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Author: Tinku Tara

Determine-all-pairs-x-y-of-integers-which-satisfy-x-2-y-2-16y-1-0-

Question Number 153763 by mathdanisur last updated on 10/Sep/21 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{pairs}\:\left(\mathrm{x};\mathrm{y}\right)\:\mathrm{of}\:\mathrm{integers} \\ $$$$\mathrm{which}\:\mathrm{satisfy} \\ $$$$\mid\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{y}^{\mathrm{2}} \mid\:-\:\sqrt{\mathrm{16y}\:+\:\mathrm{1}}\:=\:\mathrm{0} \\ $$ Answered by liberty last updated on 10/Sep/21…

S-k-1-n-sin-k-k-1-n-cos-k-where-k-k-1-n-is-an-arithmetic-progression-show-that-S-tan-where-1-n-k-1-n-k-is-the-arithmetic-mean-of-

Question Number 153758 by yeti123 last updated on 10/Sep/21 $${S}\:=\:\frac{\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}\left(\theta_{{k}} \right)}{\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\left(\theta_{{k}} \right)};\:\mathrm{where}\:\left(\theta_{{k}} \right)_{{k}\:=\:\mathrm{1}} ^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}. \\ $$$$\mathrm{show}\:\mathrm{that}\:{S}\:=\:\mathrm{tan}\left(\bar {\theta}\right) \\ $$$$\mathrm{where}\:\bar {\theta}\:=\:\frac{\mathrm{1}}{{n}}\underset{{k}\:=\:\mathrm{1}}…

Question-88218

Question Number 88218 by mathocean1 last updated on 09/Apr/20 Commented by mathocean1 last updated on 09/Apr/20 $$\mathrm{This}\:\mathrm{is}\:\mathrm{face}\:\mathrm{view}\:\mathrm{of}\:\mathrm{an}\:\mathrm{evacuation}\: \\ $$$$\mathrm{canal}.\:\mathrm{It}\:\mathrm{has}\:\mathrm{a}\:\mathrm{trapeze}\:\mathrm{form}.\:\mathrm{4m} \\ $$$$\mathrm{represents}\:\mathrm{its}\:\mathrm{small}\:\mathrm{base}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Determinate}\:\:\theta\:\in\:\left[\mathrm{60};\mathrm{90}\right]\:\mathrm{such}\:\mathrm{as}\:\mathrm{the}\: \\ $$$$\mathrm{capacity}\:\mathrm{of}\:\mathrm{canal}\:\mathrm{is}\:\mathrm{maximal}.…

Question-88214

Question Number 88214 by M±th+et£s last updated on 09/Apr/20 Answered by ajfour last updated on 09/Apr/20 $${AB}={t}\:\:,\:\:{radius}={r}\:,\:{AD}={DC}={b} \\ $$$${upper}\:{section}\:{of}\:{OB}={c} \\ $$$${t}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} \:\:\:;\:\:\:\mathrm{tan}\:\alpha=\frac{{r}}{{t}+{r}} \\ $$$$\frac{\sqrt{{r}^{\mathrm{2}}…