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 $$ \\ $$$${f}\left({g}\left({h}\left({x}\right)\right)\right)\:=\:{cos}^{\mathrm{2}}…
Question Number 225282 by fantastic last updated on 20/Oct/25 $${Q}\mathrm{225146} \\ $$$${here}\:{if}\:{the}\:{wedge}\:{and}\:{the} \\ $$$${ground}\:\:{had}\:{no}\:{friction} \\ $$$${it}\:{would}\:{start}\:{going}\:\rightarrow \\ $$$${what}\:{is}\:{acc}.\:{at}\:{time}\:{t}? \\ $$$$\mu={kx} \\ $$ Commented by mr…
Question Number 225241 by Spillover last updated on 18/Oct/25 Answered by som(math1967) last updated on 19/Oct/25 $$\:{tan}\theta=\frac{{nsin}\alpha{cos}\alpha.{sec}^{\mathrm{2}} \alpha}{\left(\mathrm{1}−{nsin}^{\mathrm{2}} \alpha\right){sec}^{\mathrm{2}} \alpha} \\ $$$$\Rightarrow{tan}\theta=\frac{{ntan}\alpha}{{sec}^{\mathrm{2}} \alpha−{ntan}^{\mathrm{2}} \alpha} \\…
Question Number 225240 by Spillover last updated on 18/Oct/25 Answered by mr W last updated on 19/Oct/25 $$\left({a}\right) \\ $$$${R}=\frac{\mid\mathrm{3}×\mathrm{5}−\mathrm{4}×\mathrm{4}+\mathrm{6}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\Rightarrow\left({x}−\mathrm{5}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}}…
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Question Number 225230 by Abdulazim last updated on 18/Oct/25 $$\:\:\:{tg}^{\mathrm{4}} \mathrm{10}°+{tg}^{\mathrm{4}} \mathrm{50}°+{tg}^{\mathrm{4}} \mathrm{70}°=? \\ $$ Commented by Frix last updated on 18/Oct/25 $$\mathrm{Claim}: \\ $$$${x}^{\mathrm{3}}…
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