Question Number 163968 by mathls last updated on 12/Jan/22 $$\left(−\mathrm{2}^{\mathrm{2}} \right)^{\mathrm{3}} =? \\ $$ Answered by cortano1 last updated on 12/Jan/22 $$\left(−\mathrm{2}^{\mathrm{2}} \right)=−\mathrm{4} \\ $$$$\left(−\mathrm{2}^{\mathrm{2}}…
Question Number 32889 by jasno91 last updated on 05/Apr/18 Commented by Rasheed.Sindhi last updated on 05/Apr/18 $$\mathrm{6}\left(\mathrm{a}\right)\:\frac{\mathrm{1}}{\mathrm{9}}\:\:,\:\frac{\mathrm{2}}{\mathrm{9}}\:\:,\frac{\mathrm{8}}{\mathrm{9}} \\ $$$$\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{They}'\mathrm{re}\:\boldsymbol{\mathrm{proper}}\:\boldsymbol{\mathrm{fractions}}. \\ $$$$\mathrm{7}\left(\mathrm{a}\right)\:\frac{\mathrm{10}}{\mathrm{9}}\:,\:\frac{\mathrm{15}}{\mathrm{9}}\:,\:\frac{\mathrm{46}}{\mathrm{9}} \\ $$$$\:\:\left(\mathrm{b}\right)\:\mathrm{They}'\mathrm{re}\:\boldsymbol{\mathrm{improper}}\:\boldsymbol{\mathrm{fractions}}. \\ $$$$\mathrm{8}.\:\:\frac{\mathrm{7}}{\mathrm{7}}\:,\:\frac{\mathrm{12}}{\mathrm{12}}\:,\:\frac{\mathrm{48}}{\mathrm{48}}\:,\:\frac{\mathrm{62}}{\mathrm{62}}\:,\:\frac{\mathrm{125}}{\mathrm{125}}…
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Question Number 163957 by HongKing last updated on 12/Jan/22 Answered by mahdipoor last updated on 12/Jan/22 $${get}\:{n}=\mathrm{3}{b}+{a}\:,\:{b}\in{N}\:\:\:\:\mathrm{0}\leqslant{a}<\mathrm{3} \\ $$$$\left[{n}\right]=\mathrm{3}{b}+\left[{a}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{{n}+\mathrm{1}}{\mathrm{3}}\right]={b}+\left[\frac{{a}+\mathrm{1}}{\mathrm{3}}\right] \\ $$$$\left[\frac{{n}}{\mathrm{3}}\right]={b}+\left[\frac{{a}}{\mathrm{3}}\right]={b}\:\:\:\:\:\:\:\:\:\:\left[\frac{{n}+\mathrm{2}}{\mathrm{3}}\right]={b}+\left[\frac{{a}+\mathrm{2}}{\mathrm{3}}\right] \\ $$$$\Rightarrow\left(−\mathrm{1}\right)^{\mathrm{3}{b}+\left[{a}\right]} +\left(−\mathrm{1}\right)^{{b}+\left[\frac{{a}+\mathrm{1}}{\mathrm{3}}\right]} =\left(−\mathrm{1}\right)^{{b}}…
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Question Number 98416 by 175 last updated on 13/Jun/20 Answered by Farruxjano last updated on 13/Jun/20 $$\boldsymbol{{i}}^{\mathrm{5}\boldsymbol{{n}}+\mathrm{1}} =\mathrm{1}\:\Rightarrow\:\mathrm{5}\boldsymbol{{n}}+\mathrm{1}=\mathrm{2}\boldsymbol{{k}}\:\left(\boldsymbol{{k}}\in\boldsymbol{{Z}}\right),\:\Rightarrow\:\boldsymbol{{n}}=\mathrm{2}\boldsymbol{{t}}+\mathrm{1}\:\left(\boldsymbol{{t}}\in\boldsymbol{{Z}}\right) \\ $$ Answered by Ramajunan last updated…
Question Number 163942 by Rasheed.Sindhi last updated on 12/Jan/22 $${f}\left(\mathrm{2}{x}\right)−\mathrm{2}\left[\:{f}\left({x}\right)\:\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$${f}\left({x}\right)=? \\ $$ Answered by mr W last updated on 12/Jan/22 $${x}\:\in\:{R} \\…
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