Question Number 157628 by mr W last updated on 25/Oct/21 $${find} \\ $$$$\left({C}_{\mathrm{0}} ^{\mathrm{100}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{2}} ^{\mathrm{100}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{4}} ^{\mathrm{100}} \right)^{\mathrm{2}} +\left({C}_{\mathrm{6}} ^{\mathrm{100}} \right)^{\mathrm{2}} +…+\left({C}_{\mathrm{100}}…
Question Number 92084 by Power last updated on 04/May/20 Answered by MJS last updated on 04/May/20 $$\mathrm{obviously}\:{x}_{\mathrm{0}} ={y}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{let}\:{x}\neq\mathrm{0}\:\wedge\:{y}={px}\:\wedge\:{p}\neq\mathrm{0} \\ $$$$\begin{cases}{{x}\left(\left({p}^{\mathrm{5}} +\mathrm{1}\right){x}^{\mathrm{4}} −\mathrm{33}{p}\right)=\mathrm{0}}\\{{x}\left(\mathrm{3}{p}^{\mathrm{2}}…
Question Number 157604 by mnjuly1970 last updated on 25/Oct/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Q}{uestion}\: \\ $$$$\:{find}\:{the}''\:{minimum}''\:{value}\:{of}: \\ $$$$ \\ $$$${f}\:\left({x}\right):=\:\mid\mathrm{1}+{x}\mid+\mid\:\mathrm{2}+{x}\mid\:+\:\mid\mathrm{4}\:+\mathrm{2}{x}\mid \\ $$$$ \\ $$ Commented by mr…
Question Number 157593 by metamorfose last updated on 25/Oct/21 $${let}\:{f}:{C}\rightarrow{R} \\ $$$$\:\:\:\:\:\:\:\:\:\:{z}\rightarrow{min}\left({y}−\left[{y}\right],\left[{y}+\mathrm{1}\right]−{y}\right)\:,\:{y}={Im}\left({z}\right) \\ $$$${let}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} ,\:{n}\in{N}^{\ast} \\ $$$${evaluate}\:{S}_{{n}} =\underset{\mathrm{0}\leqslant{k}<{n}} {\sum}{f}\left({w}^{{k}} \right) \\ $$ Terms of Service…
Question Number 157592 by metamorfose last updated on 25/Oct/21 $${let}\:{A}=\left\{\overset{.} {\mathrm{0}},\overset{.} {\mathrm{1}},\overset{.} {\mathrm{2}}\right\} \\ $$$${prove}\:{that}\:{every}\:{application}\:{from}\:{A}\: \\ $$$${to}\:{A}\:{is}\:{a}\:\mathrm{2}{nd}\:{degree}\:{polynom} \\ $$ Terms of Service Privacy Policy Contact:…
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 $${this}\:{is}\:{not}\:{a}\:{question}!…
Question Number 157565 by MathSh last updated on 24/Oct/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 157561 by MathSh last updated on 24/Oct/21 $$\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