Question Number 186002 by Shrinava last updated on 30/Jan/23 Commented by JDamian last updated on 30/Jan/23 why did you re-post the same question? Answered by mr W last updated on 30/Jan/23…
Question Number 54923 by mr W last updated on 14/Feb/19 $${Is}\:\mathrm{tan}\:\mathrm{1}°\:{rational}\:{or}\:{irrational}? \\ $$$${Give}\:{your}\:{proof}. \\ $$ Commented by Otchere Abdullai last updated on 14/Feb/19 $${tan}\mathrm{1}°\:=\mathrm{0}.\mathrm{01745506493} \\…
Question Number 120455 by udaythool last updated on 31/Oct/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185983 by mnjuly1970 last updated on 30/Jan/23 $$ \\ $$$$\:\:\:\:\:\begin{cases}{\:\:\:{f}\::\:\:\left[\:\:\mathrm{0}\:\:,\:\:\mathrm{1}\:\right]\:\rightarrow\:\mathbb{R}}\\{\:\:\:\:{f}\:\left({x}\:\right)\:=\:\frac{\:\mathrm{4}^{\:{x}} }{\mathrm{2}\:+\:\mathrm{4}^{\:{x}} }}\end{cases} \\ $$$$\:\:\:\:\:\:{is}\:\:{given}\:.\:\:{find}\:{the}\:{value}\:{of} \\ $$$$\:\:\:\:\:\:{the}\:{following}\:{expression}. \\ $$$$\:\:\:\:\:\:\mathrm{E}\:=\:{f}\:\left(\frac{\mathrm{1}}{\mathrm{20}}\:\right)+\:{f}\left(\frac{\:\mathrm{2}}{\mathrm{20}}\:\right)+…\:+{f}\:\left(\frac{\mathrm{19}}{\mathrm{20}}\right)\:−{f}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right)=? \\ $$ Answered by ARUNG_Brandon_MBU…
Question Number 185981 by mnjuly1970 last updated on 30/Jan/23 $$ \\ $$$$\:\:\:\:\:\:\:{solve}\:\:{in}\:\:\:\mathbb{R} \\ $$$$\: \\ $$$$\:\:\:\:\:\:\:\:\lfloor\:\:\mathrm{2}{x}\:\rfloor\:\:+\:\lfloor\:\:\mathrm{3}{x}\:\rfloor\:=\:\mathrm{1} \\ $$$$\: \\ $$$$\:\:\:\:\lfloor\:{x}\:\rfloor\::=\:{greatest}\:{integer}\:{number} \\ $$$$\:\:\:\:{more}\:{than}\:\:{or}\:{equal}\:\:{to}\:\:''\:{x}\:'' \\ $$ Answered…
Question Number 185982 by Shrinava last updated on 30/Jan/23 Answered by Mathspace last updated on 30/Jan/23 $${we}\:{developp}\:{first}\:\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} {C}_{\mathrm{2}{n}} ^{{k}} {x}^{{k}} \:\:{but} \\…
Question Number 185973 by normans last updated on 30/Jan/23 $$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\left[\boldsymbol{{prove}}\:\boldsymbol{{that}};\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\mathrm{2}\:=\:\mathrm{3} \\ $$$$ \\ $$ Commented by MJS_new last updated on 30/Jan/23…
Question Number 185972 by Michaelfaraday last updated on 30/Jan/23 Answered by Rasheed.Sindhi last updated on 30/Jan/23 $${x}=\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\:,\:{y}=\sqrt{{p}}\:+\frac{\mathrm{1}}{\:\sqrt{{p}}\:} \\ $$$${Show}\:{that}\:\:\:{y}^{\mathrm{2}} −\mathrm{2}={x}\left({x}^{\mathrm{2}} −\mathrm{3}\right) \\ $$$${x}^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\right)^{\mathrm{3}} ={p}+\frac{\mathrm{1}}{{p}}+\mathrm{3}\left(\sqrt[{\mathrm{3}}]{{p}}\:+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{{p}}\:}\right)…
Question Number 185939 by Rupesh123 last updated on 30/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54857 by Tawa1 last updated on 13/Feb/19 Answered by tanmay.chaudhury50@gmail.com last updated on 13/Feb/19 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\lambda{y}=\mathrm{0} \\ $$$${y}={e}^{{mx}} \\ $$$${m}^{\mathrm{2}} {e}^{{mx}} +\lambda{e}^{{mx}}…