Question Number 211502 by MathematicalUser2357 last updated on 11/Sep/24 $$\mathrm{If}\:\begin{cases}{{f}\left({x}\right)={x}^{\mathrm{2}} }\\{{g}\left({x}\right)=\mathrm{sin}\:{x}}\end{cases}, \\ $$$$\mathrm{Then}\:\mathrm{find}\:\frac{{df}}{{dg}}. \\ $$ Answered by a.lgnaoui last updated on 11/Sep/24 $$\frac{\mathrm{df}}{\mathrm{dg}}=\frac{\mathrm{df}}{\mathrm{dx}}×\frac{\mathrm{dx}}{\mathrm{dg}}.=\:\:\frac{\mathrm{f}^{'} }{\mathrm{g}'} \\…
Question Number 211496 by RojaTaniya last updated on 11/Sep/24 Answered by som(math1967) last updated on 11/Sep/24 $$\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{40}+\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{60}+\mathrm{2}{sin}\mathrm{20}{sin}\mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}−{cos}\mathrm{60}+{cos}\mathrm{40}−{cos}\mathrm{80}+{cos}\mathrm{60}−{cos}\mathrm{100}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}+{cos}\mathrm{40}−{cos}\mathrm{80}+{cos}\mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{cos}\mathrm{30}{cos}\mathrm{10} \\ $$$$={sin}\mathrm{60}{sin}\mathrm{80}…
Question Number 211492 by swargiya last updated on 11/Sep/24 Commented by MathematicalUser2357 last updated on 12/Sep/24 $$ \:\mathrm{sin}^{−\mathrm{1}} {x}−\mathrm{cos}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{6}}\: ,\: \:{x}\: \: \: \\…
Question Number 211495 by BaliramKumar last updated on 11/Sep/24 Answered by Rasheed.Sindhi last updated on 11/Sep/24 $${Let}\:{n}=\mathrm{10}{m} \\ $$$${HCF}\left(\mathrm{10}{m},\mathrm{10}\left({m}+\mathrm{1}\right)\right)=\mathrm{10} \\ $$$${HCF}\left({m},{m}+\mathrm{1}\right)=\mathrm{1}\Rightarrow{m}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},… \\ $$$$\therefore{LCM}={x}\left(\mathrm{2}-{digit}\:{numbers}\right)=\mathrm{10}{m}\left({m}+\mathrm{1}\right)\: \\ $$$$\:\:\:\:\:\:=\mathrm{20},\mathrm{60}\:\left(\mathrm{2}\:{possible}\:{values}\right)…
Question Number 211520 by a.lgnaoui last updated on 12/Sep/24 $$\mathrm{determiner}\:\mathrm{les}\:\mathrm{valeurs}\:\:\mathrm{de}\:\boldsymbol{\mathrm{p}}\mathrm{et}\:\boldsymbol{\mathrm{q}}\:\mathrm{sachant}\:\mathrm{que}\:−\mathrm{2}\:\mathrm{et}\:\mathrm{3}\:\mathrm{sont}\:\mathrm{les}\: \\ $$$$−\mathrm{2}\:\mathrm{et}\:\mathrm{3}\:\:\mathrm{sont}\:\mathrm{les}\:\mathrm{racines}\:\mathrm{de}\:\mathrm{l}\:\mathrm{equation}: \\ $$$$\mathrm{2}\boldsymbol{\mathrm{pqz}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{z}}−\mathrm{4}\left(\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}\right)=\mathrm{0} \\ $$$$ \\ $$ Commented by MrGaster last updated on…
Question Number 211485 by MATHEMATICSAM last updated on 10/Sep/24 $$\mathrm{If}\:\frac{{a}\:−\:{b}}{{c}}\:+\:\frac{{b}\:−\:{c}}{{a}}\:+\:\frac{{c}\:+\:{a}}{{b}}\:=\:\mathrm{1}\:\mathrm{and}\: \\ $$$${a}\:−\:{b}\:+\:{c}\:\neq\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}}\:=\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:. \\ $$ Answered by Frix last updated on 10/Sep/24 $$\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}+{a}}{{b}}=\mathrm{1} \\…
Question Number 211486 by mnjuly1970 last updated on 10/Sep/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{Mathematical}\:\:\:\:\:{Analysis}… \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:{f}:\mathbb{R}\:\rightarrow\:\mathbb{R}\:{is}\:{diffrentiable}\:{function}, \\ $$$$\:\:\:\:\:{f}\:\:{and}\:\:{f}\:'\:,\:{has}\:{no}\:{common}\:{zero} \\ $$$$\:\:\:\:\:{on}\:\:\mathbb{R}\:. \\ $$$$\:\:\:\:\:\:{prove}\:{that}\:{the}\:{following}\:{set}\:{is}\:{finite}. \\ $$$$ \\…
Question Number 211480 by MathematicalUser2357 last updated on 10/Sep/24 $${i}\:{logged}\:{in}\:{and}\:{i}\:{cant}\:{plot} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211482 by RojaTaniya last updated on 10/Sep/24 $$\:\mathrm{25}^{\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\:{log}_{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{27}\:+\:{log}_{\mathrm{25}} \mathrm{27}\right)} =? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 211467 by mnjuly1970 last updated on 10/Sep/24 $$ \\ $$$$\:\:\:{Solve}\:{the}\:{following}\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:+\:\sqrt{{x}\:−\sqrt{{x}}\:}\:=\:\lfloor\:{x}+\:\frac{\mathrm{1}}{{x}}\:\rfloor \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:{is}\:{floor}\:{of}\:,\:\:{x}\:\:,\: \\ $$ Answered by Frix…