Question Number 195407 by mustafazaheen last updated on 01/Aug/23 $$\mathrm{f}'\left(\mathrm{2}\right)=\mathrm{g}'\left(\mathrm{2}\right)=\mathrm{g}\left(\mathrm{2}\right)=\mathrm{2} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\left(\mathrm{fog}\right)'\left(\mathrm{2}\right)=? \\ $$ Answered by gatocomcirrose last updated on 01/Aug/23 $$\left(\mathrm{fog}\right)'\left(\mathrm{2}\right)=\mathrm{f}'\left(\mathrm{g}\left(\mathrm{2}\right)\right)\mathrm{g}'\left(\mathrm{2}\right)=\mathrm{2f}'\left(\mathrm{2}\right)=\mathrm{4} \\ $$ Answered…
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 195369 by Shlock last updated on 31/Jul/23 Answered by witcher3 last updated on 01/Aug/23 $$\mathrm{x}=−\mathrm{y} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{0}\right)\right)=\mathrm{2f}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{f}\left(\mathrm{f}\left(\mathrm{0}\right)\right)=\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\forall\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{Z}\:^{\mathrm{2}}…
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Question Number 195331 by Shlock last updated on 30/Jul/23 Answered by mr W last updated on 06/Aug/23 $${say}\:{a}={pc},\:{b}={qc} \\ $$$$\left({p}+{q}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\mathrm{1}\right)=\mathrm{10} \\ $$$$\frac{{a}+{b}}{{c}}={p}+{q} \\ $$$$\left({p}+{q}\right)_{{max}} =?…