Question Number 200589 by Ikbal last updated on 20/Nov/23 Answered by Frix last updated on 20/Nov/23 $$\mathrm{First}\:\mathrm{try}\:\mathrm{factors}\:\mathrm{of}\:\pm\mathrm{40}\:\Rightarrow \\ $$$${x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{42}{x}−\mathrm{40}= \\ $$$$=\left({x}+\mathrm{1}\right)\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{10}\right) \\…
Question Number 200586 by Calculusboy last updated on 20/Nov/23 Answered by Frix last updated on 21/Nov/23 $$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{2}+\mathrm{tan}^{\mathrm{2}} \:{x}}{dx}=\underset{\mathrm{0}} {\overset{\pi} {\int}}{xdx}−\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{x}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}…
Question Number 200579 by sonukgindia last updated on 20/Nov/23 Commented by mr W last updated on 20/Nov/23 $$\Rightarrow{Q}\mathrm{200417} \\ $$ Terms of Service Privacy Policy…
Question Number 200498 by mr W last updated on 19/Nov/23 $${solve}\:{for}\:{x}\in{R} \\ $$$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}+\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}={x} \\ $$ Commented by BaliramKumar last updated on 19/Nov/23 $$\mathrm{Q}.\:\mathrm{192958} \\ $$…
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…
Question Number 200534 by obia last updated on 19/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com