Question Number 55926 by MJS last updated on 06/Mar/19 $$\mathrm{Ancient}\:\mathrm{Roman}\:\mathrm{Number}\:\mathrm{Magic}: \\ $$$$\mathrm{take}\:\mathrm{5}\:\mathrm{matches}\:\mathrm{or}\:\mathrm{picks}\:\mathrm{and}\:\mathrm{form}\:\mathrm{a}\:\mathrm{roman}\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{−} {\overline {\mid\mid\mid}} \\ $$$$\mathrm{now}\:\mathrm{subtract}\:\mathrm{2}\:\mathrm{so}\:\mathrm{that}\:\mathrm{a}\:\mathrm{little}\:\mathrm{more}\:\mathrm{than}\:\mathrm{3}\:\mathrm{is}\:\mathrm{left} \\ $$ Answered by MJS last updated…
Question Number 55920 by ANTARES VY last updated on 06/Mar/19 $$\boldsymbol{\mathrm{Calculate}}. \\ $$$$\underset{\boldsymbol{{a}}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\boldsymbol{{b}}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{\boldsymbol{{c}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\boldsymbol{{ab}}\left(\mathrm{3}\boldsymbol{{a}}+\boldsymbol{{c}}\right)}{\mathrm{4}^{\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}} \left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{c}}+\boldsymbol{{a}}\right)}. \\ $$ Commented by mr…
Question Number 55902 by naka3546 last updated on 06/Mar/19 Answered by MJS last updated on 06/Mar/19 $${P}_{{n}} =\left({a}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} +\left({a}+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} ;\:{a}\in\mathbb{Z}^{+} \\ $$$${P}_{\mathrm{1}} =\mathrm{2}{a}+\mathrm{2}\:\in\mathbb{Z} \\ $$$${P}_{\mathrm{2}}…
Question Number 186965 by 073 last updated on 12/Feb/23 Commented by 073 last updated on 12/Feb/23 $$\mathrm{solution}\:\mathrm{please}?? \\ $$$$ \\ $$ Answered by Rasheed.Sindhi last…
Question Number 121434 by mathdave last updated on 08/Nov/20 Commented by mathdave last updated on 08/Nov/20 $${a}\:{help}\:{fom}\:{any}\:{body}\:{pls} \\ $$ Commented by Tawa11 last updated on…
Question Number 186967 by 073 last updated on 12/Feb/23 Commented by 073 last updated on 12/Feb/23 $$\mathrm{solution}\:\mathrm{plese}?? \\ $$ Answered by floor(10²Eta[1]) last updated on…
Question Number 186964 by 073 last updated on 12/Feb/23 Commented by 073 last updated on 12/Feb/23 $$\mathrm{solution}\:\mathrm{please}?? \\ $$$$\mathrm{i}\:\mathrm{need}\:\mathrm{it} \\ $$ Commented by mr W…
Question Number 186966 by 073 last updated on 12/Feb/23 Commented by 073 last updated on 12/Feb/23 $$\mathrm{solution}\:? \\ $$ Commented by 073 last updated on…
Question Number 186960 by Humble last updated on 12/Feb/23 Answered by MJS_new last updated on 12/Feb/23 $${z}={x}+{y}\mathrm{i} \\ $$$${z}=\mid{z}\mid\mathrm{e}^{\measuredangle\left({z}\right)} \\ $$$$\:\:\:\:\:\left[\mid{x}+{y}\mathrm{i}\mid=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\:\wedge\:\measuredangle\left({x}+{y}\mathrm{i}\right)=\mathrm{arctan}\:\frac{{y}}{{x}}\right] \\ $$$${z}=\sqrt{{x}^{\mathrm{2}}…
Question Number 186963 by 073 last updated on 12/Feb/23 Commented by 073 last updated on 12/Feb/23 $$\mathrm{solotion}\:\mathrm{please}?? \\ $$$$\mathrm{need}\:\mathrm{it} \\ $$ Answered by Rasheed.Sindhi last…