Question Number 100976 by M±th+et+s last updated on 29/Jun/20 $${find}\:{all}\:{possible}\:{values}\:{of}\:{x},{y},{z}\:{in}\:{terms} \\ $$$${of}\:{a},{b},{c}\:{gor}\:{a}\:{triplet}\:\left({x},{y},{z}\right)\:{that}\:{satisfy} \\ $$$$ \\ $$$${x}+\frac{\mathrm{1}}{{y}}={a} \\ $$$$ \\ $$$${y}+\frac{\mathrm{1}}{{z}}={b} \\ $$$$ \\ $$$${z}+\frac{\mathrm{1}}{{x}}={c} \\…
Question Number 166496 by HongKing last updated on 21/Feb/22 $$\mathrm{Calculas}: \\ $$$$\frac{\mathrm{2}\:\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}}\:+\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}}\:-\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:-\:\sqrt{\mathrm{2}}}\:=\:? \\ $$ Commented by cortano1 last updated on 21/Feb/22 $$\:\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}}+\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)}×\frac{\sqrt{\mathrm{5}}−\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)}{\:\sqrt{\mathrm{5}}−\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\:\frac{\mathrm{2}\sqrt{\mathrm{6}}\left(\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)}{\mathrm{5}−\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)}=−\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\…
Question Number 100960 by john santu last updated on 29/Jun/20 $$\begin{cases}{\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}}\:=\:\mathrm{3}−{x}}\\{\sqrt{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{9}}\:=\:{x}+\mathrm{3}}\end{cases}\: \\ $$ Commented by bobhans last updated on 29/Jun/20 $$\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}}\:=\:\mid\mathrm{x}−\mathrm{3}\mid\:=\:\mathrm{3}−\mathrm{x}\:\rightarrow\:\mathrm{x}\:\leqslant\:\mathrm{3} \\…
Question Number 100954 by bobhans last updated on 29/Jun/20 $$\begin{cases}{\frac{\mathrm{1}}{\mathrm{2}{x}−{y}}\:+\:\sqrt{{y}}\:=\:\mathrm{1}}\\{\frac{\sqrt{{y}}}{\mathrm{2}{x}−{y}}\:=\:−\mathrm{6}}\end{cases} \\ $$ Commented by Dwaipayan Shikari last updated on 29/Jun/20 $$\:\:\:\:\:\:\:\:\:\frac{−\mathrm{6}}{\:\sqrt{{y}}}=\frac{\mathrm{1}}{\mathrm{2}{x}−{y}}\:\:\:\:\:\:\rightarrow\left(\mathrm{1}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\frac{−\mathrm{6}}{\:\sqrt{{y}}}+\sqrt{{y}}=\mathrm{1}\:\:\:\:\Rightarrow{t}^{\mathrm{2}}…
Question Number 166468 by mathlove last updated on 20/Feb/22 $$\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2}}{x}−\mathrm{18}{x}^{\mathrm{2}} =\mathrm{6}{x}\sqrt{\mathrm{1}−\mathrm{9}{x}^{\mathrm{2}} } \\ $$$${solve}\:\:{in}\:{R} \\ $$ Answered by MJS_new last updated on 20/Feb/22 $$\mathrm{squaring}\:\&\:\mathrm{transforming}\:\Rightarrow \\…
Question Number 100908 by bemath last updated on 29/Jun/20 $$\mathrm{what}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{angle} \\ $$$$\mathrm{formed}\:\mathrm{by}\:\mathrm{a}\:\mathrm{long}\:\mathrm{needle}\:\mathrm{and}\: \\ $$$$\mathrm{short}\:\mathrm{needle}\:\mathrm{on}\:\mathrm{analog}\:\mathrm{clock}\: \\ $$$$\mathrm{that}\:\mathrm{shows}\:\mathrm{at}\:\mathrm{15}.\mathrm{50}\:? \\ $$$$\left(\mathrm{A}\right)\:\mathrm{175}^{\mathrm{o}} \:\:\:\left(\mathrm{B}\right)\:\mathrm{174}^{\mathrm{o}} \:\:\:\left(\mathrm{C}\right)\:\mathrm{173}^{\mathrm{o}} \\ $$$$\left(\mathrm{D}\right)\:\mathrm{172}^{\mathrm{o}} \:\:\:\:\left(\mathrm{E}\right)\:\mathrm{170}^{\mathrm{o}} \\ $$…
Question Number 166443 by mathls last updated on 20/Feb/22 $${fog}_{\left(\mathrm{3}\right)} =\mathrm{10} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{4} \\ $$$${g}\left({x}\right)=? \\ $$ Commented by mr W last updated on 20/Feb/22…
Question Number 166431 by BagusSetyoWibowo last updated on 20/Feb/22 $$\mathrm{2}\frac{\mathrm{2}^{{x}} }{\mathrm{16},\mathrm{0}}\:=\:\mathrm{5},\mathrm{7}^{\mathrm{2}} \:−\:\mathrm{0},\mathrm{49}\:+\:\mathrm{0} \\ $$$${How}\:{much}\:{the}\:{x}\:{is}? \\ $$ Answered by alephzero last updated on 20/Feb/22 $$\mathrm{2}\frac{\mathrm{2}^{{x}} }{\mathrm{16}}\:=\:\mathrm{32}…
Question Number 35345 by jasno91 last updated on 18/May/18 $${yes}\:\sqrt{\mathrm{2}}{x}^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166419 by ajfour last updated on 19/Feb/22 Commented by ajfour last updated on 19/Feb/22 $${Blue}\:{curve}\:\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${Red}\:{curve}\:\:\:\:{y}=\left({x}−{h}\right)−\left({x}−{h}\right)^{\mathrm{3}} +{k} \\ $$$${Find}\:{x}=\:{p}\:\:{when}\:\:{y}={c}. \\ $$…