Question Number 49468 by munnabhai455111@gmail.com last updated on 07/Dec/18 Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18 $${O}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\:{A}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:{B}\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} ,{z}_{\mathrm{2}} \right) \\ $$$${O}\overset{\rightarrow}…
Question Number 49466 by munnabhai455111@gmail.com last updated on 07/Dec/18 $${show}\:{that}\:{the}\:{area}\:{of}\:{the}\:{triangle}\:{whose}\:{vertics}\:{area}\:\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:,\:\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} ,{z}_{\mathrm{1}} \right)\:,\:\left({x}_{\mathrm{2}} ,{y}_{\mathrm{2},} {z}_{\mathrm{2}} \right)\:{is}\:\mathrm{1}/\mathrm{2}\left(\sqrt{\Sigma\left({y}_{\mathrm{1}} {z}_{\mathrm{2}} −{y}_{\mathrm{2}} {z}_{\mathrm{1}} \right)^{\mathrm{2}} \:.}\right. \\ $$ Terms of…
Question Number 49394 by ajfour last updated on 06/Dec/18 Commented by ajfour last updated on 07/Dec/18 $${Find}\:{equation}\:{of}\:{maximum}\:{area} \\ $$$${ellipse}. \\ $$ Answered by ajfour last…
Question Number 49112 by Pk1167156@gmail.com last updated on 03/Dec/18 Commented by Pk1167156@gmail.com last updated on 03/Dec/18 $$\mathrm{find}\:\theta \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 49020 by ajfour last updated on 01/Dec/18 Commented by ajfour last updated on 01/Dec/18 $${If}\:{the}\:{inscribed}\:{ellipse}\:{is}\:{of}\:{maximum} \\ $$$${area}\:{with}\:{its}\:{major}\:{axis}\:{parallel}\:{to} \\ $$$${side}\:{PQ}\:{of}\:\bigtriangleup{PQR}\:{with}\:{PR}\:=\:{q}\:{and} \\ $$$${QR}\:={p}\:,\:{find}\:\boldsymbol{{a}},\:\boldsymbol{{b}}\:{of}\:{the}\:{ellipse}. \\ $$…
Question Number 48846 by ajfour last updated on 29/Nov/18 Commented by ajfour last updated on 29/Nov/18 $${Find}\:{maximum}\:{area}\:{of}\:\bigtriangleup{ABC}\:{in} \\ $$$${terms}\:{of}\:{radii}\:{a},\:{b}\:{and}\:{R}. \\ $$ Terms of Service Privacy…
Question Number 48656 by rahul 19 last updated on 26/Nov/18 Commented by rahul 19 last updated on 26/Nov/18 $${I}'{m}\:{getting}\:\left({b}\right)\:{but}\:{ans}.\:{is}\:\left({c}\right)! \\ $$ Commented by ajfour last…
Question Number 179515 by cortano1 last updated on 30/Oct/22 Answered by Acem last updated on 30/Oct/22 Commented by Acem last updated on 30/Oct/22 $$ \\…
Question Number 179458 by cortano1 last updated on 29/Oct/22 $$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{quadrant}\:\mathrm{between}\:\mathrm{the}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{relationship}\:\mathrm{curve}\:\mathrm{y}=\frac{\mathrm{5}}{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{5}}\:\mathrm{and}\:\mathrm{x}\:\mathrm{dont} \\ $$$$\mathrm{equal}\:\mathrm{0}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{5}\right) \\ $$ Commented by Acem…
Question Number 48359 by ajfour last updated on 22/Nov/18 Answered by mr W last updated on 23/Nov/18 $$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}}{{a}^{\mathrm{2}} }+\frac{{yy}'}{{b}^{\mathrm{2}}…