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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-55902

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-186960

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}}…

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Question Number 121422 by faysal last updated on 08/Nov/20 $${a}\:{rectangle}'{s}\:{a}\:{side}'{s}\:{terminal} \\ $$$${points}\:{coordinate}\:{are}\:\left(\mathrm{2},−\mathrm{1}\right),\:\left(\mathrm{6},\mathrm{5}\right)\:{and}\:{the} \\ $$$${length}\:{of}\:{the}\:{diagonal}\:{is}\:\mathrm{8}\:{unit}. \\ $$$${what}\:{are}\:{coordinate}\:{of}\:{terminal}\:{point} \\ $$$${of}\:{paralel}\:{side}\:{that}\:{side} \\ $$ Commented by mr W last…