Question Number 144269 by alcohol last updated on 24/Jun/21 $$\int_{\mathrm{0}} ^{\:\infty} \lfloor\frac{{y}^{\mathrm{3}} }{{e}^{{y}} −\mathrm{1}}\rfloor{dy} \\ $$ Answered by mathmax by abdo last updated on 24/Jun/21…
Question Number 144226 by bramlexs22 last updated on 23/Jun/21 Commented by bramlexs22 last updated on 23/Jun/21 $$\mathrm{Find}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{yellow}\:\mathrm{circle} \\ $$ Answered by bemath last updated on…
Question Number 144209 by bramlexs22 last updated on 23/Jun/21 Answered by nimnim last updated on 23/Jun/21 $$\therefore\:\angle{ADB}=\mathrm{180}−\left(\mathrm{39}+\mathrm{180}−\mathrm{103}\right)=\mathrm{64} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 144210 by bramlexs22 last updated on 23/Jun/21 Answered by liberty last updated on 23/Jun/21 $$\mathrm{shaded}\:\mathrm{area}\:=\:\left(\frac{\mathrm{6}+\mathrm{2}}{\mathrm{2}}\right).\mathrm{4}\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{6}}.\mathrm{36}−\frac{\pi}{\mathrm{3}}.\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{16}\sqrt{\mathrm{3}}−\frac{\pi}{\mathrm{6}}\left(\mathrm{36}+\mathrm{8}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{44}\pi}{\mathrm{6}}\:=\:\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{22}\pi}{\mathrm{3}}\: \\ $$ Terms of…
Question Number 144201 by bramlexs22 last updated on 23/Jun/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles} \\ $$$$\mathrm{passing}\:\mathrm{through}\:\left(−\mathrm{4},\mathrm{3}\right)\:\mathrm{and}\:\mathrm{touching} \\ $$$$\mathrm{the}\:\mathrm{lines}\:\mathrm{x}+\mathrm{y}=\mathrm{2}\:\mathrm{and}\:\mathrm{x}−\mathrm{y}=\mathrm{2} \\ $$ Answered by benjo_mathlover last updated on 23/Jun/21 $$\mathrm{let}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{be}\: \\…
Question Number 131420 by ajfour last updated on 04/Feb/21 Commented by ajfour last updated on 04/Feb/21 $${In}\:{terms}\:{of}\:{ellipse}\:{parameters} \\ $$$${a}\:{and}\:{b},\:{find}\:{radius}\:{of}\:{the} \\ $$$${equal}\:{radii}\:{circles}. \\ $$ Answered by…
Question Number 279 by amandeep last updated on 25/Jan/15 $$\mathrm{If}\:{P}\:\left({x},{y}\right),\:{F}_{\mathrm{1}} =\left(\mathrm{3},\mathrm{0}\right),\:{F}_{\mathrm{2}} =\left(−\mathrm{3},\mathrm{0}\right)\:\mathrm{and}\: \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{25}{y}^{\mathrm{2}} =\mathrm{400}\:\mathrm{then}\:{PF}_{\mathrm{1}} +{PF}_{\mathrm{2}} =? \\ $$ Answered by 123456 last updated…
Question Number 229 by ssahoo last updated on 25/Jan/15 $$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{traingle}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{circle}\:\mid{z}\mid=\mathrm{2}\:\mathrm{and}\:\mathrm{if} \\ $$$${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} . \\ $$ Answered by 123456…
Question Number 91 by vkulkarni last updated on 25/Jan/15 $$\mathrm{A}\:\mathrm{line}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{the}\:\mathrm{mid}−\mathrm{point}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{line}\:\mathrm{joining}\:\mathrm{the}\:\mathrm{points}\:\left(−\mathrm{3},−\mathrm{4}\right)\:\mathrm{and}\:\:\left(−\mathrm{5},\mathrm{6}\right) \\ $$$$\mathrm{and}\:\mathrm{has}\:\mathrm{a}\:\mathrm{slope}\:\mathrm{of}\:\frac{\mathrm{3}}{\mathrm{4}}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}. \\ $$ Answered by ssahoo last updated on 04/Dec/14 $$\mathrm{slope}\:{m}=\frac{\mathrm{3}}{\mathrm{4}} \\…
Question Number 144074 by bobhans last updated on 21/Jun/21 $$\:\mathrm{The}\:\mathrm{parallelogram}\:\mathrm{ABCD}\:\mathrm{has} \\ $$$$\mid\mid\mathrm{AB}\mid\mid\:=\mathrm{6},\:\mid\mid\mathrm{AC}\mid\mid=\mathrm{7}\:\&\:\mathrm{d}\left(\mathrm{D},\mathrm{AC}\right)=\mathrm{2} \\ $$$$\mathrm{Find}\:\mathrm{d}\left(\mathrm{D},\mathrm{AB}\right). \\ $$ Answered by liberty last updated on 21/Jun/21 $$\mathrm{Area}_{\Delta\mathrm{ADC}} \:=\:\frac{\mathrm{2}×\mathrm{7}}{\mathrm{2}}=\mathrm{7}…