Question Number 192572 by naka3546 last updated on 21/May/23 $$\mathrm{Let}\:\:{ABCD}\:\:\mathrm{be}\:\:\mathrm{a}\:\:\mathrm{rectangle}\:\:\mathrm{having}\:\:\mathrm{an}\:\mathrm{area}\:\:\mathrm{of}\:\:\mathrm{290}. \\ $$$$\mathrm{Let}\:\:{E}\:\:\mathrm{be}\:\:\mathrm{on}\:\:{BC}\:\:\mathrm{such}\:\:\mathrm{that}\:\:{BE}\::\:{BC}\:=\:\mathrm{3}\::\:\mathrm{2}. \\ $$$$\mathrm{Let}\:\:{F}\:\:\mathrm{be}\:\:\mathrm{on}\:\:{CD}\:\:\mathrm{such}\:\:\mathrm{that}\:\:{CF}\::\:{FD}\:=\:\mathrm{3}\::\:\mathrm{1}. \\ $$$$\mathrm{If}\:\:{G}\:\:\mathrm{is}\:\:\mathrm{the}\:\:\mathrm{intersection}\:\:\mathrm{of}\:\:{AE}\:\:\mathrm{and}\:\:{BF},\:\:\mathrm{compute} \\ $$$$\mathrm{the}\:\:\mathrm{area}\:\:\mathrm{of}\:\:\bigtriangleup{BEG}. \\ $$ Terms of Service Privacy Policy…
Question Number 127012 by Algoritm last updated on 26/Dec/20 Answered by Olaf last updated on 26/Dec/20 $$\begin{cases}{{y}_{\mathrm{1}} '\:=\:{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +\:{x}\:\left(\mathrm{1}\right)}\\{{y}_{\mathrm{2}} '\:=\:{y}_{\mathrm{1}} −\mathrm{2}{y}_{\mathrm{2}} +\mathrm{2}{x}\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\::\:{y}_{\mathrm{2}}…
Question Number 192550 by mokys last updated on 20/May/23 $${how}\:{can}\:{find}\:{the}\:{sum}\:\underset{{i}=\mathrm{1}} {\overset{{r}} {\sum}}\left(\mathrm{2}{v}_{{i}} +\mathrm{1}\right)\:? \\ $$ Answered by a.lgnaoui last updated on 20/May/23 $$\mathrm{S}_{\mathrm{i}} =\sum_{\mathrm{i}=\mathrm{1}} ^{\mathrm{i}=\mathrm{r}}…
Question Number 61461 by aliesam last updated on 02/Jun/19 Answered by MJS last updated on 03/Jun/19 $$\mathrm{sin}\:{x}\:+\mathrm{sin}\:{y}\:={a} \\ $$$$\mathrm{cos}\:{x}\:+\mathrm{cos}\:{y}\:={b} \\ $$$${u}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\wedge\:{v}=\mathrm{tan}\:\frac{{y}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\left[\mathrm{sin}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} };\:\mathrm{cos}\:\mathrm{2arctan}\:{t}\:=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}}…
Question Number 126992 by Tinku Tara last updated on 26/Dec/20 $$\mathrm{App}\:\mathrm{Information}: \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{unpublished}\:\mathrm{free}\:\mathrm{versoon} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{app}\:\mathrm{from}\:\mathrm{playstore}.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}} \\ $$ Commented by Tomal last updated…
Question Number 192503 by Skabetix last updated on 19/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 126963 by ZiYangLee last updated on 25/Dec/20 $$\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}}…
Question Number 61425 by naka3546 last updated on 02/Jun/19 Commented by alphaprime last updated on 02/Jun/19 9 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192492 by lordjonathan last updated on 19/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com