Question Number 126992 by Tinku Tara last updated on 26/Dec/20 $$\mathrm{App}\:\mathrm{Information}: \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{unpublished}\:\mathrm{free}\:\mathrm{versoon} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{app}\:\mathrm{from}\:\mathrm{playstore}. \\ $$$$\mathrm{Existing}\:\mathrm{users}\:\mathrm{can}\:\mathrm{still}\:\mathrm{see}\:\mathrm{the} \\ $$$$\mathrm{app}\:\mathrm{on}\:\mathrm{playstore}. \\ $$$$\mathrm{A}\:\mathrm{paid}\:\mathrm{version}\:\mathrm{will}\:\mathrm{soon}\:\mathrm{be}\:\mathrm{available}. \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{update}\:\mathrm{pinned}\:\mathrm{message}\:\mathrm{once} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{available}.…
Question Number 192514 by Tomal last updated on 19/May/23 Commented by Skabetix last updated on 19/May/23 $$\frac{\mathrm{2}{log}_{\mathrm{2}} \left(\mathrm{7}\right)}{\mathrm{3}} \\ $$ Commented by Tomal last updated…
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Question Number 126963 by ZiYangLee last updated on 25/Dec/20 $$\mathrm{If}\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{5}×\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}×\mathrm{9}}+\ldots+\frac{\mathrm{1}}{\mathrm{2017}×\mathrm{2019}}=\frac{{x}}{{y}} \\ $$$$\mathrm{where}\:\frac{{x}}{{y}}\:\mathrm{is}\:\mathrm{in}\:\mathrm{its}\:\mathrm{lower}\:\mathrm{terms},\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}+{y}. \\ $$ Answered by Olaf last updated on 25/Dec/20 $$\mathrm{S}\:=\:\underset{{n}=\mathrm{0}} {\overset{\mathrm{1008}}…
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Question Number 61391 by Prithwish sen last updated on 02/Jun/19 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{best}\:\mathrm{book}\:\mathrm{for}\:\mathrm{learning}\:\mathrm{advanced} \\ $$$$\mathrm{calculus}\:? \\ $$ Commented by ajfour last updated on 02/Jun/19 $${Erwin}\:{Kreyszig}\:\left({wiley}\:{publication}\right) \\ $$…
Question Number 126916 by mohammad17 last updated on 25/Dec/20 Commented by mr W last updated on 25/Dec/20 $${p}=\frac{\mathrm{3}!}{\mathrm{4}!}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy…
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Question Number 126911 by bramlexs22 last updated on 25/Dec/20 Commented by bramlexs22 last updated on 25/Dec/20 $${thanks}\: \\ $$ Commented by bramlexs22 last updated on…