Question Number 13647 by chux last updated on 22/May/17 $$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{5}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{3x}^{\mathrm{2}} +\mathrm{xy}+\mathrm{y}^{\mathrm{2}} =\mathrm{1}…..\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}\:\mathrm{find}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$ Commented…
Question Number 13645 by chux last updated on 21/May/17 $$\mathrm{please}\:\mathrm{is}\:\mathrm{factor}\:\mathrm{theorem}\:\mathrm{and}\:\mathrm{error} \\ $$$$\mathrm{and}\:\mathrm{trial}\:\mathrm{the}\:\mathrm{same}?\:\mathrm{please}\:\mathrm{help}\: \\ $$$$\mathrm{cause}\:\mathrm{i}\:\mathrm{think}\:\mathrm{theres}\:\mathrm{a}\:\mathrm{difference} \\ $$$$\mathrm{but}\:\mathrm{i}\:\mathrm{cant}\:\mathrm{explain}\:\mathrm{it}. \\ $$$$ \\ $$$$\mathrm{Thankz}. \\ $$ Terms of Service…
Question Number 144708 by mathdanisur last updated on 28/Jun/21 $${Simplify}:\:\:\left(\frac{\mathrm{1}}{\left({x}-\mathrm{2}\right)!}\:-\:\frac{\mathrm{1}}{\left({x}-\mathrm{1}\right)!}\right)\centerdot{x}! \\ $$ Answered by MJS_new last updated on 28/Jun/21 $$\left(\frac{\left({x}−\mathrm{1}\right){x}}{{x}!}−\frac{{x}}{{x}!}\right){x}!={x}^{\mathrm{2}} −\mathrm{2}{x}={x}\left({x}−\mathrm{2}\right) \\ $$ Commented by…
Question Number 144693 by mathdanisur last updated on 27/Jun/21 Answered by Rasheed.Sindhi last updated on 28/Jun/21 $${All}\:{natural}\:{numbers}\:{fall}\:{into}\:{the} \\ $$$${following}\:\mathrm{6}\:{catagories}\:{with}\:{respect} \\ $$$${to}\:{modulo}\:\mathrm{6}\:: \\ $$$$\mathrm{6}{k},\mathrm{6}{k}+\mathrm{1},\mathrm{6}{k}+\mathrm{2},\mathrm{6}{k}+\mathrm{3},\mathrm{6}{k}+\mathrm{4},\mathrm{6}{k}+\mathrm{5} \\ $$$${Obviously}\:\underset{=\mathrm{6}{k}}…
Question Number 79147 by jagoll last updated on 23/Jan/20 $$\sqrt{\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}+\sqrt{\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:}\leqslant\frac{\mathrm{2}}{\mathrm{x}} \\ $$ Answered by john santu last updated on 23/Jan/20 $$\sqrt{\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}^{\mathrm{2}} }}+\sqrt{\frac{{x}^{\mathrm{3}}…
Question Number 144682 by mathdanisur last updated on 27/Jun/21 $${Compare}:\:\:{x}=\frac{{sin}\left(\mathrm{3}\right)}{{sin}\left(\mathrm{5}\right)}\:\:{and}\:\:{y}=\frac{{cos}\left(\mathrm{3}\right)}{{cos}\left(\mathrm{5}\right)} \\ $$ Answered by mindispower last updated on 27/Jun/21 $$\mathrm{5}\:{degree} \\ $$$$\frac{\mathrm{5}\pi}{\mathrm{180}},\frac{\mathrm{3}\pi}{\mathrm{180}} \\ $$$${x}\overset{{f}} {\rightarrow}{tg}\left({x}\right),{f}'\left({x}\right)=\mathrm{1}+{tg}^{\mathrm{2}}…
Question Number 144676 by loveineq last updated on 27/Jun/21 $$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{{b}}{{ca}}\:\geqslant\:\frac{\mathrm{27}}{\mathrm{8}} \\ $$$$\left(\mathrm{Found}\:\mathrm{by}\:\mathrm{WolframAlpha}\right) \\ $$ Answered by ArielVyny last updated on 27/Jun/21 $$ \\…
Question Number 13591 by FilupS last updated on 21/May/17 $$\mathrm{for}: \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}: \\ $$$${r}=\mathrm{2sin}\left(\theta\right) \\ $$ Commented by FilupS last updated…
Question Number 144663 by mathdanisur last updated on 27/Jun/21 $${x}\in\left(\mathrm{0};\pi\right)\:{and}\:\left({a};{b}\right)\:{real}\:{numbers}\:{fixed}. \\ $$$${Find}\:{the}\:{range}\:{of}\:{function}: \\ $$$${g}\left({x}\right)=\:\frac{\left(\mathrm{1}+{a}^{\mathrm{2}} +{cot}^{\mathrm{2}} {x}\right)\centerdot\left(\mathrm{1}+{b}^{\mathrm{2}} +{cot}^{\mathrm{2}} {x}\right)}{\mathrm{1}\:+\:{cot}^{\mathrm{2}} {x}} \\ $$ Answered by mindispower last…
Question Number 144645 by mathdanisur last updated on 27/Jun/21 $$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}\:+\:{tanx}}{\mathrm{1}\:+\:{sinx}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{sinx}}}} =\:? \\ $$ Answered by liberty last updated on 27/Jun/21 $$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}}…