Question Number 136734 by mohammad17 last updated on 25/Mar/21 Answered by Ñï= last updated on 25/Mar/21 $${Q}\mathrm{5}::: \\ $$$${y}_{{p}} =\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{3}{D}+\mathrm{12}}\left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{cos}\:{x}\right) \\ $$$$={e}^{{x}} \frac{\mathrm{1}}{{D}^{\mathrm{2}}…
Question Number 136724 by mohammad17 last updated on 25/Mar/21 Commented by mohammad17 last updated on 25/Mar/21 $${please}\:{sir}\:{help}\:{me} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 136719 by mohammad17 last updated on 25/Mar/21 Commented by mohammad17 last updated on 25/Mar/21 $${how}\:{can}\:{solve}\:{this}\:{help}\:{me}\:{sir} \\ $$ Commented by mr W last updated…
Question Number 5629 by LMTV last updated on 23/May/16 $$\mathscr{L}\left(\mathrm{sin}\:^{\mathrm{2}} {t}\right)=? \\ $$ Commented by FilupSmith last updated on 23/May/16 $$\mathrm{Laplace}\:\mathrm{transform}? \\ $$ Commented by…
Question Number 71149 by naka3546 last updated on 12/Oct/19 Commented by MJS last updated on 17/Oct/19 $$\mathrm{if}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$${BE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axis} \\ $$$${M}\in\left({BE}\right) \\ $$$${A}={C}'\:\wedge\:{F}={D}' \\ $$$$\Rightarrow\:\left({AF}\right)=\left({CD}\right)'\:\mathrm{and}\:\mathrm{similar}\:\mathrm{for}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{lines}…
Question Number 71141 by Mr. K last updated on 12/Oct/19 $$\left(\mathrm{2}{x}\right)^{{x}} =\frac{\mathrm{1}}{\mathrm{16}} \\ $$ Commented by mr W last updated on 12/Oct/19 $$\left(\mathrm{2}{x}\right)^{{x}} =\sqrt{\left(\mathrm{2}{x}\right)^{\mathrm{2}{x}} }=\sqrt{{t}^{{t}}…
Question Number 71134 by Mr. K last updated on 11/Oct/19 $${if}\:{a},\:{b}\:{and}\:{c}\:{are}\:{positive},\:{find}: \\ $$$$\left(\frac{{a}}{{b}}\right)^{{log}_{\mathrm{10}} {c}} ×\left(\frac{{b}}{{c}}\right)^{{log}_{\mathrm{10}} {a}} ×\left(\frac{{c}}{{a}}\right)^{{log}_{\mathrm{10}} {b}} \\ $$$$ \\ $$ Answered by MJS…
Question Number 136656 by 0731619177 last updated on 24/Mar/21 Answered by Ñï= last updated on 24/Mar/21 $${y}={x}^{{x}^{{x}^{…} } } ={x}^{{y}} \\ $$$${y}'=\left({e}^{{ylnx}} \right)'={x}^{{y}} \left({y}'{lnx}+\frac{{y}}{{x}}\right)={x}^{{y}} {y}'{lnx}+{yx}^{{y}−\mathrm{1}}…
Question Number 136643 by Ñï= last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{sin}^{−\mathrm{1}} \sqrt{{x}}}{\:\sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} }}{dx}=\frac{\pi}{\mathrm{4}}{ln}\mathrm{3} \\ $$ Answered by Ñï= last updated on 24/Mar/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 136635 by mohammad17 last updated on 24/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com