Question Number 220114 by Spillover last updated on 05/May/25 Answered by mr W last updated on 06/May/25 Commented by mr W last updated on 06/May/25…
Question Number 220115 by Spillover last updated on 05/May/25 Answered by Spillover last updated on 06/May/25 Answered by Spillover last updated on 06/May/25 Answered by…
Question Number 220108 by Spillover last updated on 05/May/25 Answered by mr W last updated on 06/May/25 $$\frac{{xh}_{\mathrm{1}} }{\mathrm{2}}={b}\:\Rightarrow{xh}_{\mathrm{1}} =\mathrm{2}{b} \\ $$$$\left(\frac{{h}_{\mathrm{2}} }{{h}_{\mathrm{1}} }\right)^{\mathrm{2}} =\frac{{a}}{{b}}\:\Rightarrow{h}_{\mathrm{2}}…
Question Number 220110 by Spillover last updated on 05/May/25 Answered by Spillover last updated on 06/May/25 Answered by Spillover last updated on 06/May/25 Terms of…
Question Number 220111 by Spillover last updated on 05/May/25 Commented by Spillover last updated on 05/May/25 $${Determine}\:\:\:\frac{{m}}{{n}} \\ $$ Answered by Spillover last updated on…
Question Number 220104 by Spillover last updated on 05/May/25 Answered by Spillover last updated on 06/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220100 by hardmath last updated on 05/May/25 Answered by MrGaster last updated on 06/May/25 Commented by hardmath last updated on 07/May/25 $$ \\…
Question Number 220101 by hardmath last updated on 05/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220103 by Spillover last updated on 05/May/25 Answered by efronzo1 last updated on 06/May/25 $$\:\:\mathrm{tan}\:\beta\:=\:\frac{{a}−\mathrm{1}}{\mathrm{1}}\:=\:\frac{{a}}{\mathrm{1}}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\mathrm{tan}\:\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{tan}\:\mathrm{2}\beta\:=\:\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\:\:\alpha−\beta\:=\:\alpha+\beta−\mathrm{2}\beta \\ $$$$\:\:\mathrm{tan}\:\left(\alpha−\beta\right)\:=\:\frac{\mathrm{tan}\:\left(\alpha+\beta\right)−\mathrm{tan}\:\mathrm{2}\beta}{\mathrm{1}+\mathrm{tan}\:\left(\alpha+\beta\right).\:\mathrm{tan}\:\mathrm{2}\beta} \\ $$…
Question Number 220096 by Nicholas666 last updated on 05/May/25 $$ \\ $$$$\:\:\:\mathrm{let}\:{a},\:{b},\:{c},\:{d},\:{e}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{K}\:=\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:+\:{e}\:+\mathrm{1}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}\:\mathrm{that}; \\ $$$$\:\:\:\:\underset{\boldsymbol{{cyc}}} {\sum}\:\frac{\mathrm{1}}{\boldsymbol{{k}}−\boldsymbol{{a}}}\:<\:\frac{\mathrm{1}}{\mathrm{4}}\:\left(\frac{\sqrt[{\mathrm{4}\:\:\:\:}]{\boldsymbol{{e}}^{\mathrm{3}} \boldsymbol{{d}}^{\mathrm{3}} \boldsymbol{{c}}}}{\boldsymbol{{c}}^{\mathrm{3}/\mathrm{4}} \boldsymbol{{d}}^{\mathrm{1}/\mathrm{2}} \boldsymbol{{e}}^{\mathrm{1}/\mathrm{4}} \sqrt{\boldsymbol{{a}}}}\:+\:\frac{\sqrt[{\mathrm{4}\:\:\:}]{\boldsymbol{{d}}^{\:\mathrm{3}} \boldsymbol{{c}}^{\mathrm{2}}…