Question Number 216538 by CrispyXYZ last updated on 10/Feb/25 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{integer}\:\mathrm{solutions}\:\mathrm{of} \\ $$$$\mathrm{3}^{{m}} =\mathrm{2}{n}^{\mathrm{2}} +\mathrm{1}. \\ $$$$ \\ $$$${I}\:{only}\:{found}\:{m}=\mathrm{1},\:\mathrm{2},\:\mathrm{5}\:{by}\:{computer} \\ $$$${from}\:{m}=\mathrm{1}\:{to}\:{m}=\mathrm{30000}. \\ $$$${Is}\:{there}\:{any}\:{greater}\:{solutions}? \\ $$ Commented…
Question Number 216532 by sniper237 last updated on 11/Feb/25 $${Let}\:{f}\::\mathbb{R}_{+} \rightarrow\mathbb{R}\:{such}\:{as}\:{f}\left({xy}\right)={f}\left({x}\right)+{f}\left({y}\right) \\ $$$$\left.\mathrm{1}\right)\:{Prove}\:{that}\:\:{f}\:{is}\:{derivable}\:{iff}\:\: \\ $$$${f}\:{is}\:{derivable}\:{at}\:{x}=\mathrm{1}. \\ $$$$\left.\mathrm{2}\left.\right)\:{Prove}\:{that}\:{if}\:{so},\:{f}\left({x}\right)={Log}_{{a}} {x}\right)\: \\ $$$${where}\:{a}\:{is}\:{positive}\:{value}\:{to}\:{precise} \\ $$ Answered by maths2…
Question Number 216534 by Tawa11 last updated on 10/Feb/25 Commented by Tawa11 last updated on 10/Feb/25 In the figure, the block of mass M =…
Question Number 216560 by Tawa11 last updated on 10/Feb/25 An object of mass M, initially at rest at the coordinate origin, explodes into three parts.…
Question Number 216525 by Mingma last updated on 10/Feb/25 Answered by Rasheed.Sindhi last updated on 10/Feb/25 $$\mathrm{11}.\mathrm{1}\:\mathrm{A}^{\mathrm{2}} =\mathrm{A}\:,\:\mathrm{B}^{\mathrm{2}} =\mathrm{B}\:,\:\mathrm{AB}=\mathrm{BA} \\ $$$$\mathrm{AB}=\mathrm{BA}\Rightarrow\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{A}^{\mathrm{2}} \mathrm{B}^{\mathrm{2}} \\ $$$$\left(\mathrm{AB}\right)^{\mathrm{2}}…
Question Number 216526 by Mingma last updated on 10/Feb/25 Answered by Wuji last updated on 10/Feb/25 $$\mathrm{let}\:\mathrm{B}=\begin{bmatrix}{\mathrm{a}\:\:\:\:\:\:\:\mathrm{b}}\\{\mathrm{c}\:\:\:\:\:\:\:\:\mathrm{d}}\\{\mathrm{e}\:\:\:\:\:\:\:\:\mathrm{f}}\end{bmatrix}\:,\:\mathrm{C}=\begin{bmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{4}}\\{−\mathrm{1}\:\:\:\:\mathrm{0}}\end{bmatrix}\:: \\ $$$$\mathrm{AB}=\begin{bmatrix}{\mathrm{1}\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{2}}\\{\mathrm{0}\:\:−\mathrm{1}\:\:\:\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{a}\:\:\:\:\mathrm{b}}\\{\mathrm{c}\:\:\:\:\:\mathrm{d}}\\{\mathrm{e}\:\:\:\:\:\mathrm{f}}\end{bmatrix}=\begin{bmatrix}{\mathrm{a}+\mathrm{3c}+\mathrm{2e}\:\:\:\:\mathrm{b}+\mathrm{3d}+\mathrm{2f}}\\{−\mathrm{c}\:\:−\mathrm{e}\:\:\:\:\:\:\:\:\:−\mathrm{d}\:\:+\mathrm{f}\:\:\:\:}\end{bmatrix} \\ $$$$\mathrm{AB}=\mathrm{C}\:\:\Rightarrow\:\mathrm{a}+\mathrm{3c}+\mathrm{2e}=\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{b}+\mathrm{3d}+\mathrm{2f}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:−\mathrm{c}\:+\mathrm{e}=−\mathrm{1}\:\:\Rightarrow\mathrm{e}=\mathrm{c}−\mathrm{1}…
Question Number 216507 by sachipra last updated on 09/Feb/25 Commented by MathematicalUser2357 last updated on 10/Feb/25 Distance formula with no explanation Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216493 by issac last updated on 09/Feb/25 $$\mathrm{Res}_{{z}={c}} \left\{{f}\left({z}\right)\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:\mathrm{C}} \:{f}\left({z}\right)\mathrm{d}{z} \\ $$$$\mathrm{Res}_{{z}=\mathrm{1}} \left\{\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{3}} }\right\}=\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\frac{{z}^{\mathrm{21}} +{z}^{\mathrm{2}} +{z}+\mathrm{1}}{\left({z}−\mathrm{1}\right)^{\mathrm{3}} }\mathrm{d}{z} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\:\oint_{\:{C}} \:\:\frac{\frac{{z}^{\mathrm{21}}…
Question Number 216489 by manxsol last updated on 09/Feb/25 $${find}\:\:{residuo} \\ $$$$\:\:\:\:\:\:\:\:\frac{{x}^{\mathrm{21}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$ Commented by issac last updated on 09/Feb/25 $$\mathrm{Q216493}…
Question Number 216491 by Jubr last updated on 09/Feb/25 Answered by MrGaster last updated on 09/Feb/25 $$\left(\mathrm{1}\right): \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{1}−{t}^{\mathrm{2}} +\frac{{t}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{t}^{\mathrm{6}} }{\mathrm{3}!}+\ldots\right){dt}}{\:\sqrt{\mathrm{1}−{e}^{−{x}^{\mathrm{2}}…