Question Number 157585 by MathSh last updated on 24/Oct/21 $$\mathrm{if}\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b}\leqslant\mathrm{c}<\frac{\pi}{\mathrm{2}}\:\:\:\mathrm{then}: \\ $$$$\frac{\mathrm{5}}{\mathrm{tan}\boldsymbol{\mathrm{a}}}\:+\:\frac{\mathrm{3}}{\mathrm{tan}\boldsymbol{\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\mathrm{tan}\boldsymbol{\mathrm{c}}}\:\geqslant\:\frac{\mathrm{27}}{\mathrm{tan}\boldsymbol{\mathrm{a}}\:+\:\mathrm{tan}\boldsymbol{\mathrm{b}}\:+\:\mathrm{tan}\boldsymbol{\mathrm{c}}} \\ $$ Answered by ghimisi last updated on 25/Oct/21 $$\mathrm{3}\left(\frac{\mathrm{1}}{{tga}}+\frac{\mathrm{1}}{{tgb}}+\frac{\mathrm{1}}{{tgc}}\right)+\mathrm{2}\left(\frac{\mathrm{1}}{{tga}}−\frac{\mathrm{1}}{{tgc}}\right)\geqslant \\ $$$$\mathrm{3}\centerdot\frac{\mathrm{9}}{{tga}+{tgb}+{tgc}}+\frac{\mathrm{2}\left({tgc}−{tga}\right)}{{tgatgc}}\geqslant\frac{\mathrm{27}}{{tga}+{tgb}+{tgc}} \\…
Question Number 157576 by amin96 last updated on 24/Oct/21 Commented by quvonch3737 last updated on 24/Oct/21 $$\mathrm{1}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }+\frac{\mathrm{16}}{{x}^{\mathrm{3}} }+…={S}\:\:/×\frac{\mathrm{1}}{{x}} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }+\frac{\mathrm{9}}{{x}^{\mathrm{3}} }+\frac{\mathrm{16}}{{x}^{\mathrm{4}} }+…=\frac{{S}}{{x}} \\…
Question Number 157575 by quvonch3737 last updated on 24/Oct/21 $$ \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\boldsymbol{\alpha}}} \\ $$$$\left(\Pi<\boldsymbol{\alpha}<\mathrm{2}\Pi\right) \\ $$ Commented by mr W last updated on 24/Oct/21 $${this}\:{is}\:{not}\:{a}\:{question}!…
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Question Number 157561 by MathSh last updated on 24/Oct/21 $$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}\:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}\geqslant\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Sigma\:\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{b}\:+\:\mathrm{kbc}}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{1}\:+\:\mathrm{k}}\:\:;\:\:\mathrm{k}>\mathrm{0}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157562 by Ar Brandon last updated on 24/Oct/21 Answered by FongXD last updated on 24/Oct/21 $$\mathrm{Given}:\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{3} \\ $$$$\mathrm{square}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\Leftrightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{2yz}+\mathrm{2xz}=\mathrm{9}…
Question Number 92013 by I want to learn more last updated on 04/May/20 $$\mathrm{Solve}: \\ $$$$\:\:\:\mathrm{2}^{\mathrm{x}} \:\:+\:\:\mathrm{3}^{\mathrm{y}} \:\:\:=\:\:\mathrm{72}\:\:\:\:\:…..\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\mathrm{2}^{\mathrm{y}} \:\:+\:\:\mathrm{3}^{\mathrm{x}} \:\:\:=\:\:\mathrm{108}\:\:\:\:\:…..\:\left(\mathrm{ii}\right) \\ $$ Commented…
Question Number 92008 by I want to learn more last updated on 04/May/20 $$\mathrm{Sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series}:\:\:\mathrm{1}\:\:+\:\:\frac{\mathrm{3}}{\mathrm{4}}\:\:+\:\:\frac{\mathrm{3}.\mathrm{5}}{\mathrm{4}.\mathrm{8}}\:\:+\:\:\frac{\mathrm{3}.\mathrm{5}.\mathrm{7}}{\mathrm{4}.\mathrm{8}.\mathrm{12}}\:\:+\:\:…\:\:\:\:\mathrm{is}\:? \\ $$ Commented by Prithwish Sen 1 last updated on 04/May/20…
Question Number 92000 by student work last updated on 04/May/20 $$\mathrm{proof}\:\mathrm{0}!!=\mathrm{1} \\ $$ Answered by $@ty@m123 last updated on 04/May/20 $${n}!={n}.\left({n}−\mathrm{1}\right)! \\ $$$$\Rightarrow\left({n}−\mathrm{1}\right)!=\frac{{n}!}{{n}} \\ $$$${Put}\:{n}=\mathrm{1}…
Question Number 26461 by ajfour last updated on 25/Dec/17 Commented by ajfour last updated on 25/Dec/17 $${A}\:{layer}\:{is}\:{cut}\:{from}\:{the}\:{back}\:{of}\: \\ $$$${upper}\:{large}\:{cuboid}\:\left({of}\:{front}\:\right. \\ $$$$\left.{surface}\:{area}\:\boldsymbol{{p}}\:{and}\:{volume}\:−\boldsymbol{{q}}\:\right). \\ $$$${This}\:{layer}\:{is}\:{of}\:{volume}\:\boldsymbol{{x}}^{\mathrm{3}} ,\:{where} \\…