Question Number 134987 by 0731619177 last updated on 09/Mar/21 Answered by Olaf last updated on 09/Mar/21 $$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\mathrm{sin}{x}\mathrm{cos}{x}}{\:\sqrt{\mathrm{1}−{k}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}}\:{dx} \\ $$$$\mathrm{I}_{\mathrm{2}} \left({k}\right)\:=\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}}…
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Question Number 69436 by azizullah last updated on 23/Sep/19 Answered by $@ty@m123 last updated on 23/Sep/19 $$\left({x}+\mathrm{4}\right)=\mathrm{2}\left({x}−\mathrm{6}\right) \\ $$$${x}=\mathrm{16} \\ $$ Commented by azizullah last…
Question Number 134977 by 0731619177 last updated on 09/Mar/21 Answered by Ñï= last updated on 09/Mar/21 $${tanhx}=\frac{{x}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}+\frac{{x}^{\mathrm{2}} }{\mathrm{5}+\frac{{x}^{\mathrm{2}} }{\mathrm{7}+…}}}} \\ $$$$\int_{{a}} ^{{b}} {tanhxdx}={ln}\left(\mathrm{cosh}\:{x}\right)\mid_{{a}} ^{{b}}…
Question Number 69431 by azizullah last updated on 23/Sep/19 Answered by Kunal12588 last updated on 23/Sep/19 $$\mathrm{120}={x}+{y}…\left(\mathrm{1}\right) \\ $$$${let}\:{x}>{y} \\ $$$$\mathrm{3}{y}=\mathrm{2}{x}…\left(\mathrm{2}\right) \\ $$$${from}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{240}=\mathrm{2}{x}+\mathrm{2}{y}…
Question Number 69433 by azizullah last updated on 23/Sep/19 Answered by Kunal12588 last updated on 23/Sep/19 $$\mathrm{630}−\mathrm{7}{x}=\mathrm{3}{x} \\ $$$$\Rightarrow{x}=\mathrm{63} \\ $$ Commented by azizullah last…
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Question Number 134934 by Ahmed1hamouda last updated on 08/Mar/21 Commented by mr W last updated on 08/Mar/21 $$\pi{a}^{\mathrm{2}} \\ $$ Commented by mr W last…
Question Number 134928 by Abdoulaye last updated on 08/Mar/21 $$ \\ $$$$\mathrm{f}\:\mathrm{derivable}\:\mathrm{in}\:\mathrm{0}\:\mathrm{and}\:\mathrm{f}\:\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${determined}\:\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{f}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$$ \\ $$$${help}\:{me} \\ $$ Terms…