Question Number 110318 by Study last updated on 28/Aug/20 $$\left({a}+{b}−{c}\right)^{\mathrm{2}} =?? \\ $$ Answered by john santu last updated on 28/Aug/20 $$={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}−\mathrm{2}{ac}−\mathrm{2}{bc}…
Question Number 175849 by Linton last updated on 08/Sep/22 $$\mathrm{2}^{\mathrm{2}{a}} +\mathrm{2}^{{a}} =\:\mathrm{10} \\ $$$${what}\:{is}\:{a}? \\ $$ Answered by MJS_new last updated on 08/Sep/22 $$\mathrm{2}^{\mathrm{2}{a}} +\mathrm{2}^{{a}}…
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Question Number 175839 by Rasheed.Sindhi last updated on 08/Sep/22 $${If}\:{w},\:{x},\:{y}\:{and}\:{z}\:{be}\:{four}\:{consecutive} \\ $$$${terms}\:{of}\:{any}\:{AP},\:{then}\:{show}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{w}^{\mathrm{2}} −{z}^{\mathrm{2}} =\mathrm{3}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right). \\ $$ Answered by mr W last…
Question Number 175838 by Rasheed.Sindhi last updated on 08/Sep/22 $${If}\:{x},{y}\:{and}\:{z}\:{be}\:{the}\:{pth},\:{qth}\:{and}\:{rth} \\ $$$${terms}\:{of}\:{an}\:{AP},\:{show}\:{that} \\ $$$$\begin{vmatrix}{{p}}&{{q}}&{{r}}\\{{x}}&{{y}}&{{z}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$ Answered by Rasheed.Sindhi last updated on 08/Sep/22 $$\begin{vmatrix}{{p}}&{{q}}&{{r}}\\{{x}}&{{y}}&{{z}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{vmatrix}=\mathrm{0} \\…
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Question Number 110299 by Lekhraj last updated on 28/Aug/20 Answered by Her_Majesty last updated on 28/Aug/20 $$\mathrm{5}{x}+\mathrm{5}{y}=\mathrm{2}{xy} \\ $$$$\Rightarrow\:\mathrm{5}\mid{xy}\:\Rightarrow\:\mathrm{5}\mid{x}\vee\mathrm{5}\mid{y} \\ $$$${let}\:{x}=\mathrm{5}{k} \\ $$$$\mathrm{25}{k}+\mathrm{5}{y}=\mathrm{10}{ky} \\ $$$${y}=\frac{\mathrm{5}{k}}{\mathrm{2}{k}−\mathrm{1}}…
Question Number 44763 by ajfour last updated on 04/Oct/18 Commented by ajfour last updated on 04/Oct/18 $${Equation}\:{of}\:{ellipse}\::\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}. \\ $$$${Find}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\:\boldsymbol{{b}}. \\ $$…
Question Number 175834 by Linton last updated on 08/Sep/22 $${xe}^{\overset{\mathrm{1}/} {{x}}} =\:{e} \\ $$$${solve}\:{for}\:{x} \\ $$ Commented by mr W last updated on 08/Sep/22 $${put}\:{your}\:{question}\:{in}\:{order}!…
Question Number 110294 by bemath last updated on 28/Aug/20 $$\mid\mathrm{2}{x}+\mathrm{1}\mid−\mid{x}−\mathrm{2}\mid\:<\:\mathrm{4}\: \\ $$$${find}\:{the}\:{solution}\:{set} \\ $$ Answered by Rio Michael last updated on 28/Aug/20 $$\mid\mathrm{2}{x}\:+\:\mathrm{1}\mid\:=\:\begin{cases}{\mathrm{2}{x}\:+\:\mathrm{1}\:,\:\:{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{−\left(\mathrm{2}{x}\:+\:\mathrm{1}\right),\:{x}\:<\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\mid{x}−\mathrm{2}\mid\:=\:\begin{cases}{{x}−\mathrm{2},\:\:{x}\:\geqslant\:\mathrm{2}}\\{−\left({x}−\mathrm{2}\right)\:,\:{x}\:<\:\mathrm{2}}\end{cases}…