Question Number 57332 by problem solverd last updated on 02/Apr/19 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{such}\:\mathrm{that} \\ $$$${k}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\left({y}−\mathrm{2}{x}+\mathrm{1}\right)\left({y}+\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{hence}\:\mathrm{obtain}\: \\ $$$$\mathrm{the}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{resulting}\:\mathrm{circle}. \\ $$ Commented by…
Question Number 57329 by naka3546 last updated on 02/Apr/19 $$\underset{−\mathrm{1}} {\int}\overset{\mathrm{2}} {\:}\:\mid{x}\mid\:\lfloor{x}\rfloor\:{dx}\:\:=\:\:\:? \\ $$ Commented by turbo msup by abdo last updated on 02/Apr/19 $${I}=\int_{−\mathrm{1}}…
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Question Number 188392 by yawa7373 last updated on 28/Feb/23 Commented by Rasheed.Sindhi last updated on 28/Feb/23 $$\boldsymbol{{sir}}\:{mr}\:{W}\:,\:{please}\:{see}\:{my}\:{third}\:{answer} \\ $$$${to}\:{Q}#\mathrm{188327}\:{in}\:{which}\:{I}'{ve}\:{used} \\ $$$${derivative}. \\ $$ Commented by…
Question Number 122848 by ZiYangLee last updated on 20/Nov/20 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }+……+\frac{{n}}{{n}^{\mathrm{2}} }\right)=? \\ $$ Answered by nico last updated on 20/Nov/20 $$=\underset{{x}\rightarrow\mathrm{0}}…
Question Number 122843 by ZiYangLee last updated on 20/Nov/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{y}=\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{3cos}\:{x}+\mathrm{5} \\ $$ Commented by Dwaipayan Shikari last updated on 20/Nov/20 $$−\mathrm{1}\leqslant{cosx}\leqslant\mathrm{1} \\ $$$$−\mathrm{3}\leqslant−\mathrm{3}{cosx}\leqslant\mathrm{3}\:\:\:\:\:\:\:\:\:\: \\…
Question Number 122836 by sandy_delta last updated on 20/Nov/20 Commented by sandy_delta last updated on 20/Nov/20 $$\mathrm{please}\:\mathrm{help} \\ $$ Answered by Snail last updated on…
Question Number 188360 by 073 last updated on 28/Feb/23 Answered by Rasheed.Sindhi last updated on 28/Feb/23 $$\:\:\mathrm{S}_{\mathrm{N}} =\frac{\mathrm{N}}{\mathrm{2}}\left({a}+{l}\right) \\ $$$$\:\:\:\mathrm{A}=\frac{{n}−\mathrm{2}}{\mathrm{2}}\left(\mathrm{1}+{n}−\mathrm{2}\right)=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\:\:\:\mathrm{B}=\frac{{n}−\mathrm{14}}{\mathrm{2}}\left(\mathrm{15}+{n}\right)=\frac{\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}} \\ $$$$\mathrm{A}−\mathrm{B}=\frac{\left({n}−\mathrm{2}\right)\left({n}−\mathrm{1}\right)}{\mathrm{2}}−\frac{\left({n}−\mathrm{14}\right)\left({n}+\mathrm{15}\right)}{\mathrm{2}}=\mathrm{42} \\…
Question Number 188359 by 073 last updated on 28/Feb/23 Answered by Rasheed.Sindhi last updated on 28/Feb/23 $$\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+…+\mathrm{48}!\equiv{x}\left({mod}\:\mathrm{6}\right) \\ $$$${For}\:{k}\geqslant\mathrm{3},\:\mathrm{6}\mid{k}!\: \\ $$$$\Rightarrow\mathrm{1}!+\mathrm{2}!\equiv{x}\left({mod}\:\mathrm{6}\right) \\ $$$$\:\:\:\:\:\mathrm{1}+\mathrm{2}\equiv{x}\left({mod}\:\mathrm{6}\right. \\ $$$${x}=\mathrm{3}…
Question Number 122799 by aurpeyz last updated on 19/Nov/20 Answered by liberty last updated on 19/Nov/20 $${letting}\:{u}\:=\:\mathrm{1}+\mathrm{3}{x}^{−\mathrm{1}} \:\Rightarrow{du}=−\mathrm{3}{x}^{−\mathrm{2}} {dx} \\ $$$${L}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\int\:{u}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{du}\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{5}}{u}^{\frac{\mathrm{5}}{\mathrm{3}}} +\:{c} \\ $$$${thus}\:\underset{\mathrm{1}}…