Question Number 164712 by cortano1 last updated on 21/Jan/22 Answered by mr W last updated on 21/Jan/22 Commented by mr W last updated on 21/Jan/22…
Question Number 98907 by Ramajunan last updated on 17/Jun/20 Answered by mr W last updated on 17/Jun/20 $$\frac{\left(\mathrm{5}+{x}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{5}+{x}\right){x}}=−\frac{\mathrm{5}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}{x}} \\…
Question Number 164437 by SLVR last updated on 17/Jan/22 $${Find}\:{the}\:{max}.\:{area}\:{of}\:\Delta{le} \\ $$$${witth}\:{sides}\:{a},{b},{c}\:{such}\:{as} \\ $$$$\mathrm{0}<{a}\leqslant\mathrm{1};\mathrm{1}\leqslant{b}\leqslant\mathrm{2};\mathrm{2}\leqslant{c}\leqslant\mathrm{3}\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\:\:\:\:\:\:\:\:{b}\right)\mathrm{1}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:{c}\right)\mathrm{2}\:\:\:\:\:\:\:\:{d}\right)\mathrm{3}/\mathrm{2} \\ $$ Answered by mahdipoor last updated on 17/Jan/22…
Question Number 98728 by Tinku Tara last updated on 15/Jun/20 $$\mathrm{Currently}\:\mathrm{working}\:\mathrm{on}\:\mathrm{enhancing} \\ $$$$\mathrm{this}\:\mathrm{app}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{shapes}. \\ $$$$\mathrm{So}\:\mathrm{posting}\:\mathrm{a}\:\:\mathrm{math}\:\mathrm{problem}\:\mathrm{realted} \\ $$$$\mathrm{to}\:\mathrm{drawing}\bar {.} \\ $$$$\mathrm{Ref}.\:\mathrm{Frame1}\:\mathrm{X}-\mathrm{Y} \\ $$$$\mathrm{Frame}\:\mathrm{2}: \\ $$$$\mathrm{Axis}\:\mathrm{translated}\:\mathrm{by}\:\left({h},{k}\right)\:\mathrm{and} \\…
Question Number 98598 by MWSuSon last updated on 15/Jun/20 $$\boldsymbol{{let}}\:\boldsymbol{{A}}=\left(\mathrm{3},\mathrm{4}\right)\:\boldsymbol{{and}}\:\boldsymbol{{B}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{variable}}\:\boldsymbol{{point}} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{the}}\:\boldsymbol{{line}}\:\mid\boldsymbol{{x}}\mid=\mathrm{6}.\:\boldsymbol{{if}}\:\overline {\boldsymbol{{AB}}}<\mathrm{4},\:\boldsymbol{{then}}\:\boldsymbol{{the}} \\ $$$$\boldsymbol{{number}}\:\boldsymbol{{of}}\:\boldsymbol{{position}}\:\boldsymbol{{of}}\:\boldsymbol{{B}}\:\boldsymbol{{with}}\:\boldsymbol{{integral}} \\ $$$$\boldsymbol{{coordinates}}\:\boldsymbol{{is}}? \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}}! \\ $$ Commented by bobhans last…
Question Number 33051 by Tinkutara last updated on 09/Apr/18 Commented by MJS last updated on 09/Apr/18 $$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{eccentric}\:\mathrm{angle}? \\ $$$$\mathrm{is}\:\mathrm{it}\:\angle{F}_{\mathrm{1}} {AF}_{\mathrm{2}} ? \\ $$ Commented by…
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Question Number 32679 by rahul 19 last updated on 31/Mar/18 $${If}\:\mathrm{4}{a}^{\mathrm{2}} −\mathrm{5}{b}^{\mathrm{2}} +\mathrm{6}{a}+\mathrm{1}=\mathrm{0}\:{and}\:{the}\:{line} \\ $$$${ax}+{by}+\mathrm{1}=\mathrm{0}\:{touches}\:{a}\:{fixed}\:{circle} \\ $$$${then}\:{the}\:{correct}\:{statement}\:{is}\:: \\ $$$$\left.\mathrm{1}\right)\:{centre}\:{of}\:{circle}\:{is}\:{at}\:\left(\mathrm{3},\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{radius}\:{of}\:{circle}\:{is}\:\sqrt{\mathrm{3}}. \\ $$$$\left.\mathrm{3}\right)\:{circle}\:{passes}\:{through}\:\left(\mathrm{1},\mathrm{0}\right). \\ $$$$\left.\mathrm{4}\right)\:{none}\:{of}\:{these}.…
Question Number 98185 by abdomathmax last updated on 12/Jun/20 $$\mathrm{developp}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$$$\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{3}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$ Answered by mathmax by abdo last updated on 12/Jun/20 $$\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{3}+\mathrm{sin}^{\mathrm{2}}…
Question Number 32508 by Tinkutara last updated on 26/Mar/18 Answered by MJS last updated on 27/Mar/18 $$\mathrm{there}\:\mathrm{are}\:\mathrm{2}\:\mathrm{tangents}\:\mathrm{of}\:\mathrm{different} \\ $$$$\mathrm{lengths}\:\mathrm{but}\:\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{answers} \\ $$$$\mathrm{fit}\:\mathrm{any}\:\mathrm{of}\:\mathrm{them}… \\ $$$$\mathrm{ellipse}:\:{y}=\pm\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }…