Question Number 71533 by pete last updated on 16/Oct/19 $$\mathrm{Three}\:\mathrm{interior}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{nonagon}\:\mathrm{are} \\ $$$$\mathrm{equal}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{six}\:\mathrm{is}\:\mathrm{1050}° \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{size}\:\mathrm{of}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equal}\:\mathrm{angles}. \\ $$ Answered by MJS last updated on 17/Oct/19 $$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{angles}\:\mathrm{is}\:\mathrm{1260}° \\…
Question Number 137065 by mr W last updated on 29/Mar/21 Commented by mr W last updated on 29/Mar/21 $${which}\:{number}\:{is}\:{larger}? \\ $$ Answered by mr W…
Question Number 5991 by sanusihammed last updated on 08/Jun/16 $${Solve}\:{the}\:{differential}\:{equation}\: \\ $$$$ \\ $$$${cos}^{\mathrm{2}} \left({x}\right)\:\frac{{dy}}{{dx}}\:+\:{y}\:=\:{tan}\left({x}\right) \\ $$ Answered by prakash jain last updated on 09/Jun/16…
Question Number 5989 by sanusihammed last updated on 08/Jun/16 $${Evaluate}\:\:\:\mathrm{10}\:×\mathrm{12}\:×\:\mathrm{14}\:×\:\mathrm{16}\:×\:\mathrm{18}\:×\:\mathrm{20}\:\:{into}\:{factorial}\:{form} \\ $$ Answered by prakash jain last updated on 08/Jun/16 $$\mathrm{10}×\mathrm{12}×\mathrm{14}×\mathrm{16}×\mathrm{18}×\mathrm{20} \\ $$$$=\mathrm{2}\left(\mathrm{5}×\mathrm{6}×\mathrm{7}×\mathrm{8}×\mathrm{9}×\mathrm{10}\right)=\frac{\mathrm{2}!\mathrm{10}!}{\mathrm{4}!} \\ $$…
Question Number 5988 by sanusihammed last updated on 08/Jun/16 $${If}\:\:\:\:{ur}\:=\:{log}\:{r} \\ $$$${show}\:{that}\:\:\:\sum_{{r}\:=\:\mathrm{1}} ^{\mathrm{10}} \:{ur}\:=\:{log}\mathrm{3628800} \\ $$$$ \\ $$$${please}\:{help}. \\ $$ Answered by Yozzii last updated…
Question Number 137056 by Ar Brandon last updated on 29/Mar/21 $$\:\:\:\:\mathrm{On}\:\mathrm{this}\:\mathrm{same}\:\mathrm{day}\:\mathrm{last}\:\mathrm{year}\:\mathrm{I}\:\mathrm{made}\:\mathrm{my}\:\mathrm{first}\:\mathrm{on}\:\mathrm{this}\:\mathrm{forum}\:\left(\mathrm{Q86484}\right) \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{was}\:\mathrm{very}\:\mathrm{astonished}\:\mathrm{by}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{which}\:\mathrm{I}\:\mathrm{received}. \\ $$$$\mathrm{I}\:\mathrm{underestimated}\:\mathrm{the}\:\mathrm{place}\:\mathrm{at}\:\mathrm{first}\:\mathrm{before}\:\mathrm{getting}\:\mathrm{to}\:\mathrm{know}\:\mathrm{it} \\ $$$$\mathrm{better}.\:\mathrm{And}\:\mathrm{I}\:\mathrm{realised}\:\mathrm{I}\:\mathrm{knew}\:\mathrm{nothing}.\: \\ $$$$\:\:\:\:\mathrm{I}'\mathrm{m}\:\mathrm{grateful}\:\mathrm{with}\:\mathrm{all}\:\mathrm{the}\:\mathrm{teachings}\:\mathrm{which}\:\mathrm{I}'\mathrm{ve}\:\mathrm{acquired}\:\mathrm{from}\: \\ $$$$\mathrm{you}\:\mathrm{all}.\:\mathrm{You}\:\mathrm{guys}\:\mathrm{are}\:\mathrm{just}\:\mathrm{so}\:\mathrm{amazing}.\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{once}\:\mathrm{more}. \\ $$$$\mathrm{Special}\:\mathrm{thanks}\:\mathrm{to}\:\mathrm{you}\:\mathrm{too}\:\mathrm{Mr}\:\mathrm{Tinku}-\mathrm{Tara}.\: \\ $$😉…
Question Number 137059 by Dwaipayan Shikari last updated on 29/Mar/21 $$\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+…}{\mathrm{4}^{\mathrm{2}} }+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{3}} }+..}{\mathrm{4}^{\mathrm{3}} }+\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }+…}{\mathrm{4}^{\mathrm{4}} }+..=\frac{\mathrm{3}}{\mathrm{4}}{log}\left(\mathrm{2}\right)−\frac{\pi}{\mathrm{8}} \\ $$ Terms of…
Question Number 71518 by TawaTawa last updated on 16/Oct/19 $$\mathrm{Express}\:\:\sqrt{\mathrm{28}}\:\:\mathrm{as}\:\mathrm{continued}\:\mathrm{fraction} \\ $$ Commented by Prithwish sen last updated on 16/Oct/19 $$\sqrt{\mathrm{28}}\:=\:\mathrm{1}+\:\sqrt{\mathrm{28}}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:=\:\mathrm{1}+\:\:\frac{\mathrm{27}}{\:\sqrt{\mathrm{28}}+\mathrm{1}}\:=\:\mathrm{1}+\frac{\mathrm{27}}{\mathrm{2}+\sqrt{\mathrm{28}}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{1}+\frac{\mathrm{27}}{\mathrm{2}+\frac{\mathrm{27}}{\:\sqrt{\mathrm{28}}+\mathrm{1}}}\:=\:\mathrm{1}+\frac{\mathrm{27}}{\mathrm{2}+\frac{\mathrm{27}}{\mathrm{2}+\frac{\mathrm{27}}{\mathrm{2}+\frac{\mathrm{27}}{\mathrm{2}+._{._{..}…
Question Number 5982 by Kasih last updated on 08/Jun/16 $${the}\:{center}\:{of}\:{circle}\:{in}\:\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0} \\ $$$${determine}\:{the}\:{equation}\:{of}\:{circle}\:{that} \\ $$$${passing}\:{through}\:\left(−\mathrm{1},\mathrm{3}\right),\left(\mathrm{7},−\mathrm{1}\right) \\ $$ Commented by Rasheed Soomro last updated on 08/Jun/16 $${the}\:{center}\:{of}\:{circle}\:{in}\:\mathrm{2}{x}+{y}−\mathrm{11}=\mathrm{0}…
Question Number 137055 by mathmax by abdo last updated on 29/Mar/21 $$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{dx}}{\left(\mathrm{1}+\mathrm{cosx}\:+\mathrm{3sinx}\right)^{\mathrm{2}} } \\ $$ Commented by MJS_new last updated on 30/Mar/21 $$\mathrm{0}\leqslant\left(\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{3sin}\:{x}\right)^{\mathrm{2}}…