Question Number 195419 by CrispyXYZ last updated on 02/Aug/23 $${x}\neq\mathrm{0}.\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ix}^{{i}−\mathrm{1}} =? \\ $$ Answered by MathedUp last updated on 02/Aug/23 $${x}\neq\mathrm{0}.\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ix}^{{i}−\mathrm{1}}…
Question Number 195413 by Calculusboy last updated on 02/Aug/23 Commented by Rasheed.Sindhi last updated on 02/Aug/23 $${x}^{{y}} +{y}^{{x}} =\mathrm{8}\Rightarrow{x}={y}=\mathrm{2} \\ $$$${x}+{y}={k}^{\mathrm{2}} \Rightarrow{k}^{\mathrm{2}} =\mathrm{2}+\mathrm{2}=\mathrm{4}\Rightarrow{k}=\pm\mathrm{2} \\ $$$${k}+\mathrm{1}=\pm\mathrm{2}+\mathrm{1}=\mathrm{3},−\mathrm{1}…
Question Number 195435 by universe last updated on 02/Aug/23 $$\sqrt{\mathrm{5}+\mathrm{1}\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{7}+\mathrm{3}\sqrt{\mathrm{8}+\mathrm{4}\sqrt{…}}}}}\:\:=\:\:\:? \\ $$ Answered by mr W last updated on 02/Aug/23 $$\mathrm{3}=\sqrt{\mathrm{9}}=\sqrt{\mathrm{5}+\mathrm{4}} \\ $$$$\mathrm{4}=\mathrm{1}×\mathrm{4}=\mathrm{1}×\sqrt{\mathrm{16}}=\mathrm{1}×\sqrt{\mathrm{6}+\mathrm{10}} \\ $$$$\mathrm{10}=\mathrm{2}×\mathrm{5}=\mathrm{2}×\sqrt{\mathrm{25}}=\mathrm{2}×\sqrt{\mathrm{7}+\mathrm{18}}…
Question Number 195428 by mathlove last updated on 02/Aug/23 $$\mathrm{3}^{{a}} =\mathrm{7}^{{b}} =\mathrm{441} \\ $$$$\frac{{ab}}{{a}+{b}}=? \\ $$ Answered by Frix last updated on 02/Aug/23 $$\mathrm{441}=\mathrm{3}^{\mathrm{2}} \mathrm{7}^{\mathrm{2}}…
Question Number 195404 by universe last updated on 01/Aug/23 Answered by AST last updated on 01/Aug/23 $$\left(\frac{\mathrm{127}^{\mathrm{2}} }{{xy}}+\frac{{x}}{{y}}+\frac{{y}}{{x}}\right)\left(\mathrm{127}+{x}+{y}\right) \\ $$$$=\frac{\mathrm{127}^{\mathrm{3}} }{{xy}}+\frac{\mathrm{127}^{\mathrm{2}} }{{y}}+\frac{\mathrm{127}^{\mathrm{2}} }{{x}}+\frac{\mathrm{127}{x}}{{y}}+\frac{{x}^{\mathrm{2}} }{{y}}+{x}+\frac{\mathrm{127}{y}}{{x}}+{y}+\frac{{y}^{\mathrm{2}} }{{x}}…
Question Number 195407 by mustafazaheen last updated on 01/Aug/23 $$\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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