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
Question Number 202716 by Rasheed.Sindhi last updated on 01/Jan/24 $$\overline {\:\:{abcd}\:\:}{is}\:{a}\:{four}\:{digit}\:{number} \\ $$$${such}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\overline {\:{cd}\:} \\ $$$${and}\:\overline {\:{cd}\:}−\overline {\:{d}\:}=\overline {\:{ab}\:}. \\ $$$$\mathcal{F}{ind}\:{the}\:{number}.…
Question Number 202715 by Rasheed.Sindhi last updated on 01/Jan/24 $$\begin{array}{|c|}{\:\:\underset{\mathrm{is}\:\mathrm{2025}} {\mathrm{Square}\:\mathrm{of}\:\mathrm{mean}\:\mathrm{of}\:\mathrm{two}\:\mathrm{numbers}}\:\:}\\\hline\end{array}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\& \\ $$$$\begin{array}{|c|}{\:\:\underset{\:\mathrm{is}\:\mathrm{2026}} {\mathrm{mean}\:\mathrm{of}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}}\:\:}\\\hline\end{array} \\ $$$$\:\:\:\:\: \\ $$$$\begin{array}{|c|}{\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}.}\\\hline\end{array}\: \\ $$ Answered by JDamian…
Question Number 202726 by ajfour last updated on 02/Jan/24 $${pk}={b} \\ $$$${pq}+{sk}={c} \\ $$$${sq}={d} \\ $$$$\sqrt{{t}}=\frac{{t}−{q}}{{k}}=\frac{{t}−{s}}{{p}} \\ $$$${Given}\:{are}\:{b},\:{c},\:{d}.\:{Find}\:{t}. \\ $$ Commented by MATHEMATICSAM last updated…