Question Number 132295 by shaker last updated on 13/Feb/21 Commented by mathmax by abdo last updated on 13/Feb/21 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{logx}−\sqrt{\mathrm{5x}−\mathrm{5}}\:\:\:\mathrm{f}\:\mathrm{defined}\:\mathrm{on}\:\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\mathrm{5}}{\mathrm{2}\sqrt{\mathrm{5x}−\mathrm{5}}}\:=\frac{\mathrm{1}}{\mathrm{x}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}}\:=\frac{\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}} \\ $$$$=\frac{\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}−\sqrt{\mathrm{5}}\mathrm{x}\right)\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}{\mathrm{2x}\sqrt{\mathrm{x}−\mathrm{1}}\left(\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}+\sqrt{\mathrm{5}}\mathrm{x}\right)}\:=\frac{\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{5x}^{\mathrm{2}} }{\left(…\right)}=\frac{−\mathrm{5x}^{\mathrm{2}}…
Question Number 132285 by bemath last updated on 13/Feb/21 $$\mathrm{Simplify}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\: \\ $$$$\frac{\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} −\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{x}\right)\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{2}}} +\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)}{\left(\mathrm{x}^{\frac{\mathrm{4}}{\mathrm{3}}} −\mathrm{x}\right)\left(\mathrm{x}+\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} +\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} \right)} \\ $$$$\mathrm{with}\:\mathrm{x}\:\neq\:\mathrm{0} \\ $$$$…
Question Number 66743 by pete last updated on 19/Aug/19 $$\mathrm{Each}\:\mathrm{month}\:\mathrm{a}\:\mathrm{store}\:\mathrm{owner}\:\mathrm{can}\:\mathrm{spend}\:\mathrm{at} \\ $$$$\mathrm{most}\:\$\mathrm{100},\mathrm{000}\:\mathrm{on}\:\mathrm{PC}'\mathrm{s}\:\mathrm{and}\:\mathrm{laptops}.\:\mathrm{A} \\ $$$$\mathrm{PC}\:\mathrm{costs}\:\mathrm{the}\:\mathrm{store}\:\mathrm{owner}\:\$\mathrm{1000}\:\mathrm{and}\:\mathrm{a} \\ $$$$\mathrm{laptop}\:\mathrm{costs}\:\mathrm{him}\:\$\mathrm{1500}.\:\mathrm{Each}\:\mathrm{PC}\:\mathrm{is}\:\mathrm{sold} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{400}\:\mathrm{while}\:\mathrm{a}\:\mathrm{laptop}\:\mathrm{is}\:\mathrm{sold} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{700}.\:\mathrm{The}\:\mathrm{store}\:\mathrm{owner}\:\mathrm{estimates} \\ $$$$\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{15}\:\mathrm{PC}'\mathrm{s}\:\mathrm{but}\:\mathrm{no}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{80}\:\mathrm{are}\:\mathrm{sold}\:\mathrm{each}\:\mathrm{month}.\:\mathrm{He}\:\mathrm{also}\:\mathrm{estimates} \\…
Question Number 132260 by Salman_Abir last updated on 12/Feb/21 Answered by Olaf last updated on 13/Feb/21 $$\left({x}\sqrt{{x}}\right)^{{x}} \:=\:{x}^{{x}\sqrt{{x}}} \\ $$$$\left({x}^{\mathrm{3}/\mathrm{2}} \right)^{{x}} \:=\:{x}^{{x}^{\mathrm{3}/\mathrm{2}} } \\ $$$${x}\mathrm{ln}{x}^{\mathrm{3}/\mathrm{2}}…
Question Number 132257 by Salman_Abir last updated on 12/Feb/21 Answered by bemath last updated on 12/Feb/21 $$\left(\mathrm{i}\right)\mathrm{E}\left(\mathrm{0},\mathrm{0},\mathrm{8}\right),\:\mathrm{C}\left(\mathrm{3},\mathrm{3},\mathrm{0}\right)\:\mathrm{vector}\: \\ $$$$\:\mathrm{EC}\:=\:\mathrm{3}\hat {\mathrm{i}}+\mathrm{3}\hat {\mathrm{j}}−\mathrm{8}\hat {\mathrm{k}}\:;\:\mid\mathrm{EC}\mid=\sqrt{\mathrm{9}+\mathrm{9}+\mathrm{64}}\:=\sqrt{\mathrm{82}} \\ $$$$\:\mathrm{H}\left(\mathrm{0},\mathrm{3},\mathrm{8}\right)\:,\mathrm{B}\left(\mathrm{3},\mathrm{0},\mathrm{0}\right)\:\mathrm{vector}\: \\…
Question Number 66715 by mr W last updated on 18/Aug/19 $${if}\:{f}\left({x}\right)=\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$ Commented by Tony Lin last updated on…
Question Number 1147 by prakash jain last updated on 04/Jul/15 $${f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −{x}+\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=? \\ $$ Commented by prakash jain last updated on 04/Jul/15 $${f}\left({f}\left(\mathrm{0}\right)\right)=\mathrm{1}…
Question Number 132198 by benjo_mathlover last updated on 12/Feb/21 $$\mathrm{If}\:\begin{cases}{\mathrm{16}^{{a}+{b}} \:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{16}^{{b}+{c}} \:=\:\mathrm{2}}\\{\mathrm{16}^{{a}+{c}} \:=\:\mathrm{2}\sqrt{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{then}\:\mathrm{c}\:=\:\_\_\: \\ $$ Answered by EDWIN88 last updated on 12/Feb/21 $$\mathrm{eq}\left(\mathrm{1}\right)\:\mathrm{16}^{\mathrm{2}\left(\mathrm{a}+\mathrm{b}\right)}…
Question Number 132180 by bounhome last updated on 12/Feb/21 $${solve}\:: \\ $$$$\mathrm{2}{sec}^{\mathrm{2}} {x}+\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right){secx}−\mathrm{3}\sqrt{\mathrm{2}}=\mathrm{0}\:;\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}} \\ $$ Answered by benjo_mathlover last updated on 12/Feb/21 $$\:\mathrm{3}\sqrt{\mathrm{2}}\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{3}\right)\mathrm{cos}\:\mathrm{x}−\mathrm{2}=\mathrm{0} \\…
Question Number 66619 by mr W last updated on 17/Aug/19 $${solve}\:{for}\:{x},{y}\in{R} \\ $$$$\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{\mathrm{ln}\:\left({y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$ Answered by Smail last updated on…