Question Number 8514 by Basant007 last updated on 14/Oct/16 $$\mathrm{show}\:\mathrm{that}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{4}} }}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$ Commented by swapnil last updated on 18/Oct/16 $$\mathrm{what}\:\mathrm{is}\:\beta\:\mathrm{here} \\…
Question Number 139587 by mathlove last updated on 29/Apr/21 Answered by qaz last updated on 29/Apr/21 $${f}\left(\mathrm{2}\right)=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{xdx}+{f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \right)\mid_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{5}=\frac{\mathrm{71}}{\mathrm{6}} \\…
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Question Number 139582 by mr W last updated on 29/Apr/21 $${Bug}\:{in}\:{version}\:\mathrm{2}.\mathrm{265} \\ $$$$\mathrm{1}.\:{the}\:{cursor}\:{keys}\:\uparrow\downarrow\:{seem}\:{not}\:{to}\:{work}. \\ $$$$\mathrm{2}.\:{when}\:{inserting}\:{new}\:\:{lines}\:{with} \\ $$$${Enter}\:{key},\:{the}\:{cursor}\:{doesn}'{t}\:{move}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 139577 by bemath last updated on 29/Apr/21 $$\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{number}\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(\mathrm{z}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{6}}+\mathrm{3}=\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{y}−\mathrm{2}\mid+\mid\mathrm{z}−\mathrm{3}\mid\: \\ $$ Answered by mr W last updated on 29/Apr/21 $${a}=\mid{x}−\mathrm{1}\mid\geqslant\mathrm{0},\:{b}=\mid{y}−\mathrm{2}\mid\geqslant\mathrm{0},\:{c}=\mid{z}−\mathrm{3}\mid\geqslant\mathrm{0} \\…
Question Number 74042 by liki last updated on 18/Nov/19 Commented by $@ty@m123 last updated on 18/Nov/19 $${I}\:{tried}\:{to}\:{send}\:{you}\:{a}\:{book}\:{but}\:{the}\:{above} \\ $$$${number}\:{is}\:{not}\:{recognised}\:{by}\:{whatsapp}. \\ $$$${Are}\:{you}\:{from}\:{Ivory}\:{coast}? \\ $$ Commented by…
Question Number 74040 by Learner-123 last updated on 18/Nov/19 $${Find}\:{orthogonal}\:{trajectories}\:{of}\:{the} \\ $$$${curves}:\:\left({x}−{c}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={c}^{\mathrm{2}} . \\ $$ Commented by Learner-123 last updated on 18/Nov/19 $${please}\:{help}……
Question Number 74041 by FCB last updated on 18/Nov/19 Commented by mathmax by abdo last updated on 18/Nov/19 $${let}\:{A}_{{n}} =\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\right)^{\frac{\mathrm{1}}{{sin}\left(\pi\sqrt{\mathrm{1}+{n}^{\mathrm{2}} }\right)}} \:\Rightarrow \\ $$$$\left.{ln}\left({A}_{{n}}…
Question Number 8503 by MNG last updated on 13/Oct/16 $${Q}.\:\mathrm{1}\:\:{sinA}+{sinB}+{sinC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos} \\ $$$$\frac{{B}}{\mathrm{2}}\:{cos}\:\frac{{C}}{\mathrm{2}}\:. \\ $$$${Q}.\mathrm{2}\:\:{cosA}\:{cosB}\:−\:{cosC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos}\frac{{B}}{\mathrm{2}} \\ $$$${cos}\frac{{C}}{\mathrm{2}}\:−\mathrm{1} \\ $$$$ \\ $$$${Q}.\mathrm{3}\:\:\frac{{sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}}{{sinA}+{sinB}+{sinC}}\:=\mathrm{8}{sin}\:\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{B}}{\mathrm{2}}\:{sin}\frac{{C}}{\mathrm{2}} \\ $$$$ \\…