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Author: Tinku Tara

Tinku-Tara-sir-it-could-be-a-new-issue-when-i-cancel-a-comment-or-an-answer-by-tapping-the-back-button-it-should-popup-a-window-asking-if-discard-or-save-now-this-popup-window-doesn-t-appear-an

Question Number 225322 by mr W last updated on 21/Oct/25 $${Tinku}\:{Tara}\:{sir}: \\ $$$${it}\:{could}\:{be}\:{a}\:{new}\:{issue}: \\ $$$${when}\:{i}\:{cancel}\:{a}\:{comment}\:{or}\:{an}\:{answer}, \\ $$$${by}\:{tapping}\:{the}\:{back}\:{button},\:{it}\:{should} \\ $$$${popup}\:{a}\:{window}\:{asking}\:{if}\:{discard}\: \\ $$$${or}\:{save}.\:{now}\:{this}\:{popup}\:{window}\: \\ $$$${doesn}'{t}\:{appear}\:{and}\:{the}\:{work}\:{made}\: \\ $$$${goes}\:{automatically}\:{lost}.…

Find-3-2-5-1-4-3-2-5-1-4-1-4-5-1-4-1-5-1-4-1-

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 $$ \\ $$$$\sqrt[{\mathrm{4}}]{\frac{\mathrm{3}+\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}−\mathrm{2}\sqrt[{\mathrm{4}}]{\mathrm{5}}}}×\sqrt[{\mathrm{4}}]{\frac{\mathrm{6}\sqrt{\mathrm{5}}\:+\:\mathrm{6}\:−\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}\:−\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{6}\sqrt{\mathrm{5}}\:+\:\mathrm{6}\:+\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}\:+\:\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{5}}}} \\ $$$$…

if-fogoh-x-cos-2-x-9-then-f-x-g-x-h-x-

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}}…

Let-S-n-x-r-1-n-sin-2r-3-2-r-1-x-cos-10r-1-2-r-1-x-sin-6r-1-2-r-1-x-cos-2r-5-2-r-1-x-2-r-1-sin-r2-3-r-x-sin-2-2-r-x-then-find-the-value-of-lim-

Question Number 225303 by fantastic last updated on 20/Oct/25 $${Let} \\ $$$${S}_{{n}} \left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{sin}\:\left(\left(\mathrm{2}{r}−\mathrm{3}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{10}{r}+\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)−\mathrm{sin}\left(\:\left(\mathrm{6}{r}−\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{2}{r}+\mathrm{5}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)}{\mathrm{2}^{{r}−\mathrm{1}} \left(\mathrm{sin}\:\left({r}\mathrm{2}^{\mathrm{3}−{r}} −{x}\right)\mathrm{sin}\:\left(\mathrm{2}^{\mathrm{2}−{r}} −{x}\right)\right)} \\ $$$${then}\:{find}\:{the}\:{value}\:{of} \\…

We-have-integrated-artificial-intelligence-images-are-checked-during-upload-and-will-be-rejected-if-not-related-to-science-daily-processing-of-other-posts-A-few-senior-users-are-exempted-fr

Question Number 225299 by Tinku Tara last updated on 20/Oct/25 $$\mathrm{We}\:\mathrm{have}\:\mathrm{integrated}\:\mathrm{artificial} \\ $$$$\mathrm{intelligence}: \\ $$$$-\:\mathrm{images}\:\mathrm{are}\:\mathrm{checked}\:\mathrm{during}\:\mathrm{upload} \\ $$$$\:\:\:\mathrm{and}\:\mathrm{will}\:\mathrm{be}\:\mathrm{rejected}\:\mathrm{if}\:\mathrm{not}\:\mathrm{related} \\ $$$$\:\:\:\mathrm{to}\:\mathrm{science} \\ $$$$-\:\mathrm{daily}\:\mathrm{processing}\:\mathrm{of}\:\mathrm{other}\:\mathrm{posts} \\ $$$$\mathrm{A}\:\mathrm{few}\:\mathrm{senior}\:\mathrm{users}\:\mathrm{are}\:\mathrm{exempted}\:\mathrm{from} \\ $$$$\mathrm{these}\:\mathrm{checks}.…

Question-225240

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}}…