Question Number 204384 by Abdullahrussell last updated on 15/Feb/24 Commented by Frix last updated on 15/Feb/24 $$\mathrm{35}^{\mathrm{36}} =\mathrm{10}{m}+\mathrm{5} \\ $$$$\mathrm{36}^{\mathrm{35}} =\mathrm{10}{n}+\mathrm{6} \\ $$$$\mathrm{4}^{\mathrm{10}{m}+\mathrm{5}} =\mathrm{100}{k}+\mathrm{24} \\…
Question Number 204349 by Abdullahrussell last updated on 14/Feb/24 Commented by mr W last updated on 14/Feb/24 $${you}\:{can}'{t}\:{prove}\:{something}\:{wrong}! \\ $$ Commented by mr W last…
Question Number 204350 by universe last updated on 14/Feb/24 Answered by mr W last updated on 14/Feb/24 $${P}\left({a}\right)={a}^{\mathrm{4}} +{a}^{\mathrm{2}} {b}+{ac}+{d}=\mathrm{1}\:\:\:\:…\left({i}\right) \\ $$$${P}\left({b}\right)={ab}^{\mathrm{3}} +{b}^{\mathrm{3}} +{bc}+{d}=−\mathrm{1}\:\:\:…\left({ii}\right) \\…
Question Number 204334 by universe last updated on 13/Feb/24 Answered by Frix last updated on 13/Feb/24 $${n}\rightarrow\infty\:\Rightarrow\:{a}_{{n}} =\sqrt{\mathrm{2}{a}_{{n}} }\:\Rightarrow\:{a}_{{n}} =\mathrm{2} \\ $$ Answered by MM42…
Question Number 204329 by mr W last updated on 13/Feb/24 $${solve}\:\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}=\left\{{x}\right\}+\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Answered by AST last updated on 13/Feb/24 $${x}\:{cannot}\:{be}\:{negative},{otherwise},{L}.{H}.{S}\:{and}\:{R}.{H}.{S} \\ $$$${will}\:{have}\:{opposite}\:{signs}.\:{x}\:{cannot}\:{also}\:{grow} \\ $$$${arbitrarily}\:{large},{otherwise}\:\frac{\mathrm{1}}{\left[{x}\right]}+\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\ll\frac{\mathrm{1}}{\mathrm{3}}…
Question Number 204330 by universe last updated on 13/Feb/24 Answered by AST last updated on 13/Feb/24 $$\mathrm{3}{a}+\mathrm{2}{b}+\mathrm{15}=\mathrm{0}\:\:\:\:\wedge\:\:\:\mathrm{6}{a}+\mathrm{2}{b}\leqslant\mathrm{0}\Rightarrow{b}\leqslant−\mathrm{3}{a} \\ $$$$\Rightarrow{b}=\frac{−\mathrm{15}−\mathrm{3}{a}}{\mathrm{2}}\leqslant−\mathrm{3}{a}\Rightarrow−\mathrm{15}−\mathrm{3}{a}\leqslant−\mathrm{6}{a}\Rightarrow\mathrm{3}{a}\leqslant\mathrm{15} \\ $$$$\Rightarrow{a}\leqslant\mathrm{5}\Rightarrow{max}\left({a}\mid{a}\in\mathbb{Z}\right)=\mathrm{5} \\ $$ Commented by…
Question Number 204270 by universe last updated on 10/Feb/24 Answered by witcher3 last updated on 10/Feb/24 $$\left(\mathrm{1}\right) \\ $$$$\mathrm{withe}\:\mathrm{recursion}\:\mathrm{We}\:\mathrm{can}\:\mathrm{easly}\:\mathrm{proov}\:\mathrm{that}\:\mathrm{a}_{\mathrm{n}} >\mathrm{0} \\ $$$$\mathrm{a}_{\mathrm{n}+\mathrm{1}} =\mathrm{f}\left(\mathrm{a}_{\mathrm{n}} \right);\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\:\mathrm{increase}\:\mathrm{function} \\…
Question Number 204249 by universe last updated on 10/Feb/24 Commented by Frix last updated on 10/Feb/24 $${P}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{7}} }{\mathrm{16}}−\frac{\mathrm{21}{x}^{\mathrm{5}} }{\mathrm{16}}+\frac{\mathrm{35}{x}^{\mathrm{3}} }{\mathrm{16}}−\frac{\mathrm{35}{x}}{\mathrm{16}} \\ $$$${Q}\left({x}\right)=\frac{\mathrm{5}{x}^{\mathrm{3}} }{\mathrm{16}}+\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{29}{x}}{\mathrm{16}}+\mathrm{1} \\…
Question Number 204187 by mnjuly1970 last updated on 08/Feb/24 $$ \\ $$$$\:\:\:\:\:\:{If}\:\:\:{p}\:,\:{q}\:,\:{r}\:>\mathrm{0}\:,\:\:{pqr}=\:\mathrm{1} \\ $$$$\:\:\:\:\: \\ $$$$\:\Rightarrow\:\:{min}\left(\frac{\:{p}^{\mathrm{3}} }{{q}+{r}}\:+\:\frac{{q}^{\:\mathrm{3}} }{{p}+{r}}\:+\frac{{r}^{\mathrm{3}} }{{p}+{q}}\:\right)=? \\ $$$$\:\:\:\ast\:{give}\:{a}\:{reason}\:\ast \\ $$ Answered by…
Question Number 204218 by mnjuly1970 last updated on 08/Feb/24 Answered by AST last updated on 08/Feb/24 $$\begin{pmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\begin{pmatrix}{\:{x}}\\{\:{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{ax}+{by}}\\{{cx}+{dy}}\end{pmatrix}=\begin{pmatrix}{\left(\mathrm{3}{a}+\mathrm{2}{c}\right){x}+\left(\mathrm{3}{b}+\mathrm{2}{d}\right){y}}\\{\left(\mathrm{4}{a}+{c}\right){x}+\left(\mathrm{4}{b}+{d}\right){y}}\end{pmatrix} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{3}{a}+\mathrm{2}{c}}&{\mathrm{3}{b}+\mathrm{2}{d}}\\{\mathrm{4}{a}+{c}}&{\mathrm{4}{b}+{d}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\Rightarrow\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}=\begin{pmatrix}{\frac{−\mathrm{1}}{\mathrm{5}}}&{\frac{\mathrm{2}}{\mathrm{5}}}\\{\frac{\mathrm{4}}{\mathrm{5}}}&{\frac{−\mathrm{3}}{\mathrm{5}}}\end{pmatrix} \\ $$$$\Rightarrow{A}^{−\mathrm{1}} =\mathrm{5}^{−\mathrm{1}} \begin{pmatrix}{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{4}}&{−\mathrm{3}}\end{pmatrix}=\begin{pmatrix}{−\mathrm{3}}&{\mathrm{6}}\\{\mathrm{12}}&{−\mathrm{9}}\end{pmatrix}=\begin{pmatrix}{\mathrm{4}}&{\mathrm{6}}\\{\mathrm{5}}&{\mathrm{5}}\end{pmatrix}…