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} \\ $$$$\mathrm{11392}={b}\mathrm{2}^{{n}} =\mathrm{2}^{\mathrm{7}} ×\mathrm{89}\Rightarrow{n}=\mathrm{7};{b}=\mathrm{89} \\…
Question Number 202203 by SEKRET last updated on 22/Dec/23 $$\:\: \\ $$$$ \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2023}} =\:\overset{\_\_\_\_\_\_\_\_\_\_\_\_\_\_} {\boldsymbol{\mathrm{abc}}………….} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\:=\:? \\ $$$$ \\ $$$$ \\…
Question Number 202198 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:−\:{b}\:=\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\gamma,\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{c}\:=\:\mathrm{0}\: \\ $$$$\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\frac{\alpha\:−\:\gamma}{\beta\:−\:\gamma}\:=\:\frac{\beta\:−\:\delta}{\alpha\:−\:\delta}\:\:. \\ $$ Answered by MM42 last updated on 22/Dec/23…
Question Number 202167 by tri26112004 last updated on 22/Dec/23 $$\underset{\mathrm{0}} {\int}^{\mathrm{1}} \underset{{x}} {\int}^{\mathrm{1}} {sin}\left({y}^{\mathrm{2}} \right){dydx}\:=\:¿ \\ $$ Answered by mnjuly1970 last updated on 22/Dec/23 $$\:{answer}:=\:{sin}^{\mathrm{2}}…
Question Number 202160 by Calculusboy last updated on 22/Dec/23 Answered by MathematicalUser2357 last updated on 22/Dec/23 $$\frac{\mathrm{1}}{\mathrm{7}!}\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{d}^{\mathrm{7}} }{{dx}^{\mathrm{7}} }\:\left(\mathrm{sin}\left(\mathrm{tan}\:{x}\right)−\mathrm{tan}\left(\mathrm{sin}\:{x}\right)\right) \\ $$ Terms of Service…
Question Number 202193 by MATHEMATICSAM last updated on 22/Dec/23 $$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\alpha\:+\:\delta,\:\beta\:+\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$${a}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \:=\:\mathrm{4}\left({b}\:−\:{q}\right). \\ $$ Answered by Rasheed.Sindhi…
Question Number 202161 by Calculusboy last updated on 22/Dec/23 Answered by som(math1967) last updated on 22/Dec/23 $$\frac{\left({m}+\mathrm{1}\right)!\left(\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\mathrm{2}{m}+\mathrm{3}\right.}{\mathrm{2}{m}\left({m}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{m}+\mathrm{1}\right)} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)!\frac{{m}+\mathrm{2}}{\mathrm{2}}×\mathrm{2}\left\{\mathrm{1}+\left({m}+\mathrm{2}−\mathrm{1}\right)\right\}}{\mathrm{2}{m}\left({m}+\mathrm{2}\right)\frac{\left({m}+\mathrm{1}\right)\left({m}+\mathrm{2}\right)}{\mathrm{2}}} \\ $$$$=\frac{\left({m}+\mathrm{1}\right)!\left({m}+\mathrm{2}\right)^{\mathrm{2}} }{{m}\left({m}+\mathrm{2}\right)^{\mathrm{2}} \left({m}+\mathrm{1}\right)} \\ $$$$=\frac{{m}\left({m}+\mathrm{1}\right)×\left({m}−\mathrm{1}\right)!}{{m}\left({m}+\mathrm{1}\right)}…