Question Number 50226 by Cheyboy last updated on 14/Dec/18 $$\mathrm{Three}\:\mathrm{prizes}\:\mathrm{are}\:\mathrm{awarded}\:\mathrm{each}\:\mathrm{for} \\ $$$$\mathrm{getting}\:\mathrm{more}\:\mathrm{than}\:\mathrm{80\%marks}, \\ $$$$\mathrm{98\%}\:\mathrm{attendance}\:\mathrm{and}\:\mathrm{good} \\ $$$$\mathrm{behaviour}\:\mathrm{in}\:\mathrm{the}\:\mathrm{college}.\mathrm{In}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{ways}\:\mathrm{the}\:\mathrm{prozes}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{awarded}\:\mathrm{if}\:\mathrm{15}\:\mathrm{student}\:\mathrm{of}\:\mathrm{the}\:\mathrm{college} \\ $$$$\mathrm{are}\:\mathrm{eligible}\:\mathrm{for}\:\mathrm{the}\:\mathrm{three}\:\mathrm{prizes}? \\ $$ Commented…
Question Number 181279 by mnjuly1970 last updated on 23/Nov/22 Answered by Rasheed.Sindhi last updated on 25/Nov/22 $$\sqrt[{\mathrm{3}}]{{x}+\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{{x}+\mathrm{3}}\:=\mathrm{0} \\ $$$$\begin{array}{|c|}{\underset{\:\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc}=\mathrm{0}} {{a}+{b}+{c}=\mathrm{0}}\:\:\:\:\:\:\:\:}\\\hline\end{array} \\ $$$${x}+\mathrm{1}+{x}+\mathrm{2}+{x}+\mathrm{3}−\mathrm{3}\sqrt[{\mathrm{3}}]{\left({x}+\mathrm{1}\right)}\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{2}}\:\sqrt[{\mathrm{3}}]{{x}+\mathrm{3}}\:=\mathrm{0}…
Question Number 181243 by Agnibhoo98 last updated on 23/Nov/22 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\frac{{x}\:−\:{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{x}\:−\:{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{x}\:−\:{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:=\:\mathrm{4}\left({a}\:+\:{b}\:+\:{c}\right) \\ $$ Answered by Frix last updated on 23/Nov/22 $${x}=\left({a}+{b}+{c}\right)^{\mathrm{2}}…
Question Number 50163 by Meritguide1234 last updated on 14/Dec/18 Answered by peter frank last updated on 14/Dec/18 $$\mathrm{let}\:\mathrm{x}=\frac{\mathrm{b}−\mathrm{c}}{\mathrm{a}}\:\:\:\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}−\mathrm{c}}\:\: \\ $$$$\mathrm{y}=\frac{\mathrm{c}−\mathrm{a}}{\mathrm{b}} \\ $$$$\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{b}}{\mathrm{c}−\mathrm{a}} \\…
Question Number 181221 by cortano1 last updated on 23/Nov/22 Answered by mr W last updated on 23/Nov/22 Commented by cortano1 last updated on 23/Nov/22 $$\mathrm{nicesolution}…
Question Number 181196 by depressiveshrek last updated on 22/Nov/22 $${Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,\:…{a}_{\mathrm{2022}} \:{be}\:{numbers} \\ $$$${ranging}\:{from}\:\left(\mathrm{0},\:+\infty\right)\:\backslash\:\left\{\mathrm{1}\right\},\:{for}\:{which} \\ $$$${the}\:{function}\:{f}\::\:\mathbb{R}\rightarrow\mathbb{R}\:{is}\:{defined}\:{as} \\ $$$${f}\left({x}\right)={a}_{\mathrm{1}} ^{{x}} +{a}_{\mathrm{2}} ^{{x}} +{a}_{\mathrm{3}} ^{{x}}…
Question Number 181182 by depressiveshrek last updated on 22/Nov/22 $${For}\:{what}\:{values}\:{of}\:{a}\:{does}\:{the}\:{system} \\ $$$${of}\:{equations}\:{only}\:{have}\:{one}\:{solution}: \\ $$$$\begin{cases}{{a}\left({x}^{\mathrm{4}} +\mathrm{1}\right)={y}+\mathrm{2}−\mid{x}\mid}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}}\end{cases} \\ $$ Answered by mr W last updated…
Question Number 181183 by Shrinava last updated on 22/Nov/22 $$\Omega_{\boldsymbol{\mathrm{n}}} \:=\:\begin{vmatrix}{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{…}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{2}^{\mathrm{2}} }&{\mathrm{2}^{\mathrm{3}} }&{…}&{\mathrm{2}^{\boldsymbol{\mathrm{n}}} }\\{\mathrm{1}}&{\mathrm{3}^{\mathrm{2}} }&{\mathrm{3}^{\mathrm{3}} }&{…}&{\mathrm{3}^{\boldsymbol{\mathrm{n}}} }\\{…}&{…}&{…}&{…}&{…}\\{\mathrm{1}}&{\mathrm{n}^{\mathrm{2}} }&{\mathrm{n}^{\mathrm{3}} }&{…}&{\mathrm{n}^{\boldsymbol{\mathrm{n}}} }\end{vmatrix}\:\:,\:\:\:\mathrm{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\boldsymbol{\mathrm{n}}}]{\frac{\Omega_{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\Omega_{\boldsymbol{\mathrm{n}}} }}\:…
Question Number 50089 by Cheyboy last updated on 13/Dec/18 $$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{16}}{\mid\mathrm{x}−\mathrm{8}\mid} \\ $$ Commented by Abdo msup. last updated on 14/Dec/18 $${let}\:{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{16}}{\mid{x}−\mathrm{8}\mid}\:\Rightarrow{f}\left({x}\right)=\frac{{x}^{\mathrm{2}} −\mathrm{8}{x}\:+\mathrm{2}{x}−\mathrm{16}}{\mid{x}−\mathrm{8}\mid}…
Question Number 181144 by Agnibhoo98 last updated on 22/Nov/22 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\left(\frac{\mathrm{3}{x}\:−\:\mathrm{28}}{\mathrm{3}{x}\:−\:\mathrm{26}}\right)^{\mathrm{3}} \:=\:\frac{{x}\:−\:\mathrm{10}}{{x}\:−\:\mathrm{8}} \\ $$ Answered by Rasheed.Sindhi last updated on 22/Nov/22 $$\left(\frac{\mathrm{3}{x}\:−\:\mathrm{28}}{\mathrm{3}{x}\:−\:\mathrm{26}}\right)^{\mathrm{3}} \:=\:\frac{{x}\:−\:\mathrm{10}}{{x}\:−\:\mathrm{8}} \\…