Question Number 202255 by SANOGO last updated on 23/Dec/23 $${calcul}\:{f}_{{n}} '\left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)={n}^{\alpha} {x}\left(\mathrm{1}−{x}\right)^{{n}\:} \: \\ $$ Answered by SANOGO last updated on 23/Dec/23…
Question Number 202249 by cortano12 last updated on 23/Dec/23 $$\:\:\:\:\mathrm{C}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{x}\:\sqrt[{\mathrm{3}}]{\mathrm{cos}\:\mathrm{x}}\:−\:\mathrm{sin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{5}} }\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202250 by MATHEMATICSAM last updated on 23/Dec/23 $$\left({a}^{\mathrm{2}} \:−\:{bc}\right){x}^{\mathrm{2}} \:+\:\mathrm{2}\left({b}^{\mathrm{2}} \:−\:{ca}\right){x}\:+\:\left({c}^{\mathrm{2}} \:−\:{ab}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{two}\:\mathrm{equal}\:\mathrm{roots}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{either}\: \\ $$$${b}\:=\:\mathrm{0}\:\mathrm{or}\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:\mathrm{3}. \\ $$ Answered by…
Question Number 202251 by BaliramKumar last updated on 23/Dec/23 $$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:+\:…………..\:+\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Commented by BaliramKumar last updated on 23/Dec/23 $$\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)}\:+\:\:…….+\:\frac{\mathrm{1}}{\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\mathrm{nd}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\}}\:=\:? \\ $$ Commented by MATHEMATICSAM…
Question Number 202276 by professorleiciano last updated on 23/Dec/23 Answered by professorleiciano last updated on 23/Dec/23 $${Area}\left({retangulo}\right)=\mathrm{4}×\mathrm{6}=\mathrm{24}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{I}\right)=\mathrm{3}×\mathrm{6}=\mathrm{18}/\mathrm{2}=\mathrm{9}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{II}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({triangulo}\:{III}\right)=\mathrm{3}×\mathrm{4}=\mathrm{12}/\mathrm{2}=\mathrm{6}{m}^{\mathrm{2}} \\ $$$${Area}\left({total}\right)=\mathrm{24}{m}^{\mathrm{2}}…
Question Number 202247 by SANOGO last updated on 23/Dec/23 $$\alpha>\mathrm{1} \\ $$$${calcul}\:\:\:{f}_{{n}} ^{'} \left({x}\right) \\ $$$${f}_{{n}} \left({x}\right)={n}^{\alpha} {x}\left(\mathrm{1}−{x}\right)^{{n}} \\ $$ Terms of Service Privacy Policy…
Question Number 202172 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\mathrm{log}_{\mathrm{12}} \mathrm{18}\:=\:\mathrm{A}\:\mathrm{and}\:\mathrm{log}_{\mathrm{24}} \mathrm{54}\:=\:\mathrm{B}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:\mathrm{AB}\:+\:\mathrm{5}\left(\mathrm{A}\:−\:\mathrm{B}\right)\:=\:\mathrm{1}. \\ $$ Answered by Rasheed.Sindhi last updated on 22/Dec/23 $$\mathrm{12}^{\mathrm{A}} =\mathrm{18}\:\&\:\mathrm{24}^{\mathrm{B}}…
Question Number 202205 by sonukgindia last updated on 22/Dec/23 Commented by mr W last updated on 22/Dec/23 $${people}\:{asked}\:{you}\:{questions}\:{about}\: \\ $$$${your}\:{posts},\:{but}\:{you}\:{just}\:{ignore}\:{them} \\ $$$${again}.\: \\ $$$$\Rightarrow{Q}\mathrm{202187} \\…
Question Number 202170 by sonukgindia last updated on 22/Dec/23 Answered by AST last updated on 22/Dec/23 $$\mathrm{112}={a}\mathrm{2}^{{m}} \Rightarrow{a}\mathrm{2}^{{m}−\mathrm{4}} =\mathrm{7}\Rightarrow\mathrm{2}^{{m}−\mathrm{4}} =\mathrm{1}\Rightarrow{m}=\mathrm{4};{a}=\mathrm{7} \\ $$$$\mathrm{11392}={b}\mathrm{2}^{{n}} =\mathrm{2}^{\mathrm{7}} ×\mathrm{89}\Rightarrow{n}=\mathrm{7};{b}=\mathrm{89} \\…
Question Number 202203 by SEKRET last updated on 22/Dec/23 $$\:\: \\ $$$$ \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2023}} =\:\overset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_} {\boldsymbol{\mathrm{abc}}………….} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\:=\:? \\ $$$$ \\ $$$$ \\…