Question Number 111734 by Aina Samuel Temidayo last updated on 04/Sep/20 $$\mathrm{A}\:\mathrm{chord}\:\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perpendicular} \\ $$$$\mathrm{bisector}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{length}\:\mathrm{18cm}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circle},\:\mathrm{has}\:\mathrm{length}. \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last updated…
Question Number 177253 by BaliramKumar last updated on 02/Oct/22 $$ \\ $$The base of a right pyramid is a hexagon of side 16 cm, and…
Question Number 46085 by peter frank last updated on 20/Oct/18 Commented by maxmathsup by imad last updated on 21/Oct/18 $${let}\:{S}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{cos}^{{n}} \theta\:{cos}\left({n}\theta\right)\:={Re}\left(\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{e}^{{in}\theta}…
Question Number 177092 by BaliramKumar last updated on 30/Sep/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 177045 by Ar Brandon last updated on 30/Sep/22 Commented by a.lgnaoui last updated on 30/Sep/22 $${DC}\:;\:\:{DM};\:\:{MK};{and}\:{AB} \\ $$ Commented by a.lgnaoui last updated…
Question Number 177044 by Ar Brandon last updated on 30/Sep/22 Answered by a.lgnaoui last updated on 30/Sep/22 $$\Sigma\:{angles}\:{interieures}=\Sigma{angles}\:{exterieures} \\ $$$$\frac{\measuredangle{B}}{\mathrm{2}}+\measuredangle{KPD}+\frac{\measuredangle{D}}{\mathrm{2}}=\mathrm{90} \\ $$$$\measuredangle{EDD}=\frac{\measuredangle{D}}{\mathrm{2}}\:\:\:\left({PK}\mid\mid\:{ED}\right) \\ $$$$\:\frac{\measuredangle{B}}{\mathrm{2}}+\measuredangle{D}=\mathrm{90} \\…
Question Number 177047 by Ar Brandon last updated on 30/Sep/22 Answered by som(math1967) last updated on 30/Sep/22 $${cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DC}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }{\mathrm{2}×\mathrm{7}{DC}}=\frac{\mathrm{24}+{DC}^{\mathrm{2}} }{\mathrm{14}{DC}} \\ $$$${again}\:\:{cosD}=\frac{\mathrm{7}^{\mathrm{2}} +{DB}^{\mathrm{2}}…
Question Number 177046 by Ar Brandon last updated on 30/Sep/22 Answered by som(math1967) last updated on 30/Sep/22 $${Area}\left(\bigtriangleup{ADE}+\bigtriangleup{BEC}\right)=\frac{\mathrm{1}}{\mathrm{2}}{AreaABCD} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}}{\mathrm{2}}\:\left({let}\right) \\ $$$${Area}\bigtriangleup{AEB}=\frac{\mathrm{1}}{\mathrm{2}}{ABCD}=\frac{{x}}{\mathrm{2}} \\ $$$${Area}\left({ADEP}+\bigtriangleup{ADE}\right) \\…
Question Number 177043 by Ar Brandon last updated on 30/Sep/22 Answered by a.lgnaoui last updated on 30/Sep/22 $$\left.\mathrm{A}\right){only}\:\mathrm{I} \\ $$ Commented by Rasheed.Sindhi last updated…
Question Number 176973 by Ar Brandon last updated on 28/Sep/22 Commented by Ar Brandon last updated on 28/Sep/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{triangle}. \\ $$ Terms of Service Privacy…