Question Number 176847 by Ar Brandon last updated on 27/Sep/22 $$\mathrm{If}\:{x},\:{y}\:\in\mathbb{Z} \\ $$$$\mathrm{23}!=\mathrm{2}^{{x}} \mathrm{5}^{{y}} {k} \\ $$$$\mathrm{and}\:{k}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{number},\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\: \\ $$$$\mathrm{max}\left({x}+{y}\right)\:? \\ $$ Answered by BaliramKumar last…
Question Number 176841 by Ar Brandon last updated on 27/Sep/22 $$\mathrm{If}\:\mathrm{3}\:\mathrm{of}\:\mathrm{10}\:\mathrm{elective}\:\mathrm{courses}\:\mathrm{are}\:\mathrm{been}\:\mathrm{delivered}\: \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{same}\:\mathrm{time}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{possibilities}\: \\ $$$$\mathrm{are}\:\mathrm{there}\:\mathrm{to}\:\mathrm{take}\:\mathrm{5}\:\mathrm{courses}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 176839 by Ar Brandon last updated on 27/Sep/22 $$\mathrm{If}\:{a}=\mathrm{2}+\sqrt{\mathrm{15}}\:\:\mathrm{Express} \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{15}}+\frac{\mathrm{196}}{\mathrm{54}}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a} \\ $$ Answered by Rasheed.Sindhi last updated on 27/Sep/22 $$\mathrm{If}\:{a}=\mathrm{2}+\sqrt{\mathrm{15}}\:\:\mathrm{Express} \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt{\mathrm{15}}+\frac{\mathrm{196}}{\mathrm{54}}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a}…
Question Number 176836 by Ar Brandon last updated on 27/Sep/22 $$\mathrm{s}\left(\mathrm{B}'−\mathrm{A}'\right)=\mathrm{4} \\ $$$$\mathrm{s}\left(\mathrm{B}−\mathrm{A}\right)=\mathrm{6} \\ $$$$\mathrm{s}\left(\mathrm{A}\right)=\mathrm{9} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{subsets}\:\mathrm{of}\:\mathrm{B}\:\mathrm{with} \\ $$$$\mathrm{at}\:\mathrm{most}\:\mathrm{2}\:\mathrm{elements}\:? \\ $$ Commented by Rasheed.Sindhi last…
Question Number 45720 by Tawa1 last updated on 15/Oct/18 $$\underset{\boldsymbol{\mathrm{r}}\:=\:\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}} {\boldsymbol{\sum}}}\:\:\left(\boldsymbol{\mathrm{r}}\:+\:\mathrm{1}\right)^{\mathrm{3}} \\ $$ Commented by math khazana by abdo last updated on 15/Oct/18 $$=\sum_{{k}=\mathrm{2}}…
Question Number 176785 by Ar Brandon last updated on 26/Sep/22 $$\frac{\mathrm{6}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }={y} \\ $$$$\mathrm{Express}\:\:\frac{\mathrm{11}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y} \\ $$ Commented by Ar Brandon last updated…
Question Number 45712 by malwaan last updated on 15/Oct/18 $$\mathrm{prove}\:\mathrm{that}\:: \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{is}\:\mathrm{irrationl}\:\mathrm{number} \\ $$ Commented by maxmathsup by imad last updated on 15/Oct/18 $${let}\:{suppose}\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\:{with}\:{p}\:{and}\:{q}\:{integrsn}\:{natural}\:{and}\:\Delta\left({p},{q}\right)=\mathrm{1}\:\Rightarrow \\…
Question Number 176774 by Ar Brandon last updated on 26/Sep/22 $$\mathrm{How}\:\mathrm{many}\:\mathrm{distinct}\:{y}\:\mathrm{exist}\:\mathrm{satisfying} \\ $$$$\mid{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{18}\mid+\mid{y}−\mathrm{3}\mid=\mathrm{5} \\ $$ Answered by mr W last updated on 26/Sep/22 $${x},{y}\in\mathbb{Z}…
Question Number 111218 by Lekhraj last updated on 02/Sep/20 Answered by nimnim last updated on 02/Sep/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{1}\:\:\:\:\:\mathrm{1}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\mathrm{4}\:\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{3}\:\:\:\:\:\mathrm{5}\:\:\:\:\mathrm{9}\:\:\:\:\:\mathrm{7}…
Question Number 45585 by pieroo last updated on 14/Oct/18 $$\mathrm{prove}\:\mathrm{that1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+\mathrm{n}^{\mathrm{3}} =\frac{\mathrm{n}^{\mathrm{2}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}\:\mathrm{for}\:\mathrm{every} \\ $$$$\mathrm{natural}\:\mathrm{number}\:\boldsymbol{\mathrm{n}} \\ $$ Commented by maxmathsup by imad…