Question Number 128698 by I want to learn more last updated on 09/Jan/21 Commented by I want to learn more last updated on 09/Jan/21 $$\mathrm{Altitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}.…
Question Number 128684 by Study last updated on 09/Jan/21 $${how}\:{we}\:{can}\:{convert}\:\mathrm{0}.\overset{−} {\mathrm{9}}\:{to}\:\frac{{q}}{{p}}? \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\mathrm{let}\:\mathrm{r}\:=\:\mathrm{0},\mathrm{99999999999}… \\ $$$$\:\:\:\:\:\:\:\mathrm{10r}\:=\:\mathrm{9}.\mathrm{9999999999}… \\ $$$$\mathrm{substract}\:\Rightarrow\:\mathrm{9r}\:=\:\mathrm{9}\:\Rightarrow\:\mathrm{r}\:=\:\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}}\:=\:\frac{\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{n}}{\mathrm{n}}\:;\:\mathrm{n}\neq\:\mathrm{0}…
Question Number 128636 by john_santu last updated on 09/Jan/21 $$\mathrm{Solve}\:\mathrm{diopthantine}\:\mathrm{equation}\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}\:=\:\frac{\mathrm{2}}{\mathrm{17}}. \\ $$ Answered by liberty last updated on 09/Jan/21 $$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{2}}{\mathrm{17}}\:;\:\mathrm{2ab}\:=\:\mathrm{17}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{consider}\:\left(\mathrm{2a}−\mathrm{17}\right)\left(\mathrm{2b}−\mathrm{17}\right)=\mathrm{4ab}−\mathrm{34}\left(\mathrm{a}+\mathrm{b}\right)+\mathrm{17}^{\mathrm{2}} \\…
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Question Number 63095 by aliesam last updated on 28/Jun/19 Answered by MJS last updated on 29/Jun/19 $$\sqrt[{\mathrm{3}}]{\mathrm{4}{x}^{\mathrm{4}} −\mathrm{40}{x}^{\mathrm{2}} +\mathrm{100}}−\sqrt[{\mathrm{3}}]{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{10}}=\mathrm{20}−\mathrm{2}{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} ={s} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{4}{s}^{\mathrm{2}}…
Question Number 63090 by ajfour last updated on 28/Jun/19 $${s}=\sqrt{{a}^{\mathrm{2}} +\left({a}^{\mathrm{2}} −{d}\right)^{\mathrm{2}} }+\sqrt{\left({b}−{a}\right)^{\mathrm{2}} +\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\sqrt{{b}^{\mathrm{2}} +\left({c}−{b}^{\mathrm{2}} \right)^{\mathrm{2}} }+{c}−{d} \\ $$$$\:{p}=\:{a}\left({a}^{\mathrm{2}} −{d}\right)+\left({a}+{b}\right)\left({b}^{\mathrm{2}}…
Question Number 128617 by ajfour last updated on 08/Jan/21 Commented by ajfour last updated on 08/Jan/21 $${Express}\:{x}\:{in}\:{terms}\:{of}\:{t},\:{then} \\ $$$${use}\:{it}\:{to}\:{solve}\:{x}^{\mathrm{3}} ={x}+{c}. \\ $$ Answered by ajfour…
Question Number 63076 by YSN last updated on 28/Jun/19 $${show}\:{that}\:{f}:{A}\rightarrow{B}\:{is}\:{bijection}\:{then}\:{f}\left({A}_{\mathrm{1}} ^{{c}} \right)=\left[{f}\left({A}_{\mathrm{1}} \right)\right]^{{c}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128589 by mnjuly1970 last updated on 08/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:…{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:{a}\in\mathbb{R}^{+} \:\: \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\&\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\right\}\Rightarrow\:\sqrt{{a}−\sqrt{{a}}\:}\:=? \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} −\mathrm{17}{a}=\mathrm{16}\sqrt{{a}}\: \\ $$$$\:\:\:\:\:\:\: \\ $$ Answered by snipers237…
Question Number 63021 by kaivan.ahmadi last updated on 27/Jun/19 $${solve}\:{this}\:{equation} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{{y}} ={y}^{{x}} \\ $$$$ \\ $$$$ \\ $$$${x},{y}\in\mathbb{R}. \\ $$ Commented by…