Question Number 196035 by sonukgindia last updated on 16/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195996 by sonukgindia last updated on 15/Aug/23 Answered by Rasheed.Sindhi last updated on 15/Aug/23 $${Let}\:{radius}\:{of}\:{blue}\:{quarter}={r} \\ $$$${and}\:{of}\:{green}={R} \\ $$$${R}={r}+\mathrm{1} \\ $$$${Diagonal}={R}+{r} \\ $$$$\left({R}+{r}\right)^{\mathrm{2}}…
Question Number 196014 by Rodier97 last updated on 16/Aug/23 $$\:\:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{t}.\sqrt{{tan}\left({t}\right)}\:{dt}\:\:=\:??? \\ $$ Commented by Frix last updated on 15/Aug/23 $$\mathrm{Question}\:\mathrm{195895}\:!!! \\ $$ Commented…
Question Number 196007 by sonukgindia last updated on 15/Aug/23 Answered by mr W last updated on 15/Aug/23 $${AC}=\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\overset{\frown} {{BC}}=\left(\pi−\alpha\right){R} \\ $$$$\left(\pi−\alpha\right){R}=\frac{{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}} \\ $$$$\left(\pi−\alpha\right)\:\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\mathrm{1}…
Question Number 196000 by Rodier97 last updated on 15/Aug/23 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{lim}_{{x}\rightarrow+\infty} \:\left({lnx}\right)^{\mathrm{2}} −\sqrt{{x}}\:\:\:\:\:???? \\ $$$$ \\ $$$$ \\…
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Question Number 195953 by sonukgindia last updated on 13/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195954 by sonukgindia last updated on 13/Aug/23 Answered by mahdipoor last updated on 13/Aug/23 $$\frac{\mathrm{1}}{\mathrm{4}}×\frac{\left(\mathrm{4}−{a}\right)−{x}}{{x}+\left({a}−\mathrm{2}\right)}×\frac{\left(\mathrm{4}−{a}\right)+{x}}{−{x}+\left({a}−\mathrm{2}\right)}={c} \\ $$$${c}=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:\begin{cases}{\mathrm{4}−{a}=\mathrm{2}−{a}\Rightarrow\nexists}\\{\mathrm{4}−{a}={a}−\mathrm{2}\Rightarrow{a}=\mathrm{3}}\end{cases} \\ $$ Answered by Rasheed.Sindhi last…
Question Number 195944 by sonukgindia last updated on 13/Aug/23 Answered by AST last updated on 13/Aug/23 $${a}_{\mathrm{2}} −{a}_{\mathrm{1}} ={a}_{\mathrm{3}} −{a}_{\mathrm{2}} \Rightarrow{xy}−\frac{{x}}{{y}}={x}+{y}−{xy} \\ $$$${a}_{\mathrm{4}} −{a}_{\mathrm{3}} ={a}_{\mathrm{3}}…
Question Number 195900 by Calculusboy last updated on 12/Aug/23 Answered by cortano12 last updated on 13/Aug/23 $$\:\:\:\left[\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:\right]_{−\mathrm{2}} ^{\mathrm{b}} =\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\:\:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}+\mathrm{1}\right)\:\right]_{−\mathrm{2}} ^{\mathrm{b}}…