Question Number 200476 by Rupesh123 last updated on 19/Nov/23 Answered by AST last updated on 19/Nov/23 Commented by AST last updated on 19/Nov/23 $${WLOG},{let}\:{O}\:{be}\:{the}\:{origin}\:{and}\:{a}=\mathrm{1}\: \\…
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Question Number 200474 by Rupesh123 last updated on 19/Nov/23 Answered by witcher3 last updated on 19/Nov/23 $$\mathrm{erf}\left(\mathrm{x}\right)=\frac{\mathrm{2}}{\:\sqrt{\pi}}\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt} \\ $$$$\mathrm{ln}\left(\mathrm{x}+\mathrm{ln}\left(\mathrm{x}\right)\right)=\int_{\mathrm{0}} ^{\mathrm{5}} \mathrm{e}^{−\mathrm{t}^{\mathrm{2}}…
Question Number 200475 by Rupesh123 last updated on 19/Nov/23 Answered by AST last updated on 19/Nov/23 $$\frac{\mathrm{2}{sin}\mathrm{20}}{{PC}}=\frac{\mathrm{1}}{{AC}}\Rightarrow\frac{{PC}}{{AC}}=\mathrm{2}{sin}\mathrm{20} \\ $$$${Let}\:\angle{PBC}={x};\:\frac{{sinx}}{{PC}}=\frac{{cos}\mathrm{10}}{{BC}}\Rightarrow{BC}=\frac{{PCcos}\left(\mathrm{10}°\right)}{{sin}\left({x}\right)} \\ $$$$\frac{{sin}\left(\mathrm{20}+{x}\right)}{{AC}}=\frac{{sin}\mathrm{50}}{{BC}}=\frac{{sin}\mathrm{50}{sinx}}{{PCcos}\mathrm{10}}\Rightarrow\frac{{PC}}{{AC}}=\frac{{sin}\mathrm{50}{sinx}}{{cos}\left(\mathrm{10}\right){sin}\left(\mathrm{20}+{x}\right)} \\ $$$$\Rightarrow\mathrm{2}{sin}\mathrm{20}=\frac{{sin}\mathrm{50}{sinx}}{{cos}\mathrm{10}{sin}\left(\mathrm{20}+{x}\right)}\Rightarrow\frac{{sinx}}{{sin}\left(\mathrm{20}+{x}\right)}=\frac{\mathrm{2}{sin}\mathrm{20}{cos}\mathrm{10}}{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \mathrm{20}} \\…
Question Number 200533 by ajfour last updated on 19/Nov/23 Commented by mr W last updated on 20/Nov/23 $${AB}\bot{AD}\:{is}\:{given}? \\ $$$${otherwise}\:{there}\:{is}\:{no}\:{unique}\:{solution}. \\ $$ Commented by mr…
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Question Number 200471 by Mingma last updated on 19/Nov/23 Answered by AST last updated on 19/Nov/23 $$\frac{{CB}_{\mathrm{1}} }{{B}_{\mathrm{1}} {A}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{1}\Rightarrow\frac{{CB}_{\mathrm{1}} }{{B}_{\mathrm{1}} {A}}=\mathrm{8} \\ $$$$\frac{\left[{CAB}\right]}{\left[{ADB}\right]}=\frac{{CC}_{\mathrm{1}} }{{DC}_{\mathrm{1}} }=\frac{{CD}}{{DC}_{\mathrm{1}}…
Question Number 200464 by Frix last updated on 19/Nov/23 $$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2023}} \:{x}}=??????? \\ $$ Answered by som(math1967) last updated on 19/Nov/23 $$\:{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{cos}^{\mathrm{2023}}…
Question Number 200465 by faysal last updated on 19/Nov/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fifth}\:\mathrm{powers}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} −\mathrm{2x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}=\mathrm{0}\:\mathrm{by} \\ $$$$\mathrm{applying}\:\mathrm{synthetic}\:\mathrm{division} \\ $$ Commented by mr W last updated on…
Question Number 200531 by cherokeesay last updated on 19/Nov/23 Answered by witcher3 last updated on 21/Nov/23 $$\mathrm{witch}\:\mathrm{Quation}? \\ $$ Terms of Service Privacy Policy Contact:…