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Category: Arithmetic

Find-the-square-root-of-5-12i-hence-solve-z-2-4-i-z-5-6i-0-

Question Number 47433 by Tawa1 last updated on 09/Nov/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}\:\:−\:\mathrm{5}\:−\:\mathrm{12i},\:\:\mathrm{hence}\:\mathrm{solve}:\:\:\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{4}\:+\:\mathrm{i}\right)\mathrm{z}\:+\:\left(\mathrm{5}\:+\:\mathrm{6i}\right)\:=\:\mathrm{0} \\ $$ Commented by peter frank last updated on 10/Nov/18 $$\mathrm{let}\:\mathrm{a}+\mathrm{ib}=\sqrt{-\mathrm{5}−\mathrm{12i}} \\ $$$$\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}}…

Question-112810

Question Number 112810 by Aina Samuel Temidayo last updated on 09/Sep/20 Answered by Rasheed.Sindhi last updated on 10/Sep/20 $${t}_{\mathrm{1}} ,{t}_{\mathrm{2}} ,{t}_{\mathrm{3}} ,…,{t}_{{n}} \:;\:{n}\in\mathbb{E}…………\left(\mathrm{1}\right) \\ $$$${Sum}\:{of}\:{odd}\:{numbered}\:{terms}:…

Question-47111

Question Number 47111 by peter frank last updated on 04/Nov/18 Commented by MrW3 last updated on 04/Nov/18 $$\frac{\mathrm{4}!/\mathrm{2}!}{\mathrm{7}!/\mathrm{3}!/\mathrm{2}!}=\frac{\mathrm{4}!\mathrm{3}!}{\mathrm{7}!}=\frac{\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{7}×\mathrm{6}×\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{35}} \\ $$ Commented by peter frank last…

Find-the-sum-to-infinity-whose-n-th-term-is-n-2-n-1-

Question Number 46860 by 786786AM last updated on 01/Nov/18 $$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{whose}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{is}\:\frac{\mathrm{n}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\:. \\ $$ Commented by maxmathsup by imad last updated on 01/Nov/18 $${let}\:{s}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty}…

Question-46829

Question Number 46829 by peter frank last updated on 01/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18 $${eqn}\:{of}\:{focal}\:{chord}\:{is} \\ $$$${y}={tan}\theta\left({x}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\right)\:{focaus}\left(\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:,\mathrm{0}\right) \\…