Question Number 207502 by hardmath last updated on 17/May/24 $$\frac{\mathrm{6}}{\mid\boldsymbol{\mathrm{x}}\:−\:\mathrm{4}\mid\:−\:\mathrm{3}}\:\:\geqslant\:\:\mathrm{1} \\ $$ Answered by efronzo1 last updated on 17/May/24 $$\:\:\:\mathrm{Let}\:\mid\mathrm{x}−\mathrm{4}\mid\:=\:\mathrm{y}\:\geqslant\mathrm{0} \\ $$$$\:\Rightarrow\frac{\mathrm{6}}{\mathrm{y}−\mathrm{3}}\:\geqslant\mathrm{1}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{6}−\left(\mathrm{y}−\mathrm{3}\right)}{\mathrm{y}−\mathrm{3}}\:\geqslant\:\mathrm{0}\: \\…
Question Number 207493 by hardmath last updated on 17/May/24 $$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\mathrm{n}\:−\:\mathrm{1}}{\mathrm{n}\:+\:\mathrm{2}}\right)^{\boldsymbol{\mathrm{n}}+\mathrm{3}} \:=\:\:? \\ $$ Answered by A5T last updated on 17/May/24 $$\left(\frac{{n}+\mathrm{2}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\left(\frac{{n}−\mathrm{1}+\mathrm{3}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} =\left(\mathrm{1}+\frac{\mathrm{3}}{{n}−\mathrm{1}}\right)^{{n}+\mathrm{3}} \\…
Question Number 207477 by AliJumaa last updated on 16/May/24 $${is}\:{there}\:{any}\:{generale}\:{form}\:{for}\:{this}\:{sequense}\: \\ $$$$\begin{cases}{{u}_{{n}+\mathrm{1}} =\frac{{au}_{{n}} +{b}}{{cu}_{{n}} +{d}}}\\{{u}_{{m}} ={k}}\end{cases} \\ $$$${I}\:{need}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n}\:{i}\:{have}\:{try}\:{to}\:{derrive}\:{it}\:{for}\:{a}\:{long}\:{time}\:{but}\:{i}\:{cant} \\ $$ Commented by Frix last…
Question Number 207462 by hardmath last updated on 16/May/24 $$\mathrm{4}\:\mathrm{sin}\:\frac{\boldsymbol{\mathrm{x}}}{\mathrm{2}}\:=\:\mathrm{1}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by Frix last updated on 16/May/24 $${x}=\mathrm{4}{n}\pi+\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{4}}\:\vee{x}=\mathrm{2}\left(\mathrm{2}{n}+\mathrm{1}\right)\pi−\mathrm{2sin}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Terms…
Question Number 207463 by hardmath last updated on 16/May/24 $$\mathrm{1}\:−\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{x}\:=\:\sqrt{\mathrm{2}}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 16/May/24 $$\mathrm{cos}^{\mathrm{2}} \:{x}\:=\sqrt{\mathrm{2}}>\mathrm{1}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution}\:\mathrm{for}\:{x}\in\mathbb{R} \\ $$ Terms…
Question Number 207450 by York12 last updated on 15/May/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:{m}\:\mathrm{and}\:{n}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{following}\:\:\mathrm{holds} \\ $$$$\:\frac{{d}\left({y}\right)}{{d}\left({x}\right)}\mid_{{x}={n}} =\left(\frac{{d}\left({x}\right)}{{d}\left({y}\right)}\mid_{{y}={m}} \right)^{−\mathrm{1}} \\ $$ Answered by mr W last updated on 15/May/24 Commented…
Question Number 207442 by York12 last updated on 15/May/24 $$\mathrm{If}\:{y}={f}\left({x}\right),\:\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }={e}^{{y}+\mathrm{1}} ,\:\mathrm{and}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{on}\:\mathrm{the}\:\mathrm{point} \\ $$$$\left({x}_{\mathrm{1}} ,−\mathrm{1}\right)\:\mathrm{is}\:\mathrm{paralel}\:\mathrm{to}\:\mathrm{the}\:\mathrm{straight}\:\mathrm{line}\:{g}\left({x}\right)={x}−\mathrm{3},\:\mathrm{then}\:\mathrm{find}\:{f}'\left({x}\right). \\ $$ Commented by York12 last updated on 15/May/24…
Question Number 207385 by hardmath last updated on 13/May/24 $$\mathrm{Find}: \\ $$$$\sqrt{\left(\mathrm{2},\mathrm{5}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\:−\:\sqrt[{\mathrm{3}}]{\left.\left(\mathrm{1},\mathrm{5}−\sqrt{\mathrm{5}}\right)^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$ Commented by Frix last updated on 13/May/24 $$=\frac{\mathrm{7}}{\mathrm{2}}−\sqrt{\mathrm{5}}−\frac{\sqrt{\mathrm{6}\left(−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}}−\frac{\sqrt{\mathrm{2}\left(−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right.}}{\mathrm{4}}\mathrm{i}…
Question Number 207408 by hardmath last updated on 13/May/24 $$\boldsymbol{\mathrm{z}}\:\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{z}}^{\mathrm{11}} \:=\:? \\ $$ Answered by Frix last updated on 13/May/24 $${z}=\mathrm{e}^{−\frac{\pi}{\mathrm{6}}\mathrm{i}} \\ $$$${z}^{\mathrm{11}} =\mathrm{e}^{−\frac{\mathrm{11}\pi}{\mathrm{6}}\mathrm{i}} =\mathrm{e}^{\left(\mathrm{2}\pi−\frac{\mathrm{11}\pi}{\mathrm{6}}\right)\mathrm{i}}…
Question Number 207407 by hardmath last updated on 13/May/24 Commented by Frix last updated on 13/May/24 $$\mathrm{2}.\:…=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{10}° \\ $$ Commented by Frix last updated on…