Question Number 66162 by peter frank last updated on 09/Aug/19 Commented by mr W last updated on 10/Aug/19 $${question}\:{is}\:{not}\:{clear}.\:{have}\:{you}\:{got} \\ $$$${a}\:{diagram}? \\ $$ Commented by…
Question Number 66163 by aliesam last updated on 09/Aug/19 Answered by mr W last updated on 09/Aug/19 $${y}={x}^{{x}^{{x}^{….} } } \\ $$$${y}={x}^{{y}} \\ $$$$\mathrm{ln}\:{y}={y}\:\mathrm{ln}\:{x} \\…
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Question Number 66161 by peter frank last updated on 09/Aug/19 Answered by mr W last updated on 10/Aug/19 $${if}\:{a}\:{line}\:{y}={mx}+{c}\:{tangents}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}…
Question Number 624 by 123456 last updated on 15/Feb/15 $$ \\ $$$${f}\left({x},{y}\right)=\sqrt{\mathrm{2}\left({x}+\sqrt{{x}^{\mathrm{2}} −{y}}\right)} \\ $$$${what}\:{is}\:{the}\:{domain}\:{of}\:{f}\left({x},{y}\right) \\ $$ Commented by prakash jain last updated on 12/Feb/15…
Question Number 622 by 123456 last updated on 12/Feb/15 $$\frac{{y}}{{x}}\left(\frac{{dy}}{{dx}}−\frac{{y}}{{x}}\right)=\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} } \\ $$ Commented by prakash jain last updated on 12/Feb/15 $${y}={kx}? \\ $$…
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Question Number 131689 by liberty last updated on 07/Feb/21 $$\:\mathrm{a}\:\mathrm{curve}\:\mathrm{C}\:\mathrm{passes}\:\mathrm{through}\:\left(\mathrm{2},\mathrm{0}\right) \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{slope}\:\mathrm{at}\:\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{as}\:\frac{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}−\mathrm{3}\right)}{\mathrm{x}+\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{curve}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 619 by Cheenz last updated on 11/Feb/15 $$\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:+\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:? \\ $$ Answered by 123456 last updated on 11/Feb/15 $${x}=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}+\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}}…
Question Number 131688 by liberty last updated on 07/Feb/21 $$\:\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{2}=\mathrm{0}. \\ $$$$\mathrm{If}\:{a}_{{n}} \:=\:\alpha^{{n}} −\beta^{{n}} \:\mathrm{for}\:{n}\:\geqslant\mathrm{1}\: \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{{a}_{\mathrm{10}} −\mathrm{2}{a}_{\mathrm{8}} }{\mathrm{2}{a}_{\mathrm{9}} }\:? \\ $$…