Question Number 207518 by hardmath last updated on 17/May/24 $$\mathrm{cos}\boldsymbol{\mathrm{x}}\:\mathrm{cos3}\boldsymbol{\mathrm{x}}\:\:=\:\:\mathrm{cos5}\boldsymbol{\mathrm{x}}\:\mathrm{cos7}\boldsymbol{\mathrm{x}} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Answered by Frix last updated on 17/May/24 $$\mathrm{cos}\:{x}\:\mathrm{cos}\:\mathrm{3}{x}\:=\mathrm{cos}\:\mathrm{5}{x}\:\mathrm{cos}\:\mathrm{7}{x} \\ $$$$\frac{\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{4}{x}}{\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{2}{x}\:+\mathrm{cos}\:\mathrm{12}{x}}{\mathrm{2}} \\…
Question Number 207486 by hardmath last updated on 17/May/24 Commented by som(math1967) last updated on 17/May/24 $${DE}\bot{AK}\:? \\ $$ Commented by hardmath last updated on…
Question Number 207519 by hardmath last updated on 17/May/24 $$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\:\frac{\left(\mathrm{x}\:+\:\mathrm{2}\right)\centerdot\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mid\mathrm{x}\:+\:\mathrm{2}\mid}\:\:=\:\:? \\ $$ Answered by Frix last updated on 17/May/24 $${f}\left({x}\right)=\frac{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{1}\right)}{\mid{x}+\mathrm{2}\mid}=\begin{cases}{−\left({x}+\mathrm{1}\right),\:{x}<−\mathrm{2}}\\{\left({x}+\mathrm{1}\right),\:−\mathrm{2}<{x}}\end{cases} \\ $$$$\mathrm{We}\:\mathrm{approach}\:\mathrm{2}\:\mathrm{from}\:\mathrm{the}\:\mathrm{negative}\:\mathrm{side}\:\mathrm{but} \\…
Question Number 207487 by hardmath last updated on 17/May/24 $$\mid\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{3x}\:−\:\mathrm{4}\mid\:=\:\mid\mathrm{x}\:−\:\mathrm{4}\mid \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{min}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{max}}\:\:=\:\:? \\ $$ Answered by A5T last updated on 17/May/24 $$\mid{x}−\mathrm{4}\mid\mid{x}+\mathrm{1}\mid=\mid{x}−\mathrm{4}\mid\Rightarrow\mid{x}−\mathrm{4}\mid=\mathrm{0}\:{or}\:\mid{x}+\mathrm{1}\mid=\mathrm{1} \\ $$$$\Rightarrow{x}−\mathrm{4}=\mathrm{0}\:{or}\:{x}+\mathrm{1}=\mathrm{1}\:{or}\:{x}+\mathrm{1}=−\mathrm{1}…
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Question Number 207482 by cherokeesay last updated on 17/May/24 Answered by mr W last updated on 17/May/24 $${x}\left({a}−{x}\right)={yb}={zc} \\ $$$$\Rightarrow{y}=\frac{{x}\left({a}−{x}\right)}{{b}} \\ $$$$\Rightarrow{z}=\frac{{x}\left({a}−{x}\right)}{{c}} \\ $$$${d}^{\mathrm{2}} ={y}^{\mathrm{2}}…
Question Number 207509 by 073 last updated on 17/May/24 Answered by mr W last updated on 17/May/24 $$\int_{{k}} ^{{k}+\mathrm{1}} \left[{x}\right]{x}\lceil{x}\rceil{dx} \\ $$$$={k}\left({k}+\mathrm{1}\right)\int_{{k}} ^{{k}+\mathrm{1}} {xdx} \\…
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Question Number 207502 by hardmath last updated on 17/May/24 $$\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}\: \\…