Question Number 126068 by benjo_mathlover last updated on 17/Dec/20 Answered by liberty last updated on 17/Dec/20 $${h}\left({x}\right)=\int_{\mathrm{1}} ^{\:{x}} {f}\left({t}\right){dt}\:\Leftrightarrow\:{h}\left(\mathrm{1}\right)=\int_{\mathrm{1}} ^{\:\mathrm{1}} {f}\left({t}\right){dt}\:=\:\mathrm{0} \\ $$ Terms of…
Question Number 60533 by Tawa1 last updated on 21/May/19 $$\mathrm{If}\:\:\mathrm{A},\:\mathrm{B},\:\mathrm{C}\:\:\mathrm{are}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{C}\:+\:\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}\:−\:\mathrm{B}\right)\:\:=\:\:\mathrm{2}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{A}\:\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B} \\ $$ Commented by malwaan last updated on 21/May/19 $$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}^{°} \\ $$$${cos}\left({x}+{y}\right)−{cos}\left({x}−{y}\right)=−\mathrm{2}{sin}\left({x}\right){sin}\left({y}\right) \\…
Question Number 191602 by ajfour last updated on 26/Apr/23 Commented by ajfour last updated on 26/Apr/23 $${Load}\:{is}\:{to}\:{be}\:{taken}\:{from}\:{A}\:{to}\:{B}. \\ $$$${If}\:{s}={h}\:,\:{find}\:{h}\:\:{in}\:{terms}\:{of}\:{a},{b}. \\ $$ Answered by mr W…
Question Number 126065 by Lordose last updated on 17/Dec/20 $$ \\ $$$$\mathrm{Show}\:\mathrm{that}:: \\ $$$$\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)\mathrm{log}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}}\mathrm{dx}\:=\:−\frac{\mathrm{3}}{\mathrm{16}}\zeta\left(\mathrm{4}\right) \\ $$$$\mathrm{Goodluck} \\ $$ Answered by mnjuly1970 last…
Question Number 191597 by Linton last updated on 26/Apr/23 $${Prove} \\ $$$$ \\ $$$$\mathrm{H}_{{x}} \:\rightarrow\:\mathrm{cot}\:\left(\:\mathrm{x}\:\right)\: \\ $$$$\mathrm{x}!\rightarrow\:\:\mathrm{sin}\:\left(\:\mathrm{x}\:\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 60527 by ajfour last updated on 21/May/19 Commented by ajfour last updated on 21/May/19 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{quadrilateral} \\ $$$$\mathrm{OAPB}.\:\mathrm{The}\:\mathrm{ellipse}\:\mathrm{equation}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{usual}\:\mathrm{one}. \\ $$ Answered by…
Question Number 191595 by MATHEMATICSAM last updated on 26/Apr/23 $${a}\:=\:\frac{{xy}}{{x}\:+\:{y}}\:,\:{b}\:=\:\frac{{xz}}{{x}\:+\:{z}}\:\mathrm{and}\:{c}\:=\:\frac{{yz}}{{y}\:+\:{z}}\:. \\ $$$$\mathrm{Represent}\:{x}\:\mathrm{in}\:{a},\:{b},\:{c}\:\mathrm{form}.\:\left[{x},\:{y},\:{z}\:\neq\:\mathrm{0}\right] \\ $$ Answered by mehdee42 last updated on 26/Apr/23 $$\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\:\:\:\left({i}\right)\:,\:\:\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{z}}\:\:\left({ii}\right)\:\:,\:\:\:\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\:\:\left({iii}\right) \\ $$$$\left({i}\right),\left({ii}\right)\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}\overset{\left({iii}\right)} {\Rightarrow}\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{x}}+\frac{\mathrm{1}}{{c}}…
Question Number 126056 by TITA last updated on 16/Dec/20 Commented by TITA last updated on 16/Dec/20 $$\mathrm{please}\:\mathrm{help} \\ $$ Answered by Olaf last updated on…
Question Number 191589 by MATHEMATICSAM last updated on 28/Apr/23 $${a}^{{x}} \:=\:{bc},\:{b}^{{y}} \:=\:{ca},\:{c}^{{z}} \:=\:{ab}. \\ $$$$\mathrm{Prove}\:\mathrm{that},\:\frac{{x}}{\mathrm{1}\:+\:{x}}\:+\:\frac{{y}}{\mathrm{1}\:+\:{y}}\:+\:\frac{{z}}{\mathrm{1}\:+\:{z}}\:=\:\mathrm{2}. \\ $$$$\left(\mathrm{Without}\:\mathrm{using}\:\mathrm{log}\right) \\ $$$${a}\:\neq\:{b}\:\neq\:{c} \\ $$ Commented by mr W…
Question Number 126052 by fajri last updated on 16/Dec/20 $$ \\ $$$$\mathrm{on}\:\mathrm{any}\:\mathrm{trapezoid}\:\mathrm{ABCD}\:\mathrm{points}\:\mathrm{E}\:\mathrm{and}\: \\ $$$$\mathrm{F}\:\mathrm{are}\:\mathrm{located}\:\mathrm{on}\:\mathrm{CD}\:\mathrm{so}\:\mathrm{that}\:\mathrm{AD}\:\mathrm{is} \\ $$$$\mathrm{paral}{l}\mathrm{el}\:\mathrm{to}\:\mathrm{BE}\:\mathrm{and}\:\mathrm{AF}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\: \\ $$$$\mathrm{BC}.\mathrm{Point}\:\mathrm{H}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{of}\:\mathrm{AF}\: \\ $$$$\mathrm{an}{d}\:\mathrm{BE}\:\mathrm{point}\:\mathrm{G}\:\mathrm{is}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{of}\: \\ $$$$\mathrm{AC}\:\mathrm{and}\:\mathrm{BE}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{4}\:\mathrm{cm} \\ $$$${an}\mathrm{d}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{CD}\:\mathrm{is}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{what}\:\mathrm{is} \\…