Question Number 147103 by mnjuly1970 last updated on 18/Jul/21 Answered by mr W last updated on 18/Jul/21 $${x}={n}+{f},\:{n}\in{Z},\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\left({n}+{f}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +{f}^{\mathrm{2}} +\mathrm{2}{nf} \\…
Question Number 81552 by ajfour last updated on 14/Feb/20 Commented by jagoll last updated on 14/Feb/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{z}\:?? \\ $$ Commented by jagoll last updated on…
Question Number 16000 by tawa tawa last updated on 16/Jun/17 Answered by mrW1 last updated on 16/Jun/17 $$\mathrm{I}\:\mathrm{assume}\:\mathrm{you}\:\mathrm{mean}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{tan}\:\theta\:+\:\mathrm{b}\:\mathrm{sec}\:\theta\:=\:\mathrm{c} \\ $$$$ \\ $$$$\mathrm{a}\:\mathrm{tan}\:\theta\:+\:\mathrm{b}\:\mathrm{sec}\:\theta\:=\:\mathrm{c} \\…
Question Number 147076 by mathdanisur last updated on 17/Jul/21 $${a}+{b}+{c}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{1}\:\:\:\:\:\Rightarrow\:\:\:{abc}=? \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{1} \\ $$ Answered by gsk2684…
Question Number 15999 by tawa tawa last updated on 16/Jun/17 $$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y} \\ $$$$\mathrm{5xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x} \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z} \\ $$ Answered by mrW1 last updated on…
Question Number 81519 by ajfour last updated on 13/Feb/20 $$\mid\mid\mid{x}\mid−\mathrm{3}\mid−\mathrm{2}\mid=\mathrm{1} \\ $$$${solve}\:{for}\:{real}\:{x}. \\ $$ Commented by mr W last updated on 13/Feb/20 $$\mid\mid\mid{x}\mid−\mathrm{3}\mid−\mathrm{2}\mid=\mathrm{1} \\ $$$$\mid\mid{x}\mid−\mathrm{3}\mid−\mathrm{2}=\pm\mathrm{1}…
Question Number 81523 by Power last updated on 13/Feb/20 Commented by mr W last updated on 14/Feb/20 $$=\mathrm{100} \\ $$ Commented by Power last updated…
Question Number 81506 by Power last updated on 13/Feb/20 Commented by Power last updated on 13/Feb/20 $$\left.\mathrm{A}\left.\right)\left.\mathrm{1}\left.−\frac{\mathrm{1}}{\mathrm{50}!}\:\:\:\:\:\:\:\mathrm{B}\right)\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}\right)\mathrm{1}+\frac{\mathrm{1}}{\mathrm{49}!}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}\right)\mathrm{2} \\ $$ Commented by mr W last updated…
Question Number 81507 by Power last updated on 13/Feb/20 Commented by mr W last updated on 13/Feb/20 $${we}\:{have}\:\frac{\mathrm{2}{n}+\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\…
Question Number 147022 by ajfour last updated on 17/Jul/21 Terms of Service Privacy Policy Contact: info@tinkutara.com