Question Number 134287 by EDWIN88 last updated on 02/Mar/21 $$\mathrm{The}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{x}+\mathrm{y}−\mathrm{1}=\mathrm{0}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{cicle} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{6x}−\mathrm{8y}\:=\:\mathrm{0}\:\mathrm{at}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:.\:\mathrm{Then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{which}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{diameter} \\ $$ Answered by bramlexs22 last updated on 02/Mar/21…
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Question Number 68740 by ahmadshah last updated on 15/Sep/19 Commented by Prithwish sen last updated on 15/Sep/19 $$\sqrt{\mathrm{30}\sqrt{\mathrm{30}\sqrt{\mathrm{30}\sqrt{\mathrm{30}}\sqrt{\mathrm{30}\left(\mathrm{30}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }}}} \\ $$$$=\sqrt{\mathrm{30}\sqrt{\mathrm{30}\sqrt{\mathrm{30}\sqrt{\mathrm{30}}\sqrt{\left(\mathrm{30}\right)^{\mathrm{2}} }}}} \\ $$$$=\sqrt{\mathrm{30}\sqrt{\mathrm{30}\sqrt{\left(\mathrm{30}\right)^{\frac{\mathrm{5}}{\mathrm{2}}}…
Question Number 3205 by Rasheed Soomro last updated on 07/Dec/15 $$\mathcal{S}{uggest}\:{minimum}\:{number}\:{of}\:\:{weights}\:,{two}\:{peices}\:{of}\:{each},\: \\ $$$${to}\:{weigh}\:{upto}\:{at}\:{least}\:\mathrm{60}\:{kg}\left({in}\:{whole}\:{kg}'{s}\right)\:{in}\:{a}\:{common} \\ $$$${balance}. \\ $$ Commented by prakash jain last updated on 07/Dec/15…
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Question Number 3204 by Filup last updated on 07/Dec/15 $$\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{i}!=\mathrm{1}×\left(\mathrm{1}×\mathrm{2}\right)×…×\left(\mathrm{1}×\mathrm{2}×…×{n}\right) \\ $$$$=\mathrm{1}^{{n}} \mathrm{2}^{{n}−\mathrm{1}} \mathrm{3}^{{n}−\mathrm{2}} …\left({n}−\mathrm{1}\right)^{\mathrm{2}} {n} \\ $$$$\therefore\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{i}!=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\left({n}−{i}+\mathrm{1}\right)^{{i}} \\…
Question Number 134272 by mnjuly1970 last updated on 01/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nice}\:\:\:\mathrm{calculus} \\ $$$$\:\:\:\:\mathrm{prove}\:\mathrm{that}:\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}\:::\:\: =\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\pi{x}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}\:{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}\:}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{compute}:\:\:\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}\:=? \\…
Question Number 134269 by I want to learn more last updated on 01/Mar/21 $$\underset{\mathrm{k}\:\:\:=\:\:\:\:\mathrm{2}} {\overset{\mathrm{99}} {\sum}}\:\:\:\frac{\mathrm{k}^{\mathrm{2}} \:\:+\:\:\mathrm{1}}{\mathrm{k}^{\mathrm{2}} \:\:−\:\:\mathrm{1}} \\ $$ Answered by mr W last…
Question Number 3198 by Filup last updated on 07/Dec/15 $$\underset{{i}={a}} {\overset{{n}} {\prod}}{i}=?\:\:\:\:\:\mathrm{for}\:\mid{i}\mid<\mathrm{1},\:\:{a}\in\mathbb{R} \\ $$ Commented by Filup last updated on 07/Dec/15 $$=\frac{\mathrm{1}}{{a}}×\frac{\mathrm{1}}{{a}+\mathrm{1}}×…×\frac{\mathrm{1}}{{n}} \\ $$$$=\frac{\mathrm{1}}{\left(\frac{{n}!}{\left({a}−\mathrm{1}\right)!}\right)} \\…