Question Number 148868 by DELETED last updated on 01/Aug/21 Answered by DELETED last updated on 01/Aug/21 $$\mathrm{given}\:\mathrm{that},\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{AB}=\mathrm{AC}=\mathrm{BC}=\mathrm{6}\:\mathrm{cm}, \\ $$$$\mathrm{and}\:\mathrm{it}'\mathrm{s}\:\mathrm{10}\:\mathrm{cm}\:\mathrm{hight},\:\mathrm{calculate}\:\mathrm{it}'\mathrm{s}\:\mathrm{volume}. \\ $$ Commented by DELETED last…
Question Number 148864 by jlewis last updated on 31/Jul/21 $$\mathrm{A}\:\mathrm{random}\:\mathrm{variable}\:\mathrm{K}\:\mathrm{has}\:\mathrm{a}\:\mathrm{pdf}\: \\ $$$$\mathrm{f}\left(\mathrm{k}\right)=\left\{_{\mathrm{0}\:\:\:\:\:\:,\:\:\:\:\:\:\:\mathrm{otherwise}} ^{\mathrm{e}^{−\mathrm{k}\:\:\:,\:\:\:\:\:\:\:\:\mathrm{0}>\mathrm{k}} } \:\:\:\:\mathrm{find}\:\mathrm{E}\left(\mathrm{K}\right)\:\mathrm{and}\:\right. \\ $$$$\mathrm{the}\:\mathrm{cdf} \\ $$ Answered by Olaf_Thorendsen last updated on…
Question Number 83331 by oyemi kemewari last updated on 01/Mar/20 Commented by jagoll last updated on 01/Mar/20 $$\left(\mathrm{13}\right)\:\int\:\mathrm{sec}\:\mathrm{x}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{sec}\:\mathrm{x}\:\left(\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\right)\:\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\int\:\frac{\mathrm{d}\left(\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\right)}{\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\mid\:+\:\mathrm{c}…
Question Number 83329 by john santu last updated on 01/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{e}^{−\mathrm{2x}} −\left(\mathrm{1}+\mathrm{ax}\right)\right)}{\mathrm{x}^{\mathrm{2}} \:\left(\mathrm{1}+\mathrm{bx}\right)}\:=\:? \\ $$ Commented by jagoll last updated on 01/Mar/20 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{bx}}\:×\:\underset{{x}\rightarrow\mathrm{0}}…
Question Number 148861 by mathdanisur last updated on 31/Jul/21 Answered by dumitrel last updated on 01/Aug/21 $$\left(\mathrm{2}{x}\right)^{\mathrm{4}} +\mathrm{324}=\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{18}\right)\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{18}\right) \\ $$$${f}\left({x}\right)=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{18}\Rightarrow{f}\left({x}+\mathrm{3}\right)=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{18} \\…
Question Number 148860 by mnjuly1970 last updated on 31/Jul/21 $$ \\ $$$$\:\:\:{f}\left({x}\right):=\:\mid\:{x}^{\:\mathrm{2}} −\:\left({a}+\mathrm{1}\right){x}\:+\mathrm{2}{a}^{\:\mathrm{2}} −\mathrm{1}\mid \\ $$$$\:\:\:\:{is}\:{differentiable}\:{on}\:\left(\:\mathrm{1},\:\mathrm{2}\:\right). \\ $$$$\:\:{find}\:{the}\:{value}\left({s}\right)\:{of}\:\:'\:\:{a}\:\:'\:. \\ $$$$\:\:…….. \\ $$ Terms of Service…
Question Number 83327 by john santu last updated on 01/Mar/20 $$\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:−\mathrm{2x}\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{p}\left(\mathrm{p}+\mathrm{1}\right)\mathrm{y}\:=\:\mathrm{0}\: \\ $$$$\mathrm{in}\:\mathrm{descending}\:\mathrm{power}\:\mathrm{of}\:\mathrm{x}.\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{solution}? \\ $$ Commented by Joel578 last updated…
Question Number 148856 by gsk2684 last updated on 31/Jul/21 $${pls}\:{solve} \\ $$ Commented by gsk2684 last updated on 31/Jul/21 Commented by Kamel last updated on…
Question Number 83323 by john santu last updated on 01/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\left(\mathrm{3x}−\mathrm{2}\right)\left(\mathrm{x}−\sqrt{\mathrm{2}}\right)}\:−\:\mathrm{x}\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\:=? \\ $$$$ \\ $$ Commented by abdomathmax last updated on 01/Mar/20 $${let}\:{f}\left({x}\right)=\sqrt{\left(\mathrm{3}{x}−\mathrm{2}\right)\left({x}−\sqrt{\mathrm{2}}\right)}−{x}\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\…
Question Number 148858 by Tawa11 last updated on 31/Jul/21 Answered by EDWIN88 last updated on 01/Aug/21 Commented by EDWIN88 last updated on 01/Aug/21 $${shaded}\:{area}\:=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{2}^{\mathrm{2}} \right)\:+\mathrm{16}−\frac{\mathrm{1}}{\mathrm{4}}\pi\left(\mathrm{4}^{\mathrm{2}}…