Question Number 196066 by sniper237 last updated on 17/Aug/23 $$\begin{pmatrix}{\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}^{−\mathrm{8}} =\:\:? \\ $$ Answered by witcher3 last updated on 17/Aug/23 $$\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}=\mathrm{A} \\ $$$$\mathrm{A}^{\mathrm{2}} =\begin{pmatrix}{\mathrm{2}\:\:\:\:\:−\mathrm{3}}\\{−\mathrm{3}\:\:\:\:\mathrm{3}}\end{pmatrix}=\mathrm{3A}−\mathrm{I}_{\mathrm{2}} \\…
Question Number 196024 by York12 last updated on 16/Aug/23 $$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{{x}+\mathrm{1}} }\right)\right]=\lambda\:,\:{evalute}\:\lambda \\ $$ Answered by mr W last updated on 16/Aug/23…
Question Number 195998 by Frix last updated on 15/Aug/23 $$\mathrm{We}\:\mathrm{can}\:\mathrm{transform}\:\mathrm{to}\:\mathrm{get}\:\mathrm{rid}\:\mathrm{of}\:\mathrm{the}\:\sqrt[{\mathrm{3}}]{…} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} ={c}^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$$$\left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)^{\mathrm{3}} ={c} \\ $$$${a}+{b}+\mathrm{3}{a}^{\frac{\mathrm{1}}{\mathrm{3}}} {b}^{\frac{\mathrm{1}}{\mathrm{3}}} \left({a}^{\frac{\mathrm{1}}{\mathrm{3}}} +{b}^{\frac{\mathrm{1}}{\mathrm{3}}} \right)={c}…
Question Number 196015 by Ahmed777hamouda last updated on 15/Aug/23 Commented by Ahmed777hamouda last updated on 15/Aug/23 $$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{{i}}_{\boldsymbol{{D}}{p}} ,{i}_{{Dn}} \\ $$ Terms of Service Privacy Policy…
Question Number 195971 by mr W last updated on 14/Aug/23 $${if}\:{f}'\left({x}\right)=\frac{{f}\left({x}+{a}\right)−{f}\left({x}\right)}{{a}},\:{find}\:{f}\left({x}\right). \\ $$ Answered by jabarsing last updated on 14/Aug/23 $${hello}\:{mr}.{w}\:{dear} \\ $$$${f}\left({x}\right)={C}\:\:\:\:\left({conestant}\:{function}\right)=? \\ $$$${is}\:{true}?…
Question Number 195931 by mr W last updated on 13/Aug/23 Commented by mr W last updated on 13/Aug/23 Commented by mr W last updated on…
Question Number 195911 by Calculusboy last updated on 13/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195904 by York12 last updated on 13/Aug/23 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{2}^{{n}} \left(\mathrm{2}{n}\right)!}{\mathrm{3}^{\mathrm{2}{n}+\mathrm{1}} \left({n}+\mathrm{1}\right)!{n}!}\right]=\lambda \\ $$$${Evaluate}\:\left(\lambda\right) \\ $$ Answered by qaz last updated on 13/Aug/23…
Question Number 195898 by Calculusboy last updated on 12/Aug/23 Answered by qaz last updated on 13/Aug/23 $${a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} −{a}_{{n}+\mathrm{1}} {a}_{{n}} =\mathrm{2}\:\:\:\:\Rightarrow{a}_{{n}+\mathrm{2}} {a}_{{n}+\mathrm{1}} =\mathrm{2}{n}+{a}_{\mathrm{1}} {a}_{\mathrm{2}} =\mathrm{2}\left({n}+\mathrm{1}\right)…
Question Number 195820 by York12 last updated on 11/Aug/23 $${a},{b},{c}>\mathrm{0}\:\&{abc}=\mathrm{1},{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}+\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}+\frac{\mathrm{1}}{\mathrm{1}+{c}+{a}}\leqslant\mathrm{1} \\ $$ Answered by CrispyXYZ last updated on 12/Aug/23 $$\mathrm{let}\:{a}={x}^{\mathrm{3}} ,\:{b}={y}^{\mathrm{3}} ,\:{c}={z}^{\mathrm{3}} ,\:\mathrm{then}\:{xyz}=\mathrm{1}.…