Question Number 69075 by Aditya789 last updated on 19/Sep/19 Commented by Prithwish sen last updated on 19/Sep/19 $$\mathrm{Divide}\:\mathrm{both}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator} \\ $$$$\mathrm{by}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1} \\ $$ Commented by…
Question Number 69072 by TawaTawa last updated on 19/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 69073 by necxxx last updated on 19/Sep/19 $${A}\:{man}\:{wants}\:{to}\:{shear}\:\mathrm{17}\:{cars}\:{between} \\ $$$${his}\:\mathrm{3}\:{children}\:{in}\:{the}\:{ratio}\:\mathrm{1}:\mathrm{2},\:\mathrm{1}:\mathrm{3},\:\mathrm{1}:\mathrm{9} \\ $$$${respectively}.{How}\:{will}\:{he}\:{go}\:{about}\:{it}? \\ $$ Commented by necxxx last updated on 19/Sep/19 $${please}\:{help} \\…
Question Number 3536 by Syaka last updated on 14/Dec/15 $${prove} \\ $$$$\frac{\mathrm{1}\:+\:{cos}\:{x}\:−\:{cos}\:{y}\:+\:{cos}\:{z}}{\mathrm{1}\:+\:{cos}\:{x}\:+\:{cos}\:{y}\:−\:{cos}\:{z}}\:=\:{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{z}\:.\:{tan}\:\frac{\mathrm{1}}{\mathrm{2}}{y} \\ $$$$ \\ $$ Commented by Yozzii last updated on 14/Dec/15 $${x}={y}=\mathrm{0}\:,\:{z}=\pi/\mathrm{2} \\…
Question Number 134605 by Khalmohmmad last updated on 05/Mar/21 Answered by Ñï= last updated on 05/Mar/21 $${f}\left(\mathrm{2}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{2}} {\sum}}{k}!=\mathrm{1}!+\mathrm{2}!=\mathrm{3} \\ $$$${g}\left({f}\left(\mathrm{2}\right)\right)={g}\left(\mathrm{3}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left({k}+\mathrm{1}\right)=\mathrm{2}+\mathrm{3}+\mathrm{4}=\mathrm{9} \\ $$…
Question Number 134604 by mohammad17 last updated on 05/Mar/21 Commented by mohammad17 last updated on 05/Mar/21 $${how}\:{can}\:{solve}\:{this} \\ $$ Commented by mohammad17 last updated on…
Question Number 134601 by abdullahquwatan last updated on 05/Mar/21 $$\int\mathrm{3}\left(\sqrt{\mathrm{x}\:+\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{x}}}}\right)\:\mathrm{dx} \\ $$ Answered by bobhans last updated on 06/Mar/21 $$\sqrt{\frac{\mathrm{x}\sqrt{\mathrm{x}}+\mathrm{4}}{\:\sqrt{\mathrm{x}}}}\:=\:\frac{\sqrt{\mathrm{x}\sqrt{\mathrm{x}}+\mathrm{4}}}{\:\sqrt[{\mathrm{4}}]{\mathrm{x}}}\: \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{t}^{\mathrm{4}} \:\Rightarrow\mathrm{dx}\:=\:\mathrm{4t}^{\mathrm{3}} \:\mathrm{dt} \\…
Question Number 69064 by Sayantan chakraborty last updated on 18/Sep/19 Commented by Sayantan chakraborty last updated on 18/Sep/19 $$\mathrm{PLEASE}\:\mathrm{HELP} \\ $$ Commented by Sayantan chakraborty…
Question Number 134596 by mohammad17 last updated on 05/Mar/21 Commented by mohammad17 last updated on 05/Mar/21 $${help}\:{me}\:{sir} \\ $$ Answered by Olaf last updated on…
Question Number 134599 by mohammad17 last updated on 05/Mar/21 Answered by Olaf last updated on 05/Mar/21 $$\left.{a}\right) \\ $$$$\int_{\mathrm{C}} \mid{z}\mid^{\mathrm{2}} {dz}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \mid{re}^{{i}\frac{\pi}{\mathrm{4}}} \mid^{\mathrm{2}} {dr}\:=\:\left[\frac{{r}^{\mathrm{3}}…