Question Number 65359 by 237174/13/1 last updated on 29/Jul/19 $$\phi\phi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 130886 by Khalmohmmad last updated on 30/Jan/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}+\mathrm{2}^{{n}} }{\mathrm{3}^{{n}} } \\ $$ Answered by EDWIN88 last updated on 30/Jan/21 $$\:\underset{{n}=\mathrm{1}} {\overset{\infty}…
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Question Number 130876 by 0731619177 last updated on 30/Jan/21 Answered by Dwaipayan Shikari last updated on 30/Jan/21 $$\int_{−\infty} ^{\infty} {re}^{−{a}^{\mathrm{2}} \left({r}−\frac{\mathrm{6}}{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{36}}{{a}^{\mathrm{2}} }} {dr}…
Question Number 65301 by LPM last updated on 28/Jul/19 Commented by MJS last updated on 28/Jul/19 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{should}\:\mathrm{ve}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{30}\:\mathrm{because}\:\mathrm{then} \\ $$$$\frac{{ab}}{\mathrm{3}}=−\mathrm{3}\:\mathrm{but}\:\mathrm{with}\:\mathrm{tbe}\:\mathrm{given}\:\mathrm{equations}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{values}\:\mathrm{different}\:\mathrm{from}\:\mathrm{each}\:\mathrm{other} \\ $$…
Question Number 65289 by naka3546 last updated on 27/Jul/19 Commented by Prithwish sen last updated on 27/Jul/19 $$\mathrm{let} \\ $$$$\boldsymbol{\Sigma\mathrm{pqr}}=\boldsymbol{\mathrm{A}},\boldsymbol{\Sigma\mathrm{pq}}=\boldsymbol{\mathrm{B}},\boldsymbol{\Sigma\mathrm{p}}=\boldsymbol{\mathrm{C}} \\ $$$$\mathrm{Then}\:\mathrm{from}\:\mathrm{the}\:\mathrm{last}\:\mathrm{three}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{5} \\…
Question Number 130818 by Study last updated on 29/Jan/21 $${if}\:\mid{z}_{\mathrm{1}} \mid=\mid{z}_{\mathrm{2}} \mid=\mid{z}_{\mathrm{3}} \mid=\mathrm{1}\:\: \\ $$$${and}\:\:\:\mid\frac{\mathrm{1}}{{z}_{\mathrm{1}} }\mid+\mid\frac{\mathrm{1}}{{z}_{\mathrm{2}} }\mid+\mid\frac{\mathrm{1}}{{z}_{\mathrm{3}} }\mid=\mathrm{1} \\ $$$$\:{find}\:\mid{z}_{\mathrm{1}} +{z}_{\mathrm{2}} +{z}_{\mathrm{3}} \mid=?\:\:\: \\ $$$${z}_{\mathrm{1}}…
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