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Category: Arithmetic

prove-that-the-locus-of-middle-point-of-the-normal-chord-of-the-parabola-y-2-4ax-is-y-2-2a-4a-3-y-2-x-2a-

Question Number 47454 by peter frank last updated on 10/Nov/18 $$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{locus} \\ $$$$\mathrm{of}\:\mathrm{middle}\:\mathrm{point}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{normal}\:\mathrm{chord}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{parabola}\:\mathrm{y}^{\mathrm{2}} =\mathrm{4ax}\:\mathrm{is} \\ $$$$\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{2a}}+\frac{\mathrm{4a}^{\mathrm{3}} }{\mathrm{y}^{\mathrm{2}} }=\mathrm{x}−\mathrm{2a} \\ $$…

Question-47447

Question Number 47447 by peter frank last updated on 10/Nov/18 Answered by …. last updated on 10/Nov/18 $$\mathrm{turtle}+\mathrm{170}=\mathrm{table}+\mathrm{cat}……\left(\mathrm{1}\right) \\ $$$$\mathrm{cat}+\mathrm{130}=\mathrm{table}+\mathrm{turtle}…..\left(\mathrm{2}\right) \\ $$$$\mathrm{from}..\left(\mathrm{1}\right)\Rightarrow\mathrm{turtle}=\mathrm{table}+\mathrm{cat}−\mathrm{170}….\left(\mathrm{3}\right) \\ $$$$\mathrm{using}\:\left(\mathrm{3}\right)\:\mathrm{in}\:\left(\mathrm{2}\right)… \\…

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…