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Question Number 139402 by mathmax by abdo last updated on 26/Apr/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{2}} \mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{x}+\mathrm{1}}\mathrm{dx} \\ $$ Answered by mathmax by abdo last updated…
Question Number 139399 by mathsuji last updated on 26/Apr/21 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{log}\left({sin}\left({x}/\mathrm{2}\right)\right)+{log}\left({cos}\left({x}/\mathrm{2}\right)\right)}{\:\sqrt{\mathrm{2}}\centerdot{cos}\left(\pi/\mathrm{4}−{x}/\mathrm{2}\right)}\:{dx} \\ $$ Answered by Ar Brandon last updated on 26/Apr/21 $$\mathrm{Let}\:{u}=\frac{\pi}{\mathrm{2}}−{x} \\ $$…
Question Number 8325 by lepan last updated on 08/Oct/16 $${In}\:\Delta\:{ABC},\angle{A}=\mathrm{90}°\:,\:{AD}\bot{BC},\:{DE}\bot{AC}, \\ $$$${AF}\bot{FG},{GH}\bot{FC}. \\ $$$$\left({a}\right){How}\:{many}\:{triangles}\:{are}\:{there}? \\ $$$$\left({b}\right){If}\:{AB}=\frac{\mathrm{16}}{\mathrm{9}}\:,\:\angle{B}=\mathrm{60}°,{find}\:{the}\:{length} \\ $$$$\:\:\:\:\:\:{of}\:{GH}. \\ $$ Terms of Service Privacy Policy…
Question Number 139393 by BHOOPENDRA last updated on 26/Apr/21 Answered by Ar Brandon last updated on 26/Apr/21 $$\mathrm{I}=\int\frac{\mathrm{cos}{x}}{\mathrm{2cos}{x}+\mathrm{sin}{x}+\mathrm{3}}{dx},\:{t}=\mathrm{tan}\frac{{x}}{\mathrm{2}}\Rightarrow{dt}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}}{dx} \\ $$$$\:\:=\int\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 139394 by EnterUsername last updated on 26/Apr/21 $$\mathrm{If}\:\mid{z}−{i}\mid\leqslant\mathrm{2}\:\mathrm{and}\:{z}_{\mathrm{0}} =\mathrm{5}+\mathrm{3}{i},\:\mathrm{then}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mid{z}_{\mathrm{0}} +{iz}\mid\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{7}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{7}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{5}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{9} \\ $$ Answered by mr W last updated on…
Question Number 139388 by mnjuly1970 last updated on 26/Apr/21 Answered by mr W last updated on 26/Apr/21 Commented by mr W last updated on 26/Apr/21…
Question Number 139390 by abenarhodym last updated on 26/Apr/21 $${a}\:{uniform}\:{pole}\:\mathrm{30}{m}\:{and}\:{weight}\:\mathrm{40}{kg}\:{is}\:{carried}\:{at}\:{P}\:\:{by}\:{John}\:{and}\:{Ama}\:\mathrm{8}{m}\:{away}\:{from}\:{Q}.\:{Find}\:{the}\:{distance}\:{from}\:{P}\:{where}\:{a}\:{mass}\:{of}\:\mathrm{20}{kg}\:{should}\:{be}\:{attached}\:{so}\:{that}\:{Amas}\:{support}\:{is}\:{twice}\:{that}\:{of}\:{John}\:{if}\:{tbe}\:{system}\:{is}\:{in}\:{equilibrim} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 8314 by tawakalitu last updated on 07/Oct/16 Commented by Tinku Tara last updated on 08/Oct/16 $$\mathrm{For}\:\mathrm{othercases}\:\mathrm{can}\:\mathrm{u}\:\mathrm{please}\:\mathrm{email} \\ $$$$\mathrm{picture}\:\mathrm{to}\:\mathrm{us}\:\mathrm{at}\:\mathrm{infoattinkutara}.\mathrm{com} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{will}\:\mathrm{troubleshoot}\:\mathrm{the}\:\mathrm{issue}. \\ $$$$\mathrm{Alternatively}\:\mathrm{try}\:\mathrm{to}\:\mathrm{post}\:\mathrm{image}\:\mathrm{as}\:\mathrm{a}\:\mathrm{new} \\…
Question Number 8311 by lepan last updated on 07/Oct/16 $${Solve}\:{the}\:{equation}\:\mathrm{6}{cos}\mathrm{2}{a}−\mathrm{5}{sin}\mathrm{2}{a}=\mathrm{1}.\mathrm{8} \\ $$$${for}\mathrm{0}°\leqslant{a}\leqslant\mathrm{180}°. \\ $$ Commented by 123456 last updated on 07/Oct/16 $$\mathrm{6cos2}{a}−\mathrm{5sin2}{a}=\mathrm{1}.\mathrm{8} \\ $$$$\mathrm{cos}^{\mathrm{2}} \mathrm{2}{a}+\mathrm{sin}^{\mathrm{2}}…