Question Number 18748 by khamizan833@yahoo.com last updated on 29/Jul/17 Answered by Joel577 last updated on 29/Jul/17 $${f}\left({x}\right)\:=\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{3}} \:−\:\mathrm{6}{x} \\ $$$${f}\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)\:=\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)^{\mathrm{3}} \:−\:\mathrm{6}\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{2}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{16}}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{32}}\:+\:\mathrm{4}\right)\:−\:\left(\mathrm{6}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{6}\:−\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}…
Question Number 149817 by ajfour last updated on 07/Aug/21 $$\:\:\:{x}^{\mathrm{3}} −{x}={c} \\ $$$$\:\:{let}\:\:{x}={t}+{h} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{t}\left({t}^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={p} \\ $$$$\mathrm{3}{ht}^{\mathrm{2}}…
Question Number 18747 by khamizan833@yahoo.com last updated on 29/Jul/17 Answered by Joel577 last updated on 29/Jul/17 $$\left(−\mathrm{2}^{−\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:=\:\left(−\mathrm{2}\right)^{−\mathrm{6}} \:=\:\mathrm{0},\mathrm{015625} \\ $$ Commented by Joel577…
Question Number 84283 by jagoll last updated on 11/Mar/20 $$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{2}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{x}−\mathrm{a}}\:=\:\mathrm{k}\:,\:\mathrm{a}>\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{a}} {\mathrm{lim}}\:\frac{\left(\mid\mathrm{x}\mid−\mathrm{1}\right)^{\mathrm{3}} −\left(\mid\mathrm{a}\mid−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }\:=\: \\ $$$$ \\ $$ Answered…
Question Number 18746 by khamizan833@yahoo.com last updated on 29/Jul/17 Answered by daffa22 last updated on 30/Jul/17 $$\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{q}}\right)+{p}}+\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{r}}\right)+{q}}+\frac{\mathrm{1}}{\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)+{r}} \\ $$$${misal}, \\ $$$$\bullet\left(\mathrm{1}+\frac{\mathrm{1}}{{q}}\right)={a} \\ $$$$\bullet\left(\mathrm{1}+\frac{\mathrm{1}}{{r}}\right)={b} \\ $$$$\bullet\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)={c}…
Question Number 18744 by aplus last updated on 29/Jul/17 Commented by aplus last updated on 29/Jul/17 $$\mathrm{help} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 149813 by mathdanisur last updated on 07/Aug/21 $$\int\left(−\mathrm{1}\right)^{\left[\boldsymbol{\mathrm{x}}\right]} \:\mathrm{dx}\:=\:? \\ $$ Commented by DonQuichote last updated on 07/Aug/21 $${deal}:\:{you}\:{show}\:{me}\:{your}\:{solution}\:{idea}\:{first} \\ $$$${then}\:{I}'{ll}\:{show}\:{you}\:{mine} \\ $$…
Question Number 18742 by Tinkutara last updated on 31/Jul/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{cot16}°\mathrm{cot44}°\:+\:\mathrm{cot44}°\mathrm{cot76}° \\ $$$$−\:\mathrm{cot76}°\mathrm{cot16}°\:\mathrm{is} \\ $$ Answered by Tinkutara last updated on 31/Jul/17 $$\frac{\mathrm{3}\:+\:\mathrm{cot}\:\mathrm{76}°\:\mathrm{cot}\:\mathrm{16}°}{\mathrm{cot}\:\mathrm{76}°\:+\:\mathrm{cot}\:\mathrm{16}°} \\ $$$$=\:\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{76}°\:\mathrm{sin}\:\mathrm{16}°\:+\:\mathrm{cos}\:\mathrm{76}°\:\mathrm{cos}\:\mathrm{16}°}{\mathrm{sin}\:\left(\mathrm{76}°\:+\:\mathrm{16}°\right)} \\…
Question Number 149809 by 0731619 last updated on 07/Aug/21 Answered by Ar Brandon last updated on 07/Aug/21 $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }} \mathrm{cos}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy}…
Question Number 84275 by john santu last updated on 11/Mar/20 $$\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{b}}−\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{a}}×\left(\boldsymbol{\mathrm{c}}×\boldsymbol{\mathrm{b}}\right)\:? \\ $$ Commented by mr W last updated on 11/Mar/20 $$−\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{b}}+\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{a}}×\left(\boldsymbol{\mathrm{c}}×\boldsymbol{\mathrm{b}}\right) \\ $$$$\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{b}}−\left(\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{c}}\:=\:\boldsymbol{\mathrm{a}}×\left(\boldsymbol{{b}}×\boldsymbol{{c}}\right) \\…