Question Number 214479 by Frangg last updated on 09/Dec/24 Answered by mehdee7396 last updated on 10/Dec/24 $$\frac{\mathrm{5}{x}−\mathrm{6}}{\mathrm{3}{x}}=\frac{\mathrm{26}}{\mathrm{18}}=\frac{\mathrm{13}}{\mathrm{9}} \\ $$$$\mathrm{45}{x}−\mathrm{54}=\mathrm{39}{x} \\ $$$$\mathrm{6}{x}=\mathrm{54}\Rightarrow{x}=\mathrm{9} \\ $$$$ \\ $$$$…
Question Number 214176 by a.lgnaoui last updated on 30/Nov/24 $$\mathrm{find}\:\mathrm{the}\:\mathrm{vslue}\:\mathrm{of}\:\boldsymbol{\mathrm{R}}\left(\mathrm{radius}\:\boldsymbol{\mathrm{o}}\mathrm{f}\:\mathrm{circle}\:\mathrm{betwen}\:\mathrm{two}\:\right. \\ $$$$\:\mathrm{diagrame}\:\mathrm{curves}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\mathrm{and}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{x}}\right)\:\mathrm{as}\:\mathrm{showen}. \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathrm{the}\:\mathrm{programer}\:\mathrm{calculator}. \\ $$ Commented by a.lgnaoui last updated on 30/Nov/24 Commented by…
Question Number 214065 by a.lgnaoui last updated on 28/Nov/24 $$\boldsymbol{{Exercice}}\:\boldsymbol{{pratique}}: \\ $$$$\mathrm{determiner}\:\mathrm{Volume}\:\mathrm{de}\:\mathrm{la}\:\mathrm{forme} \\ $$$$\mathrm{AMBN}\::\mathrm{est}\:\mathrm{dans}\:\mathrm{un}\:\mathrm{plan}\:\mathrm{horisontale} \\ $$$$\mathrm{ci}−\mathrm{dessous}.\left(\mathrm{OI}=\mathrm{12}\right) \\ $$$$\left(\mathrm{ACB}\right)\:\mathrm{assimile}\:\mathrm{a}\:\mathrm{une}\:\mathrm{chainette} \\ $$ Commented by a.lgnaoui last updated…
Question Number 213887 by efronzo1 last updated on 20/Nov/24 $$\:\:\:\mathrm{Find}\:\mathrm{amplitude},\:\mathrm{period},\:\mathrm{maximum}\: \\ $$$$\:\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{for}\:\mathrm{function} \\ $$$$\:\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{6}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{5}}\mathrm{x}\right)−\mathrm{8}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213861 by BaliramKumar last updated on 19/Nov/24 Answered by A5T last updated on 19/Nov/24 $${cotA}+{cotB}=\frac{\mathrm{1}}{{tanA}}+\frac{\mathrm{1}}{{tanB}}=\frac{{p}}{{tanAtanB}}={q} \\ $$$$\Rightarrow{tanAtanB}=\frac{{p}}{{q}} \\ $$$${cot}\left({A}+{B}\right)=\frac{\mathrm{1}}{{tan}\left({A}+{B}\right)}=\frac{\mathrm{1}−{tanAtanB}}{{tanA}+{tanB}}=\frac{\mathrm{1}−\frac{{p}}{{q}}}{{p}} \\ $$$$\Rightarrow{cot}\left({A}+{B}\right)=\frac{{q}−{p}}{{pq}} \\ $$…
Question Number 213741 by bourasi last updated on 15/Nov/24 $$\sqrt{\mathrm{1}−\mathrm{sin}} \\ $$ Commented by Frix last updated on 15/Nov/24 $$\mathrm{Is}\:\mathrm{sin}\:\mathrm{a}\:\mathrm{variable}\:\mathrm{name}\:\mathrm{or}\:\mathrm{s}×\mathrm{i}×\mathrm{n}? \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{case},\:\mathrm{is}\:\mathrm{i}=\sqrt{−\mathrm{1}}? \\ $$…
Question Number 213643 by Abdullahrussell last updated on 12/Nov/24 $$\:{Find}\:{the}\:{maximum}\:{value}\:{of} \\ $$$$\:\mathrm{3}{sin}^{\mathrm{2}} {x}−\mathrm{8}{cosx}+\mathrm{5}=? \\ $$ Answered by Berbere last updated on 12/Nov/24 $${sin}^{\mathrm{2}} \left({x}\right)=\mathrm{1}−{cos}^{\mathrm{2}} \left({x}\right)…
Question Number 213606 by CrispyXYZ last updated on 10/Nov/24 $$\bigtriangleup{ABC}.\:\mathrm{2}{a}+{b}=\mathrm{2}{c}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{of} \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:{C}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\:{A}}. \\ $$ Answered by Ghisom last updated on 10/Nov/24 $$\mathrm{wlog}\:{b}=\mathrm{1}\:\Rightarrow\:{c}={a}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 213519 by RojaTaniya last updated on 07/Nov/24 Answered by mr W last updated on 07/Nov/24 $$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{3}}{{x}} \\ $$$$\mathrm{tan}\:\left(\alpha+\alpha+\beta\right)=\frac{\mathrm{6}}{{x}} \\ $$$$\frac{\mathrm{6}}{{x}}=\frac{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{3}}{{x}}}{\mathrm{1}−\frac{\mathrm{1}}{{x}}×\frac{\mathrm{3}}{{x}}}=\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} −\mathrm{3}}…
Question Number 213504 by Spillover last updated on 06/Nov/24 Answered by A5T last updated on 06/Nov/24 $${psin}^{\mathrm{2}} {x}−{q}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)={p}−{q} \\ $$$${psin}^{\mathrm{2}} {x}−{q}+{qsin}^{\mathrm{2}} {x}={p}−{q} \\ $$$${sin}^{\mathrm{2}}…