Question Number 222299 by fantastic last updated on 22/Jun/25 $${For}\:{what}\:{value}\:{of}\:\:{k}\:\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${will}\:{have}\:{same}\:{value}\:{but}\:\:{with}\:{opposite}\:{symbol}\left({like}\:{x}={a}\:{and}\:−{a}\right) \\ $$$${i}\:{mean}\:{the}\:{two}\:{valuea}\:{of}\:{x}\:{will}\:{be}\:{this}\:{type} \\ $$$${x}=\mathrm{2}\:{and}\:−\mathrm{2}\left({both}\:\mathrm{2}\:{but}\:{opposite}\:{symbols}\right) \\ $$ Answered by mr W…
Question Number 222292 by Nicholas666 last updated on 22/Jun/25 $$ \\ $$$$\:\:\:\int_{−\infty} ^{\infty} \mathrm{sech}\left({z}\right)\:\mathrm{sech}\left({z}−{a}\right)\:{dz} \\ $$$$ \\ $$ Answered by Frix last updated on 22/Jun/25…
Question Number 222295 by alvan545 last updated on 22/Jun/25 Answered by MrGaster last updated on 22/Jun/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222323 by MrGaster last updated on 22/Jun/25 $${a}+\mathrm{3}^{{b}} ={b},\mathrm{3}^{{b}} \centerdot{a}^{{b}+\mathrm{1}} \:\mathrm{max}=? \\ $$ Answered by Frix last updated on 22/Jun/25 $${a}={b}−\mathrm{3}^{{b}} \\ $$$${f}\left({b}\right)=\mathrm{3}^{{b}}…
Question Number 222317 by wewji12 last updated on 22/Jun/25 $$\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({z}\right)\mathrm{d}{z}=\frac{\pi}{\mathrm{2}}\:,\:\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{g}\left({z}\right)\mathrm{d}{z}=\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\:\infty} \:{f}\left({z}\right)\mathrm{g}\left({z}\right)\mathrm{d}{z}=?? \\ $$ Answered by MrGaster last updated…
Question Number 222312 by ajfour last updated on 22/Jun/25 Answered by mr W last updated on 22/Jun/25 $${QC}={DC}={AB}=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{a}}=\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${a}^{\mathrm{4}} −{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}…
Question Number 222309 by fantastic last updated on 22/Jun/25 $$\begin{array}{|c|}{\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} }\\\hline\end{array} \\ $$ Answered by Rasheed.Sindhi last updated on 22/Jun/25 $$\frac{\mathrm{4}^{{x}} }{\mathrm{6}^{{x}} }+\mathrm{1}=\frac{\mathrm{9}^{{x}}…
Question Number 222336 by Nicholas666 last updated on 22/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\int\int\int\:\:\frac{−{y}\:\pm\:\sqrt{{y}^{\mathrm{2}} \:+\:\mathrm{4}{xy}}}{\mathrm{2}{x}}\:{dxdydz} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222339 by Tawa11 last updated on 22/Jun/25 $$\mathrm{Solve}:\:\:\:\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{36}}\:\:\:+\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{24}}\:\:\:\:=\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{16}} \\ $$ Answered by fantastic last updated on 22/Jun/25 $$\left(\mathrm{36}\right)^{\frac{\mathrm{1}}{{x}}} +\left(\mathrm{24}\right)^{\frac{\mathrm{1}}{{x}}} =\left(\mathrm{16}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$$${or}\:\left(\frac{\mathrm{36}}{\mathrm{16}}\right)^{\frac{\mathrm{1}}{{x}}} +\left(\frac{\mathrm{24}}{\mathrm{16}}\right)^{\frac{\mathrm{1}}{{x}}}…
Question Number 222271 by MrGaster last updated on 21/Jun/25 $${y}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{10}} ,\mathrm{Prove}:{y}=\frac{\mathrm{123}+\mathrm{55}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ Answered by Rasheed.Sindhi last updated on 21/Jun/25 $${let}\:{x}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}\:}{\mathrm{2}}\:\Rightarrow{y}={x}^{\mathrm{10}} \\ $$$$\:\:\:\:\:\mathrm{2}{x}−\mathrm{1}=\sqrt{\mathrm{5}}\: \\ $$$$\:\:\:\mathrm{4}{x}^{\mathrm{2}}…