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…
Question Number 62998 by mr W last updated on 27/Jun/19 $${Solve}\:{for}\:{x}:\:\:\mathrm{5}^{\boldsymbol{{x}}} +\mathrm{6}\boldsymbol{{x}}=\mathrm{7} \\ $$ Commented by mr W last updated on 27/Jun/19 $${Equation}\:{like}\:{this}\:{can}\:{be}\:{solved}\:{using} \\ $$$${Lambert}\:{W}\:{function}\:{as}\:{following}:…
Question Number 128508 by Lian last updated on 08/Jan/21 $${if}\:\:{x}^{{x}^{\mathrm{3}} } =\mathrm{3} \\ $$$${find}\:{x} \\ $$ Answered by mr W last updated on 08/Jan/21 $${if}\:{x}^{\mathrm{3}}…
Question Number 128509 by benjo_mathlover last updated on 08/Jan/21 Answered by liberty last updated on 08/Jan/21 $$\left(\Leftrightarrow\right)\:\frac{\mathrm{W}}{\mathrm{y}+\mathrm{x}}\:=\:\mathrm{6}\:\mathrm{days}\:\mathrm{where}\:\frac{\mathrm{W}}{\mathrm{y}}=\mathrm{10}\:\mathrm{days}\: \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{find}\:\frac{\mathrm{W}}{\frac{\mathrm{W}}{\mathrm{10}}+\mathrm{x}}=\mathrm{6}\:\Rightarrow\mathrm{W}=\frac{\mathrm{6W}}{\mathrm{10}}+\mathrm{6x} \\ $$$$\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{4W}}{\mathrm{60}}\:=\:\frac{\mathrm{W}}{\mathrm{15}}\:,\:\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{Mrs}\:\mathrm{x}\:\mathrm{can} \\ $$$$\mathrm{finish}\:\mathrm{the}\:\mathrm{job}\:\mathrm{by}\:\mathrm{herself}\:\mathrm{in}\:\mathrm{15}\:\mathrm{days} \\ $$…
Question Number 62945 by Tawa1 last updated on 27/Jun/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\:\:\left(\mathrm{6}\:−\:\mathrm{4x}\right)^{−\mathrm{3}} \\ $$ Answered by mr W last updated on 27/Jun/19 $$\left(\mathrm{6}−\mathrm{4}{x}\right)^{−\mathrm{3}} \\ $$$$=\mathrm{6}^{−\mathrm{3}} \left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}}\right)^{−\mathrm{3}} \\…