Question Number 225938 by Spillover last updated on 16/Nov/25 Answered by Spillover last updated on 17/Nov/25 Answered by Spillover last updated on 17/Nov/25 Answered by…
Question Number 225939 by Spillover last updated on 16/Nov/25 Answered by Ghisom_ last updated on 16/Nov/25 $$\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{tan}\:{x}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{\mathrm{cot}\:{x}}\:\rightarrow\:{dx}=−\mathrm{2sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cot}\:{x}}\right] \\ $$$$=−\mathrm{2}\underset{\infty}…
Question Number 225837 by Rojarani last updated on 14/Nov/25 $${Show}\:{that},\:{log}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}….\alpha}}}}\:=\mathrm{1} \\ $$ Answered by fantastic last updated on 14/Nov/25 $${let}\:\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}={x} \\ $$$${x}^{\mathrm{2}} =\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}} \\ $$$$\mathrm{7}{x}=\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}…
Question Number 225840 by hardmath last updated on 14/Nov/25 Commented by hardmath last updated on 14/Nov/25 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{MA}^{\mathrm{2}} +\mathrm{MB}^{\mathrm{2}} +\mathrm{MC}^{\mathrm{2}} +\mathrm{MD}^{\mathrm{2}} \:=\:\mathrm{4R}^{\mathrm{2}} \\ $$…
Question Number 225810 by hardmath last updated on 12/Nov/25 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{any}\:\mathrm{triangle}: \\ $$$$\frac{\mathrm{4R}}{\mathrm{r}}\:\geqslant\:\frac{\mathrm{w}_{\boldsymbol{\mathrm{a}}} \:\mathrm{w}_{\boldsymbol{\mathrm{b}}} \:\mathrm{w}_{\boldsymbol{\mathrm{c}}} }{\mathrm{h}_{\boldsymbol{\mathrm{a}}} \:\mathrm{h}_{\boldsymbol{\mathrm{b}}} \:\mathrm{h}_{\boldsymbol{\mathrm{c}}} }\:\centerdot\:\left(\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\right)\centerdot\left(\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\right)^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy…
Question Number 225613 by Jubr last updated on 04/Nov/25 Answered by mahdipoor last updated on 04/Nov/25 $$\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\mathrm{1}}}{\mathrm{2}.\mathrm{1}}=\frac{−\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$=\mathrm{cos}\left(\pm\mathrm{120}\right)+\mathrm{i}.\mathrm{sin}\left(\pm\mathrm{120}\right)=\mathrm{e}^{\mathrm{i}\left(\pm\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} \\ $$$$\mathrm{v}_{\mathrm{n}} =\mathrm{e}^{\mathrm{i}\theta\mathrm{n}} +\mathrm{e}^{\mathrm{i}\left(−\theta\right)\mathrm{n}}…
Question Number 225503 by Jyrgen last updated on 31/Oct/25 $${find}\:{x}\in\mathbb{C}\:{for}\:{r}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(−{r}\right)^{{x}} ={r} \\ $$ Answered by MrAjder last updated on 02/Nov/25 $$\left(−{r}\right)^{{x}} ={r} \\…
Question Number 225392 by Rojarani last updated on 24/Oct/25 Commented by Ghisom_ last updated on 25/Oct/25 $$\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{9}\:\mathrm{solutions}\:\mathrm{for}\:\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:\in\mathbb{C}^{\mathrm{3}} \\ $$ Answered by Kademi last updated on…
Question Number 225323 by hardmath last updated on 21/Oct/25 $$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{3}\:+\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}\:-\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} .\:\:\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\:-\:\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\mathrm{1}}\:\:=\:? \\ $$ Answered by Raphael254 last updated on 21/Oct/25 $$ \\ $$$$\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}+\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}−\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}}}×\sqrt[{\mathrm{4}}]{\frac{\mathrm{6}\sqrt{\mathrm{5}}\:+\:\mathrm{6}\:−\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}\:−\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{6}\sqrt{\mathrm{5}}\:+\:\mathrm{6}\:+\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}\:+\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{5}}}} \\ $$$$…
Question Number 225330 by mathlove last updated on 21/Oct/25 $${if}\:\:\left({fogoh}\right)\left({x}\right)={cos}^{\mathrm{2}} \left({x}+\mathrm{9}\right) \\ $$$${then}\:\:\:{f}\left({x}\right)=?\:,\:\:{g}\left({x}\right)=?\:\:,\:{h}\left({x}\right)=? \\ $$ Answered by Raphael254 last updated on 21/Oct/25 $$ \\ $$$${f}\left({g}\left({h}\left({x}\right)\right)\right)\:=\:{cos}^{\mathrm{2}}…