Question Number 195385 by Rodier97 last updated on 01/Aug/23 $$ \\ $$$$ \\ $$$$\mathrm{lim}_{{x}\Rightarrow\mathrm{a}^{+} } \:\:\:\:\frac{\sqrt{{x}}\:−\sqrt{\mathrm{a}}\:−\sqrt{{x}−\mathrm{a}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }}\:\:;\:\:\mathrm{a}\:>\:\mathrm{0} \\ $$ Answered by cortano12 last updated…
Question Number 195393 by Erico last updated on 01/Aug/23 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\sqrt[{\:\:\boldsymbol{{x}}\:\:}]{\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)^{{x}} }{{n}}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\sqrt[{\boldsymbol{{n}}}]{\mathrm{C}_{\mathrm{2}\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{n}}} } \\ $$ Answered by witcher3 last updated…
Question Number 195395 by cortano12 last updated on 01/Aug/23 Answered by kapoorshah last updated on 01/Aug/23 $${Any}\:{number}\:{power}\:\mathrm{0}\:{is}\:\mathrm{1} \\ $$$$ \\ $$$$\left(\mathrm{1}\:+\:\infty\:+\:\infty\:+\:…\:+\:\infty\right)^{\mathrm{0}} \:=\:\mathrm{1} \\ $$ Terms…
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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