Question Number 7234 by WagST last updated on 17/Aug/16 $${need}\:{help}\:{with}\:{step}\:{by}\:{step}.\:{thanks}. \\ $$$$ \\ $$$${x}\left({x}+\mathrm{4}\right)\left({x}−\mathrm{1}\right)=\mathrm{2}{x}\left({x}+\mathrm{4}\right) \\ $$$$ \\ $$$$ \\ $$ Answered by Yozzia last updated…
Question Number 7230 by Yozzia last updated on 17/Aug/16 $${An}\:{Interesting}\:{App}: \\ $$$${I}'{ve}\:{recently}\:{downloaded}\:{the}\:{app}\:{called} \\ $$$$'{Math}\:{Tricks}'\:{from}\:{the}\:{Google}\:{Play}\:{Store}. \\ $$$${If}\:{you}'{d}\:{like}\:{to}\:{improve}\:{your}\:{speed} \\ $$$${and}\:{skill}\:{in}\:{the}\:{mental}\:{calculations} \\ $$$${arena},\:{I}\:{think}\:{this}\:{app}\:{should}\:{be}\:{of} \\ $$$${interest}\:{to}\:{you}.\:{The}\:{app}\:{can}\:{throw}\:{you}\:{simple} \\ $$$${calculations}\:{that}\:{include}\:{the}\:{basic}\:{operations} \\…
Question Number 138302 by mathdave last updated on 12/Apr/21 $${if}\:\:\mathrm{tan}^{\mathrm{2}} {x}=\mathrm{1}+\mathrm{2tan}^{\mathrm{2}} {y} \\ $$$${show}\:{that} \\ $$$$\mathrm{cos2}{x}+\mathrm{sin}^{\mathrm{2}} {y}=\mathrm{0} \\ $$ Answered by Ñï= last updated on…
Question Number 7227 by FilupSmith last updated on 17/Aug/16 $${S}=\underset{{t}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{t}} {t}^{{k}} {k}^{−{n}} ,\:{k}>\mathrm{1} \\ $$$$\mathrm{Solve}\:\mathrm{for}\:{S} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 138296 by bobhans last updated on 12/Apr/21 $$\int\:\frac{{dx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}\:=? \\ $$ Answered by bemath last updated on 12/Apr/21 Answered by mathmax…
Question Number 138299 by cherokeesay last updated on 12/Apr/21 Answered by bemath last updated on 12/Apr/21 $${BD}\:=\:{r}_{\mathrm{1}} +\mathrm{2}\:\Leftrightarrow\:{r}_{\mathrm{1}} \left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\mathrm{2} \\ $$$${r}_{\mathrm{1}} =\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}−\mathrm{1}}=\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$${shaded}\:{area}=\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{2}\right)^{\mathrm{2}}…
Question Number 7222 by Tawakalitu. last updated on 16/Aug/16 $${What}\:{are}\:{the}\:{possible}\:{solution}\:{satisfying} \\ $$$${x}^{{y}} \:=\:{y}^{{x}} \\ $$ Commented by Tawakalitu. last updated on 17/Aug/16 $${I}\:{appreciate}\:{your}\:{effort}.\:{thanks}\:{sir}. \\ $$…
Question Number 138295 by greg_ed last updated on 12/Apr/21 $$\boldsymbol{\mathrm{hi}}\:! \\ $$$$\boldsymbol{\mathrm{Simplify}}\::\:{for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\:{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\:\frac{\mathrm{3}{k}+\mathrm{8}}{{k}\left({k}+\mathrm{2}\right)\mathrm{2}^{{k}} }\:. \\ $$ Answered by mr W last updated…
Question Number 7216 by peter james last updated on 16/Aug/16 Answered by Yozzia last updated on 16/Aug/16 $${ln}\left({x}+{y}\right)−{ln}\left({x}+\mathrm{3}\right)+{ln}\mid\frac{\mathrm{3}−{y}}{{x}+\mathrm{3}}\mid={B}−{ln}\mid{x}+\mathrm{3}\mid \\ $$$${Let}\:{u}=\frac{{x}+{y}}{{x}+\mathrm{3}}\Rightarrow{y}={u}\left({x}+\mathrm{3}\right)−{x} \\ $$$$\therefore\:{y}'={u}'\left({x}+\mathrm{3}\right)+{u}−\mathrm{1} \\ $$$$\Rightarrow{y}'+\mathrm{1}={u}+{u}'\left({x}+\mathrm{3}\right). \\…
Question Number 7213 by Yozzia last updated on 16/Aug/16 $${Let}\:{a}\:{and}\:{b}\:{be}\:{positive}\:{integers}\:{such} \\ $$$${that}\:{ab}+\mathrm{1}\:{divides}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} .\:{Show}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}+\mathrm{1}}\:{is}\:{the}\:{square}\:{of}\:{an}\:{integer}. \\ $$$$\left({IMO}\:\mathrm{1988}\:{Qu}.\mathrm{6}\right) \\ $$ Terms of Service…