Question Number 42698 by Rio Michael last updated on 31/Aug/18 $${A}\:{boy}\:{lying}\:{flat}\:{on}\:{level}\:{ground}\:{sees}\:{a}\:{bird}\:\:{on}\:{a}\:{tree}\:{and}\:{the} \\ $$$${angle}\:{of}\:{Elevation}\:{from}\:{the}\:{boy}\:{to}\:{the}\:{birth}\:{is}\:\mathrm{42}°,{if}\:{the}\:{boy} \\ $$$${is}\:\mathrm{6}{m}\:{from}\:{the}\:{tree}.{find}\:{the}\:{hieght}\:{of}\:{the}\:{tree}\:{if}\:{the}\:{bird} \\ $$$${is}\:{at}\:{the}\:{top}\:{of}\:{the}\:{tree} \\ $$ Answered by math1967 last updated on…
Question Number 42624 by ajfour last updated on 29/Aug/18 Commented by ajfour last updated on 29/Aug/18 $${The}\:{smaller}\:{circle}\:{has}\:{radius}\:\boldsymbol{{r}}. \\ $$$${The}\:{two}\:{parabolas}\:{are}\:{reflection} \\ $$$${of}\:{one}\:{another}\:{in}\:{x}-{axis}. \\ $$$${Equation}\:{of}\:{upper}\:{parabola}\:{is} \\ $$$$\:{y}={ax}^{\mathrm{2}}…
Question Number 42340 by ajfour last updated on 23/Aug/18 Commented by ajfour last updated on 23/Aug/18 $${Find}\:\boldsymbol{{r}}\:{in}\:{terms}\:{of}\:\boldsymbol{{R}}\:{and}\:\boldsymbol{{a}}. \\ $$ Commented by MJS last updated on…
Question Number 173352 by cortano1 last updated on 10/Jul/22 Answered by mr W last updated on 10/Jul/22 $${x}_{{n}+\mathrm{1}} ^{\mathrm{2}} =\frac{\mathrm{2}{x}_{{n}} }{{x}_{{n}} +\mathrm{1}} \\ $$$$\left(\frac{\mathrm{1}}{{x}_{{n}+\mathrm{1}} }\right)^{\mathrm{2}}…
Question Number 107696 by ajfour last updated on 12/Aug/20 Commented by ajfour last updated on 12/Aug/20 $${If}\:{both}\:{circles}\:{have}\:{the}\:{equal}\:{radius}, \\ $$$${find}\:{the}\:{radius}. \\ $$ Answered by mr W…
Question Number 107594 by Rio Michael last updated on 11/Aug/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{common}\:\mathrm{divisor}\:\mathrm{of}\:\mathrm{1122}\:\mathrm{and}\:\mathrm{1001}\:\mathrm{and}\: \\ $$$$\mathrm{express}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{common}\:\mathrm{divisor}\:{d}\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}. \\ $$$$\:\:{d}\:=\:\mathrm{1122}{x}\:+\:\mathrm{1001}{y} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{above}\:\mathrm{result}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{congruence}\:\mathrm{equation} \\ $$$$\:\mathrm{37}{x}\:\equiv\:\mathrm{11}\:\left(\mathrm{mod}\:\mathrm{33}\right) \\ $$ Answered by 1549442205PVT last…
Question Number 173033 by cortano1 last updated on 05/Jul/22 Answered by greougoury555 last updated on 05/Jul/22 $$\:{f}\:'\left({x}\right)=\:\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\:−\frac{\mathrm{4}{k}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\:=\mathrm{0} \\ $$$$\:\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\left(\mathrm{1}−\frac{\mathrm{4}{k}\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{2}{x}}\right)=\mathrm{0} \\ $$$$\left(°\right)\:\mathrm{tan}\:{x}=\mathrm{1}\Rightarrow{f}\:=\sqrt{\mathrm{2}}\:+\mathrm{2}{k}\: \\ $$$$\left(°°\right)\:\mathrm{1}\:=\frac{\mathrm{4}{k}\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)}{\mathrm{sin}\:^{\mathrm{2}}…
Question Number 173002 by Mikenice last updated on 04/Jul/22 Commented by kaivan.ahmadi last updated on 05/Jul/22 $$\mathrm{1}−{tg}^{\mathrm{2}} \theta=\mathrm{1}−\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}=\frac{{cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}= \\ $$$$\frac{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}}…
Question Number 172878 by Mikenice last updated on 02/Jul/22 Commented by Mikenice last updated on 02/Jul/22 $${please}\:{if}\:{anyone}\:{want}\:{to}\left[{solve}\:{this}\right. \\ $$$${problem}\:{should}\:{please}\:{show}\:{the} \\ $$$${diagram} \\ $$ Answered by…
Question Number 41686 by ajfour last updated on 11/Aug/18 Commented by ajfour last updated on 11/Aug/18 $${Find}\:{radius}\:{of}\:{circle}\:{in}\:{terms}\:{of}\:{a}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on…