Question Number 120091 by ZiYangLee last updated on 29/Oct/20 $$\mathrm{When}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left({x}−\mathrm{1}\right)\left({x}+\mathrm{2}\right),\: \\ $$$$\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\left({x}+\mathrm{3}\right),\:\mathrm{and}\:\mathrm{when}\:{f}\left({x}\right) \\ $$$$\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right),\:\mathrm{the}\:\mathrm{remainder} \\ $$$$\mathrm{is}\:\left(\mathrm{2}{x}+\mathrm{1}\right). \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right). \\ $$ Answered…
Question Number 120089 by ZiYangLee last updated on 29/Oct/20 $$\mathrm{If}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{satisfies} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {xf}\left({x}\right){dx}=\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({f}\left({x}\right)\right)^{\mathrm{2}} {dx}\geqslant\mathrm{4} \\ $$ Terms of…
Question Number 120071 by SOMEDAVONG last updated on 29/Oct/20 Answered by bramlexs22 last updated on 29/Oct/20 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\mathrm{1}+\frac{\mathrm{1}}{{x}}}{\mathrm{2}\left({x}−\mathrm{1}\right)}\:=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{1}−{x}}{\mathrm{2}{x}\left({x}−\mathrm{1}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{−\mathrm{1}}{\mathrm{2}{x}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Terms…
Question Number 185601 by MCH last updated on 24/Jan/23 $${Hello}\:{please}\:{I}\:{experienced}\:{a}\:{little} \\ $$$$\:{difficulty}\:{with}\:{this}\:{question}.\: \\ $$$$ \\ $$$${A}\:{charge}\:{Qa}\:=−\mathrm{20}{mC}\:\:{is}\:{located}\:{at}\:\left(−\mathrm{6}\:\mathrm{4}\:\mathrm{7}\right)\:{and}\: \\ $$$${a}\:{charge}\:{Qb}=\mathrm{50}{mC}\:{at}\:\left(\mathrm{5}\:\mathrm{8}\:−\mathrm{2}\right).\: \\ $$$$\mathrm{I}{f}\:{distances}\:{are}\:{given}\:{in}\:{meters}\: \\ $$$${find}\:\left({a}\right)\overset{−} {{R}ab}\:\left({b}\right){Rab} \\ $$$$\:\left({c}\right)\:{The}\:{vector}\:{force}\:{exerted}\:{on}\:…
Question Number 54526 by Gulay last updated on 05/Feb/19 Commented by Gulay last updated on 05/Feb/19 $$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$$$ \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 120050 by bramlexs22 last updated on 29/Oct/20 $$\:\left({i}\right)\:{y}''−\mathrm{4}{y}'+\mathrm{5}{y}=\mathrm{4sin}\:^{\mathrm{2}} \mathrm{4}{x} \\ $$$$\:\left({ii}\right)\:\frac{{x}}{\mathrm{2}}+\mathrm{1}\:=\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}\: \\ $$ Answered by bemath last updated on 29/Oct/20 $$\left(\bullet\right)\:\frac{{x}+\mathrm{2}}{\mathrm{2}}\:=\:\sqrt{\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}\:{defined}\:{on}\:{x}\:\geqslant−\mathrm{2}\:…
Question Number 54517 by hassentimol last updated on 05/Feb/19 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$ \\ $$$$\frac{\sqrt{{h}+\mathrm{1}}−\mathrm{1}}{{h}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{h}+\mathrm{1}}+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{please}… \\ $$ Answered by Kunal12588 last updated…
Question Number 120037 by floor(10²Eta[1]) last updated on 28/Oct/20 $$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{R}>\mathrm{0},\:\mathrm{x}_{\mathrm{0}} >\mathrm{0},\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{R}}{\mathrm{x}_{\mathrm{n}} }+\mathrm{x}_{\mathrm{n}} \right),\:\mathrm{n}\geqslant\mathrm{0} \\ $$$$\mathrm{Prove}:\:\mathrm{For}\:\mathrm{n}\geqslant\mathrm{1},\:\mathrm{x}_{\mathrm{n}} >\mathrm{x}_{\mathrm{n}+\mathrm{1}} >\sqrt{\mathrm{R}}\:\mathrm{and} \\ $$$$\mathrm{x}_{\mathrm{n}} −\sqrt{\mathrm{R}}\leqslant\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }\:\frac{\left(\mathrm{x}_{\mathrm{0}} −\sqrt{\mathrm{R}}\right)^{\mathrm{2}}…
Question Number 185574 by mnjuly1970 last updated on 23/Jan/23 $$ \\ $$$$\:\:\:{f}\:\left({x}\:\right)=\:{cos}\left(\mathrm{2}\pi{x}\right)+\:{sin}\left(\mathrm{2}\pi{x}\right)\:+\sqrt{\:\lfloor{x}\rfloor\:+\lfloor−{x}\:\rfloor} \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:{find}\:\:\:\:\:\:\:\:{R}_{\:{f}} \:=? \\ $$ Commented by mahdipoor last updated on…
Question Number 120028 by Don08q last updated on 28/Oct/20 Answered by bramlexs22 last updated on 29/Oct/20 $$\left({a}\right)\:{say}\:{three}\:{particular}\:{women}\:{is}\:\begin{cases}{{w}_{\mathrm{1}} }\\{{w}_{\mathrm{2}} }\\{{w}_{\mathrm{3}} }\end{cases} \\ $$$$\blacksquare={woman}\:{particular} \\ $$$$\Box=\:{the}\:{other}\:{person} \\…