Question Number 195357 by SirMUTUKU last updated on 31/Jul/23 Answered by Spillover last updated on 31/Jul/23 $$\frac{{E}}{\mathrm{1}−\pi}=\sqrt{\frac{{h}−\mathrm{0}.\mathrm{5}}{\mathrm{1}−{h}}}\: \\ $$$$\left(\frac{{E}}{\mathrm{1}−\pi}\right)^{\mathrm{2}} =\frac{{h}−\mathrm{0}.\mathrm{5}}{\mathrm{1}−{h}} \\ $$$$\frac{{E}^{\mathrm{2}} }{\left(\mathrm{1}−\pi\right)^{\mathrm{2}} }=\frac{{h}−\mathrm{0}.\mathrm{5}}{\mathrm{1}−{h}} \\…
Question Number 195356 by MJS_new last updated on 31/Jul/23 $$\mathrm{remark}\:\mathrm{to}\:\mathrm{question}\:\mathrm{195301}\:\mathrm{and}\:\mathrm{similar}\:\mathrm{ones} \\ $$$${x}^{\mathrm{2}} +{y}={a} \\ $$$${x}+{y}^{\mathrm{2}} ={b} \\ $$$${a},\:{b}\:>\mathrm{0} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{solutions}\:\mathrm{depending}\:\mathrm{on}\:{a},\:{b}? \\ $$ Answered by MJS_new…
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Question Number 195344 by Mingma last updated on 31/Jul/23 Answered by kapoorshah last updated on 31/Jul/23 $${cos}\:\alpha\:=\:{cos}\:\alpha \\ $$$$\frac{{a}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{a}.\mathrm{6}}\:=\:\frac{{b}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{b}.\mathrm{6}}…
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Question Number 195343 by Matica last updated on 31/Jul/23 $$\:{hello}\:{everyone},\:{I}'{m}\:{a}\:{new}\:{student}. \\ $$$$\:{I}\:{beg}\:{your}\:{pardon}. \\ $$$$\:{Please}\:{tell}\:{me}\:{any}\:{good}\:{math}\:{book} \\ $$$$\:{you}\:{know}.\:{Thanks}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195342 by Matica last updated on 31/Jul/23 $$\:\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that}\:\:\forall{n}\:\in\:\mathbb{N}^{\ast} \:,\:\mathrm{4}^{{n}} \left({n}!\right)^{\mathrm{3}} \:<\:\left({n}+\mathrm{1}\right)^{\mathrm{3}{n}} \:. \\ $$$$\mathrm{2}.\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{in}\:\mathbb{Z}^{\mathrm{2}} \:: \\ $$$$\:\:\:\:\:{a}./\:\:\mathrm{2}{x}^{\mathrm{3}} +{xy}−\mathrm{7}=\mathrm{0}\:, \\ $$$$\:\:\:\:\:{b}./\:\:{x}\left({x}+\mathrm{1}\right)\left({x}+\mathrm{7}\right)\left({x}+\mathrm{8}\right)={y}^{\mathrm{2}} . \\ $$…
Question Number 195369 by Shlock last updated on 31/Jul/23 Answered by witcher3 last updated on 01/Aug/23 $$\mathrm{x}=−\mathrm{y} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{0}\right)\right)=\mathrm{2f}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}\left(\mathrm{f}\left(\mathrm{0}\right)\right)=\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\forall\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{Z}\:^{\mathrm{2}}…