Question Number 117827 by ajfour last updated on 13/Oct/20 Commented by ajfour last updated on 13/Oct/20 $${Find}\:{the}\:{radius}\:{of}\:{circle}\:{shown}\:{in} \\ $$$${terms}\:{of}\:\boldsymbol{{c}}\:<\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$ Commented by ajfour last…
Question Number 183353 by Spillover last updated on 25/Dec/22 $${For}\:{the}\:{function} \\ $$$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} −\mathrm{3}\:{if}\:{x}<\mathrm{4}}\\{\frac{{x}^{\mathrm{2}} }{{x}+\mathrm{4}}\:\:\:\:\:{if}\:{x}\geqslant\mathrm{4}}\end{cases} \\ $$$$\left.{Find}\:\left({i}\right)\:\underset{{x}\rightarrow−\mathrm{4}} {\mathrm{lim}}\:{f}\left({x}\right)\:\:\:\:\:\:{ii}\right)\underset{{x}\rightarrow+\mathrm{4}} {\mathrm{lim}}\:{f}\left({x}\right) \\ $$ Answered by TheSupreme last updated…
Question Number 183352 by Spillover last updated on 26/Dec/22 $${For}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{1}−{x}^{\mathrm{2}} \:{if}\:{x}<\:\mathrm{2}}\\{\mathrm{2}{x}+\mathrm{1}\:{if}\:{x}\geqslant\mathrm{2}}\end{cases} \\ $$$${Find} \\ $$$$\left({i}\right)\underset{{x}\rightarrow^{−} \mathrm{2}} {\mathrm{lim}}{f}\left({x}\right)\:\:\:\:\:\:\:\left({ii}\right)\:\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}\:{f}\left({x}\right) \\ $$$$ \\ $$…
Question Number 52282 by Tawa1 last updated on 05/Jan/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{values}\:\mathrm{of}\:\:\mathrm{x} \\ $$$$\:\:\:\mid\frac{\mathrm{x}\:−\:\mathrm{1}}{\mathrm{2x}}\mid\:\geqslant\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 05/Jan/19 Commented by tanmay.chaudhury50@gmail.com…
Question Number 183354 by Spillover last updated on 25/Dec/22 $${Given}\:{three}\:{point}.{Find}\:{the} \\ $$$${for}\:{the}\:{plane}\:\:{through}\:{the}\:{point} \\ $$$${P}\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:\:{Q}\left(\mathrm{3},\mathrm{1},\mathrm{4}\right)\:\:{R}\left(−\mathrm{1},\mathrm{0},\mathrm{1}\right) \\ $$ Commented by mr W last updated on 25/Dec/22 $${see}\:{Q}\mathrm{182984}…
Question Number 183350 by Spillover last updated on 25/Dec/22 $${Find}\: \\ $$$$\left({a}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\:\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\left({b}\right)\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\left({c}\right)\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{3}{x}+\mathrm{1}}\:−\mathrm{4}}{{x}−\mathrm{5}} \\ $$ Commented by…
Question Number 183342 by Shrinava last updated on 25/Dec/22 $$\mathrm{Find}:\:\:\:\:\:\mathrm{3}\:−\:\frac{\mathrm{2}}{\mathrm{3}\:−\:\frac{\mathrm{2}}{\mathrm{3}\:−\:\frac{\mathrm{2}}{…}}}\:=\:? \\ $$ Answered by Red1ight last updated on 25/Dec/22 $$\mathrm{3}−\frac{\mathrm{2}}{{x}}={x} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{8}}}{\mathrm{2}}…
Question Number 117781 by hmh2k20 last updated on 13/Oct/20 $$\:\:\:\:\:\:\:\:\:\:\:\mathrm{Log}\:\left(\mathrm{cos}\beta\right)\:=\:\mathrm{p}\:\:\:\:\:\Rightarrow\:\:\:\mathrm{cos}\:\beta\:=\:\mathrm{10}^{\mathrm{p}} \:\: \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\mathrm{sec}\beta\:=\:\:\frac{\mathrm{1}}{\mathrm{cos}\beta}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{p}} }\:\:=\:\mathrm{10}^{−\mathrm{p}} \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\mathrm{Log}\:\left(\mathrm{sec}\beta\right)\:=\:\:\mathrm{Log}\:\mathrm{10}^{−\mathrm{p}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{p}\:\mathrm{Log}\:\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\mathrm{p} \\ $$ Answered by $@y@m…
Question Number 117767 by sumit Singh last updated on 13/Oct/20 $${x}^{\mathrm{2}} +{y}_{} ^{\mathrm{2}} ={a}^{\mathrm{2}} \sqrt{\mathrm{2}\:} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \:\:\:\:\:\: \\ $$$$ \\ $$$${what}\:{is}\:{intersection}\:\:{Angle}=?\: \\…
Question Number 52200 by gunawan last updated on 04/Jan/19 $$\mathrm{proof}\:\mathrm{that}\: \\ $$$$\sqrt{{z}^{\mathrm{2}} }\:\neq\:{z}\:,\:{z}\:\in\:\mathbb{C} \\ $$$$\mathrm{example} \\ $$$${i}=\sqrt{−\mathrm{1}} \\ $$$${i}^{\mathrm{2}} =−\mathrm{1} \\ $$$${i}^{\mathrm{4}} =\mathrm{1} \\ $$$$\sqrt{{i}^{\mathrm{4}}…