Question Number 22729 by selestian last updated on 22/Oct/17 Commented by math solver last updated on 22/Oct/17 $$\mathrm{0} \\ $$ Commented by selestian last updated…
Question Number 22728 by selestian last updated on 22/Oct/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 88263 by M±th+et£s last updated on 09/Apr/20 $${prove}\:{that}\: \\ $$$$\mid\frac{{e}^{{z}} −{e}^{−{z}} }{\mathrm{2}}\mid^{\mathrm{2}} +{cos}^{\mathrm{2}} {y}={sinh}^{\mathrm{2}} {x}\:\:\:\:\:{when}\:{z}={x}+{iy} \\ $$$$ \\ $$ Commented by M±th+et£s last…
Question Number 22726 by math solver last updated on 22/Oct/17 Commented by math solver last updated on 24/Oct/17 $$\:{solve}\:{q}.\mathrm{8}? \\ $$ Commented by math solver…
Question Number 88261 by ar247 last updated on 09/Apr/20 Answered by Joel578 last updated on 09/Apr/20 $${u}\:\equiv\:{v}\:\left(\mathrm{mod}\:{m}\right)\:\Rightarrow\:{u}\:−\:{v}\:=\:{mx},\:{x}\:\in\:\mathbb{Z} \\ $$$$\left(\mathrm{1}\right)\:{u}\:=\:{mx}\:+\:{v} \\ $$$$\mathrm{Since}\:\mathrm{gcf}\left({v},{m}\right)\:\mathrm{divides}\:\mathrm{both}\:{v}\:\mathrm{and}\:\mathrm{m},\:\mathrm{it}\:\mathrm{also}\:\mathrm{divides}\:{u}, \\ $$$$\mathrm{hence}\:\mathrm{gcf}\left({v},{m}\right)\:\mathrm{divides}\:\mathrm{gcf}\left({u},{m}\right) \\ $$$$\Rightarrow\:\mathrm{gcf}\left({v},{m}\right)\:\mid\:\mathrm{gcf}\left({u},{m}\right)…
Question Number 88252 by A8;15: last updated on 09/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 88253 by john santu last updated on 09/Apr/20 $$\int\:\frac{\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:{dx}}{{x}+\mathrm{1}}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 153784 by saly last updated on 10/Sep/21 Commented by saly last updated on 10/Sep/21 $$\:\:\:\:\:\:\:\:{Do}\:\:{you}\:{help}\:{me} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 88251 by A8;15: last updated on 09/Apr/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22714 by selestian last updated on 22/Oct/17 Answered by ajfour last updated on 22/Oct/17 $$\lambda\:={a}+{b}+{c}\: \\ $$$${let}\:\:\:{l}={b}\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{C}}{\mathrm{2}}\right)+{c}\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{B}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{l}={b}\left(\mathrm{1}+\mathrm{cos}\:{C}\right)+{c}\left(\mathrm{1}+\mathrm{cos}\:{B}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{l}={b}+{c}+\left({b}\mathrm{cos}\:{C}+{c}\mathrm{cos}\:{B}\right)…