Question Number 173677 by naka3546 last updated on 16/Jul/22 $$\mathrm{Find}\:\:\mathrm{tangent}\:\:\mathrm{equation}\:\:\mathrm{of}\:\: \\ $$$$\mathrm{two}\:\:\mathrm{circles}\:: \\ $$$${L}_{\mathrm{1}} \::\:\left({x}−\mathrm{5}\right)^{\mathrm{2}} \:+\:\left({y}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{9} \\ $$$${L}_{\mathrm{2}} \::\:\left({x}+\mathrm{4}\right)^{\mathrm{2}} \:+\:\left({y}−\mathrm{11}\right)^{\mathrm{2}} \:=\:\mathrm{144} \\ $$ Commented…
Question Number 108143 by abony1303 last updated on 15/Aug/20 $$\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{cos}^{\mathrm{2}} \mathrm{2x}=\mathrm{sin}^{\mathrm{2}} \mathrm{3x} \\ $$$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{ASAP} \\ $$ Answered by ajfour last updated on…
Question Number 108131 by mathdave last updated on 14/Aug/20 Commented by Her_Majesty last updated on 14/Aug/20 $${just}\:{a}\:{try} \\ $$$${for}\:{x}\rightarrow+\infty\:{we}\:{have}\:\frac{{x}+\sqrt{{x}}}{{x}}=\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}\rightarrow\mathrm{1} \\ $$$$\Rightarrow\:\sqrt{{x}+\sqrt{{x}+\sqrt{{x}}}}\sim\sqrt{{x}+\mathrm{1}} \\ $$$$\Rightarrow\:{the}\:{limit}=\mathrm{1} \\ $$$${or}:\:\sqrt{{x}+\sqrt{{x}}}\sim\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{{x}}\:{when}\:{x}\rightarrow+\infty…
Question Number 173661 by aliibrahim1 last updated on 15/Jul/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 173652 by saly last updated on 15/Jul/22 Commented by saly last updated on 15/Jul/22 $$\:\:{Help}\:{me} \\ $$ Commented by JDamian last updated on…
Question Number 108118 by mathdave last updated on 14/Aug/20 Answered by abdomathmax last updated on 14/Aug/20 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{18}} \right)^{\frac{\mathrm{1}}{\mathrm{20}}} \mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{x}^{\mathrm{20}} \right)^{\frac{\mathrm{1}}{\mathrm{18}}} \mathrm{dx}\:=\mathrm{H}−\mathrm{K}…
Question Number 108114 by mathdave last updated on 14/Aug/20 Answered by mr W last updated on 14/Aug/20 $$\mathrm{sin}\:\left(\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{5}}−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{10}}\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{5}}×\frac{\mathrm{4}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{10}}−\frac{\mathrm{3}}{\mathrm{5}}×\frac{\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{3}}{\mathrm{10}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 108107 by otchereabdullai@gmail.com last updated on 14/Aug/20 Commented by rexfordattacudjoe last updated on 16/Aug/20 $${By}\:{similarities},\frac{{r}}{\mathrm{24}}=\frac{\mathrm{10}}{\mathrm{20}} \\ $$$${r}=\frac{\mathrm{10}×\mathrm{24}}{\mathrm{20}} \\ $$$${r}=\mathrm{12}{cm} \\ $$ Commented by…
Question Number 173645 by mokys last updated on 15/Jul/22 Commented by mr W last updated on 15/Jul/22 $$\frac{\mathrm{1}}{\mathrm{2}}×\pi{r}^{\mathrm{2}} {h}=\frac{\pi×\mathrm{6}^{\mathrm{2}} ×\mathrm{6}}{\mathrm{2}} \\ $$ Commented by mokys…
Question Number 42543 by solihin last updated on 27/Aug/18 Commented by solihin last updated on 27/Aug/18 $${question}\:\:\mathrm{1}\:\:{please} \\ $$ Commented by maxmathsup by imad last…