Question Number 200186 by Calculusboy last updated on 15/Nov/23 Answered by ajfour last updated on 15/Nov/23 $$\frac{{y}}{{x}}={p}\:,\:\frac{{z}}{{y}}={q}\:,\:\frac{{x}}{{z}}={r}\:=\frac{\mathrm{1}}{{pq}} \\ $$$$\mathrm{1}+{p}+{p}^{\mathrm{2}} =\left(\mathrm{2}+{p}\right)\left(\frac{\mathrm{1}}{{r}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{q}+{q}^{\mathrm{2}} =\left(\mathrm{2}+{q}\right)\left(\frac{\mathrm{1}}{{p}}\right)^{\mathrm{2}/\mathrm{3}} \\ $$$$\mathrm{1}+{r}+{r}^{\mathrm{2}}…
Question Number 200168 by hardmath last updated on 15/Nov/23 $$\mathrm{If}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{86}\:\:\mathrm{and}\:\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{3x}^{\mathrm{2}} \:+\:\mathrm{x}\:−\:\mathrm{4} \\ $$$$\mathrm{Then}\:\mathrm{find}:\:\:\mathrm{g}\left[\mathrm{f}^{−\mathrm{1}} \left(\mathrm{g}\left(\mathrm{14}\right)\right)\right]\:=\:? \\ $$ Commented by jazeee last updated on 15/Nov/23 $$…
Question Number 200169 by hardmath last updated on 15/Nov/23 $$\mathrm{Rationalise}\:\mathrm{the}\:\mathrm{deniminator}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{fraction}: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}\:−\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}}\:=\:? \\ $$ Answered by Sutrisno last updated on 15/Nov/23 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}}×\frac{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}\right)−\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\left(\sqrt{\mathrm{6}}−\sqrt{\mathrm{3}}\right)−\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)} \\…
Question Number 200167 by hardmath last updated on 15/Nov/23 $$\mathrm{Given}\:\:\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{polynomial} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)\:=\:\mathrm{1}\:,\:\mathrm{f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{and}\:\:\mathrm{f}\left(\mathrm{3}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Find}:\:\:\mathrm{f}\left(\mathrm{4}\right)\:=\:? \\ $$ Answered by witcher3 last updated on 15/Nov/23 $$\Leftrightarrow\mathrm{f}\left(\mathrm{1}\right)−\mathrm{1}=\mathrm{2f}\left(\mathrm{2}\right)−\mathrm{1}=\mathrm{3f}\left(\mathrm{3}\right)−\mathrm{1}=\mathrm{0} \\…
Question Number 200139 by hardmath last updated on 14/Nov/23 $$\mathrm{If} \\ $$$$\mathrm{x}\::\:\mathrm{y}\::\:\mathrm{z}\:=\:\frac{\mathrm{1}}{\mathrm{7}}\::\:\frac{\mathrm{1}}{\mathrm{3}}\::\:\frac{\mathrm{1}}{\mathrm{21}} \\ $$$$\mathrm{5x}\:−\:\mathrm{2y}\:+\:\mathrm{z}\:=\:\mathrm{16} \\ $$$$ \\ $$$$\mathrm{Find}:\:\:\:\mathrm{y}\:=\:? \\ $$ Answered by ajfour last updated…
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Question Number 200087 by universe last updated on 13/Nov/23 $$\:\:\mathrm{if}\:\omega\:\neq\:\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{aand}\:\mathrm{z}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mid{z}\mid\:=\:\mathrm{1}\:\mathrm{then} \\ $$$$\:\:\mid\frac{\mathrm{2}+\mathrm{3}\omega+\mathrm{4}{z}\omega^{\mathrm{2}} }{\mathrm{4}\omega+\mathrm{3}\omega^{\mathrm{2}} {z}+\mathrm{2}{z}}\mid=\:? \\ $$ Commented by Frix last updated on 13/Nov/23…
Question Number 200066 by hardmath last updated on 13/Nov/23 $$\mathrm{If}\:\:\:\frac{\mathrm{1}\:+\:\mathrm{x}}{\:\sqrt{\mathrm{3}}}\:=\:\mathrm{3}\:\:\:\mathrm{find}\:\:\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\:−\:\mathrm{1}\:=\:? \\ $$ Answered by Frix last updated on 13/Nov/23 $${x}=−\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}\:\Rightarrow\:\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{26}}\:\Rightarrow\:{x}+\frac{\mathrm{1}}{{x}}−\mathrm{1}=\frac{−\mathrm{51}+\mathrm{81}\sqrt{\mathrm{3}}}{\mathrm{26}} \\ $$ Commented by hardmath…
Question Number 200035 by ajfour last updated on 12/Nov/23 Commented by ajfour last updated on 12/Nov/23 $${Find}\:{equation}\:{of}\:{parabola}\:{having}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{same}\:{curvature}\:{as}\:\mathrm{sin}\:{x}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{at}\:{shown}\:{point} \\ $$ Commented by…
Question Number 200025 by cortano12 last updated on 12/Nov/23 Answered by Frix last updated on 12/Nov/23 $$\left(\mathrm{1}\right)\:\:\:\:\:\sqrt{{t}+\mathrm{8}}+\sqrt{{t}}=\mathrm{4}\:\Rightarrow\:{t}=\mathrm{1}\:\Rightarrow\:{y}=\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{Transforming}\:\left(\mathrm{2}\right)\:\mathrm{to} \\ $$$${x}^{\mathrm{3}} \left({x}+\mathrm{13}\right)\sqrt{{x}+\mathrm{1}}=\mathrm{6}{x}^{\mathrm{4}} +\mathrm{14}{x}^{\mathrm{3}} +\mathrm{8} \\…