Question Number 129607 by HHHHH last updated on 16/Jan/21 $${salom} \\ $$ Answered by prakash jain last updated on 16/Jan/21 $$\mathrm{welcome} \\ $$ Answered by…
Question Number 64068 by mathmax by abdo last updated on 12/Jul/19 $${calculate}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{tsint}}{\mathrm{3}+{sin}^{\mathrm{2}} {t}}\:{dt}\: \\ $$ Commented by mathmax by abdo last updated on…
Question Number 64066 by mathmax by abdo last updated on 12/Jul/19 $${let}\:\alpha\:,\beta\:{and}\:\lambda\:{the}\:{roots}\:{of}\:{x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{1}\:=\mathrm{0}\:{find}\:{the}\:{value}\:{of} \\ $$$${A}\:=\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \:{and}\:\:{B}\:=\alpha^{\mathrm{3}} \:+\beta^{\mathrm{3}} \:+\lambda^{\mathrm{3}} \:. \\ $$ Answered by…
Question Number 64065 by mathmax by abdo last updated on 12/Jul/19 $${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 64061 by mmkkmm000m last updated on 12/Jul/19 $$\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{25}/\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}} \\ $$ Answered by MJS last updated on 12/Jul/19 $$\mathrm{no}\:\mathrm{real}\:\mathrm{solutions} \\ $$$$\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\…
Question Number 64060 by mmkkmm000m last updated on 12/Jul/19 $$\left(\mathrm{2}{x}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{25}/\left({x}+\mathrm{3}\right)^{\mathrm{2}} =\sqrt{\mathrm{2}} \\ $$ Commented by MJS last updated on 12/Jul/19 $$\mathrm{Sir}\:\mathrm{would}\:\mathrm{you}\:\mathrm{mind}\:\mathrm{using}\:\mathrm{smaller}\:\mathrm{font}\:\mathrm{size}\:\mathrm{as}\:\mathrm{it}'\mathrm{s}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{overview}\:\mathrm{the}\:\mathrm{forum}\:\mathrm{like}\:\mathrm{this} \\ $$ Terms…
Question Number 129594 by bagjagunawan last updated on 16/Jan/21 Commented by bemath last updated on 17/Jan/21 $$\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}\:\mathrm{sin}\:\mathrm{x}}\:=\: \\ $$$$\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:+\frac{\mathrm{2cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left(\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}{\mathrm{2sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\mathrm{cos}\:\mathrm{x}}\:= \\ $$$$\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\left(\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}= \\ $$$$\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\:+\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{x}−\mathrm{1}+\mathrm{cos}\:\mathrm{x}}= \\ $$$$\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\:\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}−\mathrm{1}}=\frac{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}}…
Question Number 129595 by Adel last updated on 16/Jan/21 $$\mathrm{y}=\sqrt{\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{sin}\:\mathrm{x}+………….\infty}}} \\ $$$$ \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=? \\ $$ Answered by ajfour last updated on 16/Jan/21 $${y}^{\mathrm{2}} −\mathrm{sin}\:{x}={y}…
Question Number 129591 by mohammad17 last updated on 16/Jan/21 Answered by mr W last updated on 16/Jan/21 $${AC}=\left(\mathrm{2},\mathrm{4}\right) \\ $$$${BD}=\left(\mathrm{4},−\mathrm{2}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{{AC}\centerdot{BD}}{\mid{AC}\mid×\mid{BD}\mid}=\frac{\mathrm{2}×\mathrm{4}−\mathrm{4}×\mathrm{2}}{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\frac{\mathrm{0}}{\mathrm{20}}=\mathrm{0} \\…
Question Number 129587 by Adel last updated on 16/Jan/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}} =? \\ $$ Answered by Dwaipayan Shikari last updated on 16/Jan/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)^{{x}} =\underset{{x}\rightarrow\infty}…