Question Number 61511 by Kunal12588 last updated on 03/Jun/19 $${cos}^{−\mathrm{1}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\:{cos}^{−\mathrm{1}} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} }=\mathrm{120}°=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{Find}\:\:{x} \\ $$ Answered by MJS last updated…
Question Number 192580 by Frix last updated on 21/May/23 $$\mathrm{This}\:\mathrm{might}\:\mathrm{be}\:\mathrm{helpful}: \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{easily}\:\mathrm{calculate}\:{z}_{\mathrm{1}} ^{{z}_{\mathrm{2}} } \:\mathrm{with}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\in\mathbb{C}: \\ $$$$\mathrm{Transform}\:{z}_{\mathrm{1}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{and}\:{z}_{\mathrm{2}} ={a}+{b}\mathrm{i}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} ^{{z}_{\mathrm{2}}…
Question Number 127047 by mohammad17 last updated on 26/Dec/20 $${how}\:{can}\:{graph}\:{this}\: \\ $$$$\left[{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{9}\:\right]\:{pleas}\:{sir}\:{help}\:{me}\:{with}\:{details}\:? \\ $$ Answered by benjo_mathlover last updated on 26/Dec/20 $${if}\:{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 192582 by mathdave last updated on 21/May/23 Commented by mathdave last updated on 21/May/23 $${aAnyone}\:{to}\:{help}\:{me}\:{with}\:{this} \\ $$ Commented by Frix last updated on…
Question Number 192572 by naka3546 last updated on 21/May/23 $$\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…