Question Number 22515 by Tinkutara last updated on 19/Oct/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:{ABCD},\:\mathrm{it}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{that}\:{AB}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{CD}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{diagonals}\:{AC}\:\mathrm{and}\:{BD}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\mathrm{Show}\:\mathrm{that} \\ $$$$\left(\mathrm{a}\right)\:{AD}.{BC}\:\geqslant\:{AB}.{CD}; \\ $$$$\left(\mathrm{b}\right)\:{AD}\:+\:{BC}\:\geqslant\:{AB}\:+\:{CD}. \\ $$ Terms…
Question Number 22478 by 123 45 polytechnicien last updated on 19/Oct/17 $$\begin{cases}{{x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} =\mathrm{3}}\\{{y}+{z}^{\mathrm{2}} +{x}^{\mathrm{3}} =\mathrm{3}}\\{{z}+{x}^{\mathrm{2}} +{z}^{\mathrm{3}} =\mathrm{3}}\end{cases} \\ $$$${trouver}\:{les}\:{solutions}\:{positives} \\ $$$$ \\ $$ Commented…
Question Number 87991 by necxxx last updated on 07/Apr/20 Commented by necxxx last updated on 07/Apr/20 $${please}\:{help} \\ $$ Answered by mr W last updated…
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Question Number 153226 by ajfour last updated on 05/Sep/21 Commented by mr W last updated on 05/Sep/21 $${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0} \\ $$$$\Rightarrow{x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{c}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{c}}{\mathrm{2}}} \\ $$$${any}\:{other}\:{ways}\:{to}\:{solve}?…
Question Number 87671 by TawaTawa1 last updated on 05/Apr/20 Commented by TawaTawa1 last updated on 05/Apr/20 $$\mathrm{I}\:\mathrm{got}\:\:\mathrm{10}.\mathrm{5}\:\mathrm{m}^{\mathrm{2}\:} \:\:\mathrm{but}\:\mathrm{am}\:\mathrm{not}\:\mathrm{sure}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}. \\ $$ Commented by Tony…
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Question Number 22112 by tawa tawa last updated on 11/Oct/17 $$\mathrm{A}\:\mathrm{boy}\:\mathrm{ran}\:\mathrm{around}\:\mathrm{a}\:\mathrm{circular}\:\mathrm{part}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{14m}\:\mathrm{in}\:\mathrm{15s}.\:\mathrm{Calculate}\:\mathrm{the}\: \\ $$$$\mathrm{average}\:\mathrm{velocity}\:\mathrm{and}\:\mathrm{the}\:\mathrm{average}\:\mathrm{speed}. \\ $$ Answered by $@ty@m last updated on 11/Oct/17 $${Average}\:{Verlocity}=\frac{{Displacement}}{{time}} \\ $$$$=\frac{\mathrm{0}}{\mathrm{15}}=\mathrm{0}…
Question Number 87630 by mr W last updated on 05/Apr/20 Commented by mr W last updated on 05/Apr/20 $${ajfour}\:{sir}\:{has}\:{put}\:{this}\:{question}: \\ $$$${find}\:{the}\:{radius}\:{r}\:{of}\:{the}\:{small}\:{circles} \\ $$$${in}\:{terms}\:{of}\:{the}\:{parameters}\:{a}\:{and}\:{b} \\ $$$${of}\:{the}\:{ellipse}\:\left({a}\geqslant{b}\right).…
Question Number 22079 by Tinkutara last updated on 10/Oct/17 $$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{and}\:{h}_{{a}} \:\mathrm{the} \\ $$$$\mathrm{altitude}\:\mathrm{through}\:{A}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left({b}\:+\:{c}\right)^{\mathrm{2}} \:\geqslant\:{a}^{\mathrm{2}} \:+\:\mathrm{4}{h}_{{a}} ^{\mathrm{2}} . \\ $$$$\left(\mathrm{As}\:\mathrm{usual}\:{a},\:{b},\:{c}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{sides}\:{BC},\right. \\ $$$$\left.{CA},\:{AB}\:\mathrm{respectively}.\right) \\ $$…