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Author: Tinku Tara

Find-21-22-23-24-1-

Question Number 212219 by hardmath last updated on 06/Oct/24 $$\mathrm{Find}: \\ $$$$\sqrt{\mathrm{21}\centerdot\mathrm{22}\centerdot\mathrm{23}\centerdot\mathrm{24}\:+\:\mathrm{1}}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 06/Oct/24 $$\sqrt{\left(\mathrm{22}.\mathrm{5}−\mathrm{1}.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}−.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+.\mathrm{5}\right)\left(\mathrm{22}.\mathrm{5}+\mathrm{1}.\mathrm{5}\right)+\mathrm{1}}\: \\ $$$$\sqrt{\left(\mathrm{22}.\mathrm{5}^{\mathrm{2}} −.\mathrm{5}^{\mathrm{2}}…

Can-some-please-explain-is-there-any-use-of-imaginary-numbers-in-real-life-I-have-heard-that-this-number-has-been-used-in-many-different-fields-Why-use-a-number-which-is-not-real-

Question Number 212213 by mathkun last updated on 06/Oct/24 $$\mathrm{Can}\:\mathrm{some}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{use}\:\mathrm{of}\:\mathrm{imaginary}\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{real}\:\mathrm{life}? \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{heard}\:\mathrm{that}\:\mathrm{this}\:\mathrm{number}\:\mathrm{has}\:\mathrm{been}\:\mathrm{used}\:\mathrm{in}\:\mathrm{many}\:\mathrm{different}\:\mathrm{fields}.\: \\ $$$$\mathrm{Why}\:\mathrm{use}\:\mathrm{a}\:\mathrm{number}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{real}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Answered by Frix…

Question-212209

Question Number 212209 by efronzo1 last updated on 06/Oct/24 $$\:\:\:\cancel{\underbrace{\gtrdot}}\: \\ $$ Answered by som(math1967) last updated on 06/Oct/24 $$\:{let}\:{no}\:{of}\:{solders}\:{in}\:{A},{B},{C}\:{are} \\ $$$${x},{y},{z} \\ $$$$\therefore\mathrm{37}{x}+\mathrm{23}{y}=\mathrm{29}{x}+\mathrm{29}{y} \\…

Question-212175

Question Number 212175 by RojaTaniya last updated on 05/Oct/24 Answered by A5T last updated on 05/Oct/24 $${Let}\:{a}+{b}+{c}={x};{ab}+{bc}+{ca}={y};{abc}={z} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{2}{y}=\mathrm{20}…\left({i}\right) \\ $$$${a}^{\mathrm{3}}…

Question-212181

Question Number 212181 by RojaTaniya last updated on 05/Oct/24 Answered by A5T last updated on 05/Oct/24 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right)=\mathrm{5} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{2} \\ $$$${a}^{\mathrm{3}}…

f-x-arctan-1-x-1-x-ask-f-2023-0-

Question Number 212177 by MrGaster last updated on 05/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\mathrm{arctan}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ask}:{f}^{\mathrm{2023}} \left(\mathrm{0}\right) \\ $$ Answered by a.lgnaoui last updated on 05/Oct/24 $$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\Rightarrow\:\:\:\mathrm{f}^{\mathrm{2023}} \left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2023}} }{\mathrm{4}^{\mathrm{2023}}…