Question Number 28319 by NECx last updated on 24/Jan/18 $${If}\:{the}\:{roots}\:{of}\:{x}^{\mathrm{2}} +{px}+{q}=\mathrm{0},\:{q}\neq\mathrm{0} \\ $$$${are}\:\alpha\:{and}\:\beta.{Find}\:{the}\:{roots}\:{of} \\ $$$${qx}^{\mathrm{2}} +\left(\mathrm{2}{q}−{p}^{\mathrm{2}} \right){x}+{q}=\mathrm{0}\:{in}\:{terms}\:{of} \\ $$$$\alpha\:{and}\:\beta. \\ $$ Answered by Rasheed.Sindhi last…
Question Number 159390 by pete last updated on 16/Nov/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:{x}\:\mathrm{and}\:{y}\:\mathrm{if} \\ $$$$\mathrm{log}_{\mathrm{4}} {x}\:+\mathrm{3}=\mathrm{log}_{\mathrm{27}} {y} \\ $$ Answered by abdullah_ff last updated on 16/Nov/21 $$\mathrm{log}_{\mathrm{4}} {x}+\mathrm{3}=\mathrm{log}_{\mathrm{27}}…
Question Number 159386 by HongKing last updated on 16/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 28313 by abdo imad last updated on 23/Jan/18 $${let}\:{give}\:\:{P}_{{n}} \left({x}\right)=\:\mathrm{1}−{x}^{\mathrm{2}^{{n}+\mathrm{1}} } \:\:\:{and}\:\:{Q}_{{n}} \left({x}\right)=\:\prod_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}+{x}^{\mathrm{2}^{{k}} } \right) \\ $$$${prove}\:{that}\:{Q}_{{n}\:} \:{divide}\:{P}_{{n}} . \\ $$…
Question Number 28311 by abdo imad last updated on 23/Jan/18 $${let}\:{give}\:{P}_{{n}} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{\mathrm{2}{n}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+…+\frac{\mathrm{1}}{{k}+\mathrm{1}}\right){x}^{{k}} \:\:{and} \\ $$$${Q}_{{n}} \left({x}\right)=\:\mathrm{1}+\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\:+…\frac{{x}^{{n}} }{{n}+\mathrm{1}}\:\:.{prove}\:{that}\:{Q}_{{n}\:} \:{divide}\:{P}_{{n}} . \\ $$ Terms…
Question Number 28312 by abdo imad last updated on 23/Jan/18 $${let}\:{give}\:\:{P}_{{n}} \left({x}\right)=\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}} \:+\left({x}+\mathrm{2}\right)^{{n}} −\mathrm{1}\:{and} \\ $$$${Q}\left({x}\right)=\:{x}^{\mathrm{2}} \:+\mathrm{3}{x}\:+\mathrm{2}\:\:{find}\:{R}\left({x}\right)\:/{P}_{{n}} \left({x}\right)={R}\left({x}\right)\:{Q}\left({x}\right)\:. \\ $$ Answered by sma3l2996 last updated…
Question Number 159379 by HongKing last updated on 16/Nov/21 $$\mathrm{let}\:\:\boldsymbol{\mathrm{S}}\left(\mathrm{x}\right)\:=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\mathrm{3x}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{2}} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{above}\:\mathrm{find}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{1}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \left(\mathrm{n}\:+\:\mathrm{3}\right)}\: \\ $$ Answered by Ar…
Question Number 159378 by HongKing last updated on 16/Nov/21 $$\mathrm{let}\:\:\mathrm{x};\mathrm{y}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{expression}: \\ $$$$\mathrm{P}\:=\:\mathrm{2020}\boldsymbol{\mathrm{x}}\:+\:\mathrm{2021}\boldsymbol{\mathrm{y}} \\ $$ Commented by mr W last…
Question Number 28288 by Mr eaay last updated on 23/Jan/18 Answered by Rasheed.Sindhi last updated on 23/Jan/18 $$\:^{\bullet} \left(\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}×\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\:\:^{\bullet} \left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}}…
Question Number 159353 by cortano last updated on 16/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com