Question Number 199891 by Mathspace last updated on 10/Nov/23 $${solve}\:{by}\:{laplce}\:{transform} \\ $$$${y}^{''} −{y}^{'} +{y}\:=\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$ Commented by Mathspace last updated on 10/Nov/23 $${with}\:{y}\left(\mathrm{0}\right)=\mathrm{1}\:{and}\:{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1}…
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Question Number 199774 by Mingma last updated on 09/Nov/23 Answered by ajfour last updated on 09/Nov/23 Commented by ajfour last updated on 09/Nov/23 $${P}\:\left({perimeter}\:{of}\:\bigtriangleup\:{ABC}\right) \\…
Question Number 199775 by Mingma last updated on 09/Nov/23 Answered by mathfreak01 last updated on 09/Nov/23 $$ \\ $$$$ \\ $$Assuming when have a triangle…
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Question Number 199801 by Calculusboy last updated on 09/Nov/23 Answered by Frix last updated on 09/Nov/23 $${x}={a}\sqrt{{b}−{a}\sqrt{{b}−{a}\sqrt{{b}−…}}} \\ $$$${x}={a}\sqrt{{b}−{x}} \\ $$$${x}^{\mathrm{2}} ={a}^{\mathrm{2}} \left({b}−{x}\right) \\ $$$${x}^{\mathrm{2}}…
Question Number 199769 by linlu last updated on 09/Nov/23 $$\mathrm{62}{n}\mathrm{7}{z}\mathrm{7} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199770 by Rupesh123 last updated on 09/Nov/23 Answered by aleks041103 last updated on 09/Nov/23 $${C}=\pi−\mathrm{3}\alpha \\ $$$$\Rightarrow\frac{\mathrm{5}}{{sin}\left(\mathrm{2}\alpha\right)}=\frac{\mathrm{6}}{{sin}\left(\pi−\mathrm{3}\alpha\right)}=\frac{\mathrm{6}}{{sin}\left(\mathrm{3}\alpha\right)} \\ $$$$\Rightarrow\mathrm{6}{sin}\left(\mathrm{2}\alpha\right)=\mathrm{5}{sin}\left(\mathrm{3}\alpha\right) \\ $$$${sin}\left(\mathrm{2}\alpha\right)=\mathrm{2}{sin}\left(\alpha\right){cos}\left(\alpha\right) \\ $$$${sin}\left(\mathrm{3}\alpha\right)={sin}\left(\mathrm{2}\alpha\right){cos}\left(\alpha\right)+{sin}\left(\alpha\right){cos}\left(\mathrm{2}\alpha\right)=…
Question Number 199771 by Rupesh123 last updated on 09/Nov/23 Answered by AST last updated on 09/Nov/23 $$\frac{{sin}\left(\mathrm{4}{x}\right)=\mathrm{2}{sin}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{2}{x}\right)}{{a}}=\frac{{sin}\left(\mathrm{2}{x}\right)}{{b}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{x}\right)=\frac{{a}}{\mathrm{2}{b}}=\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} {x}=\frac{{a}+\mathrm{2}{b}}{\mathrm{4}{b}} \\ $$$$\frac{{sin}\left(\mathrm{2}{x}\right)=\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right)}{{b}}=\frac{{sin}\left({x}\right)}{{c}}\Rightarrow{cos}\left({x}\right)=\frac{{b}}{\mathrm{2}{c}} \\ $$$$\Rightarrow\frac{{a}+\mathrm{2}{b}}{{b}}=\frac{{b}^{\mathrm{2}}…