Question Number 191324 by Mingma last updated on 23/Apr/23 Answered by HeferH last updated on 23/Apr/23 $$\mathrm{r}\:=\:\mathrm{20} \\ $$$$\mathrm{S}\:=\:\frac{\mathrm{15}×\mathrm{20}}{\mathrm{2}}\:+\:\frac{\mathrm{7}×\mathrm{24}}{\mathrm{2}}\:=\:\mathrm{150}+\mathrm{84}\:=\:\mathrm{234}\:\mathrm{u}^{\mathrm{2}} \: \\ $$ Commented by Mingma…
Question Number 191302 by Mingma last updated on 22/Apr/23 Answered by a.lgnaoui last updated on 22/Apr/23 $$\boldsymbol{\mathrm{S}}_{\mathrm{1}} =\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{OAB}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{h}_{\mathrm{1}} \:\:\:\:=\frac{\mathrm{CD}×\mathrm{h}_{\mathrm{1}} }{\mathrm{4}}\:\:\:\left(\mathrm{h}_{\mathrm{1}} =\mathrm{OH}_{\mathrm{1}} \right) \\ $$$$\:\:\:\frac{\mathrm{CD}×\mathrm{h}_{\mathrm{1}} }{\mathrm{4}}=\mathrm{8}\:\:\:\Rightarrow\boldsymbol{\mathrm{CD}}.\boldsymbol{\mathrm{h}}_{\mathrm{1}}…
Question Number 191276 by Mingma last updated on 22/Apr/23 Answered by mr W last updated on 22/Apr/23 $$\frac{{c}}{{a}}=\frac{{a}+{b}}{{b}−{a}} \\ $$$$\Rightarrow{c}=\frac{{a}\left({a}+{b}\right)}{{b}−{a}} \\ $$ Commented by Mingma…
Question Number 191274 by Mingma last updated on 22/Apr/23 Commented by mr W last updated on 22/Apr/23 $${it}\:{can}\:{be}\:{proved}\:{that} \\ $$$${radius}\:{of}\:{red}\:{circle}\:{R}={blue}\:{length}\:{l} \\ $$$$\Rightarrow\:{area}\:{of}\:{red}\:{circle}\:=\pi{l}^{\mathrm{2}} \\ $$ Commented…
Question Number 60088 by ajfour last updated on 17/May/19 Commented by ajfour last updated on 17/May/19 $$\mathrm{Only}\:\mathrm{hemisphere}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{R}\:\mathrm{is}\:\mathrm{grassy}, \\ $$$$\mathrm{and}\:\mathrm{occupies}\:\mathrm{central}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{field}\:\mathrm{of}\:\mathrm{side}\:\mathrm{a}>\:\mathrm{2R}. \\ $$$$\mathrm{A}\:\mathrm{goat}\:\mathrm{tied}\:\mathrm{at}\:\mathrm{front}\:\mathrm{right}\:\mathrm{corner} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{rope}\:\mathrm{of}\:\mathrm{length}\:{l}=\:\frac{{a}}{\mathrm{2}}\:,…
Question Number 191143 by Mingma last updated on 19/Apr/23 Answered by HeferH last updated on 21/Apr/23 $$\mathrm{Say}\:\mathrm{CD}\:=\:\mathrm{a},\:\mathrm{DE}\:=\:\mathrm{b},\:\mathrm{AC}\:=\:\mathrm{c},\:\mathrm{AE}\:=\:\mathrm{d} \\ $$$$\:\ast\:\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow \\ $$$$\:\mathrm{5}\left(\mathrm{b}−\mathrm{a}\right)=\:\mathrm{ab} \\ $$$$\:\ast\:\mathrm{cd}\:=\:\mathrm{ab}+\mathrm{64} \\ $$$$\:\mathrm{Since}\:\mathrm{AD}\:\mathrm{is}\:\mathrm{bisector}:…
Question Number 191128 by Mingma last updated on 18/Apr/23 Answered by mr W last updated on 18/Apr/23 Commented by mr W last updated on 19/Apr/23…
Question Number 125593 by shaker last updated on 12/Dec/20 Answered by MJS_new last updated on 12/Dec/20 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\left({search}\:{the}\:{www}\:{for}\:{it}\right)\right] \\ $$$$=\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}=…
Question Number 191126 by Mingma last updated on 18/Apr/23 Commented by mahdipoor last updated on 18/Apr/23 $$\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} −\frac{\mathrm{4}\left(\mathrm{4}+\mathrm{3}+\mathrm{2}+\mathrm{1}\right)}{\mathrm{2}} \\ $$ Terms of…
Question Number 191076 by Mingma last updated on 17/Apr/23 Answered by mr W last updated on 18/Apr/23 $${touching}\:{point}\:{P}\left({p},\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right) \\ $$$$\mathrm{tan}\:\theta=−\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{{p}^{\mathrm{3}} } \\ $$$${r}={p}−{r}\:\mathrm{sin}\:\theta={p}−\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{4}+{p}^{\mathrm{6}} }}…