Question Number 80405 by TawaTawa last updated on 02/Feb/20 $$\mathrm{Solve}: \\ $$$$\left(\mathrm{a}\right)\:\:\:\:\:\:\:\:\:\left(\mathrm{x}\:−\:\mathrm{3}\right)^{\mathrm{2}} \:\:>\:\:−\:\mathrm{5} \\ $$$$\left(\mathrm{b}\right)\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3x}^{\mathrm{2}} \:\:>\:\:−\:\mathrm{12} \\ $$ Commented by MJS last updated on 02/Feb/20…
Question Number 145918 by mathdanisur last updated on 09/Jul/21 Answered by Olaf_Thorendsen last updated on 09/Jul/21 $${l}\:=\:\mathrm{3}+\sqrt{\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{\mathrm{3}^{\mathrm{8}} +\sqrt{…}}}} \\ $$$$\left({l}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{3}^{\mathrm{2}} +\sqrt{\mathrm{3}^{\mathrm{4}} +\sqrt{\mathrm{3}^{\mathrm{8}}…
Question Number 80362 by Power last updated on 02/Feb/20 Answered by key of knowledge last updated on 02/Feb/20 $$\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+…+\mathrm{i}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)}=\frac{\left(\mathrm{1}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)\right)−\left(\mathrm{1}+…+\mathrm{i}\right)}{\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}\right)\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)\right)}= \\ $$$$\frac{\left(\mathrm{i}+\mathrm{1}\right)}{\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}\right)\left(\mathrm{1}+\mathrm{2}+…+\mathrm{i}+\left(\mathrm{i}+\mathrm{1}\right)\right)} \\ $$$$\Rightarrow\mathrm{1}−\left[\left(\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}\right)+\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\right)+…+\left(\frac{\mathrm{1}}{\mathrm{1}+…+\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{1}+…+\mathrm{100}}\right)\right]= \\ $$$$\mathrm{1}−\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+…+\mathrm{100}}\right]=\frac{\mathrm{1}}{\mathrm{5050}}…
Question Number 145886 by bramlexs22 last updated on 09/Jul/21 $$\frac{\mathrm{x}\sqrt{\frac{\mathrm{129934}}{\mathrm{14348057}}}}{\pi^{\mathrm{2}} }\:=\:\mathrm{e}^{\pi} \:\mathrm{then}\: \\ $$$$\:\frac{\sqrt{\mathrm{x}−\mathrm{96}}}{\mathrm{4}}\:=?\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 14811 by Nayon last updated on 04/Jun/17 $${why}\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid\:\:\:\:?\: \\ $$ Answered by mrW1 last updated on 04/Jun/17 $${if}\:{x}\geqslant\mathrm{0},\:\sqrt{{x}^{\mathrm{2}} }={x}=\mid{x}\mid \\ $$$${if}\:{x}<\mathrm{0},\:\sqrt{{x}^{\mathrm{2}} }=\sqrt{\left(−{x}\right)^{\mathrm{2}}…
Question Number 80347 by jagoll last updated on 02/Feb/20 $${let}\:{x}\:{and}\:{y}\:{be}\:{positif}\:{real}\:{number} \\ $$$${such}\:{that}\:\mathrm{1}\leqslant{x}+{y}\leqslant\mathrm{9}\:{and} \\ $$$${x}\leqslant\mathrm{2}{y}\leqslant\mathrm{3}{x}.\:{what}\:{is}\:{the}\: \\ $$$${largest}\:{value}\:{of}\:\:\:\frac{\mathrm{9}−{y}}{\mathrm{9}−{x}} \\ $$$$ \\ $$ Commented by john santu last…
Question Number 145879 by mathdanisur last updated on 09/Jul/21 Answered by mr W last updated on 09/Jul/21 $${let}\:{u}={f}\left({f}\left({x}\right)\right) \\ $$$${f}\left({u}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}−{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\leqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−{u}\leqslant\mathrm{0}…
Question Number 145852 by mathdanisur last updated on 08/Jul/21 $${sin}\frac{\pi}{\mathrm{24}}\centerdot{cos}\frac{\pi}{\mathrm{24}}\centerdot{cos}\frac{\pi}{\mathrm{12}}=? \\ $$ Answered by waiphyoemaung last updated on 08/Jul/21 $$\mathrm{solution} \\ $$$$\mathrm{sin}\frac{\pi}{\mathrm{24}}.\mathrm{cos}\frac{\pi}{\mathrm{24}}.\mathrm{cos}\frac{\pi}{\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\left(\mathrm{2sin}\frac{\pi}{\mathrm{24}}\mathrm{cos}\frac{\pi}{\mathrm{24}}\right)\mathrm{cos}\frac{\pi}{\mathrm{12}} \\…
Question Number 145856 by mathdanisur last updated on 08/Jul/21 Commented by mr W last updated on 08/Jul/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \left(…\right){dv}={C}={constant} \\ $$$${f}\left({x}\right)=\frac{{d}}{{dx}}\left({C}\right)=\mathrm{0} \\ $$$${f}'\left({x}\right)=\mathrm{0} \\…