Question Number 7478 by Yozzia last updated on 31/Aug/16 $${Find}\:{the}\:{value}\:{of}\:\:{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} −\mathrm{18}\right)\:\:{if}\:\:{x}\left({x}−\mathrm{3}\right)=−\mathrm{1}. \\ $$ Answered by Rasheed Soomro last updated on 31/Aug/16 $$\:\:\:{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} −\mathrm{18}\right)\:\:{if}\:\:{x}\left({x}−\mathrm{3}\right)=−\mathrm{1}…
Question Number 7472 by FilupSmith last updated on 31/Aug/16 $$\mathrm{Show}\:\mathrm{why}: \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{x}^{{x}} \:=\:\mathrm{0} \\ $$ Commented by FilupSmith last updated on 31/Aug/16 $$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:{e}^{{x}\mathrm{ln}\left({x}\right)}…
Question Number 138546 by mnjuly1970 last updated on 14/Apr/21 Answered by ajfour last updated on 14/Apr/21 $$\left({i}\right)\:\:{a}^{\mathrm{3}} ={p},\:{b}^{\mathrm{3}} ={q},\:{c}^{\mathrm{3}} ={r} \\ $$$$\:{a}+{b}+{c}=\:\boldsymbol{{v}}\:=\sqrt[{\mathrm{3}}]{{p}}+\sqrt[{\mathrm{3}}]{{q}}+\sqrt[{\mathrm{3}}]{{r}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={v}^{\mathrm{3}}…
Question Number 138532 by mathlove last updated on 14/Apr/21 $${x}+\sqrt{{y}}=\mathrm{7} \\ $$$$\sqrt{{x}}+{y}=\mathrm{11} \\ $$$${faind}\:\:{x}=?\:\:{and}\:\:{y}=? \\ $$ Answered by henderson last updated on 14/Apr/21 $$\begin{cases}{{x}\:+\:\sqrt{{y}\:}\:=\:\mathrm{7}}\\{\sqrt{{x}}\:+\:{y}\:=\:\mathrm{11}}\end{cases}\:\Leftrightarrow\:\begin{cases}{{x}\:=\:\mathrm{7}\:−\:\sqrt{{y}}}\\{\sqrt{\mathrm{7}−\sqrt{{y}}}\:+\:{y}\:=\:\mathrm{11}}\end{cases}\:\Leftrightarrow\:\begin{cases}{{x}\:=\:\mathrm{7}−\sqrt{{y}}}\\{\sqrt{\mathrm{7}−\sqrt{{y}}}\:=\:\mathrm{11}−{y}}\end{cases}\Leftrightarrow \\…
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Question Number 7433 by Tawakalitu. last updated on 28/Aug/16 Commented by Yozzia last updated on 28/Aug/16 $$ \\ $$$${u}=\frac{\underset{{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}+\sqrt{{i}}}}{\underset{{i}=\mathrm{1}} {\overset{\mathrm{99}} {\sum}}\sqrt{\mathrm{10}−\sqrt{{i}}}} \\ $$$$\Rightarrow{u}\underset{{i}=\mathrm{1}}…
Question Number 7407 by Tawakalitu. last updated on 27/Aug/16 Commented by sou1618 last updated on 27/Aug/16 $${another}\:{solution} \\ $$$$\mathrm{999}^{\mathrm{2}} <\mathrm{1024}^{\mathrm{2}} =\mathrm{2}^{\mathrm{10}×\mathrm{2}} =\mathrm{2}^{\mathrm{20}} \\ $$$${so}\:\mathrm{999}^{\mathrm{2}} <\mathrm{2}^{\mathrm{20}}…
Question Number 7413 by Tawakalitu. last updated on 28/Aug/16 Answered by sandy_suhendra last updated on 28/Aug/16 $$\left.\mathrm{9}\right)\:{the}\:{curve}\:{through}\:{the}\:{point}\:\left(−\mathrm{1},\mathrm{0}\right),\:\left(\mathrm{3},\mathrm{0}\right)\:{and}\:\left(−\mathrm{2},−\mathrm{10}\right) \\ $$$$\:\:\:\:\:{f}\left({x}\right)={a}\left({x}−{x}_{\mathrm{1}} \right)\left({x}−{x}_{\mathrm{2}} \right)\:{which}\:{x}_{\mathrm{1}} =−\mathrm{1}\:{and}\:{x}_{\mathrm{2}} =\mathrm{3} \\ $$$$\:\:\:\:\:\:−\mathrm{10}={a}\left(−\mathrm{2}+\mathrm{1}\right)\left(−\mathrm{2}−\mathrm{3}\right)…
Question Number 138474 by henderson last updated on 14/Apr/21 $$\boldsymbol{\mathrm{hi}}\:!\: \\ $$$$\boldsymbol{\mathrm{find}}\:{x}\:\boldsymbol{\mathrm{such}}\:\boldsymbol{\mathrm{that}}\::\:\forall\:{y}\:\in\:\left[\mathrm{0},\mathrm{1}\right],\:{x}\:\geqslant\:{y}\:\Rightarrow\:{x}\:\geqslant\:\mathrm{2}{y}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 72900 by aliesam last updated on 04/Nov/19 $${prove}\:{that}\: \\ $$$$ \\ $$$$−\mid{a}\mid\leqslant{a}\leqslant\mid{a}\mid \\ $$$$ \\ $$$${a}\:{is}\:{a}\:{real}\:{number} \\ $$ Answered by MJS last updated…