Question Number 72863 by ajfour last updated on 04/Nov/19 Commented by ajfour last updated on 04/Nov/19 $${Find}\:{equations}\:{of}\:{AB}\:{dividing} \\ $$$${the}\:{composite}\:{area}\:{in}\:{two}\:{equal} \\ $$$${parts},\:{when}\:\:{segment}\:{AB}\:{has} \\ $$$${least}\:{length}\:{and}\:{maximum}\:{length}. \\ $$…
Question Number 7258 by Tawakalitu. last updated on 19/Aug/16 Commented by Yozzia last updated on 19/Aug/16 $${y}'=\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}\Rightarrow{y}=\int\frac{{x}}{\:\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}{dx}. \\ $$$${Let}\:{u}=\mathrm{5}−{x}^{\mathrm{2}} \Rightarrow{du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=\frac{−\mathrm{1}}{\mathrm{2}}{du}. \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{du}}{\:\sqrt{{u}}}=\frac{−\mathrm{1}}{\mathrm{2}}\int{u}^{−\mathrm{1}/\mathrm{2}} {du}=\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}}…
Question Number 72721 by mr W last updated on 01/Nov/19 Commented by mr W last updated on 01/Nov/19 $${find}\:{the}\:{correlation}\:{of}\:{A}\:{and}\:{B}\:{such} \\ $$$${that}\:{the}\:{circle}\:{is}\:{inscribed}\:{between}\:{the} \\ $$$${parabola}\:{and}\:{the}\:{x}−{axis}\:{as}\:{shown}. \\ $$…
Question Number 138203 by liberty last updated on 11/Apr/21 $${Three}\:{circles}\:{each}\:{radius}\:\mathrm{1},\:{touch}\:{one} \\ $$$${another}\:{externally}\:{and}\:{they}\:{lie} \\ $$$${between}\:{two}\:{parallel}\:{line}.\:{The}\: \\ $$$${minimum}\:{possible}\:{distance}\:{between}\: \\ $$$${the}\:{lines}\:{is}\:\_\: \\ $$ Answered by bobhans last updated…
Question Number 6716 by Tawakalitu. last updated on 15/Jul/16 $${Prove}\:{that}\:{the}\:{locus}\:{of}\:{a}\:{point}\:{which}\:{moves}\:{its}\:{distance}\:{from}\: \\ $$$${the}\:{point}\:\left(−{b},\:\mathrm{0}\right)\:{is}\:{p}\:{times}\:{its}\:{distance}\:{from}\:{the}\:{point}\:\left({b},\:\mathrm{0}\right)\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({p}^{\mathrm{2}} \:−\:\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \right)\:−\:\mathrm{2}{b}\left({p}^{\mathrm{2}} \:+\:\mathrm{1}\right){x}\:=\:\mathrm{0} \\ $$$${Show}\:{that}\:{this}\:{locus}\:{is}\:{a}\:{circle}\:{and}\:{find}\:{its}\:{radius}. \\ $$ Answered by…
Question Number 137768 by mr W last updated on 06/Apr/21 Commented by mr W last updated on 06/Apr/21 $${find}\:{the}\:{maximal}\:{area}\:{of}\:{the}\:{shadow} \\ $$$${in}\:{terms}\:{of}\:{r},\:{h},\:{H}\:{and}\:{d}. \\ $$ Answered by…
Question Number 6681 by Tawakalitu. last updated on 11/Jul/16 Answered by Rasheed Soomro last updated on 12/Jul/16 $${Join}\:\mathrm{O}\:{and}\:\mathrm{Y}. \\ $$$$\because\:\:\mathrm{OX}=\mathrm{OY}=\mathrm{OZ}\:\:\left[{Radii}\:{of}\:{same}\:{circle}\right] \\ $$$$\therefore\:\bigtriangleup\mathrm{XOY}\:\:{and}\:\:\bigtriangleup\mathrm{ZOY}\:{are}\:{issoscel}\:{triangles}. \\ $$$$\therefore\:\:\angle\mathrm{XYO}=\angle\mathrm{OXY}=\mathrm{30}\:\:\:{and}\:\:\:\:\angle\mathrm{ZYO}=\angle\mathrm{OZY}=\mathrm{20} \\…
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Question Number 137412 by oustmuchiya@gmail.com last updated on 02/Apr/21 Answered by herbert last updated on 02/Apr/21 $${gradient}\:{of}\:{l}_{\mathrm{1}} \:=\:\frac{\mathrm{2}+\mathrm{4}}{\mathrm{5}+\mathrm{1}}\:=\frac{\mathrm{6}}{\mathrm{6}}=\mathrm{1} \\ $$$${but}\:{grad}\:{of}\:{l}_{\mathrm{1}} ×{l}_{\mathrm{2}} =−\mathrm{1} \\ $$$${grad}\:{of}\:{l}_{\mathrm{2}} =−\mathrm{1}…
Question Number 137363 by liberty last updated on 02/Apr/21 $$ \\ $$As illustrated, the rectangle has an area of 1, and E is the midpoint…