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Question Number 90282 by 20000193 last updated on 22/Apr/20 $$\frac{\mathrm{1}}{\mathrm{2}\alpha+\beta}+\frac{\mathrm{1}}{\alpha+\mathrm{2}\beta} \\ $$ Commented by jagoll last updated on 22/Apr/20 $$\frac{\alpha+\mathrm{2}\beta+\mathrm{2}\alpha+\beta}{\left(\mathrm{2}\alpha+\beta\right)\left(\alpha+\mathrm{2}\beta\right)}\:=\:\frac{\mathrm{3}\alpha+\mathrm{3}\beta}{\mathrm{2}\alpha^{\mathrm{2}} +\mathrm{5}\alpha\beta+\mathrm{2}\beta^{\mathrm{2}} } \\ $$ Terms…
Question Number 155806 by SANOGO last updated on 05/Oct/21 Answered by puissant last updated on 05/Oct/21 $$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} −{k}^{\mathrm{2}} }}\: \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{0}}…
Question Number 155772 by daus last updated on 04/Oct/21 Answered by mr W last updated on 04/Oct/21 Commented by mr W last updated on 04/Oct/21…
Question Number 24696 by Tinku Tara last updated on 24/Nov/17 $$\boldsymbol{\mathrm{App}}\:\boldsymbol{\mathrm{Update}} \\ $$$$\mathrm{App}\:\mathrm{has}\:\mathrm{been}\:\mathrm{updated}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{changes} \\ $$$$\bullet\:\mathrm{Long}\:\mathrm{image}\:\mathrm{upload}\:\mathrm{issue}\:\mathrm{with} \\ $$$$\:\:\:\:\mathrm{submit}\:\mathrm{button} \\ $$$$\bullet\:\mathrm{Added}\:\mathrm{an}\:\mathrm{option}\:\mathrm{to}\:\mathrm{rotate}\:\mathrm{image} \\ $$$$\:\:\:\:\mathrm{while}\:\mathrm{uploading} \\ $$$$\bullet\:\mathrm{Added}\:\mathrm{ability}\:\mathrm{to}\:\mathrm{set}\:\mathrm{a}\:\mathrm{password}\:\mathrm{for}…
Question Number 155757 by aliyn last updated on 04/Oct/21 $$\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} \:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\:−\:\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} \:=\:\left(\boldsymbol{{x}}−\mathrm{1}\right)\:\boldsymbol{{y}}^{\mathrm{2}} \:\:;\:\boldsymbol{{y}}\left(\mathrm{1}\right)\:=\:\mathrm{2}\: \\ $$$$ \\ $$$$\square\:\boldsymbol{{M}}\: \\ $$ Commented by tabata last updated on…
Question Number 90217 by Maclaurin Stickker last updated on 22/Apr/20 $${let}\:{x}_{{n}\:} \:{be}\:{a}\:{sequence}\:{with}\:{x}_{\mathrm{0}} =\mathrm{2}\:{and}\:{x}_{\mathrm{1}} =\mathrm{7} \\ $$$${and}\:{x}_{{n}+\mathrm{1}} =\mathrm{7}{x}_{{n}} −\mathrm{12}{x}_{{n}−\mathrm{1}} . \\ $$$${Find}\:{the}\:{general}\:{term}\:{of}\:{x}_{{n}} . \\ $$ Commented…
Question Number 24663 by NECx last updated on 24/Nov/17 $$\boldsymbol{{solve}}\:\mid{x}−\mathrm{3}\mid=\mid\mathrm{3}{x}+\mathrm{2}\mid−\mathrm{1} \\ $$ Answered by ajfour last updated on 24/Nov/17 $${Case}\:{I}:\:\:\:\:\:\:\:{x}\:<\:−\frac{\mathrm{2}}{\mathrm{3}}\:\:,\:{then} \\ $$$$\:\:\:\:\:\mathrm{3}−{x}\:=\:−\mathrm{3}{x}−\mathrm{2}−\mathrm{1} \\ $$$${or}\:\:\:\:\mathrm{2}{x}\:=−\mathrm{6} \\…
Question Number 155710 by SANOGO last updated on 03/Oct/21 $$\mathrm{li}\underset{{x}−{oo}} {\mathrm{m}}\:\:\:\frac{\mathrm{1}}{{n}\sqrt{{n}}}\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{E}\left(\sqrt{\left.{k}\right)}\right. \\ $$$$ \\ $$ Commented by yeti123 last updated on 03/Oct/21 $$\underset{\boldsymbol{{x}}\rightarrow\infty}…
Question Number 24643 by mondodotto@gmail.com last updated on 23/Nov/17 Answered by $@ty@m last updated on 23/Nov/17 $${Let}\:{x}=\mathrm{3}×{y} \\ $$$$\mathrm{3}×\mathrm{2}×\mathrm{3}×\mathrm{4}×{y}=\mathrm{360} \\ $$$$\Rightarrow{y}=\mathrm{5} \\ $$$$\Rightarrow{x}=\mathrm{15} \\ $$$$…