Question Number 10933 by Joel576 last updated on 02/Mar/17 $$\mathrm{If}\:\:{p}\:\:\mathrm{and}\:\:{q}\:\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{for}\:\mathrm{the} \\ $$$${x}^{\mathrm{2}} \:−\:\left({a}\:+\:\mathrm{1}\right){x}\:+\:\left(−{a}\:−\:\frac{\mathrm{5}}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\:{p}^{\mathrm{2}\:} +\:{q}^{\mathrm{2}} \:\:\mathrm{is}\:… \\ $$ Answered by ridwan balatif last updated…
Question Number 10907 by j.masanja06@gmail.com last updated on 01/Mar/17 $$\mathrm{express}\:\mathrm{in}\:\mathrm{partial}\:\mathrm{fraction} \\ $$$$\frac{\mathrm{3x}+\mathrm{2}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)} \\ $$ Answered by sandy_suhendra last updated on 01/Mar/17 $$=\frac{\mathrm{3x}+\mathrm{2}}{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}+\mathrm{1}\right)}=\frac{\mathrm{3x}+\mathrm{2}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}−\mathrm{1}\right)} \\…
Question Number 76438 by john santu last updated on 27/Dec/19 $${given}\:{N}\:=\frac{\mathrm{4}}{\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{4}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{8}}} +\mathrm{1}\right)\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{16}}} +\mathrm{1}\right)} \\ $$$${find}\:{value}\:{of}\:\left({N}+\mathrm{1}\right)^{\mathrm{48}} . \\ $$ Answered by mr W last updated…
Question Number 76436 by Master last updated on 27/Dec/19 Answered by john santu last updated on 27/Dec/19 $$\frac{\mathrm{1}}{{abc}}+\frac{{a}+{c}}{{ac}}=\frac{\mathrm{1}}{{b}} \\ $$$$\frac{{ab}+{bc}+\mathrm{1}}{{abc}}=\frac{\mathrm{1}}{{b}}\rightarrow{ab}+{bc}+\mathrm{1}={ac} \\ $$$${ab}+{bc}={ac}−\mathrm{1}\:\rightarrow{b}=\frac{{ac}−\mathrm{1}}{{a}+{c}} \\ $$$${by}\:{substitusi} \\…
Question Number 76428 by TawaTawa last updated on 27/Dec/19 Answered by john santu last updated on 27/Dec/19 $$\left(\mathrm{5}\right){ratio}\:=\:\frac{{a}_{\mathrm{2}} }{{a}_{\mathrm{1}} }.\:\rightarrow{a}_{\mathrm{2}} =\mathrm{10}.{a}_{\mathrm{1}} =\mathrm{10}. \\ $$$${now}\:{a}_{{n}\:} ={a}_{\mathrm{1}}…
Question Number 76431 by john santu last updated on 27/Dec/19 $${how}\:{to}\:{find}\:{x}^{\mathrm{2048}} \:+\:{x}^{−\mathrm{2048}} \\ $$$${if}\:\:{x}+{x}^{−\mathrm{1}} =\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 10872 by Saham last updated on 28/Feb/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{cultural}\:\mathrm{gathering}\:\mathrm{of}\:\mathrm{400}\:\mathrm{people},\:\mathrm{there}\:\mathrm{are}\:\mathrm{270}\:\mathrm{men}\:\mathrm{and}\:\mathrm{200} \\ $$$$\mathrm{musicians}.\:\mathrm{Of}\:\mathrm{the}\:\mathrm{latter},\:\mathrm{80}\:\mathrm{are}\:\mathrm{singers}.\:\mathrm{60}\:\mathrm{of}\:\mathrm{the}\:\mathrm{women}\:\mathrm{are}\:\mathrm{not}\:\:\mathrm{musicians} \\ $$$$\mathrm{and}\:\mathrm{220}\:\mathrm{of}\:\mathrm{the}\:\mathrm{men}\:\mathrm{are}\:\mathrm{not}\:\mathrm{singers}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{of}\:\mathrm{the}\:\mathrm{women}\:\mathrm{are} \\ $$$$\mathrm{musicians}\:\mathrm{but}\:\mathrm{not}\:\mathrm{singers}.\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}\:\mathrm{150}\:\mathrm{singers}\:\mathrm{altogether}\:\mathrm{and}\: \\ $$$$\mathrm{40}\:\mathrm{men}\:\mathrm{are}\:\mathrm{both}\:\mathrm{musicians}\:\mathrm{and}\:\mathrm{singers}. \\ $$ Answered by Abdulhalim Ghadady last…
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Question Number 10867 by Saham last updated on 28/Feb/17 $$\left(\mathrm{1}\right) \\ $$$$\mathrm{Show}\:\mathrm{that}\:: \\ $$$$\frac{\mathrm{x}^{\mathrm{2n}\:+\:\mathrm{1}} \:−\:\mathrm{y}^{\mathrm{2n}\:+\:\mathrm{1}} }{\mathrm{x}\:−\:\mathrm{y}}\:=\:\mathrm{x}^{\mathrm{2n}\:} +\:\mathrm{x}^{\mathrm{2n}\:−\:\mathrm{1}} \mathrm{y}\:+\:…\:+\:\mathrm{xy}^{\mathrm{2n}\:−\:\mathrm{1}} \:+\:\mathrm{y}^{\mathrm{2n}} \\ $$$$\left(\mathrm{2}\right) \\ $$$$\mathrm{Show}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2n}}…
Question Number 76397 by MJS last updated on 27/Dec/19 $$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\left[{z}={a}+{b}\mathrm{i};\:\bar {{z}}={a}−{b}\mathrm{i};\:{r}\in\mathbb{R}\right] \\ $$$$\sqrt{{r}^{\mathrm{2}} −{z}^{\mathrm{2}} }=\bar {{z}} \\ $$$$\sqrt{{r}^{\mathrm{2}} +{z}^{\mathrm{2}} }=\bar {{z}} \\ $$…