Question Number 58862 by George Mark Samuel last updated on 01/May/19 $$\mathrm{2}^{\mathrm{x}+\mathrm{y}=} \mathrm{6}^{\mathrm{y}} \\ $$$$\mathrm{3}^{\mathrm{x}} =\mathrm{3}\left(\mathrm{2}^{\mathrm{y}−\mathrm{1}} \right) \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$ Answered by MJS last…
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Question Number 58860 by naka3546 last updated on 01/May/19 Commented by malwaan last updated on 01/May/19 $${I}\:{cant}\:{save}\:{the}\:{image} \\ $$$${why}\:? \\ $$ Commented by naka3546 last…
Question Number 124397 by n0y0n last updated on 03/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124395 by n0y0n last updated on 03/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 124392 by bounhome last updated on 03/Dec/20 $${y}^{'} \left(\mathrm{2}{x}^{\mathrm{2}} {y}+{x}\right)={y}−\mathrm{2}{xy}^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 124393 by n0y0n last updated on 03/Dec/20 Answered by Kunal12588 last updated on 03/Dec/20 $$\mathrm{1}.\:{l}_{\mathrm{1}} :−{i}+\mathrm{3}{j}+\mathrm{2}{k}+\nu\left(−\mathrm{3}{i}−{j}−\mathrm{4}{k}\right) \\ $$$$\left(\mathrm{5},\lambda,\mu\right)\:{lies}\:{on}\:{l}_{\mathrm{1}} \\ $$$$\Rightarrow\nu=\frac{\mathrm{5}+\mathrm{1}}{−\mathrm{3}}=−\mathrm{2} \\ $$$$\lambda=\mathrm{3}−\nu=\mathrm{5} \\…
Question Number 124390 by ELIZAR last updated on 03/Dec/20 Answered by benjo_mathlover last updated on 03/Dec/20 $$\frac{\mathrm{31}}{\mathrm{9}}=\:\mathrm{3}+\frac{\mathrm{4}}{\mathrm{9}}=\:\mathrm{4}−\frac{\mathrm{5}}{\mathrm{9}}\:=\:\mathrm{4}−\frac{\mathrm{1}}{\frac{\mathrm{9}}{\mathrm{5}}}=\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}−\frac{\mathrm{6}}{\mathrm{5}}} \\ $$$$=\:\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}−\frac{\mathrm{1}}{\frac{\mathrm{5}}{\mathrm{6}}}}\:=\:\mathrm{4}−\frac{\mathrm{1}}{\mathrm{3}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{6}}}} \\ $$ Terms of Service Privacy…
Question Number 58832 by otchereabdullai@gmail.com last updated on 30/Apr/19 $$\mathrm{find}\:\mathrm{x}\:\mathrm{if}\:\mathrm{x}^{\mathrm{2}} =\mathrm{16}^{\mathrm{x}} \\ $$ Commented by tanmay last updated on 01/May/19 $${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{16}^{{x}} \\ $$$${f}\left(\mathrm{0}\right)<\mathrm{0} \\…