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Question Number 200109 by sonukgindia last updated on 14/Nov/23 Answered by mr W last updated on 14/Nov/23 $$\mathrm{2}^{{x}} =\mathrm{5}−{x} \\ $$$$\mathrm{2}^{{x}−\mathrm{5}} =\frac{\mathrm{5}−{x}}{\mathrm{32}} \\ $$$${e}^{\left({x}−\mathrm{5}\right)\mathrm{ln}\:\mathrm{2}} =\frac{\mathrm{5}−{x}}{\mathrm{32}}…
Question Number 200110 by sonukgindia last updated on 14/Nov/23 Answered by mr W last updated on 14/Nov/23 $${x}^{{x}^{{x}} } ={x}^{\mathrm{3}} \\ $$$$\Rightarrow{x}=\pm\mathrm{1} \\ $$$$\Rightarrow{x}^{{x}} =\mathrm{3}\Rightarrow{e}^{\mathrm{ln}\:{x}}…
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Question Number 200137 by ajfour last updated on 14/Nov/23 Commented by ajfour last updated on 14/Nov/23 $${Find}\:{range}\:{of}\:{M}\:\left({x}\:{coordinate}\:{of}\right. \\ $$$$\left.{point}\:{where}\:{it}\:{hits}\:{x}\:{axis}\right),\:{if}\:\mathrm{2}{M}\:{is} \\ $$$${dropped}\:{from}\:{height}\:{h}\:{and}\:{hits} \\ $$$${left}\:{gear}\:{elastically}. \\ $$…
Question Number 200105 by cortano12 last updated on 14/Nov/23 Commented by mr W last updated on 14/Nov/23 Commented by mr W last updated on 14/Nov/23…
Question Number 200139 by hardmath last updated on 14/Nov/23 $$\mathrm{If} \\ $$$$\mathrm{x}\::\:\mathrm{y}\::\:\mathrm{z}\:=\:\frac{\mathrm{1}}{\mathrm{7}}\::\:\frac{\mathrm{1}}{\mathrm{3}}\::\:\frac{\mathrm{1}}{\mathrm{21}} \\ $$$$\mathrm{5x}\:−\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{16} \\ $$$$ \\ $$$$\mathrm{Find}:\:\:\:\mathrm{y}\:=\:? \\ $$ Answered by ajfour last updated…
Question Number 200133 by sonukgindia last updated on 14/Nov/23 Answered by Frix last updated on 14/Nov/23 $$\mathrm{You}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx\mathrm{1}.\mathrm{68421311} \\ $$ Terms of Service Privacy…
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Question Number 200103 by cortano12 last updated on 14/Nov/23 $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{sin}\:\sqrt{\mathrm{x}+\mathrm{1}}−\mathrm{sin}\:\sqrt{\mathrm{x}}\:=? \\ $$ Answered by Frix last updated on 14/Nov/23 $$\mathrm{sin}\:\alpha\:−\mathrm{sin}\:\beta\:=\mathrm{2cos}\:\frac{\alpha+\beta}{\mathrm{2}}\:\mathrm{sin}\:\frac{\alpha−\beta}{\mathrm{2}} \\ $$$$\mathrm{For}\:\mathrm{large}\:{x}:\:\frac{\sqrt{{x}+\mathrm{1}}+\sqrt{{x}}}{\mathrm{2}}\sim\sqrt{{x}}\:\wedge\:\frac{\sqrt{{x}+\mathrm{1}}−\sqrt{{x}}}{\mathrm{2}}\sim\mathrm{0} \\ $$$$\Rightarrow…