Question Number 202750 by esmaeil last updated on 02/Jan/24 $${place}\left(\mathrm{1}\rightarrow\mathrm{25}\right) \\ $$$${in}\:{table}\left(\mathrm{5}×\mathrm{5}\right){in}\:{such}\:{away}\:{that} \\ $$$${the}\:{sum}\:{is}\:{constant}\:{in}\:{all}\:{directions}. \\ $$$$\left[\:\mid\underset{−} {\overset{−} {×}}\mid\:\right]\rightarrow\left({direction}\right) \\ $$ Commented by esmaeil last updated…
Question Number 202747 by MathedUp last updated on 02/Jan/24 $$\mathrm{Series}\:\:\Sigma\:\frac{\left(−\mathrm{1}\right)^{{h}−\mathrm{1}} }{{p}_{{h}} }\:\mathrm{is}\:\mathrm{Converge}??\:{h}\in\mathbb{Z}^{+} \:,\:{p}_{{h}} \in\mathbb{P}^{+} \\ $$$$\mathbb{P}^{+} =\:\mathrm{2}\:,\:\mathrm{3}\:,\:\mathrm{5}\:,\:\mathrm{7}\:,\:\mathrm{11}\:,\:\mathrm{13}\:,\:….\: \\ $$ Answered by Mathspace last updated on…
Question Number 202740 by sonukgindia last updated on 02/Jan/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202737 by Rasheed.Sindhi last updated on 02/Jan/24 $$\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{\underset{\overset{\overset{\mathrm{2}} {\mathrm{0}}} {\mathrm{24}}} {\mathbb{H}^{\mathbb{A}^{\mathbb{P}^{\mathbb{P}^{\mathbb{Y}^{\mathbb{N}^{\mathbb{E}} \mathbb{W}} \mathbb{Y}} \mathbb{E}} \mathbb{A}} \mathbb{R}} \:!}}\\\hline\end{array} \\ $$ Answered by witcher3 last…
Question Number 202765 by Mastermind last updated on 02/Jan/24 Answered by aleks041103 last updated on 02/Jan/24 $$\frac{{x}+{x}^{\mathrm{2}} +…+{x}^{{n}} −{n}}{{x}−\mathrm{1}}= \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{x}^{{k}} −\mathrm{1}}{{x}−\mathrm{1}}= \\…
Question Number 202760 by Mastermind last updated on 02/Jan/24 Answered by ajfour last updated on 02/Jan/24 $$\frac{\pi}{\mathrm{2}} \\ $$ Commented by Mastermind last updated on…
Question Number 202761 by Mastermind last updated on 02/Jan/24 Answered by shunmisaki007 last updated on 03/Jan/24 $${g}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$${f}\left({x}\right)={g}^{−\mathrm{1}} \left({x}\right) \\ $$$${f}'\left({x}\right)=\mathrm{sin}\left({x}\right) \\ $$$${f}\left({x}\right)={c}−\mathrm{cos}\left({x}\right)\:\mathrm{where}\:{c}\:\mathrm{is}\:\mathrm{constant}.…
Question Number 202762 by Mastermind last updated on 02/Jan/24 Answered by shunmisaki007 last updated on 03/Jan/24 $$\boldsymbol{{A}}=\begin{bmatrix}{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}\end{bmatrix} \\ $$$$\mathrm{adj}\left(\boldsymbol{{A}}\right)=\begin{bmatrix}{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{vmatrix}}\end{bmatrix}=\begin{bmatrix}{\mathrm{9}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{9}}\end{bmatrix} \\ $$$$\mathrm{adj}\left(\mathrm{adj}\left(\boldsymbol{{A}}\right)\right)=\begin{bmatrix}{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{9}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}\\{\begin{vmatrix}{\mathrm{0}}&{\mathrm{9}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}\end{vmatrix}}&{\begin{vmatrix}{\mathrm{9}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{9}}\end{vmatrix}}\end{bmatrix}=\begin{bmatrix}{\mathrm{27}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{27}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{27}}\end{bmatrix} \\ $$$$\mathrm{det}\left(\mathrm{adj}\left(\mathrm{adj}\left(\boldsymbol{{A}}\right)\right)\right)=\begin{vmatrix}{\mathrm{27}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{27}}&{\mathrm{0}}\\{\mathrm{0}}&{\:\mathrm{0}}&{\mathrm{27}}\end{vmatrix}=\mathrm{19},\mathrm{683} \\ $$$$\frac{\mathrm{det}\left(\mathrm{adj}\left(\mathrm{adj}\left(\boldsymbol{{A}}\right)\right)\right)}{\mathrm{5}}=\frac{\mathrm{19},\mathrm{683}}{\mathrm{5}}=\mathrm{3},\mathrm{936}\frac{\mathrm{3}}{\mathrm{5}}…
Question Number 202731 by MrGHK last updated on 02/Jan/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\:{x}\sqrt{\mathrm{1}−{x}}}{dx} \\ $$ Answered by witcher3 last updated on 02/Jan/24 $$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{\mathrm{a}−\mathrm{1}} \left(\mathrm{1}−\mathrm{t}\right)^{\mathrm{b}−\mathrm{1}}…
Question Number 202759 by LimPorly last updated on 02/Jan/24 $${Let}\:{A}=\left\{{x}\mid{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}<\mathrm{0},{x}\in{R}\right\},{B}=\left\{{x}\mid\mathrm{2}^{\mathrm{1}−{x}} +{a}\leqslant\mathrm{0},{x}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{7}\right){x}+\mathrm{5}\leqslant\mathrm{0},{x}\in{R}\right\} \\ $$$${If}\:{A}\subseteq{B}\:{Find}\:{the}\:{range}\:{of}\:{real}\:{number}\:{a} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com