Question Number 195838 by sonukgindia last updated on 11/Aug/23 Answered by sniper237 last updated on 11/Aug/23 $${Area}=\int_{{u}} ^{{v}} {f}\left({x}\right){dx}\overset{{t}={lnx}} {=}\int_{{a}} ^{{b}} \:\frac{{dt}}{\:\sqrt{\left({t}−{a}\right)\left({b}−{t}\right)}}\: \\ $$$$\:\:\:\overset{{s}=\frac{{a}+{b}}{\mathrm{2}}−{t}} {=}\int_{−{c}}…
Question Number 195806 by Mingma last updated on 11/Aug/23 Answered by som(math1967) last updated on 11/Aug/23 Commented by som(math1967) last updated on 11/Aug/23 $$\mathrm{13}^{\mathrm{2}} −{a}^{\mathrm{2}}…
Question Number 195833 by liuxinnan last updated on 11/Aug/23 $${how}\:{does}\:\:{the}\:{d}\:{get}\:{max} \\ $$ Commented by liuxinnan last updated on 11/Aug/23 Answered by liuxinnan last updated on…
Question Number 195860 by zayyadyusuf last updated on 11/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195820 by York12 last updated on 11/Aug/23 $${a},{b},{c}>\mathrm{0}\:\&{abc}=\mathrm{1},{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}+\frac{\mathrm{1}}{\mathrm{1}+{c}+{a}}\leqslant\mathrm{1} \\ $$ Answered by CrispyXYZ last updated on 12/Aug/23 $$\mathrm{let}\:{a}={x}^{\mathrm{3}} ,\:{b}={y}^{\mathrm{3}} ,\:{c}={z}^{\mathrm{3}} ,\:\mathrm{then}\:{xyz}=\mathrm{1}.…
Question Number 195855 by jabarsing last updated on 11/Aug/23 $$\begin{cases}{\mathrm{3}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}=\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{y}}−\sqrt[{\mathrm{3}}]{{z}}}\\{{x},{y},{z}\:\in\:{N}}\end{cases}\:\:\Rightarrow\:{x},{y},{z}\:=? \\ $$$${please}\:{help}\:{me} \\ $$ Answered by Rasheed.Sindhi last updated on 13/Aug/23 $$\mathrm{Unsuccessful}\:\mathrm{Try}… \\ $$$$\begin{cases}{\mathrm{3}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}=\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{y}}−\sqrt[{\mathrm{3}}]{{z}}}\\{{x},{y},{z}\:\in\:{N}}\end{cases}\:\:\Rightarrow\:{x},{y},{z}\:=? \\…
Question Number 195823 by sonukgindia last updated on 11/Aug/23 Answered by jabarsing last updated on 11/Aug/23 $${No}\:{I}\:{can}'{t} \\ $$ Commented by sniper237 last updated on…
Question Number 195854 by sonukgindia last updated on 11/Aug/23 Commented by Rasheed.Sindhi last updated on 13/Aug/23 $${Q}#\mathrm{194637} \\ $$ Commented by Rasheed.Sindhi last updated on…
Question Number 195848 by pete last updated on 11/Aug/23 $$\mathrm{Please}\:\mathrm{how}\:\mathrm{did}\:\mid\mathrm{z}−\mathrm{a}\mid=\mathrm{r}\:\mathrm{became}\: \\ $$$$\mathrm{z}=\:\mathrm{a}\:+\:\mathrm{re}^{\mathrm{i}\theta} ? \\ $$ Answered by Frix last updated on 11/Aug/23 $$\mid{z}−{a}\mid={r}\:\Rightarrow\:{z}−{a}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{This}\:\mathrm{step}\:\mathrm{is}\:\mathrm{always}\:\mathrm{true},\:\mathrm{even}\:\mathrm{for}\:{z}−{a}\in\mathbb{R}…
Question Number 195813 by jabarsing last updated on 11/Aug/23 $$\begin{cases}{\mathrm{3}\sqrt{\sqrt[{\mathrm{3}}]{\mathrm{12}}−\sqrt[{\mathrm{3}}]{\mathrm{3}}}\:\:=\:\sqrt[{\mathrm{3}}]{{x}}\:+\:\sqrt[{\mathrm{3}}]{{y}}\:−\sqrt[{\mathrm{3}}]{{z}}}\\{{x},{y},{z}\:\in\:{N}}\end{cases}\:\Rightarrow\:{x},{y},{z}\:=? \\ $$$${mr}.{W}\:{please}\:{help}\:{me} \\ $$$${and}\:{other}\:{my}\:{friends}\:{please}\:{help}\:{me} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com