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Category: Algebra

Question-166273

Question Number 166273 by MikeH last updated on 17/Feb/22 Commented by MikeH last updated on 17/Feb/22 $$\mathrm{The}\:\mathrm{figure}\:\mathrm{above}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circumcircle} \\ $$$$\mathrm{ABC}\:\mathrm{with}\:\mathrm{centre}\:\mathrm{O}\:\mathrm{and}\:\mathrm{angle}\: \\ $$$$\mathrm{AOB}\:=\:\mathrm{50}°.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{angles} \\ $$$$\mathrm{marked}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}. \\ $$…

2z-2-z-2-z-1-lt-1-

Question Number 100730 by john santu last updated on 28/Jun/20 $$\frac{\mathrm{2}{z}^{\mathrm{2}} }{{z}^{\mathrm{2}} +\mid{z}+\mathrm{1}\mid}\:<\:\mathrm{1}\: \\ $$ Commented by bemath last updated on 28/Jun/20 $$\mathrm{since}\:\mathrm{z}^{\mathrm{2}} +\mid\mathrm{z}+\mathrm{1}\mid\:>\:\mathrm{0}\:\mathrm{then}\:\mathrm{2z}^{\mathrm{2}} \:<\:\mathrm{z}^{\mathrm{2}}…

If-four-men-can-dig-a-piece-of-land-in-2days-at-eight-hours-everyday-How-many-days-can-two-men-dig-in-every-six-hours-perday-

Question Number 166261 by gadafi last updated on 16/Feb/22 $${If}\:{four}\:{men}\:{can}\:{dig}\:{a}\:{piece}\:{of}\:{land}\:{in}\:\mathrm{2}{days}\:{at}\:{eight}\:{hours}\:{everyday}.{How}\:{many}\:{days}\:{can}\:{two}\:{men}\:{dig}\:{in}\:{every}\:{six}\:{hours}\:{perday}? \\ $$ Commented by 1549442205PVT last updated on 17/Feb/22 $${you}\:{should}\:{write}\:{your}\:{post}\:{like}\:{this} \\ $$$${so}\:{it}\:{is}\:{easy}\:{for}\:{people}\:{to}\:{read}. \\ $$$${If}\:{four}\:{men}\:{can}\:{dig}\:{a}\:{piece}\:{of}\:{land}\: \\…

Question-100703

Question Number 100703 by bramlex last updated on 28/Jun/20 Commented by bramlex last updated on 28/Jun/20 $$\left(\mathrm{1}\right)\:\mathrm{2}{y}^{\mathrm{2}} \:=\:\mathrm{100}−\mathrm{40}{a}+\mathrm{4}{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{16}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} \:\Rightarrow{y}^{\mathrm{2}} =\:\mathrm{32} \\ $$$$\left(\mathrm{1}\right)\wedge\left(\mathrm{2}\right)\:\Rightarrow\mathrm{4}{a}^{\mathrm{2}} −\mathrm{40}{a}+\mathrm{100}=\mathrm{64}\:…

Question-35152

Question Number 35152 by Victor31926 last updated on 16/May/18 Commented by Rasheed.Sindhi last updated on 16/May/18 $$\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{1}……………\mathrm{I} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{35}……….\mathrm{II} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}}…

Question-35153

Question Number 35153 by Victor31926 last updated on 16/May/18 Answered by Rasheed.Sindhi last updated on 16/May/18 $$\mathrm{2}=\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+….}}} \\ $$$$\mathrm{2}−\mathrm{x}=\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+….}}} \\ $$$$\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+….}}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}−\mathrm{4x}+\mathrm{x}^{\mathrm{2}} =\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+….}}}…

Question-166215

Question Number 166215 by mathlove last updated on 15/Feb/22 Answered by nurtani last updated on 15/Feb/22 $$\bullet\:\mathrm{27}^{{x}} −\mathrm{27}^{{y}} \:=\:\mathrm{24}\:\Leftrightarrow\:\mathrm{3}^{\mathrm{3}{x}} −\mathrm{3}^{\mathrm{3}{y}} \:=\:\mathrm{24}\:\Leftrightarrow\:\left(\mathrm{3}^{{x}} \right)^{\mathrm{3}} −\left(\mathrm{3}^{{y}} \right)^{\mathrm{3}} =\:\mathrm{24}…