty last updated on 16/Dec/20 $${The}\:{area}\:{of}\:{a}\:{tringle}\:{is}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}\right)\left(\mathrm{6}\right)\mathrm{sin}\:\theta \\ $$$${A}\:=\:\mathrm{24}\:\mathrm{sin}\:\theta \\ $$$$\frac{{dA}}{{dt}}\:=\:\left(\mathrm{24}\:\mathrm{cos}\:\theta\right)\:\frac{{d}\theta}{{dt}}\:;\:{where}\:\frac{{d}\theta}{{dt}}\:=\:\mathrm{0}.\mathrm{12}\:{rad}/{sec} \\ $$$$\Leftrightarrow\:\frac{{dA}}{{dt}}\:=\:\mathrm{24}×\mathrm{0}.\mathrm{12}×\mathrm{cos}\:\frac{\pi}{\mathrm{6}}=\:\mathrm{2}.\mathrm{49}\:{m}^{\mathrm{2}} /{sec}\: \\…
Question Number 126008 by bramlexs22 last updated on 16/Dec/20 $$\:{Find}\:{all}\:{asymptotes}\:{of}\:{the}\: \\ $$$${function}\:{y}=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:. \\ $$ Answered by liberty last updated on 16/Dec/20 $${Horizontal}\:{asymptote}\::\:{y}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:=\:\underset{{x}\rightarrow\infty}…
Question Number 126003 by mnjuly1970 last updated on 16/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{integral}… \\ $$$$\:\:\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\phi=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {tan}\left({x}\right){ln}\left({sin}\left({x}\right)\right){ln}\left({cos}\left({x}\right)\right){dx}=\frac{\zeta\left(\mathrm{3}\right.}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{G}{ood}\:{luck} \\ $$ Answered by Olaf last updated…
Question Number 125996 by bramlexs22 last updated on 16/Dec/20 $$\:{If}\:{z}^{\mathrm{3}} ={x}^{\mathrm{2}} −{y}^{\mathrm{2}} \:,\:\rightarrow\begin{cases}{\frac{{dx}}{{dt}}=\mathrm{3}}\\{\frac{{dy}}{{dt}}=\mathrm{2}}\end{cases} \\ $$$${find}\:\frac{{dz}}{{dt}}\:{when}\:{x}=\mathrm{4}\:{and}\:{y}=\mathrm{1} \\ $$ Answered by Olaf last updated on 16/Dec/20 $${z}^{\mathrm{3}}…
Question Number 125995 by liberty last updated on 16/Dec/20 Commented by benjo_mathlover last updated on 16/Dec/20 $${f}\left({x}\right)=\:\begin{cases}{{x}\:;\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1}}\\{−{x}\:;\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{0}}\end{cases} \\ $$$$\:{f}\:'\left(−\mathrm{1}\right)\:=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{f}\left(−\mathrm{1}+{h}\right)−{f}\left(−\mathrm{1}\right)}{{h}} \\ $$$${f}\:'\left(−\mathrm{1}\right)=\:\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\left(−\mathrm{1}+{h}\right)−\left(\mathrm{1}\right)}{{h}} \\ $$$$\:{f}\:'\left(−\mathrm{1}\right)=\:\underset{{h}\rightarrow\mathrm{0}}…