Question Number 153764 by mathdanisur last updated on 10/Sep/21 $$\mathrm{Prove}\:\mathrm{without}\:\mathrm{any}\:\mathrm{software}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\sqrt{\mathrm{1}\:-\:\left(\frac{\mathrm{x}\:+\:\mathrm{y}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\mathrm{dxdy}\:>\:\frac{\pi}{\mathrm{4}} \\ $$ Commented by alisiao last updated on…
Question Number 153763 by mathdanisur last updated on 10/Sep/21 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{pairs}\:\left(\mathrm{x};\mathrm{y}\right)\:\mathrm{of}\:\mathrm{integers} \\ $$$$\mathrm{which}\:\mathrm{satisfy} \\ $$$$\mid\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{y}^{\mathrm{2}} \mid\:-\:\sqrt{\mathrm{16y}\:+\:\mathrm{1}}\:=\:\mathrm{0} \\ $$ Answered by liberty last updated on 10/Sep/21…
Question Number 153733 by mathdanisur last updated on 09/Sep/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 88203 by behi83417@gmail.com last updated on 09/Apr/20 $$\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{ax}}^{−\mathrm{3}} ,\:\boldsymbol{\mathrm{meets}}:\:\:\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{y}}=−\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{x}}} \:\boldsymbol{\mathrm{at}}: \\ $$$$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{B}},\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}:\:\boldsymbol{\mathrm{AB}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{minimum}}. \\ $$$$\boldsymbol{\mathrm{find}}:\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\left(\boldsymbol{\mathrm{s}}\right)\:\boldsymbol{\mathrm{of}}:\:\boldsymbol{\mathrm{a}}\:\:\:\boldsymbol{\mathrm{and}}\:\:\boldsymbol{\mathrm{min}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{AB}}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 22640 by Tinkutara last updated on 21/Oct/17 $${With}\:{usual}\:{notation},\:{show}\:{that} \\ $$$$\frac{{C}_{\mathrm{0}} }{{x}}\:−\:\frac{{C}_{\mathrm{1}} }{{x}+\mathrm{1}}\:+\:\frac{{C}_{\mathrm{2}} }{{x}+\mathrm{2}}\:−\:….\:+\:\left(−\mathrm{1}\right)^{{n}} \frac{{C}_{{n}} }{{x}+{n}}= \\ $$$$\frac{{n}!}{{x}\left({x}\:+\:\mathrm{1}\right)\left({x}\:+\:\mathrm{2}\right)….\left({x}\:+\:{n}\right)} \\ $$ Terms of Service Privacy…
Question Number 153704 by mathdanisur last updated on 09/Sep/21 Commented by mathdanisur last updated on 09/Sep/21 $$\mathrm{p}_{\boldsymbol{\mathrm{k}}} \:=\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} +…+\mathrm{x}^{\mathrm{2021}} }\:=\:\frac{\mathrm{1}\:-\:\mathrm{x}}{\mathrm{1}\:-\:\mathrm{x}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\ $$$$\:\:\:\:\:\:=\:\frac{\mathrm{1}\:-\:\mathrm{x}}{\mathrm{1}\:-\:\mathrm{x}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\…
Question Number 88160 by john santu last updated on 08/Apr/20 $${solve}\::\:{x}^{\mathrm{2}} \:=\:\mathrm{3}{x}\:+\:\mathrm{6}{y}\:;\:{xy}\:=\:\mathrm{5}{x}\:+\:\mathrm{4}{y} \\ $$ Commented by john santu last updated on 08/Apr/20 $${y}\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}}{\mathrm{6}}\:\:;\:{y}\left({x}−\mathrm{4}\right)\:=\:\mathrm{5}{x} \\…
Question Number 22618 by Tinkutara last updated on 21/Oct/17 $${If}\:\left(\mathrm{1}\:+\:{x}\right)^{{n}} \:=\:{C}_{\mathrm{0}} \:+\:{C}_{\mathrm{1}} {x}\:+\:{C}_{\mathrm{2}} {x}^{\mathrm{2}} \:+\:{C}_{\mathrm{3}} {x}^{\mathrm{3}} \\ $$$$+\:…\:+\:{C}_{{n}} {x}^{{n}} ,\:{prove}\:{that} \\ $$$$\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{1}.\mathrm{2}}{C}_{\mathrm{0}} \:+\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{2}.\mathrm{3}}{C}_{\mathrm{1}}…
Question Number 22612 by Tinkutara last updated on 21/Oct/17 $${In}\:{the}\:{binomial}\:{expasion}\:{of}\:\left({a}\:−\:{b}\right)^{\mathrm{5}} , \\ $$$${the}\:{sum}\:{of}\:\mathrm{2}^{{nd}} \:{and}\:\mathrm{3}^{{rd}} \:{term}\:{is}\:{zero}, \\ $$$${then}\:\frac{{a}}{{b}}\:{is} \\ $$ Answered by ajfour last updated on…
Question Number 153682 by EDWIN88 last updated on 09/Sep/21 $$\:\left(\sqrt[{{i}}]{{i}}\:\right)^{{xi}} \:=\:{i}^{{x}} \: \\ $$$$\:\:{x}=?\: \\ $$ Answered by MJS_new last updated on 09/Sep/21 $$\mathrm{lhs}\:\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{2}}} \:\Rightarrow\:\sqrt[{\mathrm{i}}]{\mathrm{i}}=\mathrm{e}^{\frac{\pi}{\mathrm{2}}}…