Question Number 154274 by mathdanisur last updated on 16/Sep/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{x}^{\mathrm{2}} \centerdot\mathrm{2}^{\boldsymbol{\mathrm{x}}} \left(\mathrm{x}^{\mathrm{4}} \centerdot\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{3}\centerdot\mathrm{15}^{\boldsymbol{\mathrm{x}}} \right)\:=\:\mathrm{125}^{\boldsymbol{\mathrm{x}}} \:-\:\mathrm{27}^{\boldsymbol{\mathrm{x}}} \\ $$ Commented by MJS_new last updated…
Question Number 154268 by liberty last updated on 16/Sep/21 Commented by mr W last updated on 16/Sep/21 $$\mathrm{8}!=\mathrm{40320} \\ $$ Commented by liberty last updated…
Question Number 154270 by amin96 last updated on 16/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 88731 by I want to learn more last updated on 12/Apr/20 Commented by john santu last updated on 12/Apr/20 $$\left.\mathrm{2}\left.{a}\right)\:\frac{{dy}}{{dx}}\:=\:{kx}−\mathrm{48}{x}^{−\mathrm{3}} \:\right]_{\left(−\mathrm{2},\mathrm{14}\right)} \:=\:\mathrm{0} \\…
Question Number 154267 by mathdanisur last updated on 16/Sep/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\mathrm{and}\:\mathrm{xy}+\mathrm{yz}+\mathrm{zx}=\mathrm{3xyz}\:\:\mathrm{then}: \\ $$$$\left(\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\right)\left(\underset{\boldsymbol{\mathrm{cyc}}} {\sum}\:\frac{\mathrm{1}}{\mathrm{2x}+\mathrm{y}+\mathrm{z}}\right)\:\leqslant\:\frac{\mathrm{9}}{\mathrm{8}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 23190 by ANTARES_VY last updated on 27/Oct/17 $$\underset{\boldsymbol{\mathrm{x}}\rightarrow\infty} {\boldsymbol{\mathrm{lim}}}\left(\frac{\boldsymbol{\mathrm{x}}−\mathrm{2}}{\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{10}}\right)^{\mathrm{5}\boldsymbol{\mathrm{x}}} \\ $$ Commented by ANTARES_VY last updated on 27/Oct/17 $$\boldsymbol{\mathrm{calculate}} \\ $$ Answered by…
Question Number 154262 by mathlove last updated on 16/Sep/21 Answered by som(math1967) last updated on 16/Sep/21 $${lntan}\left(\frac{\pi}{\mathrm{180}}\right)+{lntan}\left(\frac{\mathrm{2}\pi}{\mathrm{180}}\right)+…{lntan}\left(\frac{\mathrm{89}\pi}{\mathrm{180}}\right) \\ $$$${lntan}\mathrm{1}°+{lntan}\mathrm{2}°+…{lntan}\mathrm{89}°\bigstar \\ $$$${lntan}\mathrm{1}×{tan}\mathrm{2}×…{tan}\mathrm{89} \\ $$$${lntan}\mathrm{1}×{cot}\left(\mathrm{90}−\mathrm{89}\right)×{tan}\mathrm{2}×{cot}\left(\mathrm{90}−\mathrm{88}\right)…×{tan}\mathrm{45} \\ $$$${lntan}\mathrm{1}×{cot}\mathrm{1}×{tan}\mathrm{2}×{cot}\mathrm{2}×…×\mathrm{1}…
Question Number 23187 by lizan 123 last updated on 27/Oct/17 $${plz}\:\:\:{anyone}\:{answer}\:{the}\:{question}\: \\ $$$$\mathrm{23181}…{plz}\:{plz}\:{plz}… \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 88723 by M±th+et£s last updated on 12/Apr/20 $${prove}\:{that} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({k}+\mathrm{2}\right)^{\mathrm{2}} {x}^{{k}} }{\left({k}+\mathrm{3}\right)!}=\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)−\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$…
Question Number 154258 by liberty last updated on 16/Sep/21 Answered by som(math1967) last updated on 16/Sep/21 $$\frac{{a}}{{b}+{c}}\:+\mathrm{1}+\frac{{b}}{{c}+{a}}+\mathrm{1}+\frac{{c}}{{a}+{b}}+\mathrm{1}=\mathrm{76}+\mathrm{3} \\ $$$$\:{or}\:\frac{{a}+{b}+{c}}{{b}+{c}}\:+\frac{{a}+{b}+{c}}{{c}+{a}}\:+\frac{{a}+{b}+{c}}{{a}+{b}}=\mathrm{79} \\ $$$${or}.\:\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}+\frac{\mathrm{1}}{{a}+{b}}\right)=\mathrm{79} \\ $$$$\therefore\left(\frac{\mathrm{1}}{{a}+{b}}\:+\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}\right)=\frac{\mathrm{79}}{\mathrm{1580}}=\frac{\mathrm{1}}{\mathrm{20}}\bigstar \\ $$$$\bigstar{a}+{b}+{c}=\mathrm{1580}…