Question Number 145954 by mathdanisur last updated on 09/Jul/21 $$\mathrm{1}+{i}+{i}^{\mathrm{2}} +{i}^{\mathrm{3}} +…+{i}^{\mathrm{99}} =? \\ $$ Answered by mr W last updated on 09/Jul/21 $$=\frac{\mathrm{1}−{i}^{\mathrm{100}} }{\mathrm{1}−{i}}=\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{50}}…
Question Number 145947 by Khalmohmmad last updated on 09/Jul/21 Commented by hknkrc46 last updated on 09/Jul/21 $$\:\left.\begin{matrix}{\int\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}\:=\:\boldsymbol{{F}}\left(\boldsymbol{{x}}\right)\:+\:\boldsymbol{{c}}}\\{\int\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}\:=\:\boldsymbol{{G}}\left(\boldsymbol{{x}}\right)\:+\:\boldsymbol{{c}}}\end{matrix}\right\}\:\begin{matrix}{\boldsymbol{{G}}\left(\mathrm{2}\right)\:=\:\mathrm{3}}\\{\boldsymbol{{f}}\left(\mathrm{5}\right)\:=\:\mathrm{1}}\end{matrix} \\ $$$$ \\ $$$$\:\:\:\bullet\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:=\:\boldsymbol{{F}}\:'\left(\boldsymbol{{x}}\right)\:\wedge\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)\:=\:\boldsymbol{{G}}\:'\left(\boldsymbol{{x}}\right) \\ $$$$\:\:\:\bullet\:\frac{\boldsymbol{{d}}\left[\left(\boldsymbol{{x}}^{\mathrm{2}} \:+\:\mathrm{2}\right)\boldsymbol{{G}}\left(\boldsymbol{{x}}\right)\right]}{\boldsymbol{{dx}}}\:=\:\frac{\boldsymbol{{d}}\left[\boldsymbol{{F}}\left(\mathrm{3}\boldsymbol{{x}}\:−\:\mathrm{1}\right)\right]}{\boldsymbol{{dx}}} \\…
Question Number 80404 by peter frank last updated on 02/Feb/20 Answered by MJS last updated on 02/Feb/20 $${f}\left({x}\right):\:{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{5}}\:\Rightarrow\:{f}^{−\mathrm{1}} \left({y}\right):\:{x}=\sqrt{\mathrm{5}{y}} \\ $$$${x}'=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{{y}}} \\ $$$$\mathrm{surface}=\mathrm{2}\pi\underset{{a}} {\overset{{b}}…
Question Number 80405 by TawaTawa last updated on 02/Feb/20 $$\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…