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Category: Arithmetic

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

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}}…

Question-43350

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}…

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-

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}}…

Question-108679

Question Number 108679 by Rasikh last updated on 18/Aug/20 Answered by Dwaipayan Shikari last updated on 18/Aug/20 $$\sqrt{\mathrm{1}+\mathrm{8}}\:=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}.\mathrm{4}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}.\mathrm{15}}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}}}….}}=\mathrm{3} \\…

Question-108594

Question Number 108594 by Rasikh last updated on 18/Aug/20 Answered by 1549442205PVT last updated on 18/Aug/20 $$\mathrm{From}\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{1we}\:\mathrm{have} \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\mathrm{abc}\left(\mathrm{1}\right).\mathrm{Apply}\:\mathrm{Cauchy}'\mathrm{s}\: \\ $$$$\mathrm{inequality}\:\mathrm{for}\:\mathrm{three}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\mathrm{we}\:\mathrm{have}\:: \\ $$$$\mathrm{abc}=\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\geqslant\mathrm{3}\:^{\mathrm{3}}…

Question-43036

Question Number 43036 by Rio Michael last updated on 06/Sep/18 Commented by Rio Michael last updated on 06/Sep/18 $${Find}\:{the}\:{area}\:{of}\:{one}\:{square}\:{of}\:{the}\:{squares}\left(\frac{\mathrm{1}}{\mathrm{4}}\:{of}\:{the}\:{big}\:{square}\right) \\ $$ Answered by $@ty@m last…