Question Number 21232 by mondodotto@gmail.com last updated on 17/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152299 by saly last updated on 27/Aug/21 Commented by saly last updated on 27/Aug/21 $$\:\:\: \\ $$$$\:{Do}\:{you}\:\:{help}\:{me}\:? \\ $$$$ \\ $$ Commented by…
Question Number 21210 by mondodotto@gmail.com last updated on 16/Sep/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 152247 by mathocean1 last updated on 26/Aug/21 $${show}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:−\mathrm{1}}{{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mr W last updated on 26/Aug/21 $$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+……
Question Number 152241 by ZiYangLee last updated on 26/Aug/21 Answered by Olaf_Thorendsen last updated on 26/Aug/21 $$\left({i}\right)\:\mathrm{and}\:\left({ii}\right)\:: \\ $$$${p}\left({x}\right)\:=\:{a}\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{px}+{q}\right),\:{a}\neq\mathrm{0} \\ $$$$\left({iii}\right)\:: \\ $$$${p}\left(\mathrm{0}\right)\:=\:\mathrm{4}\:\Leftrightarrow\:−{aq}\:=\:\mathrm{4}\:\Rightarrow\:{q}\:=\:−\frac{\mathrm{4}}{{a}} \\…
Question Number 152208 by SOMEDAVONG last updated on 26/Aug/21 $$\mathrm{1}.\mathrm{for}\:\forall\mathrm{x}>\mathrm{0}.\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{m}\:\mathrm{to}\: \\ $$$$\mathrm{1}+\mathrm{log}_{\mathrm{5}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)\geqslant\mathrm{log}_{\mathrm{5}} \left(\mathrm{mx}^{\mathrm{2}} +\mathrm{4x}+\mathrm{m}\right)\:\mathrm{verify}\:\forall\mathrm{x}. \\ $$ Answered by Rasheed.Sindhi last updated on 26/Aug/21…
Question Number 152197 by abdurehime last updated on 26/Aug/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 21116 by mondodotto@gmail.com last updated on 13/Sep/17 Answered by Tinkutara last updated on 13/Sep/17 $$\mathrm{tan}\:\mathrm{A}=\frac{\mathrm{3}}{\mathrm{4}}\Rightarrow\mathrm{sin}\:\mathrm{A}=\frac{\mathrm{3}}{\mathrm{5}},\mathrm{cos}\:\mathrm{A}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\mathrm{A}=\mathrm{2cos}^{\mathrm{2}} \:\frac{\mathrm{A}}{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{9}}{\mathrm{10}}\Rightarrow\mathrm{cos}\:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$$\therefore\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}…
Question Number 86634 by liki last updated on 29/Mar/20 Answered by MJS last updated on 29/Mar/20 $$\int\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{tan}\:\theta\:−\mathrm{1}}{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\theta\:\rightarrow\:{d}\theta=\mathrm{cos}^{\mathrm{2}} \:\theta\:{dt}\right] \\ $$$$=\int\frac{{dt}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=…
Question Number 21038 by ajfour last updated on 10/Sep/17 Commented by ajfour last updated on 10/Sep/17 $${Find}\:{angular}\:{velocity}\:{and}\:{angular} \\ $$$${acceleration}\:{of}\:{rod}\:{if}\:{end}\:{A}\:{moves} \\ $$$${to}\:{the}\:{right}\:{with}\:{constant}\:{velocity} \\ $$$${v}. \\ $$…