Question Number 164018 by cortano1 last updated on 13/Jan/22 Answered by mr W last updated on 13/Jan/22 Commented by cortano1 last updated on 13/Jan/22 $${superb}\:{solution}…
Question Number 164001 by bekzodjumayev last updated on 12/Jan/22 Commented by bekzodjumayev last updated on 12/Jan/22 $${Help} \\ $$ Answered by mahdipoor last updated on…
Question Number 163996 by mathlove last updated on 12/Jan/22 $$\begin{cases}{\mathrm{9}^{\frac{{a}+\mathrm{1}}{{b}}} =\mathrm{125}}\\{\mathrm{5}^{\frac{{b}}{{a}}} =\mathrm{3}}\end{cases} \\ $$$${then}\:\:{faind}\:{the}\:{volve}\:{of}\:\:\frac{\mathrm{25}^{{b}} }{\mathrm{2}{a}^{\mathrm{2}} }=? \\ $$ Answered by nurtani last updated on 12/Jan/22…
Question Number 98448 by Quvonchbek last updated on 14/Jun/20 $$\:\:\:\:\:\:\:\mathrm{6}^{\mathrm{273}} +\mathrm{8}^{\mathrm{273}} \:\::\mathrm{49}\:\:\:\boldsymbol{{prove}}\:\:\boldsymbol{{the}}\:\:\boldsymbol{{divi}\mathrm{s}{ion}} \\ $$ Commented by Rasheed.Sindhi last updated on 14/Jun/20 $$\:\:\:\:\:\:\mathrm{49}\:\mid\:\mathrm{2}^{\mathrm{273}} \left(\mathrm{3}^{\mathrm{273}} +\mathrm{4}^{\mathrm{273}} \right)…
Question Number 163986 by mathlove last updated on 12/Jan/22 $$\sqrt[{\mathrm{3}}]{\mathrm{1000}}{ln}^{\mathrm{80}} \left({ln}\frac{\mathrm{8}}{\mathrm{5}}−{ln}\frac{\mathrm{8}{e}}{\mathrm{5}}\right)^{\mathrm{80}} \\ $$$${is}\:\:\:\:{simplified}\:\:{form}\:\:{equl}\:\:{to}=? \\ $$$$ \\ $$ Answered by nikif99 last updated on 12/Jan/22 $$\sqrt[{\mathrm{3}}]{\mathrm{1000}\:}\mathrm{ln}\:^{\mathrm{80}}…
Question Number 163968 by mathls last updated on 12/Jan/22 $$\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}} \\ $$$$\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{They}'\mathrm{re}\:\boldsymbol{\mathrm{proper}}\:\boldsymbol{\mathrm{fractions}}. \\ $$$$\mathrm{7}\left(\mathrm{a}\right)\:\frac{\mathrm{10}}{\mathrm{9}}\:,\:\frac{\mathrm{15}}{\mathrm{9}}\:,\:\frac{\mathrm{46}}{\mathrm{9}} \\ $$$$\:\:\left(\mathrm{b}\right)\:\mathrm{They}'\mathrm{re}\:\boldsymbol{\mathrm{improper}}\:\boldsymbol{\mathrm{fractions}}. \\ $$$$\mathrm{8}.\:\:\frac{\mathrm{7}}{\mathrm{7}}\:,\:\frac{\mathrm{12}}{\mathrm{12}}\:,\:\frac{\mathrm{48}}{\mathrm{48}}\:,\:\frac{\mathrm{62}}{\mathrm{62}}\:,\:\frac{\mathrm{125}}{\mathrm{125}}…
Question Number 163956 by HongKing last updated on 12/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163957 by HongKing last updated on 12/Jan/22 Answered by mahdipoor last updated on 12/Jan/22 $${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}}…
Question Number 163955 by HongKing last updated on 12/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com