Question Number 212209 by efronzo1 last updated on 06/Oct/24 $$\:\:\:\cancel{\underbrace{\gtrdot}}\: \\ $$ Answered by som(math1967) last updated on 06/Oct/24 $$\:{let}\:{no}\:{of}\:{solders}\:{in}\:{A},{B},{C}\:{are} \\ $$$${x},{y},{z} \\ $$$$\therefore\mathrm{37}{x}+\mathrm{23}{y}=\mathrm{29}{x}+\mathrm{29}{y} \\…
Question Number 212175 by RojaTaniya last updated on 05/Oct/24 Answered by A5T last updated on 05/Oct/24 $${Let}\:{a}+{b}+{c}={x};{ab}+{bc}+{ca}={y};{abc}={z} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{2}{y}=\mathrm{20}…\left({i}\right) \\ $$$${a}^{\mathrm{3}}…
Question Number 212186 by Ismoiljon_008 last updated on 05/Oct/24 $$ \\ $$$$\:\:\:{sin}^{\mathrm{2}} \:\mathrm{1}^{{o}} \:+\:{sin}^{\mathrm{2}} \:\mathrm{5}^{{o}} \:+\:{sin}^{\mathrm{2}} \:\mathrm{9}^{{o}} \:+\:…\:{sin}^{\mathrm{2}} \:\mathrm{89}^{{o}} \:=\:{a}\:\frac{\mathrm{1}}{{b}} \\ $$$$\:\:\:{b}\:=\:? \\ $$$$\:\:\:\mathbb{H}{elp}\:{me},\:{please} \\…
Question Number 212181 by RojaTaniya last updated on 05/Oct/24 Answered by A5T last updated on 05/Oct/24 $${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} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{2} \\ $$$${a}^{\mathrm{3}}…
Question Number 212176 by MrGHK last updated on 05/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212177 by MrGaster last updated on 05/Oct/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{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 $$\mathrm{f}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{4}}\Rightarrow\:\:\:\mathrm{f}^{\mathrm{2023}} \left(\mathrm{0}\right)=\frac{\pi^{\mathrm{2023}} }{\mathrm{4}^{\mathrm{2023}}…
Question Number 212179 by cherokeesay last updated on 05/Oct/24 Answered by cemoosky last updated on 05/Oct/24 $${x}\:=\:\mathrm{6}.\mathrm{6} \\ $$ Answered by mr W last updated…
Question Number 212159 by boblosh last updated on 04/Oct/24 Answered by Rasheed.Sindhi last updated on 04/Oct/24 $$\begin{bmatrix}{\:\:\:\:\:\mathrm{2}}&{\:\:\:\:\mathrm{3}}\\{−\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}+{a}\begin{bmatrix}{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{4}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}{a}+\mathrm{2}}&{{a}+\mathrm{3}}\\{\mathrm{0}{a}−\mathrm{1}}&{\mathrm{4}{a}−\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{8}}&{{b}}\\{{c}}&{{d}}\end{bmatrix} \\ $$$$\mathrm{2}{a}+\mathrm{2}=\mathrm{8}\Rightarrow{a}=\mathrm{3} \\ $$$${c}=−\mathrm{1} \\ $$$${b}={a}+\mathrm{3}=\mathrm{3}+\mathrm{3}=\mathrm{6}…
Question Number 212168 by boblosh last updated on 04/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 212152 by RojaTaniya last updated on 04/Oct/24 Answered by som(math1967) last updated on 04/Oct/24 $$\:{x}^{\mathrm{3}} +{ax}+{b}=\left({x}+{c}\right)\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right) \\ $$$${put}\:{x}=\mathrm{1} \\ $$$$\:{a}+{b}=−\mathrm{1}\:….{case}\mathrm{1} \\ $$$${put}\:{x}=\mathrm{2} \\…