Question Number 168906 by cortano1 last updated on 21/Apr/22 Answered by mr W last updated on 21/Apr/22 $$\mathrm{tan}\:\frac{{B}}{\mathrm{2}}=\frac{\frac{{b}}{{c}}}{\mathrm{1}+\frac{{a}}{{c}}}=\frac{{b}}{{a}+{c}}=\frac{{b}}{{a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\mathrm{tan}\:\frac{{A}}{\mathrm{2}}=\frac{\frac{{a}}{{c}}}{\mathrm{1}+\frac{{b}}{{c}}}=\frac{{a}}{{b}+{c}}=\frac{{a}}{{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\…
Question Number 103171 by bemath last updated on 13/Jul/20 Commented by bobhans last updated on 13/Jul/20 $$\left(\mathrm{2}{a}\right)\:{a}\:{sector}\:{of}\:{a}\:{circle}\:{with}\:{radius}\:{r}\:{has} \\ $$$${perimeter}\:\mathrm{20}\:{units}\:{and}\:{area}\:\mathrm{21}\:{square} \\ $$$${units}\:.\:{Find}\:{all}\:{possible}\:{of}\:{r}\:{and}\:{the}\: \\ $$$${corresponding}\:{value}\:{of}\:\theta\:. \\ $$…
Question Number 103147 by ajfour last updated on 13/Jul/20 Answered by mr W last updated on 13/Jul/20 $${A}\left({a},\mathrm{0}\right) \\ $$$${B}\left(\mathrm{0},\sqrt{{p}^{\mathrm{2}} −{a}^{\mathrm{2}} }\right) \\ $$$${C}\left(\mathrm{0},{c}\right) \\…
Question Number 168558 by ajfour last updated on 13/Apr/22 Commented by ajfour last updated on 13/Apr/22 $${If}\:{the}\:{shaded}\:{area}\:{be}\:{A}.\:{Find}\:{max} \\ $$$${of}\:{y}_{{P}} \:\left({take}\:{curve}\:{to}\:{be}\:{y}={x}^{\mathrm{2}} \right). \\ $$ Answered by…
Question Number 103001 by I want to learn more last updated on 12/Jul/20 Answered by ajfour last updated on 12/Jul/20 $${let}\:{new}\:{depth}\:{be}\:\boldsymbol{{y}}. \\ $$$$\boldsymbol{{V}}=\pi{R}^{\mathrm{2}} {H}=\pi{R}^{\mathrm{2}} \left({y}−\frac{{h}}{\mathrm{2}}\right)+\pi\left(\frac{{h}}{\mathrm{2}}\right)\left({R}^{\mathrm{2}}…
Question Number 102980 by ajfour last updated on 12/Jul/20 Commented by ajfour last updated on 12/Jul/20 $${Find}\:\:{a}/{b}\:\:{if}\:{regions}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5} \\ $$$${have}\:{equal}\:{areas}. \\ $$ Commented by mr W…
Question Number 102910 by bramlex last updated on 11/Jul/20 $$\left(\mathrm{1}\right)\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$\left(\mathrm{2}\right)\:\underset{\mathrm{m}\:=\:\mathrm{1}} {\overset{\mathrm{n}} {\prod}}\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{m}\pi}{\mathrm{2n}+\mathrm{1}}\right)\: \\ $$ Answered by floor(10²Eta[1]) last updated…
Question Number 168233 by alcohol last updated on 06/Apr/22 $$\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{3}\::\:{u}_{\mathrm{1}} \:=\:\mathrm{4}}\\{{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} \:+\:\mathrm{6}{u}_{{n}−\mathrm{1}} }\end{cases} \\ $$$${Express}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n} \\ $$ Answered by mr W last…
Question Number 167520 by MikeH last updated on 18/Mar/22 $$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)\:=\:\int_{{x}} ^{\mathrm{2}{x}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{4}} }}{dt} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{state}\:\mathrm{its}\:\mathrm{domain} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{is}\:{f}\left({x}\right)\:\mathrm{even}\:\mathrm{or}\:\mathrm{odd}? \\ $$ Answered by aleks041103 last updated on…
Question Number 101851 by bemath last updated on 05/Jul/20 Answered by bobhans last updated on 05/Jul/20 $${OA}\:=\:{OB}\:=\:{radius} \\ $$$$\angle{CAO}\:=\:\angle{OBA}\:=\:\beta\:\rightarrow\angle{AOP}\:=\:\mathrm{2}\beta \\ $$$${PA}\:{is}\:{tangent}\:\rightarrow\angle{OAP}\:=\:\mathrm{90}^{{o}} \\ $$$${then}\:\angle{APO}\:=\:\mathrm{90}^{{o}} −\mathrm{2}\beta\: \\…