Question Number 200884 by cortano12 last updated on 26/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200885 by cortano12 last updated on 26/Nov/23 $$\:\:\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$ Commented by cortano12 last updated on 26/Nov/23 $$\:\:\:\boldsymbol{\mathfrak{f}} \\ $$ Answered by Frix…
Question Number 200886 by essaad last updated on 26/Nov/23 $${resoudre}\:{dans}\:{R}\:: \\ $$$$ \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{3}+{x}}\:−\sqrt[{\mathrm{3}}]{\mathrm{3}−{x}}\:=\sqrt[{\mathrm{3}}]{\mathrm{9}−{x}^{\mathrm{2}} } \\ $$ Answered by Frix last updated on 26/Nov/23 $${a}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 200942 by Mingma last updated on 26/Nov/23 Answered by witcher3 last updated on 26/Nov/23 $$\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\geqslant\mathrm{3}\left(\mathrm{abc}\right)^{−\frac{\mathrm{2}}{\mathrm{3}}} \:\:\mathrm{AM}−\mathrm{GM} \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ac}\geqslant\Leftrightarrow \\ $$$$\mathrm{abc}\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}\right)\geqslant\mathrm{1}…
Question Number 200943 by dafne last updated on 26/Nov/23 $${v}.{m}=\frac{\Delta{d}}{\Delta{t}}=\frac{{d}−{d}_{\mathrm{0}} }{{t}−{t}_{\mathrm{0}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200936 by Mingma last updated on 26/Nov/23 Commented by AST last updated on 26/Nov/23 $${Q}\mathrm{199738} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 200937 by Mingma last updated on 26/Nov/23 Answered by MM42 last updated on 26/Nov/23 $$\frac{\mathrm{25}\sqrt{\mathrm{2}}}{\mathrm{4}}\pi \\ $$ Commented by Mingma last updated on…
Question Number 200933 by Spillover last updated on 26/Nov/23 $$ \\ $$$$\int\mathrm{coth}\:\left(\mathrm{ln}\:\left[\sqrt{\mathrm{tanh}\:\left(\mathrm{ln}\:\left(\sqrt{\mathrm{sec}^{−\mathrm{1}} \:\:\sqrt[{\mathrm{4}}]{{x}}\:\:}\right)\right)}\:\right]\right) \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 200934 by sonukgindia last updated on 26/Nov/23 Answered by Frix last updated on 27/Nov/23 $$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\left(\mathrm{tan}\:\pi{x}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} {dx}\:\overset{{t}=\left(\mathrm{cot}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} } {=} \\ $$$$=−\frac{\mathrm{5}}{\pi}\underset{\infty} {\overset{\mathrm{0}}…
Question Number 200929 by Akira181 last updated on 26/Nov/23 $$\mathrm{A}\:\mathrm{ball}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{function}\:{z}={xy}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right).\:\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{the}\:{xy}−\mathrm{plane}\:\mathrm{where}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{will} \\ $$$$\mathrm{touch}\:\mathrm{it}. \\ $$$$\mathrm{Calculus}\:\mathrm{2}\:\mathrm{problem}. \\ $$ Commented by Akira181 last updated…