Question Number 171493 by Mastermind last updated on 16/Jun/22 Answered by mr W last updated on 16/Jun/22 $${A}=\mathrm{4}×{shaded}\:{triangle}+\mathrm{1}×{square} \\ $$$${A}=\mathrm{4}×\frac{\mathrm{2}×\mathrm{3}×\sqrt{\mathrm{2}}}{\mathrm{2}×\mathrm{2}}+\left(\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{2}×\mathrm{2}×\mathrm{3}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$${A}=\mathrm{13}+\mathrm{12}\sqrt{\mathrm{2}} \\…
Question Number 171491 by Mastermind last updated on 16/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171490 by Mastermind last updated on 16/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171472 by Mastermind last updated on 15/Jun/22 $${li}\underset{{a}\rightarrow\infty} {{m}}\:\underset{{n}=\mathrm{1}} {\overset{{a}} {\sum}}\frac{{e}^{{in}} .{ln}\mid\frac{\mathrm{1}}{{x}}\mid}{\pi{n}^{\mathrm{2}} }.{tan}^{−\mathrm{1}} \left({n}\sqrt{\pi}\right) \\ $$$$ \\ $$$${Mastermind} \\ $$ Terms of Service…
Question Number 171475 by solomonwells last updated on 16/Jun/22 Answered by alephzero last updated on 17/Jun/22 $${repost}\:{of}\:{Q}\mathrm{171437} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 40378 by S.HER last updated on 21/Jul/18 $$\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{1}}{\mathrm{3}!}+….+\frac{\mathrm{1}}{\mathrm{2018}!}=? \\ $$ Commented by maxmathsup by imad last updated on 21/Jul/18 $${we}\:{have}\:{e}^{{x}} \: \\ $$$${let}\:{n}>\mathrm{2018}\:{and}\:\:{S}_{{n}}…
Question Number 171437 by solomonwells last updated on 15/Jun/22 $$\mathrm{make}\:\boldsymbol{\mathrm{r}}\:\:\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formula} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{y}}=\left(\frac{\boldsymbol{\mathrm{pr}}}{\boldsymbol{\mathrm{m}}}\:\:−\:\frac{\boldsymbol{\mathrm{p}}^{\mathrm{3}} }{\mathrm{1}}\right)^{−\mathrm{3}/\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{find}\:\boldsymbol{\mathrm{r}}\:\mathrm{if}\:\mathrm{y}=−\mathrm{8}\:\:\:,\:\:\mathrm{m}=−\mathrm{1},\:\:\mathrm{p}=\mathrm{3} \\ $$ Answered by alephzero last…
Question Number 40352 by scientist last updated on 20/Jul/18 $${If}\:{A}=\begin{bmatrix}{\mathrm{4}\:\:\:\:\:−\mathrm{3}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}{use}\:{the}\:{fact}\:{that}\: \\ $$$${A}^{\mathrm{2}} =\mathrm{4}{A}−\mathrm{3}{I}_{\mathrm{2}} \:\:{and}\:{mathematical}\:{induction}\:{to}\:{prove} \\ $$$$\:\:{A}^{{n}} =\frac{\left(\mathrm{3}^{{n}} −\mathrm{1}\right)}{\mathrm{2}}{A}\:\:+\frac{\mathrm{3}−\mathrm{3}^{{n}} }{\mathrm{2}}{I}\:\:{if}\:{n}\geqslant\mathrm{1} \\ $$ Commented by prof Abdo…
Question Number 40284 by scientist last updated on 18/Jul/18 $${How}\:{many}\:{permutation}\:{can}\:{be}\:{made}\:{using}\:{the}\:{word} \\ $$$${CANADA}\:{each}\:{of}\:{the}\:{six}\:{letter}\:{being}\:{used}\:{mixed} \\ $$$${in}\:{each}\:{permutation}?\:\:{In}\:{how}\:{many}\:{of}\:{these} \\ $$$${permutation}\:{will}\:{the}\:{three}\:{A}'{s}\:{be}\:{together}?\:{In}\: \\ $$$${how}\:{many}\:{will}\:{two}\:{A}'{s}\:{be}\:{together}\:{but}\:{not}\:{the}\:{three}? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 40283 by scientist last updated on 18/Jul/18 $${There}\:{are}\:{four}\:{ladies}\:{and}\:{four}\:{gentlemen}\:{ready}\:{to}\:{play}\: \\ $$$${tennis}.{if}\:{there}\:{is}\:{only}\:{one}\:{tennis}\:{court}\:{readily} \\ $$$${available}.\:{In}\:{how}\:{many}\:{ways}\:{is}\:{it}\:{possible}\:{to}\:{arrange}\:{for}\:{one}\:{mixed}\:{double}\:{to}\:{be}\:{played} \\ $$$$\left({a}\:{lady}\:{and}\:{gentleman}\right)? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com