Question Number 133771 by bramlexs22 last updated on 24/Feb/21 $$\mathrm{A}\:\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{1}\:\mathrm{centered} \\ $$$$\mathrm{at}\:\left(\mathrm{0},\mathrm{0}\right)\:\mathrm{is}\:\mathrm{conjoined}\:\mathrm{to}\:\mathrm{a}\:\mathrm{circle} \\ $$$$\mathrm{of}\:\mathrm{radius}\:\mathrm{2}\:\mathrm{centered}\:\mathrm{at}\:\left(\mathrm{2},\mathrm{0}\right).\: \\ $$$$\mathrm{Forming}\:\mathrm{a}\:\mathrm{single}\:\mathrm{shape}\:.\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{shape}'\mathrm{s}\:\mathrm{area}? \\ $$ Commented by bramlexs22 last updated…
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Question Number 2672 by Yozzi last updated on 24/Nov/15 $${I}\:{have}\:\mathrm{4}\:{collinear}\:{points}\:{A}\left({a},\mathrm{0}\right), \\ $$$${B}\left({b},\mathrm{0}\right),\:{C}\left({c},\mathrm{0}\right)\:{and}\:{D}\left({d},\mathrm{0}\right)\:{where}\: \\ $$$$\forall{a},{b},{c},{d}>\mathrm{0}.\:{Find}\:{a}\:{point}\:{E}\left({x},{y}\right)\:{such} \\ $$$${that}\:{the}\:{following}\:{expression}\:{is} \\ $$$${minimised}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\left({AE}+{BE}+{CE}+{DE}\right). \\ $$ Commented by Rasheed…
Question Number 67983 by peter frank last updated on 03/Sep/19 Commented by Abdo msup. last updated on 03/Sep/19 $$\left.{b}\right)\:{let}\:{f}\left({x}\right)=\left(\frac{\mathrm{1}}{{x}}\right)^{{x}\:} \:\Rightarrow{f}\left({x}\right)={x}^{−{x}} \:={e}^{−{xln}\left({x}\right)} \\ $$$$\left.{f}\:{is}?{defined}\:{on}\:\right]\mathrm{0},+\infty\left[\right. \\ $$$${lim}_{{x}\rightarrow\mathrm{0}\:{and}\:{x}>\mathrm{0}}…
Question Number 133275 by mr W last updated on 20/Feb/21 Commented by mr W last updated on 20/Feb/21 $${A}\:{hall}\:{with}\:{horizontal}\:{smooth}\:{floor} \\ $$$${has}\:{a}\:{parabola}\:{shaped}\:{wall}\:{as}\:{shown}. \\ $$$${A}\:{ball}\:{is}\:{projected}\:{along}\:{the}\:{floor} \\ $$$${from}\:{point}\:{A}\:{and}\:{returns}\:{back}\:{to}\:{this}…
Question Number 2168 by Yozzi last updated on 06/Nov/15 $${The}\:{equation}\:{of}\:{an}\:{ellipse}\:{is}\:{given} \\ $$$${as}\: \\ $$$$\:\:{x}^{\mathrm{2}} +\left({y}+\mathrm{2}{xcot}\mathrm{2}\theta\right)^{\mathrm{2}} =\mathrm{1}\:\:\:\:\left(\mathrm{0}<\theta<\mathrm{0}.\mathrm{25}\pi\right). \\ $$$${Show}\:{that}\:{the}\:{minimum}\:{and}\:{maximum} \\ $$$${distances}\:{from}\:{the}\:{centre}\:{to}\:{the}\: \\ $$$${circumference}\:{of}\:{this}\:{ellipse}\:{are} \\ $$$${tan}\theta\:{and}\:{cot}\theta\:{respectively}.\: \\…
Question Number 2158 by Yozzis last updated on 05/Nov/15 $${Find}\:{the}\:{ratio},\:{over}\:{one}\:{revolution},\:{of}\:{the}\:{distance}\:{moved}\:{by} \\ $$$${a}\:{wheel}\:{rolling}\:{on}\:{a}\:{flat}\:{surface}\:{to}\:{the}\:{distance}\:{traced}\:{out}\:{by} \\ $$$${a}\:{point}\:{on}\:{its}\:{circumference}.\: \\ $$ Commented by ssahoo last updated on 06/Nov/15 $$\mathrm{Wheel}\:\mathrm{distance}=\mathrm{2}\pi{r} \\…
Question Number 132809 by mr W last updated on 17/Feb/21 Commented by mr W last updated on 18/Feb/21 $${a}\:{mountain}\:{has}\:{the}\:{shape}\:{of}\:{a}\:{right} \\ $$$${circular}\:{cone}\:{as}\:{shown}\:\left({h}>\sqrt{\mathrm{3}}{r}\right). \\ $$$${from}\:{point}\:{A}\:{to}\:{point}\:{B}\:{two} \\ $$$${sightseeing}\:{roads}\:{one}\:{time}\:{around}…
Question Number 132433 by john_santu last updated on 14/Feb/21 Answered by liberty last updated on 14/Feb/21 $$\mathrm{by}\:\mathrm{using}\:\mathrm{Cosine}\:\mathrm{of}\:\mathrm{law}\: \\ $$$$\:\left(\mathrm{r}−\mathrm{0}.\mathrm{5}\right)^{\mathrm{2}} =\mathrm{0}.\mathrm{5}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}−\mathrm{r}\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{0}.\mathrm{5}×\left(\sqrt{\mathrm{2}}−\mathrm{r}\right)\mathrm{cos}\:\mathrm{45}° \\ $$$$\cancel{\mathrm{r}^{\mathrm{2}} }−\mathrm{r}+\cancel{\frac{\mathrm{1}}{\mathrm{4}}}=\cancel{\frac{\mathrm{1}}{\mathrm{4}}}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{r}+\cancel{\mathrm{r}^{\mathrm{2}}…
Question Number 132402 by bramlexs22 last updated on 14/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{ellipse}\:\frac{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{16}}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{25}}\:=\:\mathrm{1} \\ $$ Commented by mr W last updated on 14/Feb/21 $${the}\:{perimeter}\:{of}\:{an}\:{ellipse}\:{can}\:{not}…