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Category: Arithmetic

Question-109246

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Question Number 109246 by peter frank last updated on 22/Aug/20 Answered by Ar Brandon last updated on 22/Aug/20 $$\mathrm{a}.\:\mathrm{I}=\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{2}}}=\int\frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{1}\right)\sqrt{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}}} \\ $$$$\:\:\:\:\mathrm{x}+\mathrm{2}=\sqrt{\mathrm{2}}\mathrm{ch}\theta\Rightarrow\mathrm{dx}=\sqrt{\mathrm{2}}\mathrm{sh}\theta\mathrm{d}\theta \\ $$$$\:\:\:\mathrm{I}=\int\frac{\mathrm{sh}\theta\mathrm{d}\theta}{\left(\sqrt{\mathrm{2}}\mathrm{ch}\theta−\mathrm{1}\right)\sqrt{\mathrm{sh}^{\mathrm{2}}…

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Given-the-equations-of-twe-circles-C-1-x-2-y-2-6x-4y-9-0-and-C-2-x-2-y-2-2x-6y-9-a-Find-the-equation-of-the-circle-C-3-which-passes-through-the-centre-of-C-1-and-through-the

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Question Number 109075 by Rio Michael last updated on 20/Aug/20 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{twe}\:\mathrm{circles}\: \\ $$$${C}_{\mathrm{1}} \::\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:−\mathrm{6}{x}−\mathrm{4}{y}\:+\:\mathrm{9}\:=\:\mathrm{0}\:\mathrm{and}\:{C}_{\mathrm{2}} \::\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{6}{y}\:+\:\mathrm{9}. \\ $$$$\left(\mathrm{a}\right)\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:{C}_{\mathrm{3}} \:\mathrm{which}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{centre} \\ $$$$\mathrm{of}\:{C}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{of}\:{C}_{\mathrm{1}}…

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Question-43350

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Question Number 43350 by peter frank last updated on 10/Sep/18 Commented by maxmathsup by imad last updated on 10/Sep/18 $${let}\:\:{S}\:\left({x}\right)={x}\:+\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{3}} \:+…{nx}^{{n}} \:\Rightarrow\:\frac{{S}\left({x}\right)}{{x}}=\mathrm{1}+\mathrm{2}{x}\:+\mathrm{3}{x}^{\mathrm{2}} \:+….{nx}^{{n}−\mathrm{1}} \:{but}…

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Bobhans-1-Let-n-be-a-positive-integer-and-let-x-and-y-be-positive-real-number-such-that-x-n-y-n-1-Prove-that-k-1-n-1-x-2k-1-x-4k-k-1-n-1-y-2k-1-

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Question Number 108818 by bobhans last updated on 19/Aug/20 $$\:\:\frac{\boldsymbol{\mathcal{B}}{ob}\boldsymbol{{hans}}}{\Delta} \\ $$$$\left(\mathrm{1}\right){Let}\:{n}\:{be}\:{a}\:{positive}\:{integer},\:{and}\:{let}\:{x}\:{and}\:{y}\:{be}\:{positive}\:{real}\: \\ $$$${number}\:{such}\:{that}\:{x}^{{n}} \:+\:{y}^{{n}} \:=\:\mathrm{1}\:.\:{Prove}\:{that}\: \\ $$$$\left(\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}+{x}^{\mathrm{2}{k}} }{\mathrm{1}+{x}^{\mathrm{4}{k}} }\:\right)\left(\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}+{y}^{\mathrm{2}{k}} }{\mathrm{1}+{y}^{\mathrm{4}{k}}…

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